Principles of Communication - Iit Madras Lecture Material

Principles of Communication Prof. V. Venkata Rao

Indian Institute of Technology Madras

1.1

1 CHAPTER 1

Representation of Signals

1.1 Introduction The process of (electronic) communication involves the generation,

transmission and reception of various types of signals. The communication

process becomes fairly difficult, because:

a) the transmitted signals may have to travel long distances (there by

undergoing severe attenuation) before they can reach the destination i.e.,

the receiver.

b) of imperfections of the channel over which the signals have to travel

c) of interference due to other signals sharing the same channel and

d) of noise at the receiver input1.

In quite a few situations, the desired signal strength at the receiver input

may not be significantly stronger than the disturbance component present at that

point in the communication chain. (But for the above causes, the process of

communication would have been quite easy, if not trivial). In order to come up

with appropriate signal processing techniques, which enable us to extract the

desired signal from a distorted and noisy version of the transmitted signal, we

must clearly understand the nature and properties of the desired and undesired

signals present at various stages of a communication system. In this lesson, we

begin our study of this aspect of communication theory.

1 Complete statistical characterization of the noise will be given in chapter 3, namely, Random

Signals and Noise.



1.2

Signals physically exist in the time domain and are usually expressed as a

function of the time parameter1. Because of this feature, it is not too difficult, at

least in the majority of the situations of interest to us, to visualize the signal

behavior in the Time Domain. In fact, it may even be possible to view the signals

on an oscilloscope. But equally important is the characterization of the signals in

the Frequency Domain or Spectral Domain. That is, we characterize the signal

in terms of its various frequency components (or its spectrum). Fourier analysis

(Fourier Series and Fourier Transform) helps us in arriving at the spectral

description of the pertinent signals.

1.2 Periodic Signals and Fourier Series Signals can be classified in various ways such as:

a) Power or Energy

b) Deterministic or Random

c) Real or Complex

d) Periodic or Aperiodic etc.

Our immediate concern is with periodic signals. In this section we shall

develop the spectral description of these signals.

1.2.1 Periodic signals Def. 1.1: A signal ( )px t is said to be periodic if

( ) ( )p px t x t + T= , (1.1)

for all t and some T .

( denotes the end of definition, example, etc.)

1 We will not discuss the multi-dimensional signals such as picture signals, video signals, etc.



1.3

Let T0 be the smallest value of T for which this is possible. We call T0 as

the period of ( )px t .

Fig. 1.1 shows a few examples of periodic signals.

Fig. 1.1: Some examples of periodic signals



1.4

The basic building block of Fourier analysis is the complex exponential,

namely, ( )j f tA e 2π + ϕ or ( )A j f t +exp 2π ϕ⎡ ⎤⎣ ⎦ , where

A : Amplitude (in Volts or Amperes)

f : Cyclical frequency (in Hz)

ϕ : Phase angle at t 0= (either in radians or degrees)

Both A and f are real and non-negative. As the radian frequency, ω (in

units of radians/ sec), is equal to 2 fπ , the complex exponential can also written

as ( )j tA e ω + ϕ . We use subscripts on A , f (or ω) and ϕ to denote the specific

values of these parameters.

Fourier analysis uses tcosω or sin cos2

t t⎡ π ⎤⎛ ⎞ω = ω −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ in the represen-

tation of real signals. From Euler’s relation, we have, cos sinj te t j tω = ω + ω .

As tcosω is the Re j te ω⎡ ⎤⎣ ⎦ , where [ ]xRe denotes the real part of x , we

have

( )cos

2

j t j te et

∗ω ω+

ω = (∗denotes the complex conjugate)

2

j t j te eω − ω+

=

The term j te− ω or 2j f te− π is referred to as the complex exponential at

the negative frequency ω− (or f− ).

1.2.2 Fourier series



1.5

Let ( )px t be a periodic signal with period T0 . Then 00

1fT

= is called the

fundamental frequency and 0n f is called the thn harmonic, where n is an

integer (for 0n = , we have the DC component and for the DC singal,

T0 is not defined; 1n = results in the fundamental). Fourier series

decomposes ( )px t in to DC, fundamental and its various higher harmonics,

namely,

( ) 0j 2 nf tp nn

x t x e∞ π

= − ∞= ∑ (1.2a)

The coefficients nx constitute the Fourier series and are related to

( )px t as

( ) 0

00

j 2 nf tn p

T

1x x t e dtT

− π= ∫ (1.2b)

where 0 T

∫ denotes the integral over any one period of ( )px t . Most often, we

use the interval 0 0,2 2

T T⎛ ⎞−⎜ ⎟⎝ ⎠

or ( )00 , T . Eq. 1.2(a) is referred to as the

Exponential form of the Fourier series.

The coefficients nx are in general complex; hence

njn nx x e ϕ= (1.3)

where nx denotes the magnitude of the complex number and nϕ , the argument

(or the angle). Using Eq. 1.3 in Eq. 1.2(a), we have,

( ) ( )nj n f tp nn

x t x e 02∞ π + ϕ

= − ∞= ∑

Eq. 1.2(a) states that ( )px t , in general, is composed of the frequency

components at DC, fundamental and its higher harmonics. nx is the magnitude



1.6

of the component in ( )px t at frequency 0n f and nϕ , its phase. The plot of nx

vs. n (or 0n f ) is called the magnitude spectrum, and nϕ vs. n (or 0n f ) is called

the phase spectrum. It is important to note that the spectrum of a periodic signal

exists only at discrete frequencies, namely, at 0n f , n 0, 1, 2,= ± ± ⋅ ⋅ ⋅ , etc.

Let ( )px t be real; then

( ) j n f tn p

Tx x t e d t

T0

0

2

0

1 π=− ∫

nx∗=

That is, for a real periodic signal, we have the two symmetry properties, namely,

n nx x− = (1.4a)

- n nϕ = − ϕ (1.4b)

Properties of Eq. 1.4 are part of an if and only if (iff) relationship. That is, if

( )px t is real, then Eq. 1.4 holds and if Eq. 1.4 holds, then ( )px t has to be real.

This is because the complex exponentials at ( )0n f and ( )0n f− can be combined

into a cosine term. As an example, let the only nonzero coefficients of a periodic

signal be x , x , x2 1 0± ± . *X 0=x0 implies, x0 is real and let

2 2j4

2x e xπ

∗− = = and

j3x e x1 13π

∗− = = and 0 1x = . Then,

( ) 0 0 0 04 2 2 43 34 43 1 3 2j jj jj f t j f t j f t j f t

px t 2 e e e e e e e eπ ππ π− −− π − π π π

+ + + +=

Combining the appropriate terms results in,

( )px t f t f t0 04 cos 4 6 cos 2 14 3π π⎛ ⎞ ⎛ ⎞= π − + π − +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

which is a real signal. The above form of representing ( )px t , in terms of cosines

is called the Trigonometric form of the Fourier series.



1.7

We shall illustrate the calculation of the Fourier coefficients using the

periodic rectangular pulse train (This example is to be found in almost all the

textbooks on communication theory).

Example 1.1 For the unit amplitude rectangular pulse train shown in Fig. 1.2, let us

compute the Fourier series coefficients.

Fig. 1.2: Periodic rectangular pulse train

( )px t has a period 0 4T = milliseconds and is ON for half the period and

OFF during the remaining half. The fundamental frequency 0f = 250 Hz.

From Eq. 1.2(b), we have T

j n f tn

Tx e dt

T

0

0

0

22

02

1 −

−

= ∫ π

0

0

0

42

04

1T

j nf t

Te dt

T−

− π= ∫

0 0

2sin1

n

T n f

π⎛ ⎞⎜ ⎟⎝ ⎠=π



1.8

sin

2n

n

π⎛ ⎞⎜ ⎟⎝ ⎠=π

As can be seen from the equation for nx , all the Fourier coefficients are

real but could be bipolar (+ve or –ve). Hence nϕ is either zero or ± π for all n .

Fig. 1.3 shows the plots of magnitude and phase spectrum.

Fig. 1.3: Magnitude and phase spectra for the px (t) of example 1.1

From Fig. 1.3, we observe:



1.9

i) 0x , the average or the DC value of the pulse train is 12

. For any periodic

signal, the average value is ( )0

0

2

02

1T

pT

x t dtT

−

∫ .

ii) spectrum exists only at discrete frequencies, namely, 0f n f= , with

0 250f Hz= . Such a spectrum is called the discrete spectrum (or line

spectrum). iii) the curve drawn with broken line in Fig. 1.3(a) is the envelope of the

magnitude spectrum. The envelope consists of several lobes and the

maximum value of each lobe keeps decreasing with increase in frequency.

iv) the plot of nx vs. frequency is symmetric and the plot of nϕ vs. frequency is

anti-symmetric. This is because ( )px t is real.

v) nϕ at n , 2 4= ± ± etc. is undefined as 0nx = for these n . This is indicated

with a cross on the phase spectrum plot.

One of the functions that is useful in the study of Fourier analysis is the

( )sinc function defined by

( ) ( )sinsin sinc c

πλ= λ =

πλλ (1.5)

A plot of the sinc λ vs. λ along with a table of values are given in

appendix A1.1, at the end of the chapter.



1.10

In terms of ( )sinc λ , the Fourier coefficient of example 1.1 can be written

as, 1 sin2 2n

nx c ⎛ ⎞= ⎜ ⎟⎝ ⎠

.

Spectrum analyzer is an important laboratory instrument, which can be

used to obtain the magnitude spectrum of periodic signals (frequency resolution,

frequency range over which the spectrum can be measured etc. depend on that

particular instrument). We have given below a set of four waveforms (output of a

function generator) and their line spectra, as indicated by a spectrum analyzer.

The spectral plots 1 to 4 give the values of 101

20 log nxx

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

for the

waveforms 1 to 4 respectively. The units for the above quantities are in decibels

(dB).

Exercise 1.1

For the ( )px t of Fig .1.4, show that ( )nx c n fT 0

0sin

⎛ ⎞τ= τ⎜ ⎟⎝ ⎠

Fig. 1.4: ( )px t of Exercise 1.1



1.11

1.

Waveform 1

Spectral Plot 1



1.12

Waveform 1: A cosine signal (frequency 10 kHz).

Comments on spectral plot 1: Waveform 1 has only two Fourier coefficients,

namely, 1x− and 1x . Also, we have 1 1x x− = . Hence the spectral plot has only

two lines, namely at 10± kHz, and their values are xx

110

120 log 0= dB.

2.

Waveform 2



1.13

Waveform 2: Periodic square wave with T0

15

τ= ; T0 0.1= msec.

Comments on Spectral Plot 2: Values of various spectral components are:

i) fundamental: 0 dB

ii) second harmonic: ( )( )10

sin 0.420 log

sin 0.2cc

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

1020 log 0.809 0.924 dB= = −

iii) third harmonic: ( )( )10

sin 0.620 log

sin 0.2cc

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

100.50420 log0.935⎡ ⎤= ⎢ ⎥⎣ ⎦

1020 log (0.539) dB=

5.362= −

iv) fourth Harmonic: ( )( )10

sin 0.820 log

sin 0.2cc

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

12.04 dB= −

v) fifth harmonic: ( )( )10

sin 120 log

sin 0.2c

c⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

( )1020 log 0= = −∞

because of the limitations of the instrument, we see a small spike at 60− dB.

Similarly, the values of other components can be calculated.



1.14

3.

Waveform 3

Values of Spectral Components: Exercise



1.15

4.

Waveform 4

Values of Spectral Components: Exercise



1.16

The Example 1.1 and the periodic waveforms 1 to 4 all have fundamental

as part of their spectra. Based on this, we should not surmise that every periodic

signal must necessarily have a nonzero value for its fundamental. As a counter to

the conjuncture, let ( ) ( ) ( )px t t tcos 20 cos 2000= π π .

This is periodic with period 100 msec. However, the only spectral

components that have nonzero magnitudes are at 990 Hz and 1010 Hz. That is,

the first 99 spectral components (inclusive of DC) are zero!

Let ( )px t be a periodic voltage waveform across a 1 Ω resistor or a

current waveform flowing in a 1 Ω resistor. We now define its (normalized)

average power, denoted by pxP , as

( )px p

T

P x t d tT

0

2

0

1= ∫

Parseval’s (Power) Theorem relates pxP to nx as follows:

px nn

P x 2∞

= − ∞= ∑

(The proof of this relation is left as an exercise.)

If ( )px t consists of a single complex exponential, ie,



1.17

( ) ( )j n f tp nx t x e 02π +ϕ=

then, px nP x2=

In other words, Parseval’s power theorem implies that the total average power in

( )px t is superposition of the average powers of the complex exponentials

present in it.

When the periodic signal exhibits certain symmetries, Fourier coefficients

take special forms. Let us first define some of these symmetries (We assume

( )px t to be real).

Def. 1.2(a): A periodic signal ( )px t is even, if ( ) ( )− =p px t x t (1.6a)

Def. 1.2(b): A periodic signal ( )px t is odd, if ( ) ( )− = −p px t x t (1.6b)

Def. 1.2(c): A periodic signal ( )px t has half-wave symmetry, if

( )⎛ ⎞± = −⎜ ⎟⎝ ⎠

p pTx t x t02

(1.6c)

With respect to the symmetries defined by Eq. 1.6, we have the following

special forms for the coefficients nx :

( )px t even: nx ’s are purely real and even with respect to n

( )px t odd: nx ’s are purely imaginary and odd with respect to n

( )px t half-wave symmetric: nx ’s are zero for n even.

A proof of these properties is as follows:

i) ( )px t even:



1.18

( )T

j n f tn p

Tx x t e d t

T

00

0

2

2

2

0

1 − π

−

= ∫

( )T

j n f t j n f tp p

Tx t e d t x (t)e d t

T

00 0

0

2

2

02 2

0 0

1 − π − π

−

⎡ ⎤⎢ ⎥= +⎢ ⎥⎣ ⎦∫ ∫

Changing t to - t in the first integral, and noting ( ) ( )p px t x t− = ,

( ) ( )T T

j n f t j n f tp px t e d t x t e d t

T

0 00 0

2 22 2

0 0 0

1 π − π⎡ ⎤⎢ ⎥= +⎢ ⎥⎣ ⎦∫ ∫

( ) ( )T

px t n f t d tT

0 / 2

00 0

2 cos 2⎡ ⎤⎢ ⎥= π⎢ ⎥⎣ ⎦∫

The above integral is real and as ( )n f t n f t0 0cos 2 cos(2 )⎡ ⎤π − = π⎣ ⎦ ,

n nx x− = .

ii) ( )px t odd:

( ) ( )T

j n f t j n f tn p p

Tx x t e d t x t e d t

T

00 0

0

2

2

02 2

0 0

1 − π − π

−

⎡ ⎤⎢ ⎥= +⎢ ⎥⎣ ⎦∫ ∫

Changing t to - t , in the first integral and noting that ( ) ( )p px t x t− = − ,

we have

( ) ( )T T

j n f t j n f tp px t e d t x t e d t

T

0 00 0

2 22 2

0 0 0

1 −π π⎡ ⎤⎢ ⎥= − +⎢ ⎥⎣ ⎦∫ ∫

( )T

j n f t j n f tpx t e e d t

T

00 0

22 2

0 0

1 − π π⎡ ⎤= −⎣ ⎦∫

( ) ( )T

pj x t n f t d t

T

0 2

00 0

2 sin 2= − π∫

Hence the result.

iii) ( )px t has half-wave symmetry:



1.19

( ) ( )T

j n f t j n f tn p p

Tx x t e d t x t e d t

T

00 0

0

0

0

/ 22 2

0 / 2

1 −− π π

−

⎡ ⎤⎢ ⎥= +⎢ ⎥⎣ ⎦∫ ∫

In the first integral, replace t by 0t + T /2 .

( ) ( )0T T

j n t n j n tn p p0x x (t + T /2) e d t x t e d t

T0

0 0

0

/ 2 / 2

0 0

1 − + − ω⎡ ⎤⎢ ⎥= +⎢ ⎥⎣ ⎦∫ ∫ω π

The result follows from the relation

1,1,

j n n odde

n even− π −⎧

= ⎨⎩

1.2.3 Convergence of Fourier series and Gibbs phenomenon

As seen from Eq. 1.2, the representation of a periodic function in terms of

Fourier series involves, in general, an infinite summation. As such, the issue of

convergence of the series is to be given some consideration.

There is a set of conditions, known as Dirichlet conditions that guarantee

convergence. These are stated below.

i) ( )pT

x t0

< ∞∫

That is, the function is absolutely integrable over any period. It is easy to

verify that the above condition results in nx < ∞ for any n .

ii) ( )px t has only a finite number of maxima and minima over any period 0T .

iii) There are only finite number of finite discontinuities over any period1.

Let ( )M

j n f tM n

n Mx t x e 02π

= −

= ∑ (1.7)

1 For examples of periodic signals that do not satisfy one or more of the conditions i) to iii), the

reader is referred to [1, 2] listed at the end of this chapter.



1.20

and ( ) ( ) ( )M p Me t x t x t= −

then ( )MMx tlim

→ ∞ converges uniformly to ( )px t wherever ( )px t is continuous;

that is ( )MMe tlim 0

→ ∞= for all t.

Dirichlet conditions are sufficient but not necessary. Later on, we shall

have examples of Fourier series for functions that voilate some of the Dirichet

conditions.

If ( )px t is not absolutely integrable but square integrable, that is,

( )pT

x t dt0

2< ∞∫ , then the series converges in the mean. That is,

( )MMT

e t d t0

2lim 0→ ∞

=∫ (1.8)

Note that Eq. 1.8 does not imply that ( )MMe tlim

→∞ is zero. There could be

nonzero values in ( )MMe tlim

→∞; but they occur at isolated points, resulting in the

integral of Eq. 1.8, being equal to zero.

The limiting behavior of ( )Mx t at the points of discontinuity in ( )px t is

somewhat interesting, regardless of ( )px t being absolutely integrable or square

integrable. This is illustrated in Fig. 1.5(a). From the figure, we see that



1.21

Fig. 1.5(a): Convergence behavior of ( )Mx t at a discontinuity in ( )px t

( )Mx t passes through the mid-point of the discontinuity and has a peak

overshoot (as well as undershoot) of amplitude 0.09A (We assume M to be

sufficiently large). The period of oscillations (whose amplitudes keep decreasing

with increasing t , 0t > ) is 0

2TM

. These oscillations (with the peak overshoot as

well as the undershoot of amplitude 0. 09 A) persist even as M → ∞ . In the

limiting case, all the oscillations converge in location to the point t t1= (the point

of discontinuity) resulting in what is called as Gibbs ears as shown in Fig. 1.5(b).

Fig. 1.5(b): Gibbs ears at t = t1



1.22

(In 1898, Albert Michelson, a well-known name in the field of optics,

developed an instrument called Harmonic Analyzer (HA), which was capable of

computing the first 80 coefficients of the Fourier series. HA could also be used a

signal synthesizer. In other words, it has the ability to self-check its calculations

by synthesizing the signal using the computed coefficients. When Michelson tried

this instrument on signals with discontinuities (with continuous signals, close

agreement was found between the original signal and the synthesized signal), he

observed a strange behavior: synthesized signal, based on the 80 coefficients,

exhibited ringing with an overshoot of about 9% of the discontinuity, in the vicinity

of the discontinuity. This behavior persisted even after increasing the number of

terms beyond 80. J. W. Gibbs, professor at Yale, investigated and clarified the

above behavior by taking the saw-tooth wave as an example; hence the name

Gibbs Phenomenon.)

The convergence of the Fourier series and the corresponding

Gibbs oscillations can be seen from the animation that follows. You have been

provided with three options with respect to the number of harmonics M to be

summed. These are: M 10, 25 and 75= .

1.3 Aperiodic Signals and Fourier Transform Aperiodic (also called nonperiodic) signals can be of finite or infinite

duration. A few of the aperiodic signals occur quote often in theoretical studies.

Hence, it behooves us to introduce some notation to describe their behavior.

i) Rectangular pulse, tgaT⎛ ⎞⎜ ⎟⎝ ⎠

1,2

0,2

TttgaT Tt

⎧ <⎪⎪⎛ ⎞ = ⎨⎜ ⎟⎝ ⎠ ⎪ >

⎪⎩

(1.9a)



1.23

[Rectangular pulse is sometimes referred to as a gate pulse; hence the

symbol ( )ga ]. In view of the above tA gaT⎛ ⎞⎜ ⎟⎝ ⎠

refers to a rectangular pulse

of amplitude A and duration T , centered at 0t = .

ii) Triangular pulse, ttriT⎛ ⎞⎜ ⎟⎝ ⎠

tt t Ttri TT

outside

1 ,

0 ,

⎧− ≤⎪⎛ ⎞ = ⎨⎜ ⎟

⎝ ⎠ ⎪⎩

(1.9b)

iii) One-sided (decaying) exponential pulse, 1 texT⎛ ⎞⎜ ⎟⎝ ⎠

:

tTe t

tex tT

t

, 011 , 020 , 0

−⎧>⎪

⎪⎪⎛ ⎞ = =⎨⎜ ⎟⎝ ⎠ ⎪

<⎪⎪⎩

(1.9c)

iv) Two-sided (symmetrical) exponential pulse, 2 texT⎛ ⎞⎜ ⎟⎝ ⎠

:

tT

tT

e ttex tT

e t

, 02 1 , 0

, 0

−⎧>⎪

⎪⎛ ⎞ = =⎨⎜ ⎟⎝ ⎠ ⎪

⎪ <⎩

(1.9d)

Fig. 1.6 illustrates specific examples of these pulses.



1.24

Fig. 1.6: Examples of pulses defined by Eq. 1.9

Let ( )x t be any aperiodic signal. We define its normalized energy xE , as

( )xE x t dt2

∞

−∞

= ∫ (1.10)

An aperiodic signal with xE0 < < ∞ is said to be an energy signal.

(When no specific signal is being referred to, we use the symbol E without any

subscript to denote the energy quantity.)

1.3.1 Fourier transform Like periodic signals, aperiodic signals also can be represented in the

frequency domain. However, unlike the discrete spectrum of the periodic case,

we have a continuous spectrum for the aperiodic case; that is, the frequency

components constituting a given signal ( )x t lie in a continuous range (or

ranges), and quite often this range could be ( ),−∞ ∞ . Eq. 1.2(a) expresses ( )px t

as a sum over a discrete set of frequencies. Its counterpart for the aperiodic case

is an integral relationship given by



1.25

( ) ( ) j f tx t X f e d f2∞

π

−∞

= ∫ (1.11a)

where ( )X f is the Fourier transform of ( )x t .

Eq. 1.11(a) is given the following interpretation. Let the integral be treated

as a sum over incremental frequency ranges of width f∆ . Let ( )iX f f∆ be the

incremental complex amplitude of 2 ij f te π at the frequency if f= . If we sum a

large number of such complex exponentials, the resulting signal should be a very

good approximation to ( )x t . This argument, carried to its natural conclusion,

leads to signal representation with a sum of complex exponentials replaced by an

integral, where a continuous range of frequencies, with the appropriate complex

amplitude distribution will synthesize the given signal ( )x t .

Eq. 1.11(a) is called the synthesis relation or Inverse Fourier Transform

(IFT) relation. Quite often, we know ( )x t and would want ( )X f . The companion

relation to Eq. 1.11(a) is

( ) ( ) 2j f tX f x t e dt−∞

− π

∞

= ∫ (1.11b)

Eq. 1.11(b) is referred to as the Fourier Transform (FT) relation or, the

analysis equation, or forward transform relation. We use the notation

( ) ( )X f F x t⎡ ⎤= ⎣ ⎦ (1.12a)

( ) ( )1x t F X f− ⎡ ⎤= ⎣ ⎦ (1.12b)

Eq. 1.12(a) and Eq. 1.12(b) are combined into the abbreviated notation, namely,

( ) ( )x t X f←⎯→ . (1.12c)

( )X f is, in general, a complex quantity. That is,

( ) ( ) ( )R IX f X f j X f= +



1.26

( ) ( )j fX f e θ=

( ) ( )RX f Real part of X f= ,

( ) ( )IX f Imaginary part of X f=

( ) ( )X f magnitude of X f=

( ) ( )2 2R IX f X f= +

( ) ( ) ( )( )

arg tan I

R

X ff X f arc

X f⎡ ⎤

⎡ ⎤θ = = ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

Information in ( )X f is usually displayed by means of two plots: (a)

( ) .X f vs f , known as magnitude spectrum and (b) ( ) .f vs fθ , known as the

phase spectrum.

Example 1.2

Let ( ) tx t A gaT⎛ ⎞= ⎜ ⎟⎝ ⎠

. Let us compute and sketch ( )X f .

( )2

2

2

sin ( )T

j ft

TX f A e dt AT c f T− π

−

= =∫ ,

where

( ) ( )csin

sinπλ

λ =πλ

Appendix A1.1 contains the tabulated values of ( )csin λ . Its behavior is shown in

Fig. A1.1. Note that ( )c1 , 0

sin0 , 1, 2 etc.

λ =⎧λ = ⎨ λ = ± ±⎩

Fig. 1.7(a) sketches the magnitude spectrum of the rectangular pulse for 1AT = .



1.27

Fig. 1.7: Spectrum of the rectangular pulse

Regarding the phase plot, ( )sinc f T is real. However it could be bipolar.

During the interval, 1m mfT T

+< < , with m odd, ( )sinc f T is negative. As the

magnitude spectrum is always positive, negative values of ( )sinc f T are taken

care of by making ( ) 0180fθ = ± for the appropriate ranges, as shown in Fig.

1.7(b).

Remarkable balancing act: A serious look at the magnitude and phase plots

reveals a very charming result. From the magnitude spectrum, we find that a

rectangular pulse is composed of frequency components in the range

f−∞ < < ∞ , each with its own amplitude and phase. Each of these complex

exponentials exist for all t . But when we synthesize a signal using the complex

exponentials with the magnitudes and phases as given in Fig. 1.7, they add up to



1.28

a constant for 2Tt < and then go to zero forever. A very fascinating result

indeed!

Fig. 1.7(a) illustrates another interesting result. From the figure, we see

that most of the energy, (that is, the range of strong spectral components) of the

signal lies in the interval 1fT

< , where T is the duration of the rectangular pulse.

Hence, if T is reduced, then its spectral width increases and vice versa (As we

shall see later, this is true of other pulse types, other than the rectangular). That

is, more compact is the signal in the time-domain, the more wide-spread it would

be in the frequency domain and vice versa. This is called the phenomenon of

reciprocal spreading.

Example 1.3

Let ( ) ( )1x t ex t= . Let us find ( )X f and sketch it.

( ) 2

0

11 2

t j f tX f e e dtj f

∞− − π= =

+ π∫

Hence, ( )( )21

1 2X f

f=

+ π

( ) ( )tan 2f arc fθ = − π

A plot of the magnitude and the phase spectrum are given in Fig. 1.8.



1.29

Fig. 1.8: (a) Magnitude spectrum of the decaying exponential

(b) Phase spectrum

1.3.2 Dirichlet conditions Given ( )X f , Eq. 1.11(a) enables us to synthesize the signal ( )x t . Now

the question is: Is the synthesized signal, say ( ) ( )sx t , identical to ( )x t ? This

leads to the topic of convergence of the Fourier Integral. Analogous to the

Dirichlet conditions for the Fourier series, we have a set of sufficient conditions,

(also called Dirichlet conditions) for the existence of Fourier transform, which are

stated below:



1.30

(i) ( )x t be absolutely integrable; that is,

( )x t dt∞

−∞

< ∞∫

This ensures that ( )X f is finite for all f , because

( ) ( ) 2j f tX f x t e dt∞

− π

−∞

= ∫

( ) ( ) 2j f tX f x t e dt∞

− π

−∞

= ∫

( ) ( )2j f tx t e dt x t dt∞ ∞

− π

−∞ −∞

≤ = < ∞∫ ∫

(ii) ( )x t is single valued and has only finite number of maxima and minima

with in any finite interval.

(iii) ( )x t has a finite number of finite discontinuities with in any finite interval.

If ( ) ( ) ( ) 2limW

s j f t

WW

x t X f e dfπ

→∞−

= ∫ , then ( ) ( )sx t converges to ( )x t

uniformly wherever ( )x t is continuous.

If ( )x t is not absolutely integrable but square integrable, that is,

( ) 2x t dt

∞

−∞

< ∞∫ (Finite energy signal), then we have the convergence in the

mean, namely

( ) ( ) ( )2

0sx t x t dt∞

−∞

− =∫



1.31

Regardless of whether ( )x t is absolutely integrable or square integrable,

( ) ( )sx t exhibits Gibbs phenomenon at the points of discontinuity in ( )x t , always

passing through the midpoints of the discontinuities.

1.4 Properties of the Fourier Transform Fourier Transform has a large number of properties, which are developed

in the sequel. A thorough understanding of these properties, and the ability to

make use of them appropriately, helps a great deal in the analysis of various

signals and systems.

P1) Linearity

Let ( ) ( )1 1x t X f←⎯→ and ( ) ( )2 2x t X f←⎯→ .

Then, for all constants 1a and 2a , we have

( ) ( ) ( ) ( )1 1 2 2 1 1 2 2a x t a x t a X f a X f+ ←⎯→ +

It is very easy to see the validity of the above transform relationship. This

property will be used quite often in the development of this course material.

P2a) Area under ( )x t

If ( ) ( )x t X f←⎯→ , then

( ) ( )0x t dt X∞

−∞

=∫

The above property follows quite simply by setting 0f = in Eq. 1.11(b). As an

example of this property, we have the transform pair

( )sintga T c fTT⎛ ⎞ ←⎯→⎜ ⎟⎝ ⎠

By inspection, area of the time function is T , which is equal to ( ) 0sin | fT c f T = .

P2b) Area under ( )X f



1.32

If ( ) ( )x t X f←⎯→ , then ( ) ( )x X f df0∞

−∞

= ∫

The proof follows by making 0t = in Eq. 1.11(a).

As an illustration of this property, we have

( ) ( ) ( ) 111 2

x t ex t X fj f

= ←⎯→ =+ π

Hence ( )( )2

1 1 21 2 1 2

j fX f df df dfj f f

∞ ∞ ∞

−∞ −∞ −∞

− π= =

+ π + π∫ ∫ ∫

Noting that ( )22 0

1 2f dff

∞

−∞

π=

+ π∫ , we have

( )( )21 1

21 2X f df df

f

∞ ∞

−∞ −∞

= =+ π

∫ ∫ , which is the value of ( ) 01 |tex t =

P3) Time Scaling

If ( ) ( )x t X f←⎯→ , then ( ) 1 fx t X ⎛ ⎞α ←⎯→ ⎜ ⎟α α⎝ ⎠, where α is a real

constant.

Proof: Exercise

The value of α decides the behavior of ( )x tα . If 1α = − , ( )x tα is a time

reversed version of ( )x t . If 1α > , ( )x tα is a time compressed version of ( )x t ,

where as if 0 1< α < , we have a time expanded version of ( )x t . Let ( )x t be as

shown in Fig. 1.9(a). Then ( )2x t− would be as shown in Fig. 1.9(b).



1.33

Fig. 1.9: A triangular pulse and its time compressed and reversed version

For the special case 1α = − , we have the transform pair

( ) ( )x t X f− ←⎯→ − . That is, both the time function and its Fourier transform

undergo reversal. As an example, we know that

( ) 111 2

ex tj f

←⎯→+ π

Hence ( ) 111 2

ex tj f

− ←⎯→− π

( ) ( ) ( )1 1 2ex t ex t ex t+ − =

Using the linearity property of the Fourier transform, we obtain the

transform pair

( ) ( ) 1 12 exp1 2 1 2

ex t tj f j f

= − ←⎯→ ++ π − π

( )22

1 2 f=

+ π

Consider ( )x tα with 2α = . Then,

( ) 122 2

fx t X ⎛ ⎞←⎯→ ⎜ ⎟⎝ ⎠



1.34

Let us compare ( )X f with 12 2

fX ⎛ ⎞⎜ ⎟⎝ ⎠

by taking ( ) ( )1x t ex t= , and

( ) ( )1 2y t ex t= . That is,

( )

2 , 01 , 020 , 0

te t

y t t

t

−⎧ >⎪⎪= =⎨⎪

<⎪⎩

( ) ( )1 12 1 2 2

Y fj f

⎡ ⎤= ⎢ ⎥

+ π⎢ ⎥⎣ ⎦

12 2j f

=+ π

Let ( ) ( ) ( )x fX f X f eθ= and

( ) ( ) ( )y fY f Y f eθ= , where ( )2 2

1

2 1Y f

f=

+ π

( ) ( )tany f arc fθ = − π

Fig. 1.10 gives the plots of ( )X f and ( )Y f . In Fig. 1.10(a), we have the

plots ( ) .X f vs f and ( ) .Y f vs f . In Fig. 1.10(b), we have the plot of ( )Y f

normalized to have the maximum value of unity. This plot is denoted by ( )NY f .

Fig. 1.10(c) gives the plots of ( )x fθ and ( )y fθ . From Fig. 1.10(b), we see that

i) ( ) ( )2y t x t= has a much wider spectral width as compared to the spectrum

of the original signal. (In fact, if ( )X f is band limited to W Hz, then 2fX ⎛ ⎞

⎜ ⎟⎝ ⎠

will be band limited to W2 Hz.)



1.35

Fig 1.10: Spectral plots of ( )1ex t and ( )1 2ex t



1.36

(ii) Let ( ) ( ) ( )E f X f Y f= − . Then the value of ( )E f is dependent on f ; that

is, the original spectral magnitudes are modified by different amounts at

different frequencies (Note that ( ) ( )Y f k X f≠ where k is a constant).

In other words, ( )Y f is a distorted version of ( )X f .

(iii) From Fig. 1.10(c), we observe that ( ) ( )y xf fθ ≠ θ and their difference is a

function of frequency; that is ( )y fθ is a distorted version of ( )x fθ .

In summary, time compression would result either in the introduction of

new, higher frequency components (if the original signal is band limited) or

making the latter part of the original spectrum much more significant; the

remaining spectral components are distorted (both in amplitude and phase). On

the other hand, time expansion would result either in the loss or attenuation of

higher frequency components, and distortion of the remaining spectrum.

Let ( )x t represent an audio signal band limited to 10 kHz. Then ( )2x t will

have a spectral components upto 20 kHz. These higher frequency components

will impart shrillness to the audio, besides distorting the original signal. Similarly,

if the audio is compressed, we have loss of “sharpness” in the resulting signal

and severe distortion. This property of the FT will now be demonstrated with the

help of a recorded audio signal.

P4a) Time shift

If ( ) ( )x t X f←⎯→ then,

( ) ( )020

j f tx t t e X f− π− ←⎯→

If 0t is positive, then ( )0x t t− is a delayed version of ( )x t and if 0t is

negative, the ( )0x t t− is an advanced version of ( )x t . In any case, time shifting



1.37

will simply result in the multiplication of ( )X f by a linear phase factor. This

implies that ( )x t and ( )0x t t− have the same magnitude spectrum.

Proof: Let ( )0t tλ = − . Then,

( ) ( ) ( )020

j f tF x t t x e d∞

− π λ +

−∞

⎡ ⎤− = λ λ⎣ ⎦ ∫

( )02 2j f t j fe x e d∞

− π − π λ

−∞

= λ λ∫

( )02j f te X f− π=

P4b) Frequency Shift

If ( ) ( )x t X f←⎯→ , then

( ) ( )2 cj f tce x t X f f± π ←⎯→ ∓

where cf is a real constant. (This property is also known as modulation theorem).

Proof: Exercise As an application of the above result, let us consider the spectrum of

( ) ( ) ( )2 cos 2 cy t f t x t= π ; that is, we want the Fourier transform of

( )2 2c cj f t j f te e x tπ − π⎡ ⎤+⎣ ⎦ . From the frequency shift theorem, we have

( ) ( ) ( ) ( ) ( ) ( )2 cos 2 c c cy t f t x t Y f X f f X f f= π ←⎯→ = − + + . If ( )X f is as shown

in Fig. 1.11(a), then ( )Y f will be as shown in Fig. 1.11(b) for cf W= .

Fig.1.11: Illustration of modulation theorem



1.38

P5) Duality If we look fairly closely at the two equations constituting the Fourier

transform pair, we find that there is a great deal of similarity between them. In

Eq.1.11a, ' 'f is the variable of integration where as in Eq. 1.11b, it is the variable

' 't . The sign of the exponent is positive in Eq. 1.11a where as it is negative in

Eq. 1.11b. Both t and f are variables of the continuous type. This results in an

interesting property, namely, the duality property, which is stated below.

If ( ) ( )x t X f←⎯→ , then

( ) ( )X t x f←⎯→ − and ( ) ( )X t x f− ←⎯→

Note: This is one instance, where the variable t is associated with a function

denoted using the upper case letter and the variable f is associated with a

function denoted using a lower case letter.

Proof: ( ) ( ) 2j f tx t X f e df∞

π

−∞

= ∫

( ) ( ) 2j f tx t X f e df∞

− π

−∞

− = ∫

The result follows by interchanging the variables t and f . The proof of the

second part of the property is similar.

Duality theorem helps us in creating additional transform pairs, from the

given set. We shall illustrate the duality property with the help of a few examples.

Example 1.4

If ( ) sin 2z t A c W t= , let us use duality to find ( )Z f .



1.39

We look for a transform pair, ( ) ( )x t X f←⎯→ where in ( )X f , if we

replace f by t , we have ( ) sin 2z t A c W t= .

We know that if ( )2 2A tx t gaW W

⎛ ⎞= ⎜ ⎟⎝ ⎠

, then,

( ) ( )sin 2X f A c W f= and

( ) ( ) ( )sin 2X t A c W t z t= = ⇒

( ) ( )2 2A fZ f x f gaW W

⎛ ⎞= − = −⎜ ⎟

⎝ ⎠

As ( ) ( )ga f ga f− = , we have

( )2 2A fZ f gaW W

⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

.

Example 1.5

Find the Fourier transform of ( ) 22

1z t

t=

+.

We know that if ( ) ty t e−= , then ( )( )2

21 2

Y ff

=+ π

Let ( ) ( )x t y t= α , with 2α = π .

Then ( ) 12 2

fX f Y ⎛ ⎞= ⎜ ⎟π π⎝ ⎠

21 2

2 1 f=

π +

or ( ) 222

1X f

fπ =

+

As ( ) 22

1z t

f=

+ with f being replaced by t , we have

( ) ( )2Z f x f= π −

22 fe− π= π



1.40

Hence 22

2 21

fet

− π←⎯→ π+

P6) Conjugate functions

If ( ) ( )x t X f←⎯→ , then ( ) ( )x t X f∗ ∗←⎯→ −

Proof: ( ) ( ) 2j f tX f x t e dt∞

− π

−∞

= ∫

( ) ( ) 2j f tX f x t e dt∞

∗ ∗ π

−∞

= ∫

( ) ( ) 2j f tX f x t e dt∞

∗ ∗ − π

−∞

− = ∫

Hence the result.

From the time reversal property, we get the additional relation, namely

( ) ( )x t X f∗ ∗− ←⎯→

Def. 1.3(a): A signal ( )x t is called conjugate symmetric, if ( ) ( )x t x t∗− = .

If ( )x t is real, then ( )x t is even if ( ) ( )x t x t− = .

Exercise 1.2: Let ( ) ( ) ( )R Ix t x t j x t= +

where ( )Rx t is the real part and ( )Ix t is the imaginary part of ( )x t . Show

that

( ) ( ) ( )12Rx t X f X f∗⎡ ⎤←⎯→ + −⎣ ⎦

( ) ( ) ( )12Ix t X f X f

j∗⎡ ⎤←⎯→ − −⎣ ⎦



1.41

Def. 1.3(b): A signal ( )x t is said to be conjugate anti-symmetric if

( ) ( )x t x t∗− = − .

If ( )x t is real, then ( )x t is odd if ( ) ( )x t x t− = − .

If ( )x t is real, then ( ) ( )x t x t∗= .

As a result, ( ) ( )X f X f∗= − or ( ) ( )X f X f∗ = − .

Hence, the spectrum for the negative frequencies is the complex conjugate of the

positive part of the spectrum. This implies, that for real signals,

( ) ( )X f X f− =

( ) ( )f fθ − = − θ

Going one step ahead, we can show that if ( )x t is real and even, then

( )X f is also real and even. (Example: ( )22

1 2te

f− ←⎯→

+ π). Similarly, if ( )x t

is real and odd, its transform is purely imaginary and odd (See Example 1.7).

P7a) Multiplication in the time domain

If ( ) ( )1 1x t X f←⎯→

( ) ( )2 2x t X f←⎯→

then, ( ) ( ) ( ) ( ) ( ) ( )1 2 1 2 2 1x t x t X X f d X X f d∞ ∞

−∞ −∞

←⎯→ λ −λ λ = λ − λ λ∫ ∫

The integrals on the R.H.S represent the convolution of ( )1X f and ( )2X f . We

denote the convolution of ( )1X f and ( )2X f by ( ) ( )1 2X f X f∗ .

(Note that ∗ in between two functions represents the convolution of the two

quantities where as a superscript, it denotes the complex conjugate)



1.42

Proof: Exercise

P7b) Multiplication of Fourier transforms

Let ( ) ( )1 1x t X f←⎯→

( ) ( )2 2x t X f←⎯→

then, ( ) ( ) ( ) ( )11 2 1 2F X f X f x x t d

∞−

−∞

⎡ ⎤ = λ −λ λ⎣ ⎦ ∫

( ) ( )2 1x x t d∞

−∞

= λ −λ λ∫

As any one of the above integrals represent the convolution of ( )1x t and ( )2x t ,

we have

( ) ( ) ( ) ( )1 2 1 2x t x t X f X f∗ ←⎯→

Proof: Let ( ) ( ) ( )3 1 2x t x t x t= ∗

That is, ( ) ( ) ( )3 1 2x t x x t d∞

−∞

= λ −λ λ∫

( ) ( ) ( ) ( ) ( )j f t j f tX f F x t x t e dt x x t d e dt2 23 3 3 1 2

∞ ∞ ∞− π − π

−∞ −∞ −∞

⎡ ⎤⎡ ⎤ ⎢ ⎥= = = λ − λ λ⎣ ⎦ ⎢ ⎥⎣ ⎦

∫ ∫ ∫

Rearranging the integrals,

( ) ( ) ( ) j f tX f x x t e dt d23 1 2

∞ ∞− π

−∞ −∞

⎡ ⎤⎢ ⎥= λ − λ λ⎢ ⎥⎣ ⎦

∫ ∫

But the bracketed quantity is the Fourier transform of ( )2x t −λ . From the

property P4(a), we have

( ) ( )j fx t e X f22 2

− π λ− λ ←⎯→

Hence,



1.43

( ) ( ) ( ) 23 1 2

j fX f x X f e d∞

− π λ

−∞

= λ λ∫

( ) ( ) 22 1

j fX f x e d∞

− π λ

−∞

= λ λ∫

( ) ( )1 2X f X f=

Property P7(b), known as the Convolution theorem, is one of the very useful

properties of the Fourier transform.

The concept of convolution is very basic in the theory of signals and

systems. As will be shown later, the input and output of a linear, time- invariant

system are related by the convolution integral. For a fairly detailed treatment of

the properties of systems, convolution integral etc. the student is advised to refer

to [1 - 3].

Example 1.6

In this example, we will find the Fourier transform of tT triT⎛ ⎞⎜ ⎟⎝ ⎠

.

tT triT⎛ ⎞⎜ ⎟⎝ ⎠

can be obtained as the convolution of tgaT⎛ ⎞⎜ ⎟⎝ ⎠

with itself. That is,

t t tga ga T triT T T⎛ ⎞ ⎛ ⎞ ⎛ ⎞∗ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

As ( )sintga T c f TT⎛ ⎞ ←⎯→⎜ ⎟⎝ ⎠

, we have

( ) 2sintT tri T c f T

T⎛ ⎞ ⎡ ⎤←⎯→⎜ ⎟ ⎣ ⎦⎝ ⎠



1.44

P8) Differentiation P8a) Differentiation in the time domain

Let ( ) ( )x t X f←⎯→ ,

then, ( ) ( )2d x t j f X fdt

⎡ ⎤ ⎡ ⎤←⎯→ π⎣ ⎦ ⎣ ⎦

Generalizing, ( ) ( ) ( )2n

nn

d x tj f X f

dt←⎯→ π

Proof: We shall prove the first part; generalization follows as a consequence

this. We have,

( ) ( ) 2j f tx t X f e df∞

π

−∞

= ∫

( ) ( ) 2j f td dx t X f e dfdt dt

∞π

−∞

⎡ ⎤⎡ ⎤ ⎢ ⎥=⎣ ⎦ ⎢ ⎥⎣ ⎦

∫

Interchanging the order of differentiation and integration on the RHS,

( ) ( ) 2j f td dx t X f e dfdt dt

∞π

−∞

⎡ ⎤⎡ ⎤ =⎣ ⎦ ⎢ ⎥⎣ ⎦∫

( ) 22 j f tj f X f e df∞

π

−∞

⎡ ⎤= π⎣ ⎦∫

From the above, we see that ( )1 2F j f X f− ⎡ ⎤π⎣ ⎦ is ( )d x tdt

. Hence the property. Example 1.7

Let us find the FT of the doublet pulse ( )x t shown in Fig. 1.12 below.



1.45

Fig 1.12: ( )x t of Example 1.7

( ) d tx t T tridt T

⎡ ⎤⎛ ⎞= ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦. Hence,

( ) 2 tX f j f F T triT

⎡ ⎤⎛ ⎞= π ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

( ) ( )2 22 sinj f T c f T= π , (using the result of example 1.6)

( ) ( )( )( )

22 sin

2fT

j f TfT fT

π= π

π π

( )( ) ( )sin

2 sinfT

j T fTfTπ

= ππ

( ) ( )2 sin sinj T c fT fT= π π

As a consequence of property P8(a), we have the following interesting result.

Let ( ) ( ) ( )3 1 2x t x t x t= ∗

Then ( ) ( ) ( )3 1 2X f X f X f=

( ) ( ) ( ) ( ) ( )3 1 22 2j f X f j f X f X f⎡ ⎤π = π⎣ ⎦

( ) ( )1 22X f j f X f⎡ ⎤= π⎣ ⎦

That is,



1.46

( ) ( ) ( ) ( )1 2 1 2d dx t x t x t x tdt dt

⎡ ⎤ ⎡ ⎤∗ = ∗⎣ ⎦ ⎣ ⎦

( ) ( )1 2dx t x tdt

⎡ ⎤= ∗ ⎣ ⎦

P8b) Differentiation in the frequency domain

Let ( ) ( )x t X f←⎯→ .

Then, ( ) ( ) ( )2d X f

j t x tdf

⎡ ⎤− π ←⎯→⎣ ⎦

Generalizing, ( ) ( ) ( )2n

nn

d X fj t x t

df⎡ ⎤− π ←⎯→⎣ ⎦

Proof: Exercise

The generalized property can also be written as

( ) ( )2

n nn

n

d X fjt x tdf

⎛ ⎞←⎯→ ⎜ ⎟π⎝ ⎠

Example 1.8

Find the Fourier transform of ( ) 1 tx t t exT⎛ ⎞= ⎜ ⎟⎝ ⎠

.

We have, ( ) 111 2

ex tj f

←⎯→+ π

Hence, 111 2

tex TT j fT⎛ ⎞ ←⎯→⎜ ⎟ + π⎝ ⎠

12 1 2

t j d Tt exT df j fT

⎡ ⎤⎛ ⎞ ←⎯→⎜ ⎟ ⎢ ⎥π + π⎝ ⎠ ⎣ ⎦

( )( )2

22 1 2

j Tj Tj fT

− π←⎯→

π + π



1.47

( )

2

21 2Tj f T

⎡ ⎤⎢ ⎥←⎯→⎢ ⎥+ π⎣ ⎦

P9) Integration in time domain This property will be developed subsequently.

P10) Rayleigh’s energy theorem

This theorem states that, xE , energy of the signal ( )x t , is

( )xE X f df2

∞

−∞

= ∫

This result follows from the more general relationship, namely,

( ) ( ) ( ) ( )1 2 1 2x t x t dt X f X f df∞ ∞

∗ ∗

−∞ −∞

=∫ ∫

Proof: We have

( ) ( ) ( ) ( ) 21 2 1 2

j f tx t x t dt x t X f e df dt∗

∞ ∞ ∞∗ π

−∞ −∞ −∞

⎡ ⎤⎢ ⎥=⎢ ⎥⎣ ⎦

∫ ∫ ∫

( ) ( ) 22 1

j f tX f x t e dt df∞ ∞

∗ − π

−∞ −∞

⎡ ⎤⎢ ⎥=⎢ ⎥⎣ ⎦

∫ ∫

( ) ( )2 1X f X f df∞

∗

−∞

= ∫

If ( ) ( ) ( )1 2x t x t x t= = , then

( ) ( )xx t dt E X f df2 2∞ ∞

−∞ −∞

= =∫ ∫

Note: If ( )1x t and ( )2x t are real, then,



1.48

( ) ( ) ( ) ( ) ( ) ( )1 2 1 2 2 1x t x t d t X f X f d f X f X f d f∞ ∞ ∞

− ∞ − ∞ − ∞

= − = −∫ ∫ ∫

Property P10 enables us to compute the energy of a signal from its

magnitude spectrum. In a few situations, this may be easier than computing the

energy in the time domain. (Some authors refer to this result as Parseval’s

theorem)

Example 1.9

Let us find the energy of the signal ( ) ( )2 sin 2x t AW c W t= .

( )xE AW c W t d t2

2 sin 2∞

−∞

⎡ ⎤= ⎣ ⎦∫

In this case, it would be easier to compute xE based on ( )X f . From Example

1.4, ( )2

fX f A gaW

⎛ ⎞= ⎜ ⎟

⎝ ⎠. Hence,

W

xW

E A df W A2 22−

= =∫

More important than the calculation of the energy of the signal, Rayleigh’s

energy theorem enables to treat ( ) 2X f as the energy spectral density of ( )x t .

That is, ( ) 21X f df is the energy in the incremental frequency interval d f ,

centered at 1f f= . Let ( )W

xW

X f df E2

0.9−

=∫ . Then, 90 percent of the energy of

signal is confined to the interval f W≤ . Consider the rectangular pulse tgaT⎛ ⎞⎜ ⎟⎝ ⎠

.

The first nulls of the magnitude spectrum occur at 1fT

= ± . The evaluation of



1.49

( ) ( )1 1

2 2

1 1sin

T T

T T

X f df T c f T df− −

=∫ ∫ will yield the value 0.92 T , which is 92

percent of the total energy of tgaT⎛ ⎞⎜ ⎟⎝ ⎠

. Hence, the frequency range 1 1,T T

⎛ ⎞−⎜ ⎟⎝ ⎠

can

be taken to be the spectral width of the rectangular pulse. [The interval 2 2,T T

⎛ ⎞−⎜ ⎟⎝ ⎠

may result in about 95 percent of the total energy].

1. 5 Unified Approach to Fourier Transform So far, we have represented the periodic functions by Fourier series and

the aperiodic functions by Fourier transform. The question arises: is it possible to

unify these two approaches and talk only in terms of say, Fourier transform? The

answer is yes provided we are willing to introduce Impulse Functions both in

time and frequency domains. This would also enable us to have Fourier

transforms for signals that do not satisfy one or more of the Dirichlet’s conditions

(for the existence of the Fourier transform).

1.5.1 Unit impulse (Dirac delta function) Impulse function is not a function in its strict sense [Note that a function

( )f , takes a number y and a produces another number, ( )f y ]. It is a

distribution or generalized function. A distribution is defined in terms of its effect

on another function. The symbol ( )tδ is fairly common in the technical literature

to denote the unit impulse. We define the unit impulse as any (generalized)

function that satisfies the following conditions:

(i) ( ) 0, 0t tδ = ≠ (1.13a)

(ii) ( )t t, 0δ = ∞ = (1.13b)

(iii) Let ( )p t be any ordinary function, then



1.50

( ) ( ) ( ) ( ) ( )0 , 0p t t dt p t t dt p∞ ε

−∞ −ε

δ = δ = ε >∫ ∫ (1.13c)

( ε could be infinitesimally small)

If ( )p t t1, for= ≤ ε , then we have

( ) ( )t d t t d t 1ε ∞

−ε −∞

δ = δ =∫ ∫ (1.13d)

From Eq. 1.13(c), we see that ( )tδ operates on a function such as ( )p t

and produces the number, namely, ( )0p . As such ( )tδ falls between a function

and a transform (A transform operates on a function and produces a function).

A number of conventional functions have a limiting behavior that

approaches ( )tδ . We cite a few such functions below:

Let

(a) ( )11 tp t ga ⎛ ⎞= ⎜ ⎟ε ε⎝ ⎠

(b) ( )21 tp t tri ⎛ ⎞= ⎜ ⎟ε ε⎝ ⎠

(c) ( )31 sin tp t c ⎛ ⎞= ⎜ ⎟ε ε⎝ ⎠

Then, ( ) ( )0

lim ip t tε→

= δ , 1, 2, 3i = . ( )3p t is shown below in Fig. 1.13.



1.51

Fig. 1.13: ( )sinc with the limiting behavior of ( )tδ

( )tc ga f1Note that sin . Hence the area under the time function 1.⎛ ⎞⎛ ⎞ ←⎯→ ε =⎜ ⎟⎜ ⎟ε ε⎝ ⎠⎝ ⎠

From the above examples, we see that the shape of the function is not

very critical; its area should remain at 1 in order to approach ( )tδ in the limit.

By delaying ( )tδ by 0t and scaling it by A , we have ( )0A t tδ − . This is

normally shown as a spear (Fig. 1.14) with the weight or area of the impulse

shown in parentheses very close to it.



1.52

Fig. 1.14: Symbol for ( )0A t tδ −

Some properties of unit impulse P1) Sampling (or sifting) property

Let ( )p t be any ordinary function. Then for 0a t b< < ,

( ) ( ) ( )0 0

b

a

p t t t dt p tδ − =∫

(This is generalization of condition (iii)). Proof follows from making the

change of variable 0t t− = τ and noting ( )δ τ is zero for 0τ ≠ . Note that

for the sampling property, the values of ( )p t , 0t t≠ are of no

consequence.

P2) Replication property

Let ( )p t be any ordinary function. Then,

( ) ( ) ( )0 0p t t t p t t∗ δ − = −

The proof of this property follows from the fact, that in the process of

convolution, every value of ( )p t will be sampled and shifted by 0t

resulting in ( )0p t t− .

(Note: Some authors use this property as the operational definition of

impulse function.)



1.53

P3) Scaling Property

( ) ( )1 , 0t tδ α = δ α ≠α

Proof: (i) Let , 0tα = τ α >

( ) ( )1 1t dt d∞ ∞

−∞ −∞

δ α = δ τ τ =α α∫ ∫

( )1 t dt∞

−∞

= δα ∫

( )1 t dt∞

−∞

= δα ∫

(ii) Let 0α < ; that is α = − α , and let t− α = τ .

( ) ( )1 1t dt d∞ ∞

−∞ −∞

δ α = δ τ τ =α α∫ ∫

( )1 t dt∞

−∞

= δα ∫

It is easy to show that ( ) ( )0 01t t t t⎡ ⎤δ α − = δ −⎣ ⎦ α

.

Special Case: If 1α = − , we have the result ( ) ( )t tδ − = δ .

The above result is not surprising, especially if we look at the examples

( )1p t to ( )3p t , which are all even functions of t . Hence some authors call this

as the even sided delta function. It is also possible to come up with delta

functions as a limiting case of functions that are not even; that is, as a limiting

case of one-sided functions. In such a situation we have a left-sided delta

function or right-sided delta function etc. Left-sided delta function will prove to be

useful in the context of probability density functions of certain random variables,

subject matter of chapter 2.



1.54

Example 1.10

Find the value of

(a) ( )4

3

4

5t t dt−

δ −∫

(b) ( )5.1

3

4.9

5t t dtδ −∫

(a) ( )5tδ − is nonzero only at 5t = . The range of integration does not include

the impulse. Hence the integral is zero.

(b) As the range of integration includes the impulse, we have a nonzero value for

the product ( )3 5t tδ − . As ( )5tδ − occurs at 5t = , we can write

( ) ( )3 35 5 5t t tδ − = δ − .

Hence,

( ) ( )5.1 5.1

3 3

4.9 4.9

5 5 5 125t t dt t dtδ − = δ − =∫ ∫

Example 1.11

Let ( )4tp t tri ⎛ ⎞= ⎜ ⎟

⎝ ⎠. Find ( ) ( )2 1p t t⎡ ⎤∗ δ −⎣ ⎦ .

( ) 1 1 12 1 22 2 2

t t t⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞δ − = δ − = δ −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

( ) 1 1 1 1 1 1 22 2 2 2 2 4

tp t t p t tri⎡ ⎤ −⎛ ⎞ ⎛ ⎞ ⎛ ⎞∗ δ − = − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

Let us now compute the Fourier transform of ( )tδ . From Eq. 1.11(b), we

have,

( ) ( ) 2 1j f tF t t e dt∞

− π

−∞

⎡ ⎤δ = δ =⎣ ⎦ ∫ (1.14a)



1.55

How do we interpret this result? The spectrum of the unit impulse consists

of frequency components in the range ( ),− ∞ ∞ , all with unity magnitude and

zero phase shift, a fascinating result indeed! Hence exciting any electric network

or system with a unit impulse is equivalent to exciting the network simultaneously

with complex exponentials of all possible frequencies, all with the same

magnitude (unity in this case) and zero phase shift. That is, the unit impulse

response of a linear network is the synthesis of responses to the individual

complex exponentials and we intuitively feel that the impulse response of a

network should be able to characterize the system in the time domain. (We shall

see a little later that if the network is linear and time invariant, a simple relation

exists between the input to the network, its impulse response and the output).

The dual of the Fourier transform pair of Eq. 1.14(a) gives us

( ) ( )1 f f←⎯→ δ − = δ (1.14b)

Based on Eq. 1.14(a) and Eq. 1.14(b), we make the following observation:

a constant in one domain will transform into an impulse in the other domain.

Eq. 1.14(b) is intuitively satisfying; a constant signal has no time variations

and hence its spectral content ought to be confined to 0f = ; ( )fδ is the proper

quantity for the transform because it is zero for 0f ≠ and its inverse transform

yields the required constant in time (note that only an impulse can yield a

nonzero value when integrated over zero width).

Because of the transform pair,

( )1 f←⎯→ δ ,

we obtain another transform pair (from modulation theorem)

( )020

j f te f fπ ←⎯→ δ − (1.15a)

( )020

j f te f f− π ←⎯→ δ + (1.15b)



1.56

As 0 02 2

0cos2

j f t j f te etπ − π+

ω = , we have,

( ) ( )0 0 01cos2

t f f f f⎡ ⎤ω ←⎯→ δ − + δ +⎣ ⎦ (1.16)

Similarly,

( ) ( )0 0 01sin2

t f f f fj⎡ ⎤ω ←⎯→ δ − − δ +⎣ ⎦ (1.17)

0cosF t⎡ ⎤ω⎣ ⎦ and [ ]0sinF tω are shown in Fig. 1.15.

Fig. 1.15: Fourier transforms of (a) 0cos tω and (b) 0sin tω

Note that the impulses in Fig. 1.15(b) have weights that are complex. It is

fairly conventional to show the spectrum of 0sin tω as depicted in Fig. 1.15(b); or

else we can make two separate plots, one for magnitude and the other for phase,

where the magnitude plot is identical to that shown in Fig. 1.15(a) and the phase

plot has values of 2π

− at 0f f= and 2π

+ at 0f f= − .

In summary, we have found the Fourier transform of ( )tδ (a time function

with a discontinuity that is not finite), and using impulses in the frequency

domain, we have developed the Fourier transforms of the periodic signals such



1.57

as 02j f te± π , 0cos tω and 0sin tω , which are neither absolutely integrable nor

square integrable.

We are now in a position to present both Fourier series and Fourier

transform in a unified framework and talk only of Fourier transform whether the

signal is aperiodic or not. This is because, for a periodic signal ( )px t , we have

the Fourier series relation,

( ) 02j n f tp n

nx t x e

∞π

= −∞

= ∑

Taking the Fourier transform on both the sides,

( ) ( ) 02j n f tp p n

nF x t X f F x e

∞π

= −∞

⎡ ⎤⎡ ⎤ = = ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦

∑

02j n f tn

nx F e

∞π

= −∞

⎡ ⎤= ⎣ ⎦∑

( )0nn

x f nf∞

= −∞

= δ −∑ (1.18)

FT of ( )px t is a function of the continuous variable f , whereas, in the FS

representation of ( )px t , nx is a function of the discrete variable n . However, as

( )pX f is purely impulsive, spectral components exist only at 0f n f= , with

complex weights nx . As inversion of ( )pX f requires integration, we require

impulses in the spectrum. As such, the differences between the line spectrum of

sec. 1.1 and spectral representation given by Eq. 1.18 are only minor in nature.

They both provide the same information, differing essentially only in notation.

There is an interesting relation between nx and the Fourier transform of

one period of a periodic signal. Let,



1.58

( ) ( ) 0 0,2 2

0 ,

pT Tx t t

x toutside

⎧ − < <⎪= ⎨⎪⎩

( )0

0

0

22

02

1T

j n f tn p

Tx x t e dt

T− π

−

= ∫

( )0

0

0

22

02

1T

j n f t

Tx t e dt

T− π

−

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦∫

( ) ( ) 002

2

0 0

1 1nj t

Tj n f tx t e dt x t e dtT T

⎛ ⎞∞ ∞ − π⎜ ⎟− π ⎝ ⎠

−∞ −∞

⎡ ⎤⎢ ⎥= =⎢ ⎥⎣ ⎦∫ ∫

The bracketed quantity is ( )0

nfT

X f=

Hence, 0 0

1n

nx XT T

⎛ ⎞= ⎜ ⎟

⎝ ⎠ (1.19)

Example 1.12 Find the Fourier transform of the uniform impulse train

( ) ( )0pn

x t t nT∞

= −∞

= δ −∑ shown in Fig 1.16 below.

Fig. 1.16: Uniform impulse train



1.59

Let ( )x t be one period of ( )px t in the interval, 0 0

2 2T Tt− < < . Then,

( ) ( )x t t= δ for this example. But as ( ) 1tδ ←⎯→ , from Eq. (1.19) we have,

0

1nx

T= for all n . Hence,

( ) ( )00

1p

nX f f n f

T

∞

= −∞

= δ −∑ (1.20)

From Eq. 1.20, we have another interesting result:

A uniform periodic impulse train in either domain will transform into

another uniform impulse train in the other domain.

From the transform pair, ( ) 1tδ ←⎯→ , we have

[ ] ( )1 21 j f tF e df t∞

− π

− ∞

= = δ∫

As ( ) ( )t tδ − = δ , we have, ( )2j f te df t∞

− π

− ∞

= δ∫

That is, ( )2j f te df t∞

± π

− ∞

= δ∫ (1.21)

Using Eq. 1.21, we show that ( )x t and ( )X f constitute a transform pair.

Let ( ) ( ) 2ˆ j f tx t X f e df∞

π

− ∞

= ∫

( ) 2 2j f j f tx e d e df∞ ∞

− π λ π

− ∞ − ∞

⎡ ⎤⎢ ⎥= λ λ⎢ ⎥⎣ ⎦

∫ ∫

( ) ( )2j f tx e df d∞ ∞

π − λ

− ∞ − ∞

= λ λ∫ ∫

( ) ( )2j f tx e df d∞ ∞

π − λ

− ∞ − ∞

⎡ ⎤⎢ ⎥= λ λ⎢ ⎥⎣ ⎦

∫ ∫



1.60

( ) ( ) ( )x t d x t∞

− ∞

= λ δ − λ λ =∫

As ( ) ( )x t x t= , we have ( )X f uniquely representing ( )x t .

1.5.2 Impulse response and convolution

Let ( )x t be the input to a Linear, Time-Invariant (LTI) system resulting in

the output, ( )y t . We shall now establish a relation between ( )x t and ( )y t .

From the replication property of the impulse, we have,

( ) ( ) ( ) ( ) ( )x t x t t x t d∞

− ∞

= ∗ δ = τ δ − τ τ∫

Let ( )t⎡ ⎤δ⎣ ⎦R denote the output (response) of the LTI system, when the

input is exited by the unit impulse ( )tδ . This is generally denoted by the symbol

( )h t and is called the impulse response of the system. That is, when ( )tδ is

input to an LTI system, its output ( ) ( ) ( )y t t h t⎡ ⎤= δ =⎣ ⎦R . As the system is time

invariant, ( ) ( )t h t⎡ ⎤δ − τ = −τ⎣ ⎦R .

As the system is linear, ( ) ( ) ( ) ( )x t x h t⎡ ⎤τ δ − τ = τ − τ⎣ ⎦R

and ( ) ( ) ( ) ( ) ( ) ( )x t y t x t d x h t d∞ ∞

− ∞ − ∞

⎡ ⎤⎡ ⎤ ⎢ ⎥= τ δ − τ τ = τ − τ τ⎣ ⎦ ⎢ ⎥⎣ ⎦

∫ ∫=R R

That is, ( ) ( ) ( )y t x t h t= ∗ (1.22)

The following properties of convolution can be established:

Convolution operation

P1) is commutative

P2) is associative



1.61

P3) distributes over addition

P1) implies that ( ) ( ) ( ) ( )1 2 2 1x t x t x t x t∗ = ∗

That is, ( ) ( ) ( ) ( )1 2 2 1x x t d x x t d∞ ∞

− ∞ − ∞

τ − τ τ = τ − τ τ∫ ∫ .

P2) implies that ( ) ( ) ( ) ( ) ( ) ( )1 2 3 1 2 3x t x t x t x t x t x t⎡ ⎤ ⎡ ⎤∗ ∗ = ∗ ∗⎣ ⎦ ⎣ ⎦ , where the

bracketed convolution is performed first. Of course, we assume that every

convolution pair gives rise to bounded output.

P3) implies that ( ) ( ) ( ) ( ) ( ) ( ) ( )1 2 3 1 2 1 3x t x t x t x t x t x t x t⎡ ⎤∗ + = ∗ + ∗⎣ ⎦ .

Note that the properties P1) to P3) are valid even if the independent variable is

other than t .

Taking the Fourier transform on both sides of Eq. 1.22, we have,

( ) ( ) ( )Y f X f H f= (1.23)

where ( ) ( )h t H f←⎯→ . The quantity ( )H f is referred to (quite obviously) as

the frequency response of the system and describes the frequency domain

behavior of the system. (As ( ) ( )( )

Y fH f

X f= , it is also referred to as the transfer

function of the LTI system). As ( )H f is, in general, complex, it is normally shown

as two different plots, namely, the magnitude response: ( ) .H f vs f and the

phase response: ( )arg .H f vs f⎡ ⎤⎣ ⎦ .

If ( ) ( )1 2H f H f≠ , we then have ( ) ( )1 11 2F H f F H f− −⎡ ⎤ ⎡ ⎤≠⎣ ⎦ ⎣ ⎦ . That is,

( ) ( )1 2h t h t≠ . In other words, the impulse response of any LTI system can be

used to uniquely characterize the system in the time domain.



1.62

Example 1.13 RC-lowpass filter (RC-LPF) is one among the quite often used LTI

systems in the study of communication theory. This network is shown in Fig.

1.17. Let us find its frequency response as well as the impulse response.

Fig. 1.17: The RC-lowpass filter

One of the important properties of any LTI system is: if the input

( ) 02j f tx t e π= , then the output ( )y t is also a complex exponential given by

( ) ( ) π= j f ty t H f e 020 .

But, ( ) 020

0

12

12

j f tj f Cy t eR

j f C

ππ=

+π

or ( ) 02

0

11 2

j f ty t ej f RC

π=+ π

, when ( ) 02j f tx t e π=

Generalizing this result, we obtain,

( ) 11 2

H fj f RC

=+ π

(1.24)

That is,

( )( )2

11 2

H ff RC

=+ π

(1.25a)

( ) ( ) ( )arg tan 2f H f arc f RC⎡ ⎤θ = = − π⎣ ⎦ (1.25b)



1.63

Let 12

FRC

=π

. Then,

( ) 21

1H f

fF

=⎛ ⎞+ ⎜ ⎟⎝ ⎠

(1.26)

A plot of ( ) .H f vs f and ( )arg .H f vs f⎡ ⎤⎣ ⎦ is shown in Fig. 1.18.

Fig. 1.18 RC-LPF: (a) Magnitude response

(b) Phase response



1.64

Let us now compute ( ) ( )1h t F H f− ⎡ ⎤= ⎣ ⎦ . We have the transform pair

( )ex tj f11

1 2←⎯→

+ π

Hence, texRC RC j f RC

1 111 2

⎛ ⎞←⎯→⎜ ⎟ + π⎝ ⎠

That is,

( ) RC LPF

th t exRC RC

1 1−

⎛ ⎞⎡ ⎤ = ⎜ ⎟⎣ ⎦

⎝ ⎠ (1.27)

This is shown in Fig. 1.19.

Fig. 1.19: Impulse response of an RC-LPF

Example 1.14

The input ( )x t and the impulse response ( )h t of an LTI system are as

shown in Fig. 1.20. Let us find the output ( )y t of the system.



1.65

Fig. 1.20: The input ( )x t and the impulse response ( )h t of an LTI system

Let ( ) ( ) ( )y t h x t d∞

− ∞

= τ − τ τ∫

To compute ( )h t we have to perform the following three steps:

i) Obtain ( )h τ and ( )x t − τ for a given 1t t= .

ii) Take the product of the quantities in (i).

iii) Integrate the result of (ii) to obtain ( )1y t .

( )h τ is the same as ( )h t with the change of variable from t to τ . Note

that τ is the variable of integration. ( )x t⎡ ⎤− τ⎣ ⎦ is actually ( )x t⎡ ⎤− τ −⎣ ⎦ ; that is,

we first reverse ( )x τ to get ( )x − τ and then shift by t , the time instant for which

( )y t is desired. This completes the operations in step (i). The operations

involved in steps (ii) and (iii) are quite easy to understand.

In quite a few situations, where convolution is to be performed, it would be

of great help to have the plots of the quantities in step (i). These have been



1.66

shown in Fig. 1.21 for three different values of t , namely 0t = , 1t = − and

2t = .

From Fig.1.21(c), we see that if 1t < − , then ( )τh and ( )x t − τ do not

overlap; that ( ) 0y t = , for 1t < − . For 1 0t− < ≤ , overlap of ( )h τ and

( )x t − τ increases as t increases and the integral of the product (which is

positive) increases linearly reaching a value of 1 for 0t = . For 0 2t< ≤ , net

area of the product ( ) ( )τ − τh x t , keeps decreasing and at 2t = , we have,

( ) ( ) 0h x t d∞

− ∞

τ − τ τ =∫ .



1.67

Fig. 1.21: A few plots to implement step (i) of convolution:

(a): ( )h τ

(b), (c), (d): ( )x t − τ for 0t = , 1t = − and 2t = respectively.



1.68

Using the similar arguments, ( )y t can be computed for 2t > . The result

of the convolution is indicated in Fig.1.22.

Fig. 1.22: Complete output ( )y t of Example 1.14

(Sometimes, computing ( ) ( ) ( )y t x t h t= ∗ , could be very tricky and might

even be sticky1.)

1 Some people claim that convolution has driven many electrical engineering students to contemplate theology either for salvation or as an alternative career (IEEE Spectrum, March 1991, page 60). For an interesting cartoon expressing the student reaction to the convolution operation, see [3].

Exercise 1.3

Let ( ) , 00 ,

te tx totherwise

− α⎧ ≥⎪= ⎨⎪⎩

( ) , 0

0 ,

te th totherwise

− β⎧ ≥⎪= ⎨⎪⎩

where , 0α β > . Find ( ) ( ) ( )y t x t h t= ∗ for

(i) α = βand (ii) α ≠ β



1.69

1.5.3 Signum function and unit step function Def. 1.4: Signum Function

We denote the signum function by ( )sgn t and define it as,

( )1 , 0

sgn 0 , 01, 0

tt t

t

>⎧⎪= =⎨⎪− <⎩

(1.28)

Def. 1.5: Unit Step Function

We denote the unit step function by ( )u t , and define it as,

Exercise 1.4

Find ( ) ( ) ( )y t x t h t= ∗ where

( ) 2 , 20 ,

tx t

outside⎧ <

= ⎨⎩

( ) 2 , 00 ,

te th toutside

−⎧ ≥⎪= ⎨⎪⎩

Exercise 1.5

Let ( )

1 , 950 105051 , 1050 950

100 ,

f Hz

X f f Hz

elsewhere

⎧ < <⎪⎪⎪= − < < −⎨⎪⎪⎪⎩

(a) Compute and sketch ( ) ( ) ( )Y f X f X f= ∗

(b) Let 1 50f Hz= , 2 2050f Hz= and 3 20f Hz=

Verify that ( ) ( ) ( )1 2 32 and 1Y f Y f Y f= = = .



1.70

( )

1 , 01 , 020 , 0

t

u t t

t

>⎧⎪⎪= =⎨⎪

<⎪⎩

(1.29)

We shall now develop the Fourier transforms of ( )sgn t and ( )u t .

( )⎡ ⎤⎣ ⎦sgnF t :

Let ( ) ( ) ( )t tx t e u t e u t− α α= − − (1.30)

where α is a positive constant. Then ( ) ( )0

sgn limt x tα →

= , as can be seen from

Fig. 1.23.

Fig. 1.23: ( )sgn t as a limiting case of ( )x t of Eq. 1.30

( )( )22

1 1 42 2 2

j fX fj f j f f

− π= − =

α + π α − π α + π

( ) ( )0

1sgn limF t X fj fα →

⎡ ⎤ = =⎣ ⎦ π (1.31)

( )⎡ ⎤⎣ ⎦F u t

As ( ) ( )1 1 sgn2

u t t⎡ ⎤= +⎣ ⎦ , we have



1.71

( ) ( )1 12 2

U f fj f

= δ +π

(1.32)

We shall now state and prove the FT of the integral of a function.

Let ( ) tp t ga11 ⎛ ⎞= ⎜ ⎟ε ε⎝ ⎠

. Then, we know that ( ) ( )10lim p t tε →

= δ .

But ( ) ( )0 0

1 102

1lim2

p t dt p t dtε →

ε − ∞−

= =∫ ∫

Properties of FT continued...

P9) Integration in the time domain

Let ( ) ( )x t X f←⎯→

Then, ( ) ( ) ( ) ( )012 2

t Xx d X f f

j f− ∞

τ τ ←⎯→ + δπ∫ (1.33)

Proof:

Consider ( ) ( ) ( ) ( )x t u t x u t d∞

− ∞

∗ = τ − τ τ∫ . As ( ) 0u t − τ = for tτ > ,

( ) ( ) ( ) ( ) ( ) ( ) ( ). Butt

x t u t x d F x t u t X f U f− ∞

⎡ ⎤∗ = τ τ ∗ =⎣ ⎦∫ .

Hence, ( ) ( ) ( )12 2

t fx d X f

j f− ∞

⎡ ⎤δτ τ ←⎯→ +⎢ ⎥π⎣ ⎦

∫

Here there are two possibilities:

(i) ( )0 0X = ; then ( ) ( )2

t X fx d

j f− ∞

τ τ ←⎯→π∫

(ii) ( )0 0X ≠ ; then ( ) ( ) ( ) ( )02 2

t X f X fx d

j f− ∞

δτ τ ←⎯→ +

π∫



1.72

and ( ) ( )2

1 102

lim 1p t dt p t dtε

∞

ε →ε − ∞−

= =∫ ∫ .

That is,

( ) ( )t

d u t− ∞

δ τ τ =∫ (1.34a)

or ( ) ( )d u tt

d t= δ (1.34b)

We shall now give an alternative proof for the FT relation,

( ) ( )u t fj f

1 12 2

←⎯→ δ +π

.

As ( ) ( )d u tt

d t= δ ,

( )j f U f2 1π =

or ( )U fj f

12

=π

But this is valid only for f 0≠ because of the following argument.

( ) ( )u t u t 1+ − = . Therefore,

( ) ( ) ( )U f U f f+ − = δ

As ( )fδ is nonzero only for f 0= , we have

( ) ( ) ( ) ( )U U U f0 0 2 0+ − = = δ or

( ) ( )U f102

= δ . Hence,

( )( )f f

U ff

j f

1 , 02

1 , 02

⎧ δ =⎪⎪= ⎨⎪ ≠

π⎪⎩



1.73

Example 1.15

(a) For the scheme shown in Fig. 1.24, find the impulse response (This system is

referred to as Zero-Order-Hold, ZOH).

(b) If two such systems are cascaded, what is the overall impulse response

(cascade of two ZOHs is called a First-Order-Hold, FOH).

Fig. 1.24: Block schematic of a ZOH

a) ( )h t of ZOH:

When ( ) ( )x t t= δ , we have

( ) ( ) ( )v t t t T= δ − δ −

Hence ( ) ( ) ZOHy t h t⎡ ⎤= ⎣ ⎦ , the impulse response of the ZOH, is

( ) ( ) ( ) ( ) 2t t

Ttd T d u t u t T ga

T− ∞ − ∞

⎛ ⎞−⎜ ⎟δ τ τ − δ τ − τ = − − = ⎜ ⎟

⎜ ⎟⎝ ⎠

∫ ∫

b) Impulse response of two LTI systems in cascade is the convolution of the

impulse responses of the constituents. Hence,

( ) 2 2FOH

T Tt th t ga ga

T T

⎡ ⎤ ⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎡ ⎤ = ∗⎢ ⎥ ⎢ ⎥⎣ ⎦

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

t TT triT−⎛ ⎞= ⎜ ⎟

⎝ ⎠



1.74

Eq. 1.34(a) can also be established by working in the frequency domain.

From Eq. 1.33, with ( ) ( )x t t= δ and ( ) 1X f = ,

( ) ( ) ( )1 1 . That is,2 2

t

d f U fj f− ∞

δ τ τ ←⎯→ + δ =π∫

( ) ( )t

d u t− ∞

δ τ τ =∫

Eq. 1.34(b) helps in finding the derivatives of signals with discontinuities.

Consider the pulse ( )p t shown in Fig. 1.25(a).

Fig. 1.25: (a) A signal with discontinuities

(b) Derivative of the signal at (a)

( )p t can be written as

( ) ( ) ( ) ( )2 2 1 3p t u t u t u t= + − + + −



1.75

Hence,

( ) ( ) ( ) ( )2 2 1 3d p t

t t td t

= δ + − δ + + δ −

which is shown in Fig. 1.25(b). From this result, we note that if there is a step

discontinuity of size A at 1t t= in the signal, its derivative will have an impulse

of weight A at 1t t= .

Example 1.16

Let ( )x t be the doublet pulse of Example 1.7 (Fig.1.12). We shall find

( )X f

from ( )d x tdt

.

( ) ( ) ( ) ( )2d x t

t T t t Tdt

= δ + − δ + δ −

Taking Fourier transform on both the sides,

( ) 2 22 2j f T j f Tj f X f e eπ − ππ = − +

( )2j f T j f Te eπ − π= −

( )( ) ( )

22 2

j f T j f T j f T j f Te e e eX f jT

j f T j

π − π π − π− −=

π

( ) ( )2 sin sinjT c f T f T= π

Example 1.17

Let ( )x t , ( )h t and ( )y t denote the input, impulse response and the

output respectively of an LTI system. It is given that,

( ) ( )2tx t t e u t−= and ( ) ( )4 th t e u t−= .

Find a) ( )Y f



1.76

b) ( ) ( )1y t F Y f− ⎡ ⎤= ⎣ ⎦

c) ( ) ( ) ( )y t x t h t= ∗

a) From Eq. 1.22, we obtain

( ) ( ) ( )Y f X f H f=

If ( ) ( )2tz t e u t−= , then ( ) 12 2

Z fj f

=+ π

.

As ( ) ( )x t t z t= , we have ( ) ( )( )2

12 2 2

d Z fjX fdf j f

⎛ ⎞= =⎜ ⎟π + π⎝ ⎠

( ) 14 2

H fj f

=+ π

Hence, ( )( )2

1 14 22 2

Y fj fj f

=+ π+ π

(b) Using partial fraction expansion,

( )( )

( )21 1 14 2 4

2 2 4 22 2Y f

j f j fj f

−= + +

+ π + π+ π

Hence, ( ) ( ) ( ) ( )2 2 41 1 14 2 4

t t ty t e u t t e u t e u t− − −⎛ ⎞= − + +⎜ ⎟⎝ ⎠

(c) ( ) ( ) ( )y t x h t d∞

− ∞

= τ − τ τ∫

( ) ( ) ( )42 te u e u t d∞

− − τ− τ

− ∞

= τ τ − τ τ∫

( ) 0y t = for 0t < because for t negative, ( )u t − τ is 1 only for τ negative; but

then ( ) 0u τ = . As ( ) 0u t − τ = for tτ > , we have

( ) ( )42

0

tty t e e d− − τ− τ= τ τ∫



1.77

4 2

0

tte e d− τ= τ τ∫

2 2

4

02 2

tt e ee d

τ τ− ⎡ ⎤

= τ − τ⎢ ⎥⎣ ⎦

∫

2 2

4

02 4

ttt e ee t

τ−

⎧ ⎫⎡ ⎤⎪ ⎪= −⎨ ⎬⎢ ⎥⎣ ⎦⎪ ⎪⎩ ⎭

2 2

4 1 , 02 4

t tte et e t

−− ⎡ ⎤−

= − ≥⎢ ⎥⎣ ⎦

0= for 0t <

Exercise 1.6

Let ( )( ) 0

0

0

1 cos 2 ,2

0 ,2

Tf t tx t

Tt

⎧ + π <⎪⎪= ⎨⎪ >⎪⎩

where 0T is the period of the cosine signal and 00

1fT

= .

(a) Show that

( ) ( ) ( )32 0 0

03 22 2

d x t d x tT Tf t tdtd t

⎧ ⎫⎛ ⎞ ⎛ ⎞⎪ ⎪= π δ + − δ − −⎨ ⎬⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎪ ⎪⎩ ⎭

(b) Taking the FT of the equation in (a) above, show that

( ) ( )002 2

0

sinfX f c f Tf f

=−



1.78

1.6 Correlation Functions Two basic operations that arise in the study of communication theory are:

(i) convolution and (ii) correlation. As we have some feel for the convolution

operation by now, let us develop the required familiarity with the correlation

operation.

When we say that there is some correlation between two objects, we imply

that there is some similarity between them. We would like to quantify this intuitive

notion and come up with a formal definition for correlation so that we have a

mathematically consistent and physically meaningful measure for the correlation

of the objects of interest to us.

Our interest is in electrical signals. We may like to quantify, say, the

similarity between a transmitted signal and the received signal or between two

different transmitted or received signals etc. We shall first introduce the cross

correlation functions; this will be followed by the special case, namely, auto-

correlation function.

In the context of correlation functions, we have to distinguish between the

energy signals and power signals. Accordingly, we make the following definitions.

1.6.1 Cross-correlation functions (CCF) Let ( )x t and ( )y t be signals of the energy type. We now define their

cross-correlation functions, ( ) ( )andx y y xR Rτ τ .

Def. 1.6(a): The cross-correlation function ( )x yR τ is given by

( ) ( ) ( )x yR x t y t d t∞

∗

− ∞

τ = − τ∫ (1.35a)

Def. 1.6(b): The cross-correlation function ( )y xR τ is given by



1.79

( ) ( ) ( )y xR y t x t d t∞

∗

− ∞

τ = − τ∫ (1.35b)

In Eq. 1.35(a), ( )y t∗ − τ is a conjugated and shifted version of ( )y t , τ

accounting for the shift in ( )y t∗ . Note that the variable of integration in Eq. 1.35

is t ; hence ( )x yR as well as ( )y xR is a function of τ , the shift parameter ( τ is

also called the scanning parameter or the search parameter).

Let ( )x t and ( )y t be the signals of the power type.

Def. 1.7(a): The cross-correlation function, ( )x yR τ is given by

( ) ( ) ( )2

2

1limT

x y T TR x t y t dt

T∗

→ ∞−

τ = − τ∫ (1.36a)

Def. 1.7(b): The cross-correlation function, ( )y xR τ is given by

( ) ( ) ( )2

2

1limT

y x T TR y t x t dt

T∗

→ ∞−

τ = − τ∫ (1.36b)

The power signals that we have to deal with most often are of the periodic

variety. For periodic signals, we have the following definitions:

Def. 1.8(a): ( ) ( ) ( )0

0

2

02

1p p

T

x y p pT

R x t y t dtT

∗

−

τ = − τ∫ (1.37a)

Def. 1.8(b): ( ) ( ) ( )0

0

2

02

1p p

T

y x p pT

R y t x t dtT

∗

−

τ = − τ∫ (1.37b)

Let t − τ = λ in Eq. 1.35(a). Then, t = τ + λ and dt d= λ . Hence,

( ) ( ) ( )x yR x y d∞

∗

− ∞

τ = λ + τ λ λ∫

( )y xR∗= − τ (1.38)



1.80

As ( ) ( )x y y xR Rτ ≠ τ , cross-correlation, unlike convolution is not in

general, commutative. To understand the significance of the parameter τ ,

consider the situation shown in Fig. 1.26.

Fig. 1.26: Waveforms used to compute ( )y xR τ and ( )z xR τ

If we compute ( ) ( )x t y t dt∞

− ∞∫ , we find it to be zero as ( )x t and ( )y t do not

overlap. However, if we delay ( )x t by half a unit of time, we find that 12

x t⎛ ⎞−⎜ ⎟⎝ ⎠

and ( )y t start overlapping and for 12

τ > , we have nonzero value for the

integral. For the value of 2.75τ , we will have a positive value for



1.81

( ) ( )y t x t dt∞

− ∞

− τ∫ , which would be about the maximum of ( )y xR τ for any τ .

For values of 2.75τ > , ( )y xR τ keeps decreasing, becoming zero for 5τ > .

Similarly, if we compute ( )z xR τ , we find that ( )maxz xR τ would be much smaller

than ( )maxy xR τ (Note that for 2.75τ , the product quantity, ( ) ( )y t x t − τ , is

essentially positive for all t ). If ( )y t is the received signal of a communication

system, then we are willing to accept ( )x t as the likely transmitted signal (we can

treat the received signal as a delayed and distorted version of the transmitted

signal) whereas if ( )z t is received, it would be difficult for us to accept that ( )x t

could have been the transmitted signal. Thus the parameter τ helps us to find

time-shifted similarities present between the two signals.

From Eq. 1.35(a), we see that computing ( )x yR τ for a given τ , involves

the following steps:

(i) Shift ( )y t∗ by τ

(ii) Take the product of ( )x t and ( )y t∗ − τ

(iii) Integrate the product with respect to t .

The above steps closely resemble the operations involved in convolution. It is not

difficult to see that

( ) ( ) ( )x yR x y∗τ = τ ∗ − τ , because

( ) ( ) ( ) ( )x y x t y t dt∞

∗ ∗

− ∞

⎡ ⎤τ ∗ − τ = − τ −⎣ ⎦∫

( ) ( )x t y t dt∞

∗

− ∞

= − τ∫ ( )x yR= τ (1.39a)

Let ( ) ( )x y x yE f F R⎡ ⎤= τ⎣ ⎦ .

Then, ( ) ( ) ( )x yE f F x y∗⎡ ⎤= τ ∗ − τ⎣ ⎦ ( ) ( )X f Y f∗= (1.39b)



1.82

Example 1.18

Let ( ) ( )exp1x t t= and ( )2ty t ga ⎛ ⎞= ⎜ ⎟

⎝ ⎠.

Let us find (a) ( )x yR τ and (b) ( )y xR τ .

From the results of (a) and (b) above, let us verify Eq. 1.38.

(a) ( )τx yR :

( )x t and ( )y t are sketched below (Fig. 1.27).

Fig. 1.27: Waveforms of Example 1.18

(i) 1τ < − :

( )x t and ( )y t − τ do not overlap and the product is zero. That is,

( ) 0x yR τ = for 1τ < − .

ii) 1 1− ≤ τ < :

( ) ( )1

1

0

1tx yR e dt e

+ τ− + τ−τ = = −∫



1.83

(iii) 1τ ≥ :

( ) ( ) ( )1

1 1

1

tx yR e dt e e

τ +− τ − − τ +−

τ −

τ = = −∫

1e ee

− τ ⎛ ⎞= −⎜ ⎟⎝ ⎠

(b) ( )τy xR :

(i) For 1τ > , ( )y t and ( )x t − τ do not overlap. Hence, ( ) 0y xR τ = for 1τ > .

(ii) For 1 1− < τ ≤ ,

( ) ( )1

ty xR e dt− − τ

τ

τ = ∫

1e e eτ − τ −⎡ ⎤= −⎣ ⎦

( )11 e− − τ= −

(iii) For 1τ ≤ − ,

( ) ( )1

1

ty xR e dt− − τ

−

τ = ∫

1e ee

τ ⎛ ⎞= −⎜ ⎟⎝ ⎠

( )x yR τ and ( )y xR τ are plotted in Fig. 1.28.



1.84

Fig. 1.28: Cross correlation functions of Example 1.18

From the plots of ( )x yR τ and ( )y xR τ , it is easy to see that

( ) ( )y x x yR Rτ = − τ

Def. 1.9: Two signals ( )x t and ( )y t are said to be orthogonal if

( ) ( ) 0x t y t dt∞

∗

− ∞

=∫ (1.40)

Eq. 1.41 implies that for orthogonal signals, say ( )x t and ( )y t

( ) 0 0x yR τ =τ = (1.41a)

We have the companion relation to Eq. 1.41(a), namely

( )0 0y xR = , if ( )x t and ( )y t are orthogonal. (1.41b)

Let ( )px t and ( )py t be periodic with period 0T . Then, from Eq. 1.37(a),

for any integer n ,



1.85

( ) ( )0p p p px y x yR nT Rτ + = τ

That is, ( )p px yR τ is also periodic with the same period as ( )px t and ( )py t .

Similarly, we find ( )p py xR τ . Derivation of the FT of ( )

p px yR τ is given in appendix

A1.2.

1.6.2 Autocorrelation function (ACF) ACF can be treated as a special case of CCF. In Eq. 1.35(a), let

( ) ( )x t y t= . Then, we have

( ) ( ) ( )x xR x t x t dt∞

∗

− ∞

τ = − τ∫ (1.42a)

Instead of ( )x xR τ , we use somewhat simplified notation, namely, ( )xR τ which is

called the auto correlation function of ( )x t .

ACF compares ( )x t with a shifted and conjugated version of itself. If

( )x t and ( )x t∗ − τ are quite similar, we can expect large value for ( )xR τ ,

whereas as a value of ( )xR τ close to zero implies the orthogonality of the two

signals. Hence ( )xR τ can provide some information about the time variations of

the signal.

Let ( )t − τ = λ in Eq. 1.42(a). We then have,

( ) ( ) ( )xR x x d∞

∗

− ∞

τ = τ + λ λ λ∫

( ) ( )x t x t dt∞

∗

− ∞

= + τ∫ (1.42b)

Eq. 1.42(b) gives another relation for ( )xR τ . This is quite meaningful

because, assuming τ positive, ( )x t∗ − τ is a right shifted version of ( )x t∗ . In Eq.



1.86

1.42(a), we keep ( )x t fixed and move ( )x t∗ to the right by τ and take the

product; in Eq. 1.42(b), we keep ( )x t∗ fixed and move ( )x t to the left by τ . This

does not change the integral of Eq. 1.42(a), because if we let 1t t= and 1τ = τ ,

the product of Eq. 1.42(a) is ( ) ( )1 1 1x t x t∗ − τ . This product is obtained from Eq.

1.42(b) for 1 1t t= − τ . For a given 1τ = τ , as t is varied, all the product

quantities are obtained and hence the integral for a given 1τ remains the same.

(Note that shifting a function does not change its area.)

If ( )x t is a power signal, then the ACF is special case of Eq. 1.36(a). That

is, for the power signals, we have

( ) ( ) ( )2

2

limT

x T TR x t x t dt∗

→ ∞−

τ = − τ∫ (1.43a)

It can easily be shown that,

( ) ( ) ( )2

2

limT

x T TR x t x t dt∗

→ ∞−

τ = + τ∫ (1.43b)

For power signals that are periodic, we have

( ) ( ) ( )0

0

2

02

1T

x p pT

R x t x t dtT

∗

−

τ = − τ∫ (1.44a)

( ) ( )0

0

2

02

1T

p pT

x t x t dtT

∗

−

= + τ∫ (1.44b)

Properties of ACF (energy signals) P1) ACF exhibits conjugate symmetry. That is,

( ) ( )x xR R∗− τ = τ (1.45a)



1.87

Proof: Exercise

Eq. 1.45(a) implies that the real part of ( )xR τ is an even function of τ

where as the imaginary part is an odd function of τ .

P2) ( ) ( ) 20x xR x t d t E

∞

− ∞

= =∫ (1.45b)

where xE is the energy of the signal ( )x t (Eq. 1.10).

P3) Maximum value of ( )xR τ occurs at the origin. That is, ( ) ( )0x xR Rτ ≤ .

Proof: The proof of the above property follows from Schwarz’s inequality;

which is stated below.

Let ( )1g t and ( )2g t be two energy signals.

Then, ( ) ( ) ( ) ( )2

2 21 2 1 2g t g t dt g t dt g t dt

∞ ∞ ∞

− ∞ − ∞ − ∞

≤∫ ∫ ∫ .

Let ( ) ( )1g t x t= and ( ) ( )2g t x t∗= − τ .

From the Schwarz’s Inequality,

( ) ( ) ( ) ( )2

22x t x t dt x t dt x t dt

∞ ∞ ∞∗ ∗

− ∞ − ∞ − ∞

− τ ≤ − τ∫ ∫ ∫

( ) ( )2 20x xR R⎡ ⎤τ ≤ ⎣ ⎦ or

( ) ( )0x xR Rτ ≤ (1.45c)

P4) Let ( )xE f denote the Energy Spectral Density (ESD) of the signal ( )x t .

That is, ( )x xE f df E∞

− ∞

=∫

Then, ( ) ( )x xR E fτ ←⎯→

Proof From Eq. 1.39(b),

( ) ( ) ( )x yE f X f Y f∗=



1.88

Let ( ) ( )x t y t= ; then ( ) ( ) 2x xE f X f= . That is,

( ) ( ) 2xR X fτ ←⎯→

But ( ) 2X f is the ESD of ( )x t . That is,

( ) ( ) ( ) 2x x xE f E f X f= =

Hence,

( ) ( )x xR E fτ ←⎯→ (1.45d)

It is to be noted that ( )xE f depends only on the magnitude spectrum,

( )X f . Let ( )x t and ( )y t be two signals such that ( ) ( )X f Y f= . Then,

( ) ( )x yR Rτ = τ . Note that if ( ) ( )x yf fθ ≠ θ , ( )x t may not have any

resemblance to ( )y t ; but their ACFs will be the same. In other words, ACF does

not provide a unique description of the signal. Given ( )x t , its ACF is unique;

but given some ACF, we can find many signals that have the given ACF.

P5) Let ( ) ( ) ( )x t y t v t= + . Then,

( ) ( ) ( ) ( ) ( )x y v y v v yR R R R Rτ = τ + τ + τ + τ (1.45e)

Proof: Exercise

If ( ) ( ) 0y v v yR Rτ = τ ≡ (that is, ( )y t and ( )v t∗ − τ are orthogonal for all

τ ), then, ( ) ( ) ( )x y vR R Rτ = τ + τ (1.45f)

In such a situation, ( )xR τ is the superposition of the ACFs of the components of

( )x t . This also leads to the superposition of the ESDs; that is

( ) ( ) ( )x y vE f E f E f= + (1.45g)

Properties of ACF (periodic signals) We list below the properties of the ACF of periodic signals. Proofs of these

properties are left as an exercise.



1.89

P1) ACF exhibits conjugate symmetry

( ) ( )p px xR R∗− τ = τ (1.46a)

P2) ( ) ( )0

0

2 2

0 02

1p p

T

x p xT

R x t d t PTτ =

−

τ = =∫ (1.46b)

where pxP denotes the average power of ( )px t (Sec. 1.2.2, Pg. 1.16).

P3) ( ) ( )0px xR Rτ ≤ (1.46c)

That is, the maximum value of ( )pxR τ occurs at the origin.

P4) ( ) ( )0 , 1, 2, 3, . . .p px xR nT R nτ ± = τ = (1.46d)

where 0T is the period of ( )px t . That is, the ACF of a periodic signal is also

periodic with the same period as that of the signal.

P5) Let ( )pxP f denote the Power Spectral Density (PSD) of ( )px t . That is,

( )p px xP f df P

∞

− ∞

=∫ . Then,

( ) ( )p px xR P fτ ←⎯→ (1.46e)

As ( )pxR τ is periodic, we expect the PSD to be purely impulsive.

Example 1.19

a) Let ( )x t be the signal shown in Fig. 1.29. Compute and sketch ( )xR τ .

Exercise 1.7

Let ( ) ( ) 0 0,2 2

0 ,

pT Tx t t

x toutside

⎧ − < <⎪= ⎨⎪⎩

and ( ) ( )x t X f←⎯→ .

Show that ( ) ( )2

20 00

1p px x

n

n nF R P f X fT TT

∞

= − ∞

⎛ ⎞ ⎛ ⎞⎡ ⎤τ = = δ −⎜ ⎟ ⎜ ⎟⎣ ⎦ ⎝ ⎠ ⎝ ⎠∑



1.90

b) ( )x t of part (a) is given as the input to an LTI system with the impulse

response ( )h t . If the output ( )y t of the system is ( )3xR t − , find ( )h t .

Fig. 1.29: ( )x t of Example 1.19

a) Computation of ( )xR τ :

We know that the maximum value of the ACF occurs at the origin; that is, at

0τ = and ( )0x xR E= .

( ) 10 1. 1 . 2 1.54xR = + =

Consider the product ( ) ( )x t x t − τ for 0 1< τ ≤ . As τ increases in this range,

the overlap between the positive parts of the pulses ( )x t and ( )x t − τ (and also

between the negative parts and these pulses) decreases, which implies a

decrease in the positive value for the integral of the product. In addition, a part of

( )x t − τ that is positive overlaps with the negative part of ( )x t , there by further

reducing the positive value of the ( ) ( )x t x t − τ∫ . (The student is advised to

make a sketch of ( )x t and ( )x t − τ .) This decrease is linear (with a constant

slope) until 1τ = . The value of ( )1xR is 14

⎛ ⎞−⎜ ⎟⎝ ⎠

. For 1 2< τ < , the positive part

of ( )x t − τ fully overlaps with the negative part of ( )x t ; this makes ( )xR τ further

negative and the ACF reaches its minimum value at 2τ = . As can easily be



1.91

checked, ( ) 122xR ⎛ ⎞= −⎜ ⎟

⎝ ⎠. As τ increases beyond 2, the ( )xR τ becomes less

and less negative and becomes zero at 3τ = . As ( ) ( )x xR R− τ = τ , we have all

the information necessary to sketch ( )xR τ , which is shown in Fig. 1.30.

Fig. 1.30: ACF of the signal of Example 1.19

b) Calculating ( )h t :

We have ( ) ( ) ( )y t x t h t= ∗

( )3xR t= −

( ) ( )3xR t t= ∗ δ −

But from Eq. 1.39(a), ( ) ( ) ( )xR t x t x t= ∗ − .

Hence, ( ) ( ) ( ) ( )3y t x t x t t⎡ ⎤= ∗ − ∗ δ −⎣ ⎦

( ) ( )3x t x t⎡ ⎤= ∗ − −⎣ ⎦

That is, ( ) ( )3h t x t⎡ ⎤= − −⎣ ⎦

which is sketched in Fig. 1.31.



1.92

Fig. 1.31 Impulse response of the LTI system of Example 1.19

Example 1.20

Let ( ) ( )0cosx t A t= ω + θ . We will find ( )xR τ .

Method 1:

( ) ( ) ( )0

0

22

0 00

2

1 cos cos

T

xT

R A t t dtT

−

⎡ ⎤τ = ω + θ ω − τ + θ⎣ ⎦∫

( ) 0

0

2 2

0 0 00

2

1 cos 2 2 cos2

T

T

A t dtT

−

= ω − ω τ + θ + ω τ∫

2

0cos2

A= ω τ

We find that:

(i) ( )xR τ is periodic with the same period as ( )x t .

(ii) Its maximum value occurs at 0τ =

(iii) The maximum value is 2

2A which is the average power of the signal.

(iv) ( )xR τ is independent of θ .



1.93

Method 2:

( ) ( ) ( )0 00cos

2 2j t j tA AA t e eω + θ − ω + θ⎡ ⎤ω + θ = +⎢ ⎥⎣ ⎦

( ) ( )v t w t= +

As ( ) ( ) ( )x t v t w t= + , we have

( ) ( ) ( ) ( ) ( )x v w v w w vR R R R Rτ = τ + τ + τ + τ .

It is not difficult to see that ( ) ( ) 0v w w vR Rτ = τ = for all τ . Hence,

( ) ( ) ( )x v wR R Rτ = τ + τ

But ( ) ( )w vR R∗τ = τ . Hence,

( ) ( )2 Rex vR R⎡ ⎤τ = τ⎣ ⎦

( ) ( ) ( )0 0

00 0

0 0

2 22 2

0 02 2

14 4

T T

j tj t jv

T T

A AR e e dt e dtT T

⎡ ⎤− ω − τ + θω + θ ω τ⎣ ⎦

− −

⎡ ⎤⎢ ⎥τ = =⎢ ⎥⎢ ⎥⎣ ⎦

∫ ∫

02

4jA e ω τ=

Hence ( ) ( )2

0cos2x

AR τ = ω τ

Example 1.21

Let ( ) ( )sin 2 , 0 2x t t t= π ≤ ≤ . Let us find its ACF and sketch it.

In Fig. 1.32, we show ( )x t and ( )x t − τ for 0 2< τ < . Note that if 2τ > ,

( ) ( ) 0x t x t − τ = which implies ( ) 0xR τ = for 2τ > .



1.94

Fig. 1.32: (a) Sinusoidal pulse of Example 1.21 and (b) its shifted version

( ) ( ) ( )2

sin 2 sin 2xR t t dtτ

⎡ ⎤τ = π π − τ⎣ ⎦∫

( ) ( )2 cos 2 cos 4 22

tdt

τ

πτ − π − πτ= ∫

( ) 22 sin 4 2cos 22 4

tt

τ τ

⎡ ⎤π − πτπτ⎡ ⎤= − ⎢ ⎥⎢ ⎥ π⎣ ⎦ ⎣ ⎦

( ) ( ) ( ) ( )τ πτ π − πτ − πτ= πτ − −

πcos 2 sin 8 2 sin 2

cos 22 4

( ) cos 2 sin 2cos 2 , 0 22 2

τ πτ πτ= πτ − + ≤ τ ≤

π

As ( ) ( )x xR R− τ = τ , we have

( ) ( ) ( ) ( )

,

sin 2cos 2cos 2 , 2

2 20

xRotherwise

⎧ π ττ πτπτ − + τ ≤⎪τ = π⎨

⎪⎩

( )xR τ is plotted in Fig. 1.33.



1.95

Fig. 1.33: ( )xR τ of Example 1.21

1.7 Hilbert Transform Let ( )x t be the input of an LTI system with the impulse response ( )h t .

Then, the output ( )y t is

( ) ( ) ( )y t x t h t= ∗

or ( ) ( ) ( )Y f X f H f= ∗

That is, ( ) ( ) ( )Y f X f H f= (1.47a)

( ) ( ) ( )y x hf f fθ = θ + θ (1.47b)

From Eq. 1.47 we see that an LTI system alters, in general, both the magnitude

spectrum and the phase spectrum of the input signal. However, there are certain

networks, called all pass networks, which would alter only the (input) phase

spectrum. That is, if ( )H f is the frequency response of an all pass network, then



1.96

( ) ( )Y f X f=

( ) ( ) ( )y x hf f fθ = θ + θ

An interesting case of all-pass network is the ideal delay, with the impulse

response ( ) ( )dh t t t= δ − . Though ( ) ( )y xf fθ ≠ θ , phase shift imparted to each

input spectral component is proportional to the frequency, the proportionality

constant being 2 dtπ . Another interesting network is the Hilbert transformer. Its

output is characterized by:

(i) ( ) ( )Y f X f= , 0f ≠ and

(ii) ( )( )

( )

, 02

, 02

x

y

x

f ff

f f

π⎧− + θ >⎪⎪θ = ⎨ π⎪ + θ <⎪⎩

That is, a Hilbert transform is essentially a 2π⎛ ⎞±⎜ ⎟

⎝ ⎠ phase shifter.

Hence, we define the Hilbert transformer in the frequency domain, with the

frequency response function

( ) ( )sgnH f j f= − (1.48a)

where ( )1 , 0

sgn 0 , 01 , 0

ff f

f

>⎧⎪= =⎨⎪− <⎩

As ( ) 1sgnj ft

⎡ ⎤− ←⎯→⎣ ⎦ π,

( ) 1h tt

=π

(1.48b)

When ( )x t is the input to a Hilbert transformer, we denote its output as ( )x t

where

( ) ( ) 1x t x tt

= ∗π



1.97

( )( )

1 xd

t

∞

− ∞

τ= τ

π − τ∫ 1 (1.49a)

and ( ) ( ) ( )ˆ sgnX f j f X f= − (1.49b)

( )x t is called the Hilbert transform of ( )x t .

Note: Unlike other transforms, both ( )x t and ( )x t are functions of the same

variable ( t in our case).

Hilbert Transform (HT) will prove quite useful later on in the study of

bandpass signals and single sideband signals. For the present, let us look at

some examples of HT.

Example 1.22

Hilbert transform ( )tδ .

Let ( ) ( )x t t= δ . Let us find ( )x t .

As ( ) 1tδ ←⎯→ , we have

( ) ( )ˆ sgnF t j f⎡ ⎤δ = − ⇒⎣ ⎦

( ) 1ˆ tt

δ =π

This also establishes the relation, ( ) 1 1 !!tt t

⎡ ⎤δ ∗ =⎢ ⎥π π⎣ ⎦

Example 1.23 HT of a cosine signal.

Let ( ) ( )0cos 2x t f t= π . Let us find ( )x t .

1 This integral is actually Cauchy’s principal value.



1.98

( ) ( ) ( )0 012

X f f f f f⎡ ⎤= δ − + δ +⎣ ⎦

( ) ( ) ( )ˆ sgnX f j f X f= − . That is,

( ) ( ) ( )0 01ˆ

2X f f f f f

j⎡ ⎤= δ − − δ +⎣ ⎦

That is, ( ) ( )0ˆ sin 2x t f t= π .

Alternatively,

if ( ) 021

j f tx t e π= , then ( ) ( )02 21ˆ

j f tx t e

ππ −=

and if ( ) 022

j f tx t e− π= , then ( ) ( )02 22ˆ

j f tx t e

π− π −=

Hence,

( ) ( ) ( ) ( )⎡ ⎤= π = +⎣ ⎦x t f t x t x t0 1 21cos 22

has ( ) 0ˆ cos2

x t t π⎛ ⎞= ω −⎜ ⎟⎝ ⎠

( )0sin t= ω

Similarly, we can show that if ( ) ( )0sin 2x t f t= π , then ( ) 0ˆ cosx t t= − ω .

Example 1.24

Let ( ) 21

1x t

t=

+. Let us find ( )x t .

( ) ( ) 1x t x tt

= ∗π

( )( )2

1 11

dt

∞

− ∞

= τπ + τ − τ∫

( )2 2

1 11 1

t ddtt

∞ ∞

− ∞ − ∞

⎡ ⎤+ τ τ⎢ ⎥= τ +

π − τ+ + τ⎢ ⎥⎣ ⎦∫ ∫

As 21 0

1d d

t

∞ ∞

− ∞ − ∞

ττ = τ =

− τ+ τ∫ ∫ , we have



1.99

( ) 2 21 1ˆ

1 1tx t dt

∞

− ∞

⎡ ⎤⎢ ⎥= τ

π + + τ⎢ ⎥⎣ ⎦∫

As the bracketed integral is equal to π , we have

( ) 2ˆ

1tx tt

=+

Example 1.25

Let ( ) ( ) cos 2 cx t m t f t= π where ( )m t is a lowpass signal with

( ) 0M f = for f W> and cf W> . We will show that

( ) ( ) ( ) ( ) ( )ˆ sin 2 cos 2c cx t m t f t m t f t= π = π .

( ) ( ) ( )12 c cX f M f f M f f⎡ ⎤= − + +⎣ ⎦

( ) ( ) ( ) ( )1ˆ sgn2 c cX f M f f M f f j f⎡ ⎤ ⎡ ⎤= − + + −⎣ ⎦ ⎣ ⎦

But ( )cM f f⎡ ⎤−⎣ ⎦ is nonzero only for 0f >

and ( )cM f f⎡ ⎤+⎣ ⎦ is nonzero only for 0f < .

Hence,

( )( )

( )

2

2

1 , 02ˆ1 , 02

jc

jc

M f f e fX f

M f f e f

π−

π

⎧ − >⎪⎪= ⎨⎪ + <⎪⎩

Exercise 1.8

(a) Let ( )1sin tx t

t= . Show that ( )1

1 cosˆ tx tt

−=

(b) Let ( ) ( )2x t ga t= . Show that ( )2

11 2ˆ ln 1

2

tx t

t

−= −

π +



1.100

Consider ( ) ( ) ( ) ( ) ( )1sin 22c c cm t f t M f f f f f

j⎡ ⎤π ←⎯→ ∗ δ − − δ +⎣ ⎦

( ) ( ) ( )2 212

j jjc cM f f f e e e f f

π π− −π

⎡ ⎤←⎯→ ∗ δ − + δ +⎢ ⎥

⎢ ⎥⎣ ⎦

( ) ( ) ( )2 212

j jc cM f f f e e f f

π π−⎡ ⎤

←⎯→ ∗ δ − + δ +⎢ ⎥⎢ ⎥⎣ ⎦

That is,

( ) ( )( )

( )

2

2

1 , 02sin 21 , 02

jc

cj

c

M f f e fF m t f t

M f f e f

π−

π

⎧− >⎪

⎪⎡ ⎤π = ⎨⎣ ⎦⎪ + <⎪⎩

As ( ) ( )ˆsin 2 cF m t f t X f⎡ ⎤π =⎣ ⎦ , we have

( ) ( ) ( ) ( ) ( )ˆ sin 2 cos 2c cx t m t f t m t f t= π = π

Note: It is possible to establish even a stronger result, which is stated below.

Let ( ) ( ) ( )x t m t v t= where ( )m t is a lowpass signal with ( ) 0M f = for

f W> and ( )v t is a high-pass signal with ( ) 0V f = for f W< . Then

( ) ( ) ( )ˆ ˆx t m t v t= . [We assume that there are no impulses in either ( )M f or

( )V f .

1.7.1 Properties of Hilbert transform

Our area of application of HT is real signals. Hence, we develop the

properties of HT as applied to real signals. We assume that the signals under

consideration have no impulses in their spectra at 0f = .

P1) A signal ( )x t and its HT, ( )x t , have the same energy.

Proof: ( ) 2xE X f d f

∞

− ∞

= ∫



1.101

As ( ) ( ) ( )ˆ sgnX f j f X f= −

( ) ( )X f X f= . Hence =x xE E ˆ .

Note: Though ( )0

ˆf

X f=

is zero, it will not change the value of the integral; and

hence the energy.

P2) ( ) ( )ˆHT x t x t⎡ ⎤ = −⎣ ⎦

That is, applying the HT twice on a given signal ( )x t changes the sign of the

signal. Intuitively, this is satisfying. Each time we perform HT, we change the

phase of a spectral component in ( )X f by 090 . Hence, performing the

transformation twice results in a phase shift of 0180 .

Proof: ( ) ( ) ( )ˆ sgnx t j f X f←⎯→ −

( ) ( ) ( ) ( )ˆ sgn sgnHT x t j f j f X f⎡ ⎤ ⎡ ⎤←⎯→ − −⎣ ⎦ ⎣ ⎦

( )( ) ( ) ( )2sgnj f X f X f= − = −

Example 1.26

Let ( ) 1x tt

= . We shall find ( )x t .

From Example 1.22, we have ( ) 1ˆ tt

δ =π

.

Hence,

( ) ( ) 1ˆHT t t HTt

⎡ ⎤⎡ ⎤δ = − δ = ⎢ ⎥⎣ ⎦ π⎣ ⎦. That is,

( )1HT tt⎡ ⎤ = − πδ⎢ ⎥⎣ ⎦

P3) A signal ( )x t and its HT, ( )x t , are orthogonal.



1.102

Proof: We know that

( ) ( ) ( ) ( )ˆˆx t x t d t X f X f d f∞ ∞

− ∞ − ∞

= −∫ ∫

(See the note, property P10, Sec. 1.4). Hence,

( ) ( ) ( ) ( ) ( )ˆ sgnx t x t d t X f j f X f d f∞ ∞

− ∞ − ∞

⎡ ⎤= − − −⎣ ⎦∫ ∫

As ( )x t is real, ( ) ( )X f X f∗− = . Therefore,

( ) ( ) ( ) ( ) 2ˆ sgnx t x t d t j f X f d f∞ ∞

− ∞ − ∞

= − −∫ ∫

0= (Note that the integrand on the RHS is odd)

P4) ( )HT x t⎡ ⎤α⎣ ⎦ , where α is a nonzero constant is ( ) ( )ˆsgn x t⎡ ⎤α α⎣ ⎦ .

Proof: We will first establish that if ( ) ( ) ( )y t x t h t= ∗ and

( ) ( ) ( )z t x t h t= α ∗ α , then ( ) ( )1z t y t= αα

.

If ( ) ( ) ( )z t x t h t= α ∗ α , then

( ) 21 f fZ f X H

a a⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠α

Also, ( ) 1 fy t Y ⎛ ⎞α ←⎯→ ⎜ ⎟α α⎝ ⎠, where

f f fY X H⎛ ⎞ ⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟ ⎜ ⎟α α α⎝ ⎠ ⎝ ⎠ ⎝ ⎠

( )2 Z f= α

Hence, ( ) ( ) ( )21y t Z f Z fα ←⎯→ α = αα

That is, ( ) ( )y tz t

α=

α



1.103

( ) ( ) 1HT x t x tt

⎡ ⎤α = α ∗⎣ ⎦ π

( ) 1x tt

⎛ ⎞= α ∗ α⎜ ⎟απ⎝ ⎠

( ) 1x tt

⎡ ⎤= α α ∗⎢ ⎥απ⎣ ⎦

As ( ) ( )1 ˆx t x tt

⎡ ⎤∗ =⎢ ⎥π⎣ ⎦

, we have ( ) ( )ˆ1 x tx t

tα⎡ ⎤

α ∗ =⎢ ⎥πα α⎣ ⎦.

Hence ( ) ( ) ( ) ( )ˆ ˆsgnHT x t x t x tα⎡ ⎤α = α = α α⎣ ⎦ α

.

As a simple illustration of the property, let ( ) 21

1x t

t=

+ and 2α = . Then

( ) 21

1 4x t

tα =

+. Let us obtain 2

11 4

HTt

⎡ ⎤⎢ ⎥+⎣ ⎦

using P4.

As ( ) 2ˆ

1tx tt

=+

, ( )( )2 2 2

1 2 2sgn 21 4 1 41 2

t tHTt tt

⎡ ⎤⎡ ⎤⎢ ⎥= =⎢ ⎥+ +⎢ ⎥+⎣ ⎦ ⎣ ⎦

If 2α = − , ( ) 21

1 4x t

tα =

+ which is the same as with 2α = .

With 2α = − , ( ) ( ) 22ˆsgn

1 4tx tt

⎛ ⎞−α α = − ⎜ ⎟

+⎝ ⎠

22

1 4tt

=+

, as required

Exercise 1.9

Using the result sin 1 cost tHTt t

−⎛ ⎞ =⎜ ⎟⎝ ⎠

,

find the ( )sinHT c t⎡ ⎤⎣ ⎦ .



1.104

P5) The cross correlation function of ( )x t and ( )x t , ( )ˆx xR τ is the negative of

the HT of ( )ˆxR τ . That is,

( ) ( )ˆˆ

xx xR Rτ = − τ

Proof: ( ) ( ) ( )ˆ ˆx xF R F x x⎡ ⎤ ⎡ ⎤τ = τ ∗ − τ⎣ ⎦⎣ ⎦

( ) ( ) ( )sgnX f j f X f⎡ ⎤= − − −⎣ ⎦

( ) ( ) ( ) ( )2 2sgn sgnX f j f X f j f⎡ ⎤ ⎡ ⎤= − − =⎣ ⎦ ⎣ ⎦

That is,

( ) ( )ˆˆ

xx xR Rτ = − τ

1.8 Bandpass Signals Consider a communication system that transmits the signal

( ) ( ) cos 2 cs t m t f t= π , where ( )m t is a (lowpass) message signal and

( )cos 2 cf tπ is the (high-frequency) carrier term. Then the spectrum ( )S f of the

transmitted signal is ( ) ( ) ( )12 c cS f M f f M f f⎡ ⎤= − + +⎣ ⎦ . If ( )M f is as shown in

Fig 1.34(a), then for a carrier frequency 100cf k Hz= , ( )S f will be as shown in

Fig 1.34(b).

Exercise 1.10

Show that ( ) ( )ˆˆ

xx xR Rτ = τ .



1.105

Fig. 1.34: A typical narrowband, bandpass spectrum

We see that the spectrum of ( )s t is confined to the frequency interval

95 105f k Hz≤ ≤ . Where as ( )m t is a lowpass signal, ( )s t is a bandpass

signal. Moreover ( )s t is a narrowband, bandpass signal because the spectral

width of ( )S f , ( )105 95 10 k Hz− = , is quite small in comparison with the carrier

frequency cf of 100 k Hz . Hence, we call ( )s t as a Narrowband, Bandpass

(NBBP) signal. Fig 1.35 shows some more spectra that represent NBBP signals.



1.106

Fig. 1.35: A few more examples of NBBP spectra

NBBP signals play an important role in the communication process. Let us

assume that the rest of the communication system (channel, a part of the

receiver etc.) is also of the bandpass variety. The study of such transmission-

reception schemes becomes a little complicated because of the presence of the

carrier term in some form or the other (The term cos ctω in ( ) cos cm t tω is

meant only to “carry” the information ( )m t and is not part of the information) If

we develop tools to study bandpass signals and bandpass systems, independent

of the carrier, the analysis of the communications schemes would become

somewhat simplified. (That is, bandpass signals and bandpass systems are

studied in terms of their lowpass equivalents.) The mathematical concepts of pre-

envelope and complex envelope have been developed for this purpose. We shall

make use of these concepts in our studies on linear modulation and angle

modulation.



1.107

1.8.1 Pre-envelope Def. 1.10: Let ( )x t be any real signal with the FT, ( )X f . We define its pre-

envelope as,

( ) ( ) ( )ˆpex t x t j x t= + (1.50a)

Taking the Fourier transform of Eq. 1.54(a), we have

( ) ( ) ( ) ( )sgnpeX f X f j j f X f⎡ ⎤= + −⎣ ⎦

( ) ( ) ( )sgnX f f X f= +

That is,

( )( )( )

2 , 0

0 , 00 , 0

pe

X f f

X f X ff

⎧ >⎪

= =⎨⎪ <⎩

(1.50b)

(We assume that ( )X f has no impulse at 0f = ). That is, ( )pex t has spectrum

only for 0f ≥ , even though ( )X f is two sided (As ( )pex t has spectral

components only for 0f ≥ , some authors use the symbol ( )x t+ to denote the

pre-envelope of ( )x t ). Of course, ( )pex t∗ will have spectrum only for 0f ≤ ).

Consider the signals ( )1x t and ( )2x t whose spectra are shown in Fig. 1.36.



1.108

Fig. 1.36: Typical two-sided spectra

The corresponding ( )1, peX f and ( )2, peX f are as shown in Fig. 1.37(a) and (b)

respectively.

Fig. 1.37: Fourier transform of (a) ( )1, pex t and (b) ( )2, pex t of the signals in



1.109

Because of Eq. 1.50(b), we can write

( ) ( ) 2

0

2 j f tpex t X f e d f

∞π= ∫ (1.51)

Example 1.27

Let ( ) 21

1x t

t=

+. Let us find ( )peX f and ( )pex t .

First, let us compute ( )X f . From Example 1.5, we know that

22

11

fet

− π←⎯→ π+

.

Hence ( ) ( )22 fpeX f e u f− π= π , where ( )

1 , 01 , 020 , 0

f

u f f

f

>⎧⎪⎪= =⎨⎪

<⎪⎩

.

We require ( )1peF X f− ⎡ ⎤⎣ ⎦ .

As ( )ex tj f11

1 2←⎯→

+ π

( )ex tj f

1 11 22 1

π ←⎯→π +

.

From duality, ( ) ( )fex f e u fj t

2 12 1 2 21

− ππ π = π ←⎯→−

.

That is, ( ) 21 1

1 1pej tx t

j t t+

= =− +

2 21

1 1tj

t t= +

+ +.

Then, ( ) 2ˆ

1tx tt

=+

, which is a known result.



1.110

As can be seen from the above discussion, the spectrum of ( )pex t is still

bandpass (though one-sided) if ( )x t is a bandpass signal. Let ( )x t be a

bandpass signal with ( )peX f “centered” with respect to 102cf k Hz= as shown

in Fig. 1.38(a).

Fig 1.38: A typical ( )peX f and shifted version

(We can treat cf to be the center frequency in Fig. 1.38(a), by taking the

bandpass spectrum from 94 to 110 k Hz , though the spectrum is zero for the

frequency range 94 to100 k Hz .) From Fig 1.38(b), we see that ( )pe cX f f+ is a

lowpass spectrum, nonzero in the frequency range ( )2− to 8 k Hz .



1.111

1.8.2 Complex envelope Def. 1.11: We now define the complex envelope of ( )x t , denoted ( )c ex t as

( ) ( ) 2 cj f tc e pex t x t e− π= (1.52a)

Eq. 1.52(a) implies

( ) ( )c e pe cX f X f f= + (1.52b)

(We assume that we know the center frequency cf and it is such that

( )pe cX f f+ is lowpass in nature).

Equation 1.52(a) also implies

( ) ( ) 2 cj f tpe c ex t x t e π= (1.53)

( )c ex t is also referred to as the equivalent lowpass signal of the bandpass signal

( )x t . In general ( )c ex t is complex. Let ( )cx t be the real part and ( )sx t its

imaginary part. Then,

( ) ( ) ( )c e c sx t x t j x t= + (1.54)

We will show a little later that, both ( )cx t and ( )sx t are lowpass in nature. Using

Eq. 1.54 in Eq. 1.53, we obtain,

( ) ( ) ( ) cj f tpe c sx t x t j x t e 2π⎡ ⎤= +⎣ ⎦

As the real bandpass signal ( )x t is the real part of ( )pex t , we have

( ) ( ) ( ) ( ) ( )c c s cx t x t f t x t f tcos 2 sin 2= π − π (1.55)

Eq. 1.55 is referred to as the canonical representation of the bandpass signal.

( )cx t , which is the coefficient of the cosine term, is usually referred to as the in-

phase component and ( )sx t , the coefficient of the sine term, as the quadrature

component. Note that ( )sin 2 cf tπ is in phase quadrature to ( )cos 2 cf tπ . (It is

also common in the literature to use the symbol ( )Ix t for the in-phase

component and ( )Qx t for the quadrature component.)



1.112

To express Eq. 1.54 in polar form, let

( ) ( ) ( )c sA t x t x t2 2= + (1.56a)

( ) ( )( )

s

c

x tt

x t1tan− ⎡ ⎤

ϕ = ⎢ ⎥⎢ ⎥⎣ ⎦

(1.56b)

Then,

( ) ( ) ( )j tcex t A t e ϕ= (1.56c)

( ) ( ) ( ) 2Re Re cj f tpe c ex t x t x t e π⎡ ⎤⎡ ⎤= =⎣ ⎦ ⎣ ⎦ (1.57)

( ) ( ) 2Re cj t j f tA t e eϕ π⎡ ⎤= ⎣ ⎦ (1.58)

Eq. 1.58 resembles phasor representation of a sinusoid. We know that,

( ) 2cos 2 Re cj f tjcA f t A e e πϕ⎡ ⎤π + ϕ = ⎣ ⎦ (1.59)

jA e ϕ⎡ ⎤⎣ ⎦ is generally referred to the as the phasor associated with the sinusoidal

signal ( )cos 2 cA f tπ + ϕ . (The phasor is a complex number providing information

about the amplitude and phase (at 0t = ) of the sinusoid.) The quantity

2 cj f tjA e e πϕ⎡ ⎤⎣ ⎦ can be treated as a rotating vector as shown in Fig. 1.39(a).

Comparing Eq. 1.58 with Eq. 1.59, we find that they have a close resemblance.

Phasor of the monochromatic (single frequency) signal has constant amplitude

A and a fixed phase ϕ . In the case of the complex envelope (of a narrowband

signal) both ( )A t and ( )tϕ are, (slowly) time-varying. This is shown in Fig

1.39(b). (Note that a single frequency sinusoid is the extreme case of a

narrowband signal with zero spectral width!)



1.113

Fig. 1.39: (a) Phasor representation of ( )2 cj f tA e π + ϕ

(b) Complex envelope as a (slowly) varying amplitude and phase

In other words, complex envelope can be treated as a generalization of

the phasor representation used for single frequency sinusoids; the generalization

permits the amplitude and phase to change as a function of time. Note, however,

that for a given time 1t t= , ( ) ( )11

j tA t e ϕ is a complex number, having the

necessary information about the narrowband signal. As we shall see later,

different modulation schemes are basically different methods of controlling either

( )A t or ( )tϕ (or both) as a function of the message signal ( )m t .

If ( )x t is a NBBP signal with spectrum confined to the frequency range

cf f W± ≤ , then ( )cx t and ( )sx t are lowpass signals with spectrum confined



1.114

to f W≤ . This is because ( ) ( )c e pe cX f X f f= + is nonzero only for f W≤ .

As ( )cx t is the real part of ( )c ex t , we have

( )( ) ( )c e c e

c

x t x tx t

2

∗⎡ ⎤+⎣ ⎦=

or ( ) ( ) ( ) ( )c e c ec c

X f X fF x t X f

2

∗+ −⎡ ⎤ = =⎣ ⎦

As ( )c eX f and ( )c eX f∗ − are zero for f W> , ( )cX f is also zero for f W> .

Similarly, ( )sx t is also a lowpass signal with ( )sX f 0= for f W> .

The scheme shown below (Fig 1.40) enables us to obtain ( )cx t and ( )sx t

from ( )x t .

Fig. 1.40: Scheme for the recovery of ( )cx t and ( )sx t from ( )x t

In Fig 1.40, ( )x t a NBBP signal, with the spectrum confined to the interval

cf f W± ≤ , where cW f<< .

( ) ( )1 2 cos cv t x t t= ω

( ) ( )( )c c s c cx t t x t t t2 cos sin cos= ω − ω ω

( ) ( )c c s cx t t x t t22 cos sin 2= ω − ω



1.115

( ) ( )cc s c

tx t x t t1 cos 22 sin 22

+ ω⎡ ⎤= − ω⎢ ⎥⎣ ⎦

( ) ( ) ( ) ( )c c c s cx t x t t x t tcos 2 sin 2= + ω − ω (1.60)

In Fig. 1.40, ( )l pH f is an Ideal Lowpass Filter (ILPF) with the frequency

response.

( )1 ,0 ,l p

f WH f

outside

⎧ ≤⎪= ⎨⎪⎩

( )cx t and ( )sx t are lowpass signals, band limited to W Hz. ( ) ( )c cx t tcos 2ω

and ( ) ( )s cx t tsin 2ω have bandpass spectra centered at cf± . These quantities

will be filtered out by the ILPF and at the output of the top channel, we obtain

( )cx t . Similar analysis will show that the output of the bottom channel is ( )sx t .

From Eq. 1.58, we have,

( ) ( ) ( )cos cx t A t t t⎡ ⎤= ω + ϕ⎣ ⎦ (1.61)

Eq. 1.61 is referred to as the envelope and phase representation of ( )x t . ( )A t

is called the natural envelope (or simply the envelope) of ( )x t and ( )tϕ , its

phase. As we assume that cf is known, the information about ( )x t is contained

in either of the quantities ( ) ( )( )c sx t x t, or ( ) ( ),A t t⎡ ⎤ϕ⎣ ⎦ and these are lowpass in

nature. Note that

( ) ( ) ( )ce peA t x t x t= = (1.62)

and is always non-negative.

We shall now illustrate the concepts of ( )pex t , ( )cex t , ( )A t and ( )tϕ with

the help of a few examples.



1.116

Example 1.28

Let ( ) [ ]cos 2 cx t f t= π . Let us find ( )pex t , ( )cex t , ( )A t and ( )tϕ .

( ) ( ) ( )12 c cX f f f f f⎡ ⎤= δ − + δ +⎣ ⎦

From Eq. 1.50(b), we have

Hence ( ) ( ) ( ) 2 cj f tpe c peX f f f x t e π= δ − ⇒ =

As ( ) ( )ceX f f= δ , we obtain ( ) 1cex t =

As ( )cex t is real and positive, ( ) 0tϕ = and ( ) ( ) 1ceA t x t= = .

Example 1.29

Let ( ) cos ctx t ga tT⎛ ⎞= ω⎜ ⎟⎝ ⎠

. Assume that 1cf T >> so that ( )x t can be

taken as a NBBP signal. We shall find ( )pex t , ( )cex t and ( )A t .

Method 1 (Frequency domain):

Because of the assumption 1cf T >> , we can take ( )X f approximately as

( )( )

( )

sin , 02

sin , 02

c

c

T c f f T fX f

T c f f T f

⎧ ⎡ ⎤− >⎣ ⎦⎪⎪⎨⎪ ⎡ ⎤+ <⎣ ⎦⎪⎩

From Eq. 1.50(b),

( ) ( )sin , 0

0 ,c

peT c f f T f

X fotherwise

⎧ ⎡ ⎤− >⎪ ⎣ ⎦= ⎨⎪⎩

Hence,

( ) 2 cj f tpe

tx t ga eT

π⎛ ⎞= ⎜ ⎟⎝ ⎠



1.117

( ) ( ) 2 cj f tce pex t x t e− π=

tgaT⎛ ⎞= ⎜ ⎟⎝ ⎠

As ( )cex t is real, we have ( ) ( )= ⇒ ϕ =sx t t0 0 .

( ) ( )cetA t x t g aT⎛ ⎞= = ⎜ ⎟⎝ ⎠

Note that for Examples 1.28 and 1.29, the complex envelope is real valued and is

equal to the envelope, ( )A t .

Method 2 (Time domain):

Comparing the given ( )x t with Eq. 1.55, we find that it is already in the

canonic form with ( )ctx t gaT⎛ ⎞= ⎜ ⎟⎝ ⎠

, ( )sx t 0=

Hence, ( )cetx t gaT⎛ ⎞= ⎜ ⎟⎝ ⎠

And ( ) ( ) 2 cj f tpe cex t x t e π=

2 cj f ttga eT

π⎛ ⎞= ⎜ ⎟⎝ ⎠

( ) tA t gaT⎛ ⎞= ⎜ ⎟⎝ ⎠

and ( ) 0tϕ =

Example 1.30

( )x t is a NBBP signal with ( )X f as shown in Fig 1.41. Let us find ( )cx t

and ( )sx t .



1.118

Fig. 1.41: Spectrum of the NBBP signal of Example 1.30

From the given spectrum, we can obtain ( )ceX f , which is shown below in

Fig. 1.42.

Fig. 1.42: Spectrum of the complex envelope of the signal of Example 1.30

As shown in the Fig 1.42, we can take ( )ceX f as the sum of A and B.

Inverse Fourier transform of A : ( ) 100200 sin 100 j tc t e− π

Inverse Fourier transform of B : ( ) 5050 sin 50 j tc t e π

Hence ( ) ( ) ( )100 50200 sin 100 50 sin 50j t j tcex t c t e c t e− π π= +

( ) ( ) ( ) ( ) ( )cx t c t t c t t200 sin 100 cos 100 50 sin 50 cos 50= π + π

( ) ( ) ( ) ( ) ( )= π − πsx t c t t c t t50 sin 50 sin 50 200 sin 100 sin 100



1.119

1.9 Bandpass (BP) Systems Let a signal ( )x t be input to an LTI system with impulse response ( )h t . If

( )y t is the output, then ( ) ( ) ( )Y f X f H f= and

( ) ( ) ( )2 2Y f X f H f

2=

If ( )xE f is the energy spectral density of ( )x t , then

( ) ( ) ( ) 2y xE f E f H f= (1.63a)

Let ( ) ( )2hH f R←⎯→ τ be the Fourier transform pair where ( )hR τ is the ACF

of the impulse response ( )h t . Then,

( ) ( ) ( )y x hR R Rτ = τ ∗ τ

( ) ( ) ( )xR h h∗⎡ ⎤= τ ∗ τ ∗ − τ⎣ ⎦ (1.63b)

Exercise 1.11

Find the pre-envelope of ( ) ( )sinx t c t= .

Hint: use the result of Exercise 1.9.

Exercise 1.12

Let ( ) ( ) ( )cx t ex t t1 sin ⎡ ⎤= ω + ∆ω⎣ ⎦ where cω >> ∆ω .

Find ( )cex t .

Exercise 1.13 Find the expression for the envelope of

( ) ( ) ( )1 cos cosm cx t k t t⎡ ⎤= + ω ω⎣ ⎦ where k is a real constant and

c mf f>> . Sketch it for 0.5 and 1.5k = .



1.120

Let ( )x t be a bandpass signal with ( )X f zero for cf f W± > . A

bandpass signal is usually processed a bandpass system; that is, a system with

passband in the interval cf f B± < where B W≤ . We would like to study the

effect of a BP system on a BP input. This study is again greatly facilitated if we

were to use complex envelopes of the signals involved.

In appendix A1.3 (Eq A1.3.8), it is shown that

( ) ( ) ( )12ce ce cey t x t h t⎡ ⎤= ∗⎣ ⎦ (1.64)

We shall now give an example to illustrate the use of Eq. 1.64.

Example 1.31

Let ( )x t be a sinusoidal pulse given by

( ) ( )62 cos 2 10 , 0 1 sec

0 ,

t t mx t

outside

⎧ ⎡ ⎤π ≤ ≤⎪ ⎣ ⎦= ⎨⎪⎩

( )x t is the input to an LTI system with impulse response ( ) ( )h t x T t= − ,

where 1 secT m= . Find ( )cey t and ( )y t .

( )x t can be taken as a NBBP signal. Its complex envelope,

( ) 2 , 0 1 seccex t t m= ≤ ≤

( ) ( ) ( )62 cos 2 10h t T t⎡ ⎤= π −⎣ ⎦

( ) ( ) ( ) ( ) 6 6 6 62 cos 2 10 cos 2 10 sin 2 10 sin 2 10T t T t= π π + π π

( ) ( ) ( ) 3 6 3 62 cos 2 10 cos 2 10 sin 2 10 sin 2 10t t⎡ ⎤= π × π + π × π ×⎣ ⎦

( )62 cos 2 10 , 0 1 sect t m⎡ ⎤= π ≤ ≤⎣ ⎦

Again, we can treat ( )h t as a NBBP signal, with



1.121

( ) 2 , 0 1 sec0 ,ce

t mh t

outside≤ ≤⎧

= ⎨⎩

That is, ( ) ( ) 22ce ce

Ttx t h t ga

T

⎛ ⎞−⎜ ⎟= =⎜ ⎟⎝ ⎠

From Eq.1.64,

( ) ( ) ( ) ( )31 2 102ce ce ce

t Ty t x t h t triT

− −⎛ ⎞⎡ ⎤= ∗ = × ⎜ ⎟⎣ ⎦ ⎝ ⎠

( ) ( ) ( )cos 2ce cy t y t f t= π

Obtaining ( )y t directly as ( ) ( )x t h t⎡ ⎤∗⎣ ⎦ would be quite cumbersome.

Exercise 1.14

Let ( ) ( ) ( )6sin 200 cos 2 10x t c t t⎡ ⎤= π⎣ ⎦ be the input to an NBBP

system with ( ) ( ) ( )6 61 1

1 10 102

H f H f H fj⎡ ⎤= − − +⎣ ⎦ (1.65)

where ( )1H f is as shown in Fig. 1.43.

Fig. 1.43: ( )1H f of Eq. 1.65

Find ( )cey t , the complex envelope of the output.



1.122

Appendix A1.1

Tabulation of ( )sinc λ

λ ( )sin λc λ ( )sin λc λ ( )sin λc 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50 1.55 1.60 1.65

1.000000 0.995893 0.983631 0.963397 0.935489 0.900316 0.858393 0.810331 0.756826 0.698645 0.636619 0.571619 0.504550 0.436331 0.367882 0.300104 0.233871 0.170010 0.109291 0.052414 -0.000001 -0.047424 -0.089422 -0.125661 -0.155915 -0.180064 -0.198091 -0.210086 -0.216236 -0.216821 -0.212203 -0.202833 -0.189207 -0.171888

1.70 1.75 1.80 1.85 1.90 1.95 2.00 2.05 2.10 2.15 2.20 2.25 2.30 2.35 2.40 2.45 2.50 2.55 2.60 2.65 2.70 2.75 2.80 2.85 2.90 2.95 3.00 3.05 3.10 3.15 3.20 3.25 3.30 3.35

-0.151481 -0.128616-0.103943 -0.078113 -0.051770 -0.025536 0.000000 0.024290 0.046840 0.067214 0.085045 0.100035 0.111964 0.120688 0.126138 0.128323 0.127324 0.123291 0.116435 0.107025 0.095377 0.081847 0.066821 0.050705 0.033919 0.016880 0.000000 -0.016326 -0.031730 -0.045876 -0.058468 -0.069255 -0.078036 -0.084661

3.40 3.45 3.50 3.55 3.60 3.65 3.70 3.75 3.80 3.85 3.90 3.95 4.00 4.05 4.10 4.15 4.20 4.25 4.30 4.35 4.40 4.45 4.50 4.55 4.60 4.65 4.70 4.75 4.80 4.85 4.90 4.95 5.00

-0.089038 -0.091128 -0.090946 -0.088561 -0.084092 -0.077703 -0.069600 -0.060021 -0.049237 -0.037535 -0.025222 -0.012607 -0.000000 0.012295 0.023991 0.034821 0.044547 0.052960 0.059888 0.065199 0.068802 0.070650 0.070736 0.069097 0.065811 0.060993 0.054791 0.047385 0.038979 0.029796 0.020074 0.010059 0.000000



1.123

Fig. A1.1: Plot of ( )sinc λ for 5λ ≤



1.124

Appendix A1.2 Fourier transform of ( )

p px yR τ

As ( )p px yR τ is periodic, we expect the spectrum to be purely impulsive.

We have only to decide the weights of these impulses.

( ) ( ) ( )0

0

2

02

1p p

T

x y p pT

R x t y t dtT

∗

−

τ = − τ∫

Let ( ) ( ) 0 0,2 2

0 ,

pT Tx t t

x toutside

⎧ − < <⎪= ⎨⎪⎩

then, ( ) ( )0pn

x t x t nT∞

= − ∞= −∑

Similarly, ( ) ( )0pn

y t y t nT∞

∗ ∗

= − ∞− τ = − τ −∑

where ( ) ( ) 0 0,2 2

0 ,

pT Ty t t

y toutside

∗∗

⎧ − < <⎪= ⎨⎪⎩

( ) ( ) ( )0

0

2*

00

2

1p p

T

x ynT

R x t y t nT dtT

∞

= − ∞−

⎡ ⎤τ = − τ −⎢ ⎥

⎢ ⎥⎣ ⎦∑∫

( ) ( )0

0

/ 2*

00 / 2

1 T

n T

x t y t nT dtT

∞

= −∞ −

= − τ −∑ ∫

As ( ) 0x t = for 02

Tt > ,



1.125

( ) ( ) ( )00

1p px y

nR x t y t nT dt

T

∞∞∗

= − ∞ −∞

τ = − τ −∑ ∫

( )00

1x y

nR nT

T

∞

= − ∞= τ +∑

Taking the Fourier transform on both sides,

( ) ( )00

1Fp px y x y

nR F R nT

T

∞

= − ∞

⎡ ⎤⎡ ⎤τ = τ +⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦

∑

As ( ) ( ) ( )x yR X f Y f∗τ ←⎯→ ,

( ) ( ) ( ) 02

0

1Fp p

j n f Tx y

nR X f Y f e

T

∞π∗

= − ∞

⎡ ⎤⎡ ⎤τ = ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦

∑

( ) ( ) 02

0

1 j n f T

nX f Y f e

T

∞π∗

= − ∞

⎡ ⎤= ⎢ ⎥

⎢ ⎥⎣ ⎦∑ (A1.2.1)

But from Example 1.12, we have

( ) ( )10 0

0

1

m mt mT F f n f

T

∞ ∞−

= − ∞ = − ∞

⎡ ⎤δ − = δ −⎢ ⎥

⎢ ⎥⎣ ⎦∑ ∑ (A1.2.2)

02

0

1 j m f t

me

T

∞π

= − ∞

⎡ ⎤= ⎢ ⎥

⎢ ⎥⎣ ⎦∑

Replacing t by f , 0T by 0f , we get the dual relation

( ) 020

0

1 j m f T

m mf mf e

f

∞ ∞π

= − ∞ = − ∞

⎡ ⎤δ − = ⎢ ⎥

⎢ ⎥⎣ ⎦∑ ∑

As 00

1fT

= , we have

02

0 0

1 j m f T

m m

mf eT T

∞ ∞π

= − ∞ = − ∞

⎛ ⎞δ − =⎜ ⎟⎝ ⎠

∑ ∑ (A1.2.3)



1.126

Using Eq. A1.2.3 in Eq. A1.2.1, we obtain,

( ) 20 0 00

1Fp px y

m

m m mR X Y fT T TT

∞∗

= − ∞

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎡ ⎤τ = δ −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎣ ⎦ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦∑ (A1.2.4)

where ( )00

mfT

mX X fT =

⎛ ⎞=⎜ ⎟

⎝ ⎠

and ( )0

0mfT

mY Y fT

∗ ∗

=

⎛ ⎞=⎜ ⎟

⎝ ⎠.



1.127

Appendix A1.3 Complex envelope of the output of a BP system Let ( )x t , a BP signal, be applied as input to a BP system with impulse response

( )h t . Let the resulting output be denoted ( )y t , which is also a BP signal. We

shall derive a relation for ( )c ey t in terms of ( )c ex t and ( )c eh t .

We know that,

( ) ( ) ( )c c s cx t x t t x t tcos sin= ω − ω ,

( ) ( ) ( )ce c sx t x t j x t= + ,

( ) ( ) 2Re cj f tcex t x t e π⎡ ⎤= ⎣ ⎦ ,

( ) ( ) ( ) 12 ce c ce cX f X f f X f f∗ ⎡ ⎤= − + − +⎣ ⎦ . (A1.3.1)

Similarly, let

( ) ( ) ( )ce c sh t h t j h t= + , (A1.3.2)

( ) ( ) ( )c c s ch t h t t h t tcos sin= ω − ω , (A1.3.3)

( ) ( ) 2Re cj f tceh t h t e π⎡ ⎤= ⎣ ⎦ ,

( ) 2 22 c cj f t j f tce ceh t h e h eπ − π∗= + . (A1.3.4)

Taking the FT of Eq. A1.3.4, we have

( ) ( ) ( )2 ce c ce cH f H f f H f f∗ ⎡ ⎤= − + − +⎣ ⎦ (A1.3.5)

But ( ) ( ) 2Re cj f tcey t y t e π⎡ ⎤= ⎣ ⎦ .

Therefore,

( ) ( ) ( ) ( ) ( )2

ce c ce cY f f Y f fY f X f H f

∗ ⎡ ⎤− + − +⎣ ⎦= = (A1.3.6)

Because of Eq. A1.3.1 and A1.3.5, Eq. A1.3.6 becomes

( ) ( ) ( ) ( ) ( ) ( ) 14 ce c ce c ce c ce cX f H f H f f H f f X f f X f f∗ ∗⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤= − + − + − + − +⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

Consider the product term



1.128

( ) ( )ce c ce cH f f X f f∗ ⎡ ⎤− − +⎣ ⎦

( )ce cH f f− has spectrum confined to the range ( ),c cf B f B− + . ( )ce cX f f∗ ⎡ ⎤− +⎣ ⎦

has nonzero spectral components only in the range ( ) ( ) ,c cf W f W− + − − .

That is, the spectra ( )ce cH f f− and ( )ce cX f f∗ ⎡ ⎤− +⎣ ⎦ do not overlap; hence the

product is zero. Similarly, ( ) ( ) 0ce c ce cH f f X f f∗ ⎡ ⎤− + − =⎣ ⎦ . Hence,

( ) ( ) ( ) ( )

( ) ( )

12 4

14

ce c ce cce c ce c

ce c ce c

Y f f Y f fH f f X f f

H f f X f f

∗

∗ ∗

⎡ ⎤− + − +⎣ ⎦ = − −

⎡ ⎤ ⎡ ⎤+ − + − +⎣ ⎦ ⎣ ⎦

( )ce cY f f− has nonzero spectral components only in the range ( ),c cf B f B− + .

That is,

( ) ( ) ( )1 12 4ce c ce c ce cY f f H f f X f f⎡ ⎤− = − −⎣ ⎦ ,

and ( ) ( )( ) ( )( )1 12 4ce c ce c ce cY f f H f f X f f∗ ∗ ∗⎡ ⎤⎡ ⎤− + = − + − +⎣ ⎦ ⎣ ⎦ .

In other words,

( ) ( ) ( )12ce ce ceY f X f H f= (A1.3.7)

Therefore,

( ) ( ) ( )12ce ce cey t x t h t⎡ ⎤= ∗⎣ ⎦ (A1.3.8)

From Eq. A1.3.8, we obtain the equations for ( )cy t and ( )sy t .

( ) ( ) ( ) ( ) ( ) ce c s c sy t x t j x t h t j h t12

⎡ ⎤ ⎡ ⎤= + ∗ +⎣ ⎦ ⎣ ⎦

Therefore,

( ) ( ) ( ) ( ) ( ) c c c s sy t x t h t x t h t12

= ∗ − ∗ (A1.3.9)

( ) ( ) ( ) ( ) ( ) s c s s cy t x t h t x t h t12

= ∗ + ∗ (A1.3.10)

and, ( ) ( ) ( )ce c sy t y t j y t= + .



1.129

Exercise A1.3.1

Given the pairs ( ) ( )( )c sx t x t, and ( ) ( )c sh t h t,⎡ ⎤⎣ ⎦ suggest a scheme to

recover ( )cy t and ( )sy t .



1.130

References 1. Oppenheim, A. V, Willisky, A. S, with Hamid Nawab, S., Signals and Systems

(2nd Edition), PHI, 1997

2. Ashok Ambardar, Analog and Digital Signal Processing (2nd Edition),

Brooks/Cole Publishing Company, Thomson Asia Pvt. Ltd., Singapore, 1999

3. Lathi, B. P., Signal Processing and Linear systems, Berkeley-Cambridge

Press, 1998

Note: The above three books have a large collection of problems. The student is

advised to try to solve them.

Other Suggested Books

1. Couch II, L. W., Digital and Analog Communication Systems (6th Edition)

Pearson Asia, 2001

2. Carlson, A. B., Communication Systems (4th Edition), Mc Graw-Hill, 2003

3. Lathi, B. P., Modern Digital and Analog Communication Systems (3rd Edition),

Oxford University Press, 1998



3.1

CHAPTER 3

Random Signals and Noise

3.1 Introduction The concept of 'random variable' is adequate to deal with unpredictable

voltages; that is, it enables us to come up with the probabilistic description of the

numerical value of a random quantity, which we treat for convenience to be a

voltage quantity. In the real world, the voltages vary not only in amplitude but

also exhibit variations with respect to the time parameter. In other words, we

have to develop mathematical tools for the probabilistic characterization of

random signals. The resulting theory, which extends the mathematical model of

probability so as to incorporate the time parameter, is generally called the theory

of Random or Stochastic Processes.

Before we get started with the mathematical development of a random

process, let us consider a few practical examples of random processes and try to

justify the assertion that the concept of random process is an extension of the

concept of a random variable.

Let the variable X denote the temperature of a certain city, say, at 9 A.M.

In general, the value of X would be different on different days. In fact, the

temperature readings at 9 A.M. on two different days could be significantly

different, depending on the geographical location of the city and the time

separation between the days of observation (In a place like Delhi, on a cold

winter morning, temperature could be as low as 40 F whereas at the height of

the summer, it could have crossed 100 F even by 9 A.M.!).



3.2

To get the complete statistics of X , we need to record values of X over

many days, which might enable us to estimate ( )Xf x .

But the temperature is also a function of time. At 12 P.M., for example, the

temperature may have an entirely different distribution. Thus the random variable

X is a function of time and it would be more appropriate to denote it by ( )X t . At

least, theoretically, the PDF of ( )1X t could be very much different from that of

( )2X t for 1 2t t≠ , though in practice, they may be very much similar if 1t and 2t

are fairly closely spaced.

As a second example, think of a situation where we have a very large

number of speakers, each one of them uttering the same text into their individual

microphones of identical construction. The waveforms recorded from different

microphones would be different and the output of any given microphone would

vary with time. Here again, the random variables obtained from sampling this

collection of waveforms would depend on the sampling instants.

As a third example, imagine a large collection of resistors, each having the

same value of resistance and of identical composition and construction. Assume

that all these resistors are at room temperature. It is well known that thermal

voltage (usually referred to as thermal noise) develops across the terminals of

such a resistor. If we make a simultaneous display of these noise voltages on a

set of oscilloscopes, we find amplitude as well as time variations in these signals.

In the communications context, these thermal voltages are a source of

interference. Precisely how they limit our capacity to enjoy, say, listening to

music in an AM receiver, is our concern. The theory of random processes

enables one to come up with a quantitative answer to this kind of problem.



3.3

3.2 Definition of a Random Process Consider a sample space S (pertaining to some random experiment) with

sample points 1 2, , ..........., , ........ns s s . To every js ∈ S , let us assign a real

valued function of time, ( ),jx s t which we denote by ( )jx t . This situation is

illustrated in Fig. 3.1, which shows a sample space S with four points and four

waveforms, labeled ( ), 1, 2, 3, 4jx t j = .

Now, let us think of observing this set of waveforms at some time instant

1t t= as shown in the figure.

Since each point js of S has associated with it, a number ( )1jx t and a

probability jP , the collection of numbers, ( ) 1 , 1, 2, 3, 4jx t j = forms a random

variable. Observing the waveforms at a second time instant, say 2t , yields a

different collection of numbers, and hence a different random variable. Indeed

this set of four waveforms defines a random variable for each choice of the

observation instant. The above situation can easily be extended to the case

where there are infinite numbers of sample points and hence, the number of

waveforms associated with them are correspondingly rich.



3.4

Fig 3.1: A simple random process

The probability system composed of a sample space, an ensemble

(collection) of time functions and a probability measure is called a Random

Process (RP) and is denoted ( )X t . Strictly speaking, a random process is a

function of two variables, namely s ∈ S and ( ),t ∈ − ∞ ∞ . As such, a better

notation would be ( ),X s t . For convenience, we use the simplified notation ( )X t

to denote a random process. The individual waveforms of ( )X t are called

sample functions and the probability measure is such that it assigns a probability

to any meaningful event associated with these sample functions.

Given a random process ( )X t , we can identify the following quantities:

( )X t : The random process

( )jx t : The sample function associated with the sample point js

( )iX t : The random variable obtained by observing the process at it t=

( )j ix t : A real number, giving the value of ( )jx t at it t= .



3.5

We shall now present a few examples of random processes.

Example 3.1

Consider the experiment of tossing a fair coin. The random process ( )X t

is defined as follows: ( ) ( )sinX t t= π , if head shows up and ( ) 2X t t= , if the

toss results in a tail. Sketch the sample functions. We wish to find the expression

for the PDF of the random variables obtained from sampling the process at (a)

0t = and (b) 1t = .

There are only two sample functions for the process. Let us denote them

by ( )1x t and ( )2x t where ( ) ( )1 sinx t t= π and ( )2 2x t t= which are shown in

Fig. 3.2.

Fig. 3.2: The ensemble for the coin tossing experiment

As heads and tails are equally likely, we have ( ) ( )1 212

P x t P x t⎡ ⎤ ⎡ ⎤= =⎣ ⎦ ⎣ ⎦ .

Let 0X denote the random variable ( ) 0| tX t = and 1X correspond to ( ) 1| tX t = .

Then, we have ( ) ( )0Xf x x= δ and ( ) ( ) ( )

1

1 22Xf x x x⎡ ⎤= δ + δ −⎣ ⎦ .



3.6

Note that this is one among the simplest examples of RPs that can be used to

construe the concept.

Example 3.2 Consider the experiment of throwing a fair die. The sample space consists

of six sample points, 1 6, ........,s s corresponding to the six faces of the die. Let

the sample functions be given by ( ) ( )1 12ix t t i= + − for , 1, ......., 6is s i= = .

Let us find the mean value of the random variable ( ) 1| tX X t == .

A few of the sample functions of this random process are shown below

(Fig 3.3).

Fig. 3.3: A few sample functions of the RP of Example 3.2

The PDF of X is

( ) ( )6

1

1 1 16 2X

if x x i

=

⎡ ⎤⎛ ⎞= δ − + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∑

[ ] ( )1 1 3 5 7 9 11 3.012

E X = + + + + + =

The examples cited above have two features in common, namely (i) the

number of sample functions are finite (in fact, we could even say, quite small)



3.7

and (ii) the sample functions could be mathematically described. In quite a few

situations involving a random process, the above features may not be present.

Consider the situation where we have at our disposal N identical

resistors, N being a very large number (of the order of a million!).

Let the experiment be 'picking a resistor at random' and the sample

functions be the thermal noise voltage waveforms across these resistors. Then,

typical sample functions might look like the ones shown in Fig. 3.4.

Assuming that the probability of any resistor being picked up is 1N

, we find

that this probability becomes smaller and smaller as N becomes larger and

larger. Also, it would be an extremely difficult task to write a mathematical

expression to describe the time variation of any given voltage waveform.

However, as we shall see later on in this chapter, statistical characterization of

such noise processes is still possible which is adequate for our purposes.

Fig. 3.4: The ensemble for the experiment 'picking a resistor at random'

One fine point deserves a special mention; that is, the waveforms (sample

functions) in the ensemble are not random. They are deterministic. Randomness



3.8

in this situation is associated not with the waveforms but with the uncertainty as

to which waveform will occur on a given trial.

3.3 Stationarity By definition, a random process ( )X t implies the existence of an infinite

number of random variables, one for every time instant in the range,

t− ∞ < < ∞ . Let the process ( )X t be observed at n time instants,

1 2, , ..........., nt t t . We then have the corresponding random variables

( ) ( ) ( )1 2, , ..........., nX t X t X t . We define their Joint distribution function by

( ) ( ) ( ) ( ) ( ) 1 1 2 1 1, ..........., , , ..........., , ........

n n n nX t X tF x x x P X t x X t x= ≤ ≤ . Using a

vector notation is quite often convenient and we denote the joint distribution by

( ) ( )F xX t where the n-component random vector ( ) ( ) ( )( )1 , ..........., nX t X t=X t

and the dummy vector ( )1 2, , ..........., nx x x=x . The joint PDF of ( )X t , ( ) ( )fX t x ,

is given by

( ) ( ) ( ) ( )1 2 ........

n

n

f Fx x x

∂=

∂ ∂ ∂X t X tx x

We say a random process ( )X t is specified if and only if a rule is given or

implied for determining ( ) ( )X tF x or ( ) ( )X tf x for any finite set of observation

instants ( )1 2, , ..........., nt t t .

In application, we encounter three methods of specification. The first (and

simplest) is to state the rule directly. For this to be possible, the joint density

function must depend in a known way on the time instants. For the second

method, a time function involving one or more parameters is given. For example,

( ) ( )cos cX t A t= ω + Θ where A and cω are constants and Θ is a random

variable with a known PDF. The third method of specifying a random process is



3.9

to generate its ensemble by applying a stated operation to the sample functions

of a known process. For example, a random process ( )Y t may be the result of

linear filtering on some known process ( )X t . We shall see later on in this lesson,

examples of all these methods of specification.

Once ( ) ( )X tf x is known, it is possible for us to compute the probability of

various events. For example, we might be interested in the probability of the

random process ( )X t passing through a set of windows as shown in Fig. 3.5.

Let A be the event:

( ) ( ) ( ) 1 1 1 2 2 2 3 3 3: , ,A s a X t b a X t b a X t b= < ≤ < ≤ < ≤

That is, the event A consists of all those sample points js such that the

corresponding sample functions ( ) jx t satisfy the requirement,

( ) , 1, 2, 3i j i ia x t b i≤ ≤ = . Then the required quantity is ( )P A . A typical

sample function which would contribute to ( )P A is shown in the same figure.

( )P A can be calculated as

( ) ( ) ( )1 2 3

1 2 3

b b b

a a aP A f= ∫ ∫ ∫ X t x d x

where ( ) ( ) ( ) ( )( )1 2 3, ,X t X t X t=x t and 1 2 3d x d x d x=d x

Fig. 3.5: Set of windows and a waveform that passes through them



3.10

The above step can easily be generalized to the case of a random vector

with n-components.

Stationary random processes constitute an important subset of the

general class of random processes. We shall define stationarity here. Let the

process ( )X t be observed at time instants 1 2, , .........., nt t t and ( )X t be the

corresponding random vector.

Def. 3.1: A random process ( )X t is said to be strictly stationary or Stationary in

a Strict Sense (SSS) if the joint PDF ( ) ( )fX t x is invariant to a translation of the

time origin; that is, ( )X t is SSS, only if

( ) ( ) ( ) ( )X t T X tf x f x+ = (3.1)

where ( ) ( )1 2, , .........., nt T t T t T t T+ = + + + .

For ( )X t to be SSS, Eq. 3.1 should be valid for every finite set of time

instants , 1, 2, ..........,jt j n= , and for every time shift T and dummy vector

x . If ( )X t is not stationary, then it is called a nonstationary process.

One implication of stationarity is that the probability of the set of sample

functions of this process which passes through the windows of Fig. 3.6(a) is

equal to the probability of the set of sample functions which passes through the

corresponding time shifted windows of Fig. 3.6(b). Note, however, that it is not

necessary that these two sets consist of the same sample functions.



3.11

Fig. 3.6: (a) Original set of Windows (b) The translated set

Example 3.3 We cite below an example of a nonstationary process (It is easier to do

this than give a nontrivial example of a stationary process). Let

( ) ( )sin 2X t F t= π

where F is a random variable with the PDF

( )1 , 100 200

1000 ,

F

f Hzf f

otherwise

⎧ ≤ ≤⎪= ⎨⎪⎩

(Note that this specification of ( )X t corresponds to the second method

mentioned earlier on in this section). We now show that ( )X t is nonstationary.

( )X t consists of an infinite number of sample functions. Each sample

function is a sine wave of unit amplitude and a particular frequency f . Over the

ensemble, the random variable F takes all possible values in the range

( )100, 200 Hz . Three members of this ensemble, (with 100, 150f = and 200

Hz) are plotted in Fig. 3.7.



3.12

Fig. 3.7: A few sample functions of the process of example 3.3

To show that ( )X t is nonstationary, we need only observe that every

waveform in the ensemble is,

zero at 0t = ,

positive for 0 2.5 sect m< <

negative for ( )2.5 sec 0m t− < < .



3.13

Thus the density function of the random variable ( )1X t for 1 1 sect m= is

identically zero for negative arguments whereas the density function of the RV

( )2X t for 2 1 sect m= − is non-zero only for negative arguments (Of course, the

PDF of ( )0X is an impulse). For a process that is SSS, the one-dimensional

PDF is independent of the observation instant, which is evidently not the case for

this example. Hence ( )X t is nonstationary.

3. 4 Ensemble Averages We had mentioned earlier that a random process is completely specified,

if ( ) ( )fX t x is known. Seldom is it possible to have this information and we may

have to be content with a partial description of the process based on certain

averages. When these averages are derived from the ensemble, they are called

ensemble averages. Usually, the mean function and the autocorrelation function,

(or the auto-covariance function) of the process provide a useful description of

the processes. At times, we require the cross-correlation between two different

processes. (This situation is analogous to the random variable case, wherein we

had mentioned that even if the PDF of the variable is not available, certain

averages such as mean value, variance etc., do provide adequate or useful

information).

Def. 3.2: The Mean Function

The mean function of a process ( )X t is

( ) ( ) ( ) ( ) ( )X x tm t E X t X t x f x d x∞

− ∞

⎡ ⎤= = =⎣ ⎦ ∫ (3.2)

For example, if iX and jX are the random variables obtained by sampling

the process at it t= and jt t= respectively, then

( ) ( )ii x X iX x f x d x m t

∞

− ∞

= =∫



3.14

( ) ( )jj x X jX x f x d x m t

∞

− ∞

= =∫

In general, ( )Xm t is a function of time.

Def. 3.3: The Autocorrelation Function

The autocorrelation function of a random process ( )X t is a function of

two variables kt and it , and is given by

( ) ( ) ( ),X k i k iR t t E X t X t⎡ ⎤= ⎣ ⎦ (3.3)

Denoting the joint PDF of the random variables ( )kX t and ( )iX t by ( ), ,k iX Xf x y

we may rewrite Eq. 3.3 as

( ) ( ),, ,k iX k i X XR t t x y f x y d x d y

∞ ∞

− ∞ − ∞

= ∫ ∫

We also use ACF to denote the Auto Correlation Function.

Def. 3.4: The Auto-covariance Function

Let ( ),X k iC t t denote the auto covariance function of ( )X t . It is given by

( ) ( ) ( )( ) ( ) ( )( ),X k i k X k i X iC t t E X t m t X t m t⎡ ⎤= − −⎣ ⎦ (3.4a)

It is not too difficult to show that

( ) ( ) ( ) ( ), ,X k i X k i X k X iC t t R t t m t m t= − (3.4b)

In general, the autocorrelation and the auto-covariance would be a function of

both the arguments kt and it . If the process has a zero mean value (that is,

( ) 0Xm t = for all t ), then ( ) ( ), ,X k i X k iC t t R t t= .

For a stationary process, the ensemble averages defined above take a

simpler form. In particular, we find that mean function of the process is a

constant. That is,

( )X Xm t m= (3.5a)

Xm being a constant. In such a case, we can simply mention the mean value of

the process. Also, for a stationary process, we find that the autocorrelation and



3.15

auto-covariance functions depend only on the time difference ( )k it t− , rather

than on the actual values of kt and it . This can be shown as follows.

( ) ( ) ( ),X k i k iR t t E X t X t⎡ ⎤= ⎣ ⎦

( ) ( )k iE X t T X t T⎡ ⎤= + +⎣ ⎦ , as ( )X t is stationary.

In particular if iT t= − , then

( ) ( ) ( ), 0X k i k iR t t E X t t X⎡ ⎤= −⎣ ⎦ (3.5b)

( ) ( ), , 0X k i X k iR t t R t t= −

In order to simplify the notation, it is conventional to drop the second argument

on the RHS of Eq. 3.5(b) and write as

( ) ( ),X k i X k iR t t R t t= −

In view of Eq. 3.4(b), it is not difficult to see that for a stationary process

( ) ( ),X k i X k iC t t C t t= −

It is important to realize that for a process that is SSS, Eq. 3.5(a) and Eq.

3.5(b) hold. However, we should not infer that any process for which Eq. 3.5 is

valid, is a stationary process. In any case, the processes satisfying Eq. 3.5 are

sufficiently useful and are termed Wide Sense Stationary (WSS) processes.

Def. 3.5: Wide Sense Stationarity

A process ( )X t is WSS or stationary in a wide sense1, provided

( )X Xm t m= (3.6a)

and ( ) ( ),X k i X k iR t t R t t= − (3.6b)

Wide sense stationarity represents a weak kind of stationarity in that all

processes that are SSS are also WSS; but the converse is not necessarily true.

When we simply use the word stationary, we imply stationarity in the strict sense.

1 For definitions of other forms of stationarity (such as Nth order stationary) see [1, P302]



3.16

3.4.1 Properties of ACF The autocorrelation function of a WSS process satisfies certain properties,

which we shall presently establish. ACF is a very useful tool and a thorough

understanding of its behavior is quite essential for a further study of the random

processes of interest to us. For convenience of notation, we define the ACF of a

wide-sense stationary process ( )X t as

( ) ( ) ( )XR E X t X t⎡ ⎤τ = + τ⎣ ⎦ (3.7)

Note that τ is the difference between the two time instants at which the process

is being sampled.

P1) The mean-square value of the process is ( )0XR

τ =τ .

This follows from Eq. 3.7 because,

( ) ( )20XR E X t⎡ ⎤= ⎣ ⎦

As ( )0XR is a constant, we infer that for a WSS process, mean and mean-

square values are independent of time.

P2) The ACF is an even function of τ ; that is,

( ) ( )X XR R− τ = τ

This is because ( ) ( ) ( ) ( )E X t X t E X t X t⎡ ⎤ ⎡ ⎤+ τ = + τ⎣ ⎦ ⎣ ⎦ . That is,

( ) ( )X XR t t R t t⎡ ⎤ ⎡ ⎤+ τ − = − + τ⎣ ⎦ ⎣ ⎦

which is the desired result.

P3) ACF is maximum at the origin.

Consider the quantity, ( ) ( ) 2E X t X t⎡ ⎤± + τ⎣ ⎦

Being the expectation of a squared quantity, it is nonnegative. Expanding

the squared quantity and making use of the linearity property of the

expectation, we have

( ) ( )0 0X XR R± τ ≥

which implies ( ) ( )0X XR R≥ τ .



3.17

P4) If the sample functions of the process ( )X t are periodic with period 0T , then

the ACF is also periodic with the same period.

This property can be established as follows.

Consider ( ) ( )E X t X t⎡ ⎤+ τ⎣ ⎦ for 0Tτ ≥ . As each sample function

repeats with period 0T , the product repeats and so does the expectation of

this product.

The physical significance of ( )XR τ is that it provides a means of

describing the inter-dependence of two random variables obtained by

observing the process ( )X t at two time instants τ seconds apart. In fact if

( )X t is a zero mean process, then for any 1τ = τ , ( )( )

1

0X

X

RR

τ is the

correlation coefficient of the two random variables spaced 1τ seconds apart.

It is therefore apparent that the more rapidly ( )X t changes with time, the

more rapidly ( )XR τ decreases from its maximum value ( )0XR as τ

increases (Fig. 3.8). This decrease may be characterized by a de-

correlation time 0τ , such that for ( ), XR0τ ≥ τ τ remains below some

prescribed value, say ( )0

100XR

. We shall now take up a few examples to

compute some of the ensemble averages of interest to us.

Fig. 3.8: ACF of a slowly and rapidly fluctuating random process



3.18

Example 3.4

For the random process of Example 3.2, let us compute ( )2, 4XR .

( ) ( ) ( )2, 4 2 4XR E X X⎡ ⎤= ⎣ ⎦

( ) ( )6

12 4j j j

jP x x x

=

⎡ ⎤= ⎣ ⎦∑

A few of the sample functions of the process are shown below.

Fig. 3.9: Sampling the process of example 3.2 at 2 and 4t =

As 16jP x⎡ ⎤ =⎣ ⎦ for all j , we have

( ) ( ) ( )6

1

12, 4 2 46X j j

jR x x

=

= ∑

for 1j = , the two samples that contribute to the ACF have the values 1

(indicated by the on ( )1x t ) and 2 (indicated by × on ( )1x t ). The other sample

pairs are ( ) ( ) ( ) ( ) ( )2, 3 , 3, 4 , 4, 5 , 5, 6 , 6, 7 .



3.19

Hence ( ) [ ]12, 4 2 6 12 20 30 426XR = + + + + +

112 18.666

= =

Exercise 3.1

Let a random process ( )X t consist of 6 equally likely sample

functions, given by ( ) , 1, 2, ......, 6ix t i t i= = . Let X and Y be the

random variables obtained by sampling process at 1t = and 2t =

respectively.

Find

a) [ ]E X and [ ]E Y

b) ( ), ,X Yf x y

c) ( )1, 2XR



3.20

Example 3.5

Let ( ) ( )cos cX t A t= ω + Θ ,

where A and cω are constants and Θ is a random variable with the PDF,

( )1 , 0 2

20 ,

fotherwise

Θ

⎧ ≤ θ < π⎪ πθ = ⎨⎪⎩

Let us compute a) ( )Xm t and b) ( )1 2,XR t t .

Exercise 3.2

A random process ( )X t consists of 5 sample functions, each

occurring with probability 15

. Four of these sample functions are given

below.

( ) ( ) ( )1 cos 2 sin 2x t t t= π − π

( ) ( ) ( )2 sin 2 cos 2x t t t= π − π

( )3 2 cosx t t= −

( )4 2 sinx t t= −

a) Find the fifth sample function ( )5x t of the process ( )X t such that the

process ( )X t is

i) zero mean

ii) ( ) ( )1 2 1 2,X XR t t R t t= −

b) Let V be the random variable ( )0t

X t=

and W be the random

variable ( )4t

X t= π

. Show that, though the process is WSS,

( ) ( )V Wf v f v≠ .



3.21

(Note that the random processes is specified in terms of a random parameter,

namely Θ , another example of the second method of specification)

The ensemble of ( )X t is composed of sinusoids of amplitude A and

frequency cf but with a random initial phase. Of course, a given sample function

has a fixed value for 1θ = θ , where 10 2≤ θ < π .

Three sample functions of the process are shown in Fig. 3.10 for 610cf Hz= and 1A = .



3.22

Fig 3.10: Three sample functions of the process of example 3.5



3.23

a) ( ) ( )cosX cm t E A t⎡ ⎤= ω + Θ⎣ ⎦ . As Θ is the only random variable, we have,

( ) ( )2

0

cos2X cAm t t d

π

= ω + θ θπ ∫

0=

Note that Θ is uniform in the range 0 to 2π .

This is a reasonable answer because, for any given 1θ = θ , the

ensemble has both the waveforms, namely, ( )1cos cA tω + θ and

( ) ( )1 1cos cosc cA t A t⎡ ⎤ω + θ + π = − ω + θ⎣ ⎦ . Both the waveforms are

equally likely and sum to zero for all values of t .

b) ( ) ( ) ( )1 2 1 2, cos cosX c cR t t E A t A t⎡ ⎤= ω + Θ ω + Θ⎣ ⎦

( ) ( ) 2

1 2 1 2cos 2 cos2 c cA t t t t⎡ ⎤ ⎡ ⎤= ω + + Θ + ω −⎣ ⎦ ⎣ ⎦

As the first term on the RHS evaluates to zero, we have

( ) ( )2

1 2 1 2, cos2X cAR t t t t⎡ ⎤= ω −⎣ ⎦

As the ACF is only a function of the time difference, we can write

( ) ( ) ( ) ( )2

1 2 1 2, cos2X X X cAR t t R t t R= − = τ = ω τ

Note that ( )X t is composed of sample functions that are periodic with

period 1

cf. In accordance with property P4, we find that the ACF is also

periodic with period 1

cf. Of course, it is also an even function of τ (property

P3).



3.24

Example 3.6 In this example, we shall find the ACF of the random impulse train process

specified by

( ) ( )n dn

X t A t nT T= δ − −∑

where the amplitude nA is a random variable with ( ) ( ) 11 12n nP A P A= = =− = .

Successive amplitudes are statistically independent. The time interval dt of the

first impulse from the origin has uniform PDF in the range ( )0, T . That is,

( ) 1 , 0DT d df t t T

T= ≤ ≤ and zero elsewhere. Impulses are spaced T seconds

Exercise 3.3

For the process ( )X t of example 3.5, let 2cω = π . let Y be the

random variable obtained from sampling ( )X t and 14

t = . Find ( )Yf y .

Exercise 3.4

Let ( ) ( )= ω + Θcos cX t A t where A and cω are constants, and θ is

a random variable, uniformly distributed in the range 0 ≤ θ ≤ π . Show the

process is not WSS.

Exercise 3.5

Let ( ) ( ) ( )cos sinc cZ t X t Y t= ω + ω where X and Y are independent

Gaussian variables, each with zero mean and unit variance. Show that ( )Z t is

WSS and ( ) ( )τ = ω τZ cR cos . Let ( )1Z t V= . Show that V is ( )0, 1N .



3.25

apart and nA is independent of dT . (The symbol n∑ indicates summation with

respect to n where n , an integer, ranges from ( ),− ∞ ∞ .)

A typical sample function of the process ( )X t is shown in Fig. 3.11.

Fig. 3.11: Sample function of the random impulse train process

( ) ( ) ( ),X m d n dm n

R t t E A t mT T A t nT T⎧ ⎫⎡ ⎤ ⎡ ⎤+ τ = δ − + τ − δ − −⎨ ⎬⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎩ ⎭∑ ∑

( ) ( )m n d dm n

E A A t mT T t nT T⎡ ⎤= δ − + τ − δ − −⎢ ⎥⎣ ⎦∑∑

As nA is independent of dT , we can write (after interchanging the order of

expectation and summation),

( ) ( ) ( ),X m n d dm n

R t t A A t nT T t nT T+ τ = δ − + τ − δ − −∑∑

But, 1,0,m n

m nA A

otherwise=⎧

= ⎨⎩

This is because when m n= , ( ) ( )2 22 1 11 1 12 2m n mA A A= = + − = . If m n≠ ,

then m n m nA A A A= , as successive amplitudes are statistically independent. But

( ) ( )1 11 1 02 2m nA A= = + − = .

Using this result, we have



3.26

( ) ( ) ( ),X d dn

R t t t nT T t nT T+ τ = δ − + τ − δ − −∑

( ) ( )0

1 T

d d dn

t nT t t nT t d tT

= δ − + τ − δ − −∑ ∫

(Note that dT is uniformly distributed in the range 0 to T )

Let dt nT t x− − = . Then,

( ) ( ) ( )( )1

1,t nT

Xn t n T

R t t x x d xT

−

− +

+ τ = δ δ + τ∑ ∫

( ) ( )−∞

+ ∞

= δ δ + τ∫ x x d xT1

Letting y x= − , we have

( ) ( ) ( ) ( ) ( )1 1,XR t t y y d y y y d yT T

∞ ∞

− ∞ − ∞

+ τ = δ − δ τ − = δ δ τ −∫ ∫

( ) ( ) ( )1 1T T⎡ ⎤= δ τ ∗ δ τ = δ τ⎣ ⎦

That is, the ACF is a function of τ alone and it is an impulse!

It is to be pointed out that in the case of a random process, we can also

define time averages such as time-averaged mean value or time-averaged ACF

etc., whose calculation is based on the individual sample functions. There are

certain processes, called ergodic processes where it is possible to interchange

the corresponding ensemble and time averages. More details on ergodic

processes can be found in [2].

3.4.2 Cross-correlation Consider two random processes ( )X t and ( )Y t . We define the two

cross-correlation functions of ( )X t and ( )Y t as follows:

( ) ( ) ( ), 1 2 1 2,X YR t t E X t Y t⎡ ⎤= ⎣ ⎦



3.27

( ) ( ) ( ), 1 2 1 2,Y XR t t E Y t X t⎡ ⎤= ⎣ ⎦

where 1t and 2t are the two time instants at which the processes are observed.

Def. 3.6: Two processes are said to be jointly wide-sense stationary if,

i) ( )X t and ( )Y t are WSS and

ii) ( ) ( ) ( ), 1 2 , 1 2,X Y X Y X YR t t R t t R= − = τ

Cross-correlation is not generally an even function of τ as is the case with

ACF, nor does it have a maximum at the origin. However, it does obey a

symmetrical relationship as follows:

( ) ( )X Y Y XR Rτ = − τ (3.8)

Def. 3.7:

Two random process ( )X t and ( )Y t are called (mutually) orthogonal if

( )1 2, 0X YR t t = for every 1t and 2t .

Def. 3.8:

Two random process ( )X t and ( )Y t are uncorrelated if

( ) ( ) ( ) ( )1 2 1 2 1 2, , 0X Y X Y X YC t t R t t m t m t= − = for every 1t and 2t .

Exercise 3.6 Show that the cross-correlation function satisfies the following

inequalities.

a) ( ) ( ) ( )0 0X Y X YR R Rτ ≤

b) ( ) ( ) ( )⎡ ⎤τ ≤ +⎣ ⎦1 0 02X Y X YR R R



3.28

3.5 Systems with Random Signal Excitation In Chapter 1, we discussed the transmission of deterministic signals

through linear systems. We had developed the relations for the input-output

spectral densities. We shall now develop the analogous relationships for the case

when a linear time-invariant system is excited by random signals.

Consider the scheme shown in Fig. 3.12. ( )h t represents the (known)

impulse response of a linear time-invariant system that is excited by a random

process ( )X t , resulting in the output process ( )Y t .

Fig 3.12: Transmission of a random process through a linear filter

We shall now try to characterize the output process ( )Y t in terms of the

input process ( )X t and the impulse response ( )h t [third method of

specification]. Specifically, we would like to develop the relations for ( )Ym t and

( )1 2,YR t t when ( )X t is WSS.

Let ( )jx t be a sample function of ( )X t which is applied as input to the

linear time-invariant system. Let ( )jy t be the corresponding output where ( )jy t

belongs to ( )Y t . Then,

( ) ( ) ( )j jy t h x t d∞

− ∞

= τ − τ τ∫

As the above relation is true for every sample function of ( )X t , we can write

( ) ( ) ( )Y t h X t d∞

− ∞

= τ − τ τ∫

Consider first the mean of the output process



3.29

( ) ( ) ( ) ( )Ym t E Y t E h X t d∞

− ∞

⎡ ⎤⎡ ⎤= = τ − τ τ⎢ ⎥⎣ ⎦

⎢ ⎥⎣ ⎦∫ (3.9a)

Provided that ( )E X t⎡ ⎤⎣ ⎦ is finite for all t , and the system is stable, we may

interchange the order of the expectation and integration with respect to τ in Eq.

3.9(a) and write

( ) ( ) ( )Ym t h E X t d∞

− ∞

⎡ ⎤= τ − τ τ⎣ ⎦∫ (3.9b)

where we have used the fact that ( )h τ is deterministic and can be brought

outside the expectation. If ( )X t is WSS, then ( )E X t⎡ ⎤⎣ ⎦ is a constant Xm , so that

Eq. 3.9(b) can be simplified as

( ) ( )Y Xm t m h d∞

− ∞

= τ τ∫

( )0Xm H= (3.10)

where ( ) ( )0

0f

H H f=

= and ( )H f is the transfer function of the given system.

We note that ( )Ym t is a constant.

Let us compute ( ),YR t u , where t and u denote the time instants at

which the output process is observed. We have,

( ) ( ) ( ) ( ) ( ) ( ) ( )1 1 1 2 2 2,YR t u E Y t Y u E h X t d h X u d∞ ∞

− ∞ − ∞

⎡ ⎤⎡ ⎤= = τ − τ τ τ − τ τ⎢ ⎥⎣ ⎦

⎢ ⎥⎣ ⎦∫ ∫

Again, interchanging the order of integration and expectation, we obtain

( ) ( ) ( ) ( ) ( )1 1 2 2 1 2,YR t u d h d h E X t X u∞ ∞

− ∞ − ∞

⎡ ⎤= τ τ τ τ − τ − τ⎣ ⎦∫ ∫

( ) ( ) ( )1 1 2 2 1 2,Xd h d h R t u∞ ∞

− ∞ − ∞

= τ τ τ τ − τ − τ∫ ∫

If ( )X t is WSS, then



3.30

( ) ( ) ( ) ( )1 1 2 2 1 2,Y XR t u d h d h R∞ ∞

− ∞ − ∞

= τ τ τ τ τ − τ + τ∫ ∫ (3.11)

where t uτ = − . Eq. 3.11 implies that ( ),YR t u is only a function of t u− .

Hence, the LHS of Eq. 3.11 can be written as ( )YR τ . Eq. 3.10 and 3.11 together

imply that if ( )X t is WSS, then so is ( )Y t .

3.6 Power Spectral Density The notion of Power Spectral Density (PSD) is an important and useful

one. It provides the frequency domain description of a stationary (at least WSS)

random process. From Eq. 3.11, we have

( ) ( ) ( ) ( ) ( )21 2 2 1 1 20Y XE Y t R h h R d d

∞ ∞

− ∞ − ∞

⎡ ⎤ = = τ τ τ − τ τ τ⎣ ⎦ ∫ ∫

But, ( ) ( ) 121

j fh H f e d f∞

π τ

− ∞

τ = ∫

Hence, ( ) ( ) ( ) ( )1222 2 1 1 2

j fXE Y t H f e d f h R d d

∞ ∞ ∞π τ

− ∞ − ∞ − ∞

⎡ ⎤⎡ ⎤ = τ τ − τ τ τ⎣ ⎦ ⎣ ⎦∫ ∫ ∫

( ) ( ) ( ) 122 2 2 1 1

j fXH f d f h d R e d

∞ ∞ ∞π τ

− ∞ − ∞ − ∞

= τ τ τ − τ τ∫ ∫ ∫

Let 2 1τ − τ = λ ; that is, 1 2τ = τ − λ .

( ) ( ) ( ) ( )222 22 2

j f j fXE Y t H f d f h e d R e d

∞ ∞ ∞π τ − π λ

− ∞ − ∞ − ∞

⎡ ⎤ = τ τ λ λ⎣ ⎦ ∫ ∫ ∫

The second integral above is ( )H f∗ , the complex conjugate of ( )H f . The third

integral will be a function f , which we shall denote by ( )XS f . Then

( ) ( ) ( )22XE Y t S f H f d f

∞

− ∞

⎡ ⎤ =⎣ ⎦ ∫ (3.12)



3.31

where ( ) ( ) 2j fX XS f R e d

∞− π λ

− ∞

= λ λ∫ .

We will now justify that ( )XS f can be interpreted as the power spectral

density of the WSS process ( )X t . Suppose that the process ( )X t is passed

through an ideal narrowband, band-pass filter with the passband centered at cf

and having the amplitude response

( )11,210,2

c

c

f f fH f

f f f

⎧ ± < ∆⎪⎪= ⎨⎪ ± > ∆⎪⎩

Then, from Eq. 3.12, we find that if the filter band-width is sufficiently small

and ( )XS f is a continuous function, then the mean square value of the filter

output is approximately,

( ) ( ) ( )2 2 X cE Y t f S f⎡ ⎤ ∆⎣ ⎦

The filter however passes only those frequency components of the input random

process ( )X t that lie inside the narrow frequency band of width f∆ centered

about cf± . Thus, ( )X cS f represents the frequency density of the average power

in the process ( )X t , evaluated at cf f= . The dimensions of ( )XS f are

watts/Hz.

3.6.1 Properties of power spectral density The PSD ( )XS f and the ACF ( )XR τ of a WSS process ( )X t form a

Fourier transform pair and are given by

( ) ( ) 2j fX XS f R e d

∞− π τ

− ∞

= τ τ∫ (3.13)



3.32

( ) ( ) 2j fX XR S f e d f

∞π τ

− ∞

τ = ∫ (3.14)

Eq. 3.13 and 3.14 are popularly known as Weiner-Khinchine Relations. Using

this pair of equations, we shall derive some general properties of PSD of a WSS

random process.

P1) The zero frequency value of the PSD of a WSS process equals the total

area under the graph of ACF; that is

( ) ( )0X XS R d∞

− ∞

= τ τ∫

This property follows directly from Eq. 3.13 by putting 0f = .

P2) The mean square value of a WSS process equals the total area under the

graph of the PSD; that is,

( ) ( )2XE X t S f d f

∞

− ∞

⎡ ⎤ =⎣ ⎦ ∫

This property follows from Eq. 3.14 by putting 0τ = and noting

( ) ( )20XR E X t⎡ ⎤= ⎣ ⎦ .

P3) The PSD is real and is an even function of frequency; that is, ( )XS f is real

and ( ) ( )X XS f S f− = .

This result is due to the property of the ACF, namely, ( )XR τ is real and

even.

P4) The PSD of a WSS process is always non-negative; that is,

( ) 0XS f ≥ , for all f .

To establish this, assume that ( )XS f is negative, for a certain frequency

interval, say ( )1 1,f f f+ ∆ . Let ( )X t be the input to a narrowband filter with

the transfer function characteristic,

( ) 1 11,0,

f f f fH f

otherwise

⎧ ≤ ≤ + ∆⎪= ⎨⎪⎩



3.33

Then from Eq. 3.12, we have ( ) ( ) ( )22XE Y t S f H f d f

∞

− ∞

⎡ ⎤ =⎣ ⎦ ∫ , as a

negative quantity which is a contradiction.

We shall now derive an expression for the power spectral density, ( )YS f ,

of the output of Fig. 3.12.

Using the relation ( ) ( )1X XR F S f− ⎡ ⎤τ = ⎣ ⎦ , Eq. 3.11 can be written as

( ) ( ) ( ) ( ) ( )1 221 2 1 2

j fY XR h h S f e d f d d

∞ ∞ ∞π τ − τ + τ

− ∞ − ∞ − ∞

⎡ ⎤τ = τ τ τ τ⎢ ⎥

⎢ ⎥⎣ ⎦∫ ∫ ∫

( ) ( ) ( )1 22 2 21 1 2 2

j f j f j fXh e d h e d S f e d f

∞ ∞ ∞− π τ π τ π τ

− ∞ − ∞ − ∞

⎡ ⎤ ⎡ ⎤= τ τ τ τ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∫ ∫ ∫

( ) ( ) ( ) 2j fXH f H f S f e d f

∞∗ π τ

− ∞

= ⋅∫

( ) ( )2 2j fXH f S f e d f

∞π τ

− ∞

⎡ ⎤= ⎢ ⎥⎣ ⎦∫ (3.15)

As ( ) ( )Y YR S fτ ←⎯→ , Eq. 3.15 implies

( ) ( ) ( ) 2Y XS f S f H f= (3.16)

Note that when the input to an LTI system is deterministic, we have the input-

output FT relationship, ( ) ( ) ( )Y f X f H f= . The corresponding time domain

relationship is ( ) ( ) ( )y t x t h t= ∗ . Let ( )hR τ denote the ACF of the ( )h t . Then

( ) ( ) 2hR H fτ ←⎯→ (see P4, sec 1.6.2). Hence

( ) ( ) ( )Y x hR R Rτ = τ ∗ τ

( ) ( ) ( )xR h h∗= τ ∗ τ ∗ − τ (3.17a)

If the impulse response is real, then

( ) ( ) ( ) ( )Y xR R h hτ = τ ∗ τ ∗ − τ (3.17b)



3.34

We shall now take up a few examples involving the computation of PSD.

Example 3.7

For the random process ( )X t of example 3.5, let us find the PSD.

Since ( ) ( )2

cos2X cAR τ = ω τ , we have

( ) ( ) ( )2

4X c cAS f f f f f⎡ ⎤= δ − + δ +⎣ ⎦

Example 3.8: (Modulated Random Process)

Let ( ) ( ) ( )cos cY t X t t= ω + Θ where ( )X t is a WSS process with known

( )XR τ and ( )XS f . Θ is a uniformly distributed random variable in the range

( )0 2− π . ( )X t and Θ are independent. Let us find the ACF and PSD of ( )Y t .

( ) ( ) ( )YR E Y t Y t⎡ ⎤τ = + τ⎣ ⎦

( ) ( ) ( ) ( ) cos cosc cE X t t X t t⎡ ⎤= + τ ω + τ + Θ ω + Θ⎣ ⎦

As ( )X t and Θ are independent, we have

( ) ( ) ( ) ( )( ) ( )cos cosY c cR E X t X t E t t⎡ ⎤⎡ ⎤τ = + τ ω + τ + Θ ω + Θ⎣ ⎦ ⎣ ⎦

( ) ( )1 cos2 X cR= τ ω τ

( ) ( ) ( ) ( )14Y Y X c X cS f F R S f f S f f⎡ ⎤ ⎡ ⎤= τ = − + +⎣ ⎦ ⎣ ⎦

Example 3.9: (Random Binary Wave)

Fig. 3.13 shows the sample function ( )jx t of a process ( )X t consisting of

a random sequence of binary symbols 1 and 0. It is assumed that:



3.35

1. The symbols 1 and 0 are represented by pulses of amplitude A+ and A−

volts, respectively and duration T seconds.

2. The pulses are not synchronized, so that the starting time of the first pulse,

dt is equally likely to lie anywhere between zero and T seconds. That is, dt

is the sample value of a uniformly distributed random variable dT with its

probability density function defined by

( )1 , 0

0 ,d

dT d

t Tf t T

elsewhere

⎧ ≤ ≤⎪= ⎨⎪⎩

Fig. 3.13: Random binary wave

3. During any time interval ( )1 dn T t t nT− ≤ − ≤ , where n is an integer, the

presence of a 1 or 0 is determined by tossing a fair coin; specifically, if the

outcome is ‘heads’, we have a 1 and if the outcome is 'tails', we have a 0.

These two symbols are thus equally likely, and the presence of a 1 or 0 in

anyone interval is independent of all other intervals. We shall compute

( )XS f and ( )XR τ .

The above process can be generated by passing the random impulse train

(Example 3.6) through a filter with the impulse response,

( ) , 00 ,A t T

h totherwise

≤ ≤⎧= ⎨⎩



3.36

( ) ( )sinH f AT c f T=

Therefore, ( ) ( )2 2 21 sinXS f A T c f TT

=

( )2 2sinA T c f T=

Taking the inverse Fourier transform, we have

( )2 1 , 0

0 ,X

A TR T

otherwise

⎧ ⎛ ⎞τ− ≤ τ ≤⎪ ⎜ ⎟τ = ⎨ ⎝ ⎠

⎪⎩

The ACF of the random binary wave process can also be computed by direct

time domain arguments. The interested reader is referred to [3].

We note that the energy spectral density of a rectangular pulse ( )x t of

amplitude A and duration T , is ( ) ( )2 2 2sinxE f A T c f T=

Hence, ( ) ( )x xS f E f T= .

Exercise 3.7 In the random binary wave process of example 3.9, let 1 be

represented by a pulse of amplitude A and duration T sec. The binary zero is

indicated by the absence of any pulse. The rest of the description of the

process is the same as in the example 3.9. Show that

( ) ( ) ( )22

2 2

sin4X

f TAS f ff T

⎡ ⎤π= δ +⎢ ⎥

π⎢ ⎥⎣ ⎦

Exercise 3.8

The input voltage to an RLC series circuit is a WSS process ( )X t with

( ) 2X t = and ( ) 24XR e− ττ = + . Let ( )Y t be the voltage across the

capacitor. Find

a) ( )Y t

b) ( )YS f



3.37

3.7 Cross-Spectral Density Just as the PSD of a process provides a measure of the frequency

distribution of the given process, the cross spectral density provides a measure

of frequency interrelationship between two processes. (This will become clear

from the examples that follow). Let ( )X t and ( )Y t be two jointly WSS random

processes with their cross correlation functions ( )X YR τ and ( )Y XR τ . Then, we

define, ( )X YS f and ( )Y XS f as follows:

( ) ( ) 2j fX Y X YS f R e d

∞− π τ

− ∞

= τ τ∫ (3.18)

( ) ( ) 2j fY X Y XS f R e d

∞− π τ

− ∞

= τ τ∫ (3.19)

That is,

( ) ( )X Y X YR S fτ ←⎯→

( ) ( )Y X Y XR S fτ ←⎯→

The cross-spectral density is in general complex. Even if it is real, it need

not be positive. However, as ( ) ( )X Y Y XR Rτ = − τ , we find

( ) ( ) ( )X Y Y X Y XS f S f S f∗= − =

We shall now give a few examples that involve the cross-spectrum.

Example 3.10

Let ( ) ( ) ( )Z t X t Y t= +

where the random processes ( )X t and ( )Y t are jointly WSS and

( ) ( ) 0X t Y t= = . We will show that to find ( )ZS f , we require the cross-

spectrum.

( ) ( ) ( ),ZR t u E Z t Z u⎡ ⎤= ⎣ ⎦



3.38

( ) ( )( ) ( ) ( )( )E X t Y t X u Y u⎡ ⎤= + +⎣ ⎦

( ) ( ) ( ) ( ), , , ,X X Y Y X YR t u R t u R t u R t u= + + +

Letting t uτ = − , we have

( ) ( ) ( ) ( ) ( )Z X X Y Y X YR R R R Rτ = τ + τ + τ + τ (3.20a)

Taking the Fourier transform, we have

( ) ( ) ( ) ( ) ( )Z X X Y Y X YS f S f S f S f S f= + + + (3.20b)

If ( )X t and ( )Y t are uncorrelated, that is,

( ) ( ) ( ) 0X YR X t Y tτ = + τ =

then,

( ) ( ) ( )Z X YR R Rτ = τ + τ and

( ) ( ) ( )Z X YS f S f S f= +

Hence, we have the superposition of the autocorrelation functions as well as the

superposition of power spectral densities.

Example 3.11

Let ( )X t be the input to an LTI system with the impulse response ( )h t . If

the resulting output is ( )Y t , let us find an expression for ( )Y XS f .

( ) ( ) ( ),Y XR t u E Y t X u⎡ ⎤= ⎣ ⎦

( ) ( ) ( )1 1 1E h X t d X u∞

− ∞

⎡ ⎤= τ − τ τ⎢ ⎥

⎢ ⎥⎣ ⎦∫

Interchanging the order of expectation and integration, we have

( ) ( ) ( ) ( )1 1 1,Y XR t u h E X t X u d∞

− ∞

⎡ ⎤= τ − τ τ⎣ ⎦∫

If ( )X t is WSS, then



3.39

( ) ( ) ( )1 1 1,Y X XR t u h R t u d∞

− ∞

= τ − − τ τ∫

Letting t u− = τ gives the result

( ) ( ) ( )1 1 1,Y X XR t u h R d∞

− ∞

= τ τ − τ τ∫

( ) ( )Xh R= τ ∗ τ (3.21a)

That is, the cross-correlation between the output and input is the convolution of

the input ACF with the filter impulse response. Taking the Fourier transform,

( ) ( ) ( )Y X XS f S f H f= (3.21b)

Eq. 3.21(b) tells us that ( )X t and ( )Y t will have strong cross-correlation in

those frequency components where ( ) ( )XS f H f is large.

Example 3.12

In the scheme shown (Fig. 3.14), ( )X t and ( )Y t are jointly WSS. Let us

compute ( )V ZS f .

Fig. 3.14: Figure for the example 3.12

( ) ( ) ( ),V ZR t u E V t Z u⎡ ⎤= ⎣ ⎦

( ) ( ) ( ) ( )1 1 1 1 2 2 2 2E h X t d h Y u d∞ ∞

− ∞ − ∞

⎧ ⎫⎪ ⎪= τ − τ τ τ − τ τ⎨ ⎬⎪ ⎪⎩ ⎭∫ ∫

Interchanging the order of expectation and integration,



3.40

( ) ( ) ( ) ( ) ( )1 1 2 2 1 2 1 2,V ZR t u h h E X t Y u d d∞ ∞

− ∞ − ∞

⎡ ⎤= τ τ − τ − τ τ τ⎣ ⎦∫ ∫

As ( )X t and ( )Y t are jointly WSS, we have

( ) ( ) ( ) ( )1 1 2 2 1 2 1 2,V Z X YR t u h h R d d∞ ∞

− ∞ − ∞

= τ τ τ − τ + τ τ τ∫ ∫

where t uτ = − . That is,

( ) ( ) ( ),V Z V Z V ZR t u R t u R= − = τ

It not too difficult to show that

( ) ( ) ( ) ( )1 2V Z X YS f H f H f S f∗= (3.22)

We shall now consider some special cases of Example 3.12.

i) Let ( )1H f and ( )2H f of Fig. 3.14 have non-overlapping passbands. Then

( ) ( )1 2 0H f H f∗ = for all f . That is, ( )V ZS f 0≡ ; this implies ( )VZR 0τ ≡

and we have ( )V t and ( )Z t being orthogonal. In addition, either ( )V t or

( )Z t (or both) will be zero (note that ( ) ( )1 0 XV t H m= and

( ) ( )2 0 YZ t H m= ), because atleast one of the quantities, ( )1 0H or ( )2 0H

has to be zero. That is, the two random variables, ( )iV t and ( )jZ t obtained

from sampling the processes ( )V t at it , and ( )Z t at jt respectively, are

uncorrelated. That is, the processes ( )V t and ( )Z t are uncorrelated.

ii) Let ( ) ( )X t Y t= and ( )X t is WSS. Then we have the situation shown in

Fig. 3.15.



3.41

Fig. 3.15: Two LTI systems with a common input

Then from Eq. 3.22, we obtain,

( ) ( ) ( ) ( )1 2V Z XS f H f H f S f∗= (3.23)

iii) In the scheme of Fig. 3.15, if ( ) ( ) ( )1 2h t h t h t= = , then we have the

familiar result

( ) ( ) ( )2V XS f H f S f=

3.8 Gaussian Process Gaussian processes are of great practical and mathematical significance

in the theory of communication. They are of great practical significance because

a large number of noise processes encountered in the real world (and that

interfere with the communication process) can be characterized as Gaussian

processes. They are of great mathematical significance because Gaussian

processes possess a neat mathematical structure which makes their analytical

treatment quite feasible.

Let a random process ( )X t be sampled at time instants 1 2, , ........, nt t t .

Let ( ) ( ) ( )1 1 2 2, , ............, n nX t X X t X X t X= = = and X denote the row vector

( )1 2, , ..........., nX X X . The process ( )X t is called Gaussian, if ( )fX x is an n-

dimensional joint Gaussian density for every 1n ≥ and

( ) ( )1 2, , ..........., ,nt t t ∈ − ∞ ∞ . The n-dimensional Gaussian PDF is given by



3.42

( )( )

( ) ( )11

2 2

1 1exp22

Tx Xn

X

f CC

−⎡ ⎤= − − −⎢ ⎥⎣ ⎦πX Xx x m x m (3.24)

where xC is the covariance matrix, denotes its determinant and 1XC − is its

inverse. Xm is the mean vector, namely, ( )1 2, , ..........., nX X X and the

superscript T denotes the matrix transpose. The covariance matrix xC is given

by

11 12 1

21 22 2

1 2

n

nx

nnn n

C

λ λ λ⎛ ⎞⎜ ⎟λ λ λ⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟λλ λ⎝ ⎠

where cov ,i j i jX X⎡ ⎤λ = ⎣ ⎦

( ) ( )i i j jE X X X X⎡ ⎤= − −⎣ ⎦

Similarly, ( ) ( )1 1 2 2, , ........,X n nx X x X x X− = − − −x m

Specification of a Gaussian process as above corresponds to the first

method mentioned earlier. Note that an n-dimensional Gaussian PDF depends

only on the means and the covariance quantities of the random variables under

consideration. If the mean value of the process is constant and

( ) ( )cov ,i j i jX t X t⎡ ⎤λ = ⎣ ⎦ depends only on i jt t− , then the joint PDF is

independent of the time origin. In other words, a WSS Gaussian process is also

stationary in a strict sense. To illustrate the use of the matrix notation, let us take

an example of a joint Gaussian PDF with 2n = .

Example 3.13

1X and 2X are two jointly Gaussian variables with 1 2 0X X= = and

1 2σ = σ = σ . The correlation coefficient between 1X and 2X is ρ . Write the

joint PDF of 1X and 2X in i) the matrix form and ii) the expanded form.



3.43

i) As 211 22λ = λ = σ and 2

12 21λ = λ = ρσ , we have

( )2 2

4 22 2

, 1x XC C⎡ ⎤σ ρσ

= = σ − ρ⎢ ⎥ρσ σ⎣ ⎦

( )2 2

12 24 2

11XC− ⎡ ⎤σ − ρσ

= ⎢ ⎥− ρσ σσ − ρ ⎣ ⎦

( )2 2

1111− ρ⎡ ⎤

= ⎢ ⎥− ρσ − ρ ⎣ ⎦

Therefore,

( ) ( ) ( )1 2

1, 1 2 1 22 22 2

2

11 1, exp12 12 1

X X

xf x x x x

x

⎧ ⎫− ρ ⎛ ⎞⎡ ⎤−⎪ ⎪= ⎨ ⎬⎜ ⎟⎢ ⎥− ρσ − ρπσ − ρ ⎣ ⎦ ⎝ ⎠⎪ ⎪⎩ ⎭

ii) Taking the matrix products in the exponent above, we have the expanded

result, namely,

( ) ( )1 2

2 21 1 2 2

, 1 2 1 22 22 2

21, exp , ,2 12 1

X Xx x x xf x x x x

⎧ ⎫− ρ +⎪ ⎪= − − ∞ < < ∞⎨ ⎬σ − ρπσ − ρ ⎪ ⎪⎩ ⎭

This is the same as the bivariate Gaussian PDF of Chapter 2, section 6 with

1x x= and 2x y= , and 0X Ym m= = , and X Yσ = σ = σ .

Example 3.14

( )X t is a Gaussian process with ( ) 3Xm t = and

( ) 1 20.2 0.21 2, 4 4t t

XC t t e e− − − τ= = . Let us find, in terms of ( )Q ,

a) ( )5 2P X⎡ ⎤≤⎣ ⎦

b) ( ) ( )8 5 1P X X⎡ ⎤− ≤⎣ ⎦



3.44

a) Let Y be the random variable obtained by sampling ( )X t at 5t = . Y is

Gaussian with the mean value 3, and the variance ( )0 4XC= = . That is,

Y is ( )3, 4N .

[ ] ( )3 12 0.52 2

YP Y P Q−⎡ ⎤≤ = ≤ − =⎢ ⎥⎣ ⎦

b) Let ( ) ( )8 5Z X X= − and ( )8Y X= , and ( )5W X= . Note that

Z Y W= − is Gaussian. We have 0Z = and 2 2 2

,2Z Y W Y W Y Wσ = σ + σ − ρ σ σ

[ ]2 2 2 cov ,Y W Y W= σ + σ −

0.2 34 4 2 4 e− ×= + − ×

( )0.68 1 3.608e−= − =

[ ] [ ]11 2 0 1 2 12

P Z P Z P Z⎧ ⎫⎡ ⎤≤ = ≤ ≤ = − >⎨ ⎬⎣ ⎦ ⎩ ⎭

( )11 2 1 2 0.523.6 3.6ZP Q⎡ ⎤

= − > = −⎢ ⎥⎣ ⎦



3.45

We shall now state (without proof) some of the important properties of n

jointly Gaussian random variables, ( )1 2, , ........, nX X X=X . (These properties

are generalizations of the properties mentioned for the bivariate case in

Chapter 2).

P1) If ( )1 2, , ........, nX X X=X are jointly Gaussian, then any subset of them

are jointly Gaussian.

P2) The iX ’s are statistically independent if their covariance matrix is diagonal;

that is, 2i j i i jλ = σ δ where

Exercise 3.9

Let X be a zero-mean Gaussian vector with four components, 1 2, ,X X

3 4andX X . We can show that

1 2 3 4 1 2 3 4 1 3 2 4 1 4 2 3X X X X X X X X X X X X X X X X= ⋅ + ⋅ + ⋅

The above formula can also be used to compute the moments such as

41X , 2 2

1 2X X , etc. 41X can be computed as

= = σX X X X X4 41 1 1 1 1 13

Similarly, ( )22 2 2 21 2 1 1 2 2 1 2 1 22X X X X X X X X= = σ σ +

A zero mean stationary Gaussian process is sampled at 1t t= and 2t .

Let 1X and 2X denote the corresponding random variables. The covariance

matrix of 1X and 2X is

2 11 2XC ⎡ ⎤

= ⎢ ⎥⎣ ⎦

Show that 2 21 2 6E X X⎡ ⎤ =⎣ ⎦



3.46

1,0,i j

i ji j=⎧

δ = ⎨ ≠⎩

P3) Let ( )1 2, , .........., nY Y Y=Y be a set of vectors obtained from

( )1 2, , .........., nX X X=X by means of the linear transformation

T T TA= +Y X a

where A is any n n× matrix and ( )1 2, , .........., na a a=a is a vector of

constants ia . If X is jointly Gaussian, then so is Y .

As a consequence of P3) above, we find that if a Gaussian process

( )X t is input to a linear system, then the output ( )Y t is also Gaussian

process. We shall make use of this result in developing the properties of

narrowband noise.

3.9 Electrical Noise Electronic communication systems are made up of circuit elements such

as R , L and C , and devices like diodes, transistors, etc. All these components

give rise to what is known as internal circuit noise. It is this noise that sets the

fundamental limits on communication of acceptable quality.

Two of the important internal circuit noise varieties are: i) Thermal noise

ii) Shot noise

Exercise 3.10

Let T T TA= +Y X a

where the notation is from P3) above. Show that

TY XC AC A=



3.47

Historically, Johnson and Nyquist first studied thermal noise in metallic

resistors; hence it is also known as Johnson noise or resistance noise. The

random motion of the free electrons in a conductor caused by thermal agitation

gives rise to a voltage ( )V t at the open ends of the conductor and is present

whether or not an electrical field is applied. Consistent with the central limit

theorem, ( )V t is a Gaussian process with zero mean and the variance is a

function of R and T , where R is the value of the resistance and T is the

temperature of R . It has also been found that the spectral density of ( )V t

( )( )2in Volts Hz , denoted ( )VS f is essentially constant for 1210f ≤ Hz, if T is

290 K or 63 F (290 K is taken as the standard room temperature). 1210 Hz is

already in the infrared region of EM spectrum. This constant value of ( )VS f is

given by

( ) 22VS f R kT V Hz= (3.25)

where T : the temperature of R , in degrees Kelvin ( )K C 273= +

k : Boltzman’s constant 231.37 10 Joules K−= × .

It is to be remembered that Eq. 3.25 is valid only upto a certain frequency limit.

However this limit is much, much higher than the frequency range of interest to

us.

If this open circuit voltage is measured with the help of a true RMS

voltmeter of bandwidth B (frequency range: toB B− ), then the reading on the

instrument would be 2 2 4R kT B R kT B V⋅ = .

Thermal noise sources are also characterized in terms of available noise

PSD.

Def. 3.9: Available noise PSD is the maximum PSD that can be delivered by a

source.



3.48

Let us treat the resistor to be noise free. But it is in series with a noise

source with ( ) 2VS f R kT= . In other words, we have a noise source with the

source resistance R (Fig. 3.16(a)).

Fig. 3.16: (a) Resistor as noise source (b) The noise source driving a load

We know that the maximum power is transferred to the load, when the load is

matched to the source impedance. That is, the required load resistance is R

(Fig. 3.16(b)). The transfer function of this voltage divider network is 12

i.e. RR R

⎛ ⎞⎜ ⎟+⎝ ⎠

. Hence,

available ( ) ( ) 2VS fPSD H f

R=

2 Watts Hz4 2R kT kT

R= = (3.26)

It is to be noted that available PSD does not depend on R , though the

noise voltage is produced by R .

Because of Eq. 3.26, the available power in a bandwidth of B Hz is,

22a

kTP B kT B= ⋅ = (3.27)

In many solid state devices (and of course, in vacuum tubes!) there exists

a noise current mechanism called shot noise. In 1918, Schottky carried out the

first theoretical study of the fluctuations in the anode current of a temperature



3.49

limited vacuum tube diode (This has been termed as shot noise). He showed that

at frequencies well below the reciprocal of the electron transit time (which

extends upto a few Giga Hz), the spectral density of the mean square noise

current due to the randomness of emission of electrons from the cathode is given

by,

( ) ( )22 Amp HzIS f q I=

where q is the electronic charge and I is the average anode current. (Note that

units of spectral density could be any one of the three, namely, (a) Watts Hz ,

(b) ( )2Volts Hz and (c) ( )2Amp Hz . The circuit of Fig. 3.16(a) could be drawn

with ( )IS f for the source quantity. Then, we will have the Norton equivalent

circuit with ( ) 22 2

IR kT kTS fR R

= = in parallel with the resistance R .)

Semiconductor diodes, BJTs and JFETs have sources of shot noise in

them. Shot noise (which is non-thermal in nature), occurs whenever charged

particles cross a potential barrier. Here, we have an applied potential and there is

an average flow in some direction. However, there are going to be fluctuations

about this average flow and it is these fluctuations that contribute a noise with a

very wide spectrum.

In p-n junction devices, fluctuations of current occurs because of (i)

randomness of the transit time across space charge region separating p and n

regions, (ii) randomness in the generation and recombination of electron-hole

pairs, and (iii) randomness in the number of charged particles that diffuse etc.

Schottky’s result also holds for the semiconductor diode. The current

spectral density of shot noise in a p-n junction diode is given by

( ) ( )2 2I SS f q I I= + (3.28)

where I is the net DC current and SI is the reverse saturation current.



3.50

A BJT has two semiconductor junctions and hence two sources of shot

noise. In addition, it contributes to thermal noise because of internal ohmic

resistance (such as base resistance).

A JFET has a reverse biased junction between the gate terminal and the

semiconductor channel from the source to the drain. Hence, we have the gate

shot noise and the channel thermal noise in a JFET. (Note that gate shot noise

could get amplified when the device is used in a circuit.)

3.9.1 White noise Eq. 3.25 and Eq. 3.28 indicate that we have noise sources with a flat

spectral density with frequencies extending upto the infrared region of the EM

spectrum.

The concept of white noise is an idealization of the above. Any noise

quantity (thermal or non-thermal) which has a fiat power spectrum (that is, it

contains all frequency components in equal proportion) for f− ∞ < < ∞ is

called white noise, in analogy with white light. We denote the PSD of a white

noise process ( )W t as

( ) 0 watts Hz2W

NS f = (3.29)

where the factor 12

has been included to indicate that half the power is

associated with the positive frequencies and half with negative frequencies. In

addition to a fiat spectrum, if the process happens to be Gaussian, we describe it

as white Gaussian noise. Note that white and Gaussian are two different

attributes. White noise need not be Gaussian noise, nor Gaussian noise need be

white. Only when "whiteness" together with "Gaussianity" simultaneously exists,

the process is qualified as White Gaussian Noise (WGN) process.



3.51

White noise, whether Gaussian or not, must be fictitious (as is the case

with everything that is "ideal") because its total mean power is infinity. The utility

of the concept of white noise stems from the fact that such a noise, when passed

through a linear filter for which

( ) 2H f

∞

− ∞

< ∞∫ (3.30)

the filter output is a stationary zero mean noise process ( )N t that is meaningful

(Note that white noise process, by definition, is a zero mean process).

The condition implied by Eq.3.30 is not too stringent as we invariably deal

with systems which are essentially band limited and have finite value for ( )H f

within this band. In so far as the power spectrum at the output is concerned, it

makes little difference how the input power spectrum behaves outside of the

pass band. Hence, if the input noise spectrum is flat within the pass band of the

system, we might as well treat it to be white as this does not affect the calculation

of output noise power. However, assuming the input as ‘white’ will simplify the

calculations.

As ( ) 0

2WNS f = , we have, for the ACF of white noise

( ) ( )0

2WNR τ = δ τ , (3.31)

as shown in Fig. 3.17. This is again a nonphysical but useful result.



3.52

Fig.3.17: Characterization of white noise: a) Power spectral density

b) Auto correlation function

Eq. 3.31 implies that any two samples of white noise, no matter how closely

together in time they are taken, are uncorrelated. (Note that ( ) 0WR τ = for

0τ ≠ ). In addition, if, the noise process is Gaussian, we find that any two

samples of WGN are statistically independent. In a sense, WGN represents the

ultimate in randomness!

Imagine that white noise is being displayed on an oscilloscope. Though

the waveforms on successive sweeps are different, the display on the

oscilloscope always appears to be the same, no matter what sweep speed is

used. In the case of display of a deterministic waveform (such as a sinusoid)

changing the time-base makes the waveform to ‘expand’ or ‘contract’. In the case

of white noise, however, the waveform changes randomly from one instant to the

next, no matter what time scale is used and as such the display on the scope

appears to be the same for all time instants. If white noise drives a speaker, it



3.53

should sound monotonous because the waveform driving the speaker appears to

be the same for all time instants.

As mentioned earlier, white noise is fictitious and cannot be generated in

the laboratory. However, it is possible to generate what is known as Band Limited

White Noise (BLWN) which has a flat power spectral density for f W≤ , where

W is finite. (BLWN sources are commercially available. The cost of the

instrument depends on the required bandwidth, power level, etc.) We can give

partial demonstration (audio-visual) of the properties of white noise with the help

of these sources.

We will begin with the audio. By clicking on the speaker symbol, you will

listen to the speaker output when it is driven by a BLWN source with the

frequency range upto 110 kHz. (As the speaker response is limited only upto 15

kHz, the input to the speaker, for all practical purposes, is white.)

We now show some flash animation pictures of the spectral density and

time waveforms of BLWN.

1. Picture 1 is the display put out by a spectrum analyzer when it is fed with a

BLWN signal, band limited to 6 MHz. As can be seen from the display, the

spectrum is essentially constant upto 6 MHz and falls by 40 to 45 dB at

about 7MHz and beyond.

2. Picture 2 is time domain waveform (as seen on an oscilloscope) when the 6

MHz, BLWN signal is the input to the oscilloscope. Four sweep speeds

have been provided for you to observe the waveform. These speeds are:

100 µsec/div, 200 µsec/div, 500 µsec/div and 50 µsec/div. As you switch

from 100 µsec/div to 200 µsec/div to 500 µsec/div, you will find that the



3.54

display is just about the same, whereas when you switch from 500 µsec/div

to 50 µsec/div, there is a change, especially in the display level.

Consider a 500 kHz sine wave being displayed on an oscilloscope

whose sweep is set to 50 µsec/div. This will result in 25 cycles per division

which implies every cycle will be seen like a vertical line of appropriate

height. As the frequency increases, the display essentially becomes a band

of fluorescence enclosed between two lines. (The spikes in the display are

due to large amplitudes in noise voltage which occur with some non-zero

probability.)

Now consider a 20 kHz sine wave being displayed with the same

sweep speed, namely, 50 µsec/div. This will result in 1 cycle/div which

implies we can see the fine structure of the waveform. If the sweep speed

were to be 500 µsec/div, then a 20 kHz time will result in 10 cycles/div,

which implies that fine structure will not be seen clearly. Hence when we

observe BLWN, band limited to 6 MHz, on an oscilloscope with a sweep

speed of 50 µsec/div, fine structure of the low frequency components could

interfere with the constant envelope display of the higher frequency

components, thereby causing some reduction in the envelope level.

3. Picture 3 is the display from the spectrum analyzer when it is fed with a

BLWN signal, band limited to 50 MHz. As can be seen from the display,

PSD is essentially constant upto 50 MHz and then starts falling, going 40

dB down by about 150 MHz.

4. Picture 4 is the time domain signal, as seen on an oscilloscope, when the

input is the 50 MHz wide, BLWN signal.

You have again the choice of the same four sweep rates. This time

you will observe that when you switch from 500 µsec/div to 50 µsec/div, the



3.55

change is much less. This is because, this signal has much wider spectrum

and the power in the frequency range of 100f ≤ kHz is 0.002P where P

is the total power of the BLWN signal.

From the above, it is clear that as the BLWN tends towards white noise,

variations in the time waveform keep reducing, resulting in a steady picture on

the oscilloscope no matter what sweep speed is used for the display.

Example 3.15: (White noise through an ILPF)

White Gaussian noise with the PSD of 0

2N is applied as input to an ideal

LPF of bandwidth B . Find the ACF of the output process. Let Y be the random

variable obtained from sampling the output process at 1t = sec. Let us find

( )Yf y .

Let ( )N t denote the output noise process when the WGN process ( )W t

is the input. Then,

( )0 ,

20 ,

N

N B f BS f

otherwise

⎧ − ≤ ≤⎪= ⎨⎪⎩

Taking the inverse Fourier transform of ( )NS f , we have

( ) ( )0 sin 2NR N B c Bτ = τ

A plot of ( )NR τ is given in Fig. 3.18.



3.56

Fig 3.18: ( )NR τ of example 3.15

The filter output ( )N t is a stationary Gaussian process. Hence the PDF is

independent of the sampling time. As the variable Y is Gaussian, what we need

is the mean value and the variance of Y . Evidently [ ] 0E Y = and

( )2 200Y YY R N Bσ = = = . Hence Y is ( )00,N N B .

Note that ACF passes through zeros at 2nB

τ = where 1, 2, .....n = ± ± .

Hence any two random variables obtained by sampling the output process at

times 1t and 2t such that 1 2t t− is multiple of 12B

, are going to be statistically

independent.

Example 3.16: (White Noise through an RC-LPF)

Let ( )W t be input to an RC-LPF. Let us find the ACF of the output ( )N t .

If X and Y are two random variables obtained from sampling ( )N t with a

separation of 0.1 sec, let us find X Yρ .



3.57

The transfer function of the RC LPF is given by ( ) 11 2

H fj f RC

=+ π

.

Therefore ( )( )

02

21 2

NNS f

f RC=

+ π, and

( ) 0 exp4NNRRC RC

⎛ ⎞− ττ = ⎜ ⎟

⎝ ⎠

[ ]2

X YX Y

X Y

E XYE X

λρ = =

σ σ ⎡ ⎤⎣ ⎦

(Note that ( ) 0N t = and X Yσ = σ ).

[ ] ( )0.1 secNE X Y R

τ == τ

( )2 00 4N

NE X RRCτ =

⎡ ⎤ = τ =⎣ ⎦

Hence, ( )0.1 RCX Y e−ρ =

Exercise 3.11

( )X t is a zero mean Gaussian process with ( ) 21

1XR τ =+ τ

. Let

( ) ( )Y t X t= where ( )X t is the Hilbert transform of ( )X t . The process ( )X t

and ( )Y t are sampled at 1t = and 2t = sec. Let the corresponding random

variables be denoted by ( )1 2,X X and ( )1 2,Y Y respectively.

a) Write the covariance matrix of the four variables 1 2,X X 1 2andY Y .

b) Find the joint PDF of 2 ,X 1 2andY Y .



3.58

Let 0τ denote the de-correlation time where 0τ is defined such that if

0τ > τ , then ( ) ( )0.01 0N NR Rτ ≤ . Then for the RC-LPF, 0 4.61 RCτ = . That

is, if the output process ( )N t is sampled at 1t and 2t such that

1 2 4.61t t RC− ≥ , then the random variables ( )1N t and ( )2N t will be

Exercise 3.12 The impulse response of a filter (LTI system) is given by

( ) ( ) ( )th t t e u t− α= δ − α

where α is a positive constant. If the input to the filter is white Gaussian noise

with the PSD of 0

2N watts/Hz, find

a) the output PSD

b) Show that the ACF of the output is

( )0

2 2N e− α τα⎡ ⎤δ τ −⎢ ⎥⎣ ⎦

Exercise 3.13 White Gaussian noise process is applied as input to a zero-order-hold

circuit with a delay of T sec. The output of the ZOH circuit is sampled at

t T= . Let Y be the corresponding random variable. Find ( )Yf y .

Exercise 3.14 Noise from a 10 kΩ resistor at room temperature is passed through an

ILPF of bandwidth 2.5 MHz. The filtered noise is sampled every microsecond.

Denoting the random variables corresponding to the two adjacent samples as

1Y and 2Y , obtain the expression for the joint PDF of 1Y and 2Y .



3.59

essentially uncorrelated. If ( )N t happens to be a Gaussian process, then ( )1N t

and ( )2N t will be, for all practical purposes, independent.

In example 3.15, we found that the average output noise power is equal to

0N B where B is the bandwidth of the ILPF. Similarly, in the case of example

3.16, we find the average output noise power being equal to 0

4NRC

⎛ ⎞⎜ ⎟⎝ ⎠

where

12 RCπ

is the 3-dB bandwidth of the RC-LPF. That is, in both the cases, we find

that the average output noise power is proportional to some measure of the

bandwidth. We may generalize this statement to include all types of low pass

filters by defining the noise equivalent bandwidth as follows.

Suppose we have a source of white noise with the PSD ( ) 0

2WNS f =

connected to an arbitrary low pass filter of transfer function ( )H f . The resulting

average output noise power N is given by,

( )20

2NN H f d f

∞

− ∞

= ∫

( ) 2

00

N H f d f∞

= ∫

Consider next the same source of white noise connected to the input of an

ILPF of zero frequency response ( )0H and bandwidth B . In this case, the

average output noise power is,

( )20' 0N N B H= .

Def. 3.10: The noise equivalent bandwidth of an arbitrary filter is defined as the

bandwidth of an ideal rectangular filter that would pass as much white noise



3.60

power as the filter in question and has the same maximum gain as the filter

under consideration.

Let NB be the value of B such that 'N N= . Then NB , the noise

equivalent bandwidth of ( )H f , is given by

( )

( )

2

02 0N

H f d fB

H

∞

=∫

(3.32)

The notion of equivalent noise bandwidth is illustrated in Fig. 3.19.

Fig. 3.19: Pictorial representation of noise equivalent bandwidth

The advantage of the noise equivalent bandwidth is that if NB is known,

we can simply calculate the output noise power without worrying about the actual

shape of ( )H f . The definition of noise equivalent bandwidth can be extended to

a band pass filter.

Example 3.17 Compute the noise equivalent bandwidth of the RC-LPF.

We have ( )0 1H = . Hence,

( )2

0

141 2

Nd fB

RCf RC

∞

= =+ π∫



3.61

3.10 Narrowband Noise The communication signals that are of interest to us are generally

NarrowBand BandPass (NBBP) signals. That is, their spectrum is concentrated

around some (nominal) centre frequency, say cf , with the signal bandwidth

usually being much smaller compared to cf . Hence the receiver meant for such

signals usually consists of a cascade of narrowband filters; this means that even

if the noise process at the input to the receiver is broad-band (may even

considered to be white) the noise that may be present at the various stages of a

receiver is essentially narrowband in nature. We shall now develop the statistical

characterization of such NarrowBand Noise (NBN) processes.

Let ( )N t denote the noise process at the output of a narrowband filter

produced in response to a white noise process, ( )W t , at the input. ( )WS f is

taken as 0

2N . If ( )H f denotes the filter transfer function, we have

( ) ( ) 20

2NNS f H f= (3.33)

In fact, any narrowband noise encountered in practice could be modeled as the

output of a suitable filter. In Fig. 3.20, we show the waveforms of experimentally

generated NBBP noise. This noise process is generated by passing the BLWN

(with a flat spectrum upto 110 kHz) through a NBBP filter, whose magnitude

characteristic is shown in Fig. 3.21. This filter has a centre frequency of 101 kHz

and a 3-dB bandwidth of less than 3 kHz.



3.62

Fig. 3.20: Some oscilloscope displays of narrowband noise

X-axis: time

Y-axis: voltage



3.63

The magnitude response of the filter is obtained by sweeping the filter input from

87 kHz to 112 kHz.

Fig. 3.21: Frequency response of the filter used to generate NBN

Plot in Fig. 3.20(a) gives us the impression that it is a 101 kHz sinusoid with

slowly changing envelope. This is only partially correct. Waveforms at (b) and (c)

are expanded versions of a part of the waveform at (a). From (b) and (c), it is

clear that zero crossings are not uniform. In Fig. 3.20(c), the cycle of the

waveform shown in green fits almost into the space between two adjacent

vertical lines of the graticule, whereas for the cycle shown in red, there is a clear

offset. Hence the proper time-domain description of a NBBP signal would be: it is

a sinusoid undergoing slow amplitude and phase variations. (This will be justified

later on by expressing the NBBP noise signal in terms of its envelope and

phase.)

We shall now develop two representations for the NBBP noise signal.

These are (a) canonical (also called in-phase and quadrature component)

representation and (b) Envelope and Phase representation.



3.64

3.10.1 Representation of narrowband noise a) Canonical representation

Let ( )n t represent a sample function of a NBBP noise process ( )N t , and

( )p en t and ( )c en t , its pre-envelope and complex envelope respectively. We will

assume cf to be the (nominal) centre frequency of the noise process. Then, we

can write:

( ) ( ) ( )p en t n t j n t= + and (3.34)

( ) ( ) ( )exp 2c e p e cn t n t j f t= − π (3.35)

where ( )n t is the Hilbert transform of ( )n t .

The complex envelope ( )c en t itself can be expressed as

( ) ( ) ( )c e c sn t n t j n t= + (3.36)

With the help of Eq. 3.34 to 3.36, it is easy to show,

( ) ( ) ( ) ( ) ( )cos 2 sin 2c c cn t n t f t n t f t= π + π (3.37)

( ) ( ) ( ) ( ) ( )cos 2 sin 2s c cn t n t f t n t f t= π − π (3.38)

As ( ) ( ) 2Re cj f tc en t n t e π⎡ ⎤= ⎣ ⎦

( ) ( ) 2Re cj f tc sn t j n t e π⎡ ⎤= +⎣ ⎦

We have,

( ) ( ) ( ) ( ) ( )cos 2 sin 2c c s cn t n t f t n t f t= π − π (3.39)

As in the deterministic case, we call ( )cn t the in-phase component and ( )sn t the

quadrature component of ( )n t .

As Eq. 3.39 is valid for any sample function ( )n t of ( )N t , we shall write

( ) ( ) ( ) ( ) ( )cos 2 sin 2c c s cN t N t f t N t f t= π − π (3.40a)



3.65

Eq. 3.40(a) is referred to as the canonical representation of ( )N t . ( )cN t and

( )sN t are low pass random processes; ( )cn t and ( )sn t are sample functions of

( )cN t and ( )sN t respectively.

b) Envelope and phase representation

Let us write ( )n t as

( ) ( ) ( )cos 2 cn t r t f t t⎡ ⎤= π + ψ⎣ ⎦ (3.41a)

( ) ( ) ( ) ( ) ( )cos cos 2 sin sin 2c cr t t f t t f t⎡ ⎤= ψ π − ψ π⎣ ⎦ (3.41b)

Comparing Eq. 3.41(b) with Eq. 3.39, we find that

( ) ( ) ( )coscn t r t t= ψ

( ) ( ) ( )sinsn t r t t= ψ

or ( ) ( ) ( )1

2 2 2c sr t n t n t⎡ ⎤= +⎣ ⎦ (3.42a)

and ( ) ( )( )

tan s

c

n tt arc

n t⎡ ⎤

ψ = ⎢ ⎥⎢ ⎥⎣ ⎦

(3.42b)

( )r t is the envelope of ( )n t and ( )tψ its phase. Generalizing, we have

( ) ( ) ( )( )cos 2 cN t R t f t t⎡ ⎤= π + Ψ⎣ ⎦ (3.43)

where ( )R t is the envelope process and ( )tΨ is the phase process. Eq. 3.43

justifies the statement that the NBN waveform exhibits both amplitude and phase

variations.

Exercise 3.15

Show that ( ) ( ) ( ) ( ) ( )sin 2 cos 2c c s cN t N t f t N t f t= π + π (3.40b)



3.66

3.10.2 Properties of narrowband noise We shall now state some of the important properties of NBN. (For proofs,

refer to Appendix A3.1.)

P1) If ( )N t is zero mean, then so are ( )cN t and ( )sN t .

P2) If ( )N t is a Gaussian process, then ( )cN t and ( )sN t are jointly Gaussian.

P3) If ( )N t is WSS, then ( )cN t and ( )sN t are WSS.

P4) Both ( )cN t and ( )sN t have the same PSD which is related to ( )NS f of the

original narrowband noise as follows:

( ) ( ) ( ) ( ) ,0 ,c s

N c N cN N

S f f S f f B f BS f S f

elsewhere

⎧ − + + − ≤ ≤⎪= = ⎨⎪⎩

where it is assumed that ( )NS f occupies the frequency interval

c cf B f f B− ≤ ≤ + and cf B> .

P5) If the narrowband noise ( )N t is zero mean, then ( )cN t and ( )sN t have

the same variance as the variance of ( )N t .

P6) The cross-spectral densities of the quadrature components of narrowband

noise are purely imaginary, as shown by

( ) ( ) ( ) ( ) ,

0 ,c s s c

N c N cN N N N

j S f f S f f B f BS f S f

elsewhere

⎧ ⎡ ⎤+ − − − ≤ ≤⎪ ⎣ ⎦= − = ⎨⎪⎩

P7) If ( )N t is zero-mean Gaussian and its PSD, ( )NS f is locally symmetric

about cf± , then ( )cN t and ( )sN t are statistically independent.

Property P7) implies that if ( )NS f is locally symmetric about cf± , then

( )c sN NR τ is zero for all τ . Since ( )cN t and ( )sN t are jointly Gaussian, they

become independent. However, it is easy to see that ( )0 0c sN NR = whether or

not there is local symmetry. In other words the variables ( )1cN t and ( )1sN t , for

any sampling instant 1t , are always independent.



3.67

Note: Given the spectrum of an arbitrary band-pass signal, we are free to choose

any frequency cf as the (nominal) centre frequency. The spectral shape ( )cNS f

and ( )sNS f will depend on the cf chosen. As such, the canonical representation

of a narrowband signal is not unique. For example, for the narrowband spectra

shown in Fig. 3.22, if 1f is chosen as the representative carrier frequency, then

the noise spectrum is 12 B wide, whereas if 2f (which is actually the mid-

frequency of the given band) is chosen as the representative carrier frequency,

then the width of the spectrum is 22 B . Note for the ( )NS f of Fig. 3.22, it is not

possible for us to choose an cf such that ( )NS f exhibits local symmetry with

respect to it.

Fig. 3.22: Narrowband noise spectrum with two different centre frequencies

Example 3.18

For the narrowband noise spectrum ( )NS f shown in Fig. 3.23, sketch

( )cNS f for the two cases, namely a) 10cf k Hz= and b) 11cf k Hz= . c) What

is the variance ( )cN t ?



3.68

Fig. 3.23: ( )NS f for the example 3.18

a) From P4), we have,

( ) ( ) ( ) ,0 ,c

N c N cN

S f f S f f B f BS f

elsewhere

⎧ − + + − ≤ ≤⎪= ⎨⎪⎩

Using 10cf k Hz= , and plotting ( )N cS f f− and ( )N cS f f+ , we obtain Fig.

3.24.

Fig. 3.24: Shifted spectra: (a) ( )N cS f f+ and (b) ( )N cS f f−

Taking 2B k Hz= and extracting the relevant part of ( )N cS f f+ and

( )N cS f f− , we have ( )cNS f as shown in Fig. 3.25.



3.69

Fig. 3.25: ( )

cNS f with 10cf k Hz=

b) By repeating the above procedure, we obtain ( )cNS f (the solid line) shown

in Fig. 3.26.

Fig. 3.26: ( )

cNS f with 11cf k Hz=

c) ( )2 2 1.0cN N NS f d f Watt

∞

− ∞

σ = σ = =∫



3.70

Exercise 3.16

Assuming ( )N t to be WSS and using Eq. 3.40(a), establish

( ) ( ) ( ) ( ) ( )τ = τ π τ − τ π τc s cN N c N N cR R f R fcos 2 sin 2 (3.44)

Exercise 3.17

Let ( )N t represent a narrowband, zero-mean Gaussian process with

( )0 ,

20 ,

c cN

N f B f f BS f

otherwise

⎧ − ≤ ≤ +⎪= ⎨⎪⎩

Let X and Y be two random variables obtained by sampling ( )cN t and

( )cd N td t

at 1t t= , where ( )cN t is the in phase component of ( )N t .

a) Show that X and Y are independent.

b) Develop the expression for ( ), ,X Yf x y .

Exercise 3.18

Let ( )N t represent a NBN process with the PSD shown in Fig. 3.27

below.

Fig. 3.27: Figure for exercise 3.18

Let ( ) ( ) ( ) ( ) ( )cos sinc c s cN t N t t N t t= ω − ω with 50cf = kHz.

Show that ( ) ( ) ( )3 32 sin 10 sin 3 103s cN NR cτ = τ π τ .



3.71

We shall now establish a relation between ( )NR τ and ( )c eNR τ , where

( )c eNR τ evidently stands for the ACF of the complex envelope of the narrowband

noise. For complex signals, such as ( )c eN t , we define its ACF as

( ) ( ) ( ),c eN c e c eR t t E N t N t∗⎡ ⎤+ τ = + τ⎣ ⎦

( ) ( )( ) ( ) ( )( )c s c sE N t j N t N t j N t⎡ ⎤= + τ + + τ −⎣ ⎦

( ) ( ) ( ) ( )c s s c c sN N N N N NR R j R j R= τ + τ + τ − τ

But ( ) ( )c sN NR Rτ = τ (Eq. A3.1.7)

and ( ) ( )s c c sN N N NR Rτ = − τ (Property of cross correlation)

( )c sN NR= − τ

The last equality follows from Eq. A3.1.8. Hence,

( ) ( ) ( ) ( )( ), 2c e ce c s cN N N N NR t t R R j R+ τ = τ = τ + τ . From Eq. 3.44, we have

( ) ( ) 21 Re2

c

c e

j fN NR R e+ π τ⎡ ⎤τ = τ⎣ ⎦

( ) ( )2 214

c c

c e c e

j f j fN NR e R eπ τ − π τ∗⎡ ⎤= τ + τ⎣ ⎦ (3.45)

Taking the Fourier transform of Eq. 3.45, we obtain

( ) ( ) ( )∗⎡ ⎤= − + − −⎣ ⎦14 c e c eN N c N cS f S f f S f f (3.46)

Note: For complex signals, the ACF is conjugate symmetric; that is, if ( )X t

represents a complex random process that is WSS, then ( ) ( )X XR R∗− τ = τ . The

PSD ( ) ( )X XS f F R⎡ ⎤= τ⎣ ⎦ is real, nonnegative but not an even function of

frequency.



3.72

3.10.3 PDF of the envelope of narrowband noise From Eq. 3.43, we have,

( ) ( ) ( )( )cos 2 cN t R t f t t⎡ ⎤= π + Ψ⎣ ⎦

with ( ) ( ) ( )( )coscN t R t t= Ψ (3.47a)

and ( ) ( ) ( )( )sinsN t R t t= Ψ (3.47b)

That is,

( ) ( ) ( )1

2 2 2c sR t N t N t⎡ ⎤= +⎣ ⎦ (3.48a)

( ) ( )( )

1tan s

c

N tt

N t− ⎡ ⎤

Ψ = ⎢ ⎥⎢ ⎥⎣ ⎦

(3.48b)

Our interest is the PDF of the random variable ( )1R t , for any arbitrary

sampling instant 1t t= .

Let cN and sN represent the random variables obtained from sampling

( )cN t and ( )sN t at any time 1t t= . Assuming N(t) to be a Gaussian process,

we have cN and sN as zero mean, independent Gaussian variables with

variance 2σ where ( )2NS f d f

∞

− ∞

σ = ∫ .

Hence,

( )2 2

2 21, exp

2 2c s

c sN N c s

n nf n n⎡ ⎤+

= −⎢ ⎥πσ σ⎣ ⎦

For convenience, let ( )1R t R= and ( )1tΨ = Ψ . Then,

coscN R= Ψ and sinsN R= Ψ

and the Jacobian of the transformation is,

−ψ ψ⎡ ⎤

⎢ ⎥= =− ψ ψ⎢ ⎥⎣ ⎦

1cos sinsin cosJ rr r



3.73

Therefore,

( )2

2 2,

exp , 0, 0 2, 2 2

0 ,R

r r rf r

otherwiseΨ

⎧ ⎛ ⎞− ≥ ≤ ψ < π⎪ ⎜ ⎟ψ = πσ σ⎨ ⎝ ⎠

⎪⎩

( ) ( )2

,0

,R Rf r f r dπ

Ψ= ψ ψ∫

It is easy to verify that

( )2

2 2exp , 02

0 ,R

r r rf r

otherwise

⎧ ⎛ ⎞− ≥⎪ ⎜ ⎟= σ σ⎨ ⎝ ⎠

⎪⎩

(3.49)

Similarly, it can be shown that

( )1 , 0 2

20 ,

fotherwise

Ψ

⎧ ≤ ψ < π⎪ πψ = ⎨⎪⎩

(3.50)

As ( ) ( ) ( ), ,R Rf r f r fψ Ψψ = ψ , we have R and Ψ as independent variables.

The PDF given by Eq. 3.49 is the Rayleigh density which was introduced

in Chapter 2. From the above discussion, we have the useful result, namely, the

envelope of narrowband Gaussian noise is Rayleigh distributed.

Let us make a normalized plot of the Rayleigh PDF by defining a new

random variable V as RV =σ

(transformation by a multiplicative constant).

Then,

( )2

exp , 02

0 ,V

vv vf v

otherwise

⎧ ⎛ ⎞− ≥⎪ ⎜ ⎟= ⎨ ⎝ ⎠

⎪⎩

( )Vf v is plotted in Fig. 3.28.



3.74

Fig. 3.28: Normalized Rayleigh PDF

The peak value of the PDF occurs at 1v = and ( )1 0.607Vf = .

3.10.4 Sine wave plus narrowband noise Let a random process ( )X t be given by

( ) ( ) ( )cos 2 cX t A f t N t= π + (3.51)

where A and cf are constants, ( )N t represents the narrowband noise process

whose centre frequency is taken as cf . Our interest is to develop an expression

for the PDF of the envelope of ( )X t .

Using the canonical form for ( )N t , Eq. 3.51 can be written as

( ) ( ) ( ) ( ) ( ) ( )cos 2 cos 2 sin 2c c c s cX t A f t N t f t N t f t= π + π − π

Let ( ) ( )c c'N t A N t= + . Assume ( )N t to be a Gaussian process with zero mean

and variance 2σ . Then, for any sampling instant 1t t= , let c'N denote the

random variable ( )1c'N t and let sN denote the random variable ( )1sN t . From our

earlier discussion, we find that c'N is ( )2,N A σ , sN is ( )20,N σ and c

'N is

independent of sN . Hence,



3.75

( ) ( )22

2 2,

1, exp2 2c s

c s

c sN N'

''

n A nf n n

⎡ ⎤− +⎢ ⎥= −⎢ ⎥πσ σ⎢ ⎥⎣ ⎦

(3.52)

Let ( ) ( ) ( )c s'R t N t N t

12 22⎧ ⎫⎡ ⎤ ⎡ ⎤= +⎨ ⎬⎣ ⎦⎢ ⎥⎣ ⎦⎩ ⎭

( ) ( )( )

1tan s

c'

N tt

N t−⎡ ⎤

Ψ = ⎢ ⎥⎢ ⎥⎣ ⎦

where ( )R t and ( )tΨ are the envelope and phase, respectively of ( )X t . By a

procedure similar to that used in the computation of the PDF of the envelope of

narrowband noise, we find that

( )2 2

, 2 2

2 cos, exp , 0, 0 22 2R

r r A Arf r rΨ

⎡ ⎤+ − ψψ = − ≥ ≤ ψ ≤ π⎢ ⎥πσ σ⎣ ⎦

where ( )1R R t= and ( )1tΨ = Ψ

The quantity of interest is ( )Rf r , where

( ) ( )2

,0

,R Rf r f r dπ

Ψ= ψ ψ∫

That is,

( )22 2

2 2 20

exp exp cos2 2R

r r A Arf r dπ⎛ ⎞+ ⎛ ⎞= − ψ ψ⎜ ⎟ ⎜ ⎟πσ σ σ⎝ ⎠⎝ ⎠∫ (3.53)

The integral on the RHS of Eq. 3.53 is identified in terms of the defining

integral for the modified Bessel function of the first kind and zero order.

Let ( ) ( )2

00

1 exp cos2

I y y dπ

= ψ ψπ ∫ (3.54)

A plot of ( )0I y is shown in Fig. 3.29. In Eq. 3.54, if we let 2

Ary =σ

, Eq. 3.53 can

be rewritten as



3.76

( )2 2

02 2 2exp2R

r r A Arf r I⎛ ⎞+ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟σ σ σ⎝ ⎠⎝ ⎠

(3.55)

The PDF given by Eq. 3.55 is referred to as Rician distribution.

Fig. 3.29: Plot of ( )0I y

The graphical presentation of the Rician PDF can be simplified by introducing

two new variables, namely, RV =σ

and Aα =

σ. Then, the Rician density of Eq.

3.55 can be written in a normalized form,

( ) ( )2 2

0exp2V

vf v v I v⎛ ⎞+ α

= − α⎜ ⎟⎝ ⎠

(3.56)

which is plotted in Fig. 3.30 for various values of α .



3.77

Fig. 3.30: Normalized Rician PDF for various values of α

Based on these curves, we make the following observations:

i) From Fig. 3.28, we find ( )0 0 1I = . If 0A = , then 0α = and

( )2

exp , 02

0 ,V

vv vf v

otherwise

⎧ ⎛ ⎞− ≥⎪ ⎜ ⎟= ⎨ ⎝ ⎠

⎪⎩

which is the normalized Rayleigh PDF shown earlier in Fig. 3.27. This is

justified because if 0A = , then ( )R t is the envelope of only the

narrowband noise.

ii) For 1y >> , ( )0 2

yeI yyπ

. Using this approximation, it can be shown that

( )Vf v is approximately Gaussian in the vicinity of v = α , when α is

sufficiently large. That is, when the sine-wave amplitude A is large

compared with σ (which is the square root of the average power in ( )N t ),

( )Vf v can be approximated by a Gaussian PDF over a certain range. This

can be seen from Fig. 3.29 for the cases of 3 and 5α = .



3.78

Appendix A3.1 Properties of Narrowband Noise: Some Proofs. We shall now give proofs to some of the properties of the NBN, mentioned

in sec.3.10.2.

P1) If ( )N t is zero mean Gaussian, then so are ( )cN t and ( )sN t .

Proof: Generalizing Eq. 3.37 and 3.38, we have

( ) ( ) ( ) ( ) ( )cos 2 sin 2c c cN t N t f t N t f t= π + π (A3.1.1)

( ) ( ) ( ) ( ) ( )cos 2 sin 2s c cN t N t f t N t f t= π − π (A3.1.2)

( )N t is obtained as the output of an LTI system with ( ) 1h tt

=π

, with the

input ( )N t . This implies that if ( )N t is zero mean, then so is ( )N t . Taking

the expectation of A3.1.1, we have

( ) ( ) ( ) ( ) ( )cos 2 sin 2c c cN t N t f t N t f t= π + π

As ( ) ( ) 0N t N t= = , we obtain ( ) 0cN t = . Taking the expectation of

A3.1.2, we obtain ( ) 0sN t = .

P2) If ( )N t is a Gaussian process, then ( )cN t and ( )sN t are jointly Gaussian.

Proof: This property follows from Eq. 3.40(a) because, ( )N t is guaranteed

to be Gaussian only if ( )cN t and ( )sN t are jointly Gaussian.

P3) If ( )N t is WSS, then ( )cN t and ( )sN t are WSS.

Proof: We will first establish that

( ) ( ) ( ) ( ) ( ) ( ), cos 2 sin 2c c

NN N N c cR t t R R f R f+ τ = τ = τ π τ + τ π τ (A3.1.3)

(In Eq. A3.1.3, ( )NR τ is the Hilbert transform of ( )NR τ .)



3.79

Consider the scheme shown in Fig A3.1.1.

Fig. A3.1.1: Scheme to obtain ( )N t from ( )N t

Eq. 3.21(a) gives us

( ) ( )1NN N

R Rτ = ∗ τπτ

(A3.1.4a)

( )NR= τ (A3.1.4b)

( ) ( )N N N NR Rτ = − τ ,

( ) ( )1NR= ∗ − τ

π − τ

( )1NR= − ∗ τ

πτ

( )NR= − τ

That is, ( ) ( )N N N NR Rτ = − τ (A3.1.5)

( ) ( ) ( ),cN c cR t t E N t N t⎡ ⎤+ τ = + τ⎣ ⎦

Expressing ( )cN t + τ and ( )cN t in terms of the RHS quantities of Eq.

A3.1.1 and after some routine manipulations, we will have

( ) ( ) ( ) ( )

( ) ( ) ( )( )

( ) ( ) ( )

( ) ( ) ( )( )

1, cos 221 cos 2 221 sin 221 sin 2 22

cN N cN

N cN

cN N N N

cN N N N

R t t R R f

R R f t

R R f

R R f t

⎡ ⎤+ τ = τ + τ π τ⎣ ⎦

⎡ ⎤+ τ − τ π + τ⎣ ⎦

⎡ ⎤+ τ − τ π τ⎣ ⎦

⎡ ⎤+ τ + τ π + τ⎣ ⎦

(A3.1.6)



3.80

( )NR τ is the autocorrelation of the Hilbert transform of ( )N t . As ( )N t and

( )N t have the same PSD, we have ( ) ( )N NR Rτ = τ and from Eq. A3.1.5,

we have ( ) ( )N N N NR R Tτ = − . Using these results in Eq. A3.1.6, we obtain

Eq. A3.1.3.

P4) Both ( )cN t and ( )sN t have the same PSD which is related to ( )NS f of the

original narrowband noise as follows:

( ) ( ) ( ) ( ) ,0 ,c s

N c N cN N

S f f S f f B f BS f S f

elsewhere

⎧ − + + − ≤ ≤⎪= = ⎨⎪⎩

where it is assumed that ( )NS f occupies the frequency interval

c cf B f f B− ≤ ≤ + and cf B> .

Proof: Taking the Fourier transform of Eq. A3.1.3, we have

( ) ( ) ( )

( ) ( ) ( ) ( )

12

1 sgn sgn2

cN N c N c

N c c N c c

S f S f f S f f

S f f f f S f f f f

⎡ ⎤= − + +⎣ ⎦

⎡ ⎤− − − − + +⎣ ⎦

( ) ( ) ( ) ( )1 11 sgn 1 sgn2 2N c c N c cS f f f f S f f f f⎡ ⎤ ⎡ ⎤= − − + + + + +⎣ ⎦ ⎣ ⎦

(A3.1.9)

But ( )2 ,

1 sgn0 ,

cc

f ff f

outside<⎧

− − = ⎨⎩

(A3.1.10a)

( )2 ,

1 sgn0 ,

cc

f ff f

outside> −⎧

+ + = ⎨⎩

(A3.1.10b)

Exercise A3.1.1

Show that ( ) ( )s cN NR t R= τ (A3.1.7)

and ( ) ( ) ( ) ( ) ( )sin 2 cos 2c s

NN N N c cR R f R fτ = τ π τ − τ π τ (A3.1.8)



3.81

Using Eq. A3.1.10 in Eq. A3.1.9 completes the proof.

Exercise A3.1.2 Establish properties P5 to P7.



3.82

References

1) A. Papoulis, Probability, Random variables and Stochastic processes (3rd

ed), McGraw Hill, 1991

2) Henry Stark and John W. Woods, Probability and Random Processes with

Applications to Signal processing (3rd ed), Pearson Education Asia, 2002

3) K. Sam Shanmugam and A. M. Breiphol, Random Signals: Detection,

Estimation and data analysis, John Wiley, 1988



2.1

2 UUCHAPTER 2

Probability and Random Variables

2.1 Introduction At the start of Sec. 1.1.2, we had indicated that one of the possible ways

of classifying the signals is: deterministic or random. By random we mean

unpredictable; that is, in the case of a random signal, we cannot with certainty

predict its future value, even if the entire past history of the signal is known. If the

signal is of the deterministic type, no such uncertainty exists.

Consider the signal ( ) ( )1cos 2x t A f t= π + θ . If A , θ and 1f are known,

then (we are assuming them to be constants) we know the value of ( )x t for all t .

( A , θ and 1f can be calculated by observing the signal over a short period of

time).

Now, assume that ( )x t is the output of an oscillator with very poor

frequency stability and calibration. Though, it was set to produce a sinusoid of

frequency 1f f= , frequency actually put out maybe f1' where ( )f f f1 1 1

' ∈ ± ∆ .

Even this value may not remain constant and could vary with time. Then,

observing the output of such a source over a long period of time would not be of

much use in predicting the future values. We say that the source output varies in

a random manner.

Another example of a random signal is the voltage at the terminals of a

receiving antenna of a radio communication scheme. Even if the transmitted



2.2

(radio) signal is from a highly stable source, the voltage at the terminals of a

receiving antenna varies in an unpredictable fashion. This is because the

conditions of propagation of the radio waves are not under our control.

But randomness is the essence of communication. Communication

theory involves the assumption that the transmitter is connected to a source,

whose output, the receiver is not able to predict with certainty. If the students

know ahead of time what is the teacher (source + transmitter) is going to say

(and what jokes he is going to crack), then there is no need for the students (the

receivers) to attend the class!

Although less obvious, it is also true that there is no communication

problem unless the transmitted signal is disturbed during propagation or

reception by unwanted (random) signals, usually termed as noise and

interference. (We shall take up the statistical characterization of noise in

Chapter 3.)

However, quite a few random signals, though their exact behavior is

unpredictable, do exhibit statistical regularity. Consider again the reception of

radio signals propagating through the atmosphere. Though it would be difficult to

know the exact value of the voltage at the terminals of the receiving antenna at

any given instant, we do find that the average values of the antenna output over

two successive one minute intervals do not differ significantly. If the conditions of

propagation do not change very much, it would be true of any two averages (over

one minute) even if they are well spaced out in time. Consider even a simpler

experiment, namely, that of tossing an unbiased coin (by a person without any

magical powers). It is true that we do not know in advance whether the outcome

on a particular toss would be a head or tail (otherwise, we stop tossing the coin

at the start of a cricket match!). But, we know for sure that in a long sequence of

tosses, about half of the outcomes would be heads (If this does not happen, we

suspect either the coin or tosser (or both!)).



2.3

Statistical regularity of averages is an experimentally verifiable

phenomenon in many cases involving random quantities. Hence, we are tempted

to develop mathematical tools for the analysis and quantitative characterization

of random signals. To be able to analyze random signals, we need to understand

random variables. The resulting mathematical topics are: probability theory,

random variables and random (stochastic) processes. In this chapter, we shall

develop the probabilistic characterization of random variables. In chapter 3, we

shall extend these concepts to the characterization of random processes.

2.2 Basics of Probability We shall introduce some of the basic concepts of probability theory by

defining some terminology relating to random experiments (i.e., experiments

whose outcomes are not predictable).

2.2.1. Terminology Def. 2.1: Outcome

The end result of an experiment. For example, if the experiment consists

of throwing a die, the outcome would be anyone of the six faces, 1 6,........,F F

Def. 2.2: Random experiment An experiment whose outcomes are not known in advance. (e.g. tossing a

coin, throwing a die, measuring the noise voltage at the terminals of a resistor

etc.)

Def. 2.3: Random event A random event is an outcome or set of outcomes of a random experiment

that share a common attribute. For example, considering the experiment of

throwing a die, an event could be the 'face 1F ' or 'even indexed faces'

( 2 4 6, ,F F F ). We denote the events by upper case letters such as A , B or

A A1 2, ⋅ ⋅ ⋅ ⋅



2.4

Def. 2.4: Sample space The sample space of a random experiment is a mathematical abstraction

used to represent all possible outcomes of the experiment. We denote the

sample space by S .

Each outcome of the experiment is represented by a point in S and is

called a sample point. We use s (with or without a subscript), to denote a sample

point. An event on the sample space is represented by an appropriate collection

of sample point(s).

Def. 2.5: Mutually exclusive (disjoint) events Two events A and B are said to be mutually exclusive if they have no

common elements (or outcomes).Hence if A and B are mutually exclusive, they

cannot occur together.

Def. 2.6: Union of events

The union of two events A and B , denoted A B∪ , also written as

( )A B+ or ( A or B ) is the set of all outcomes which belong to A or B or both.

This concept can be generalized to the union of more than two events.

Def. 2.7: Intersection of events The intersection of two events, A and B , is the set of all outcomes which

belong to A as well as B . The intersection of A and B is denoted by ( )A B∩

or simply ( )AB . The intersection of A and B is also referred to as a joint event

A and B . This concept can be generalized to the case of intersection of three or

more events.

Def. 2.8: Occurrence of an event An event A of a random experiment is said to have occurred if the

experiment terminates in an outcome that belongs to A .



2.5

Def. 2.9: Complement of an event

The complement of an event A , denoted by A is the event containing all

points in S but not in A .

Def. 2.10: Null event

The null event, denoted φ , is an event with no sample points. Thus φ = S

(note that if A and B are disjoint events, then AB = φ and vice versa).

2.2.2 Probability of an Event The probability of an event has been defined in several ways. Two of the

most popular definitions are: i) the relative frequency definition, and ii) the

classical definition.

Def. 2.11: The relative frequency definition:

Suppose that a random experiment is repeated n times. If the event A

occurs An times, then the probability of A , denoted by ( )P A , is defined as

( ) lim An

nP An→ ∞

⎛ ⎞= ⎜ ⎟⎝ ⎠

(2.1)

Ann

⎛ ⎞⎜ ⎟⎝ ⎠

represents the fraction of occurrence of A in n trials.

For small values of n , it is likely that Ann

⎛ ⎞⎜ ⎟⎝ ⎠

will fluctuate quite badly. But

as n becomes larger and larger, we expect, Ann

⎛ ⎞⎜ ⎟⎝ ⎠

to tend to a definite limiting

value. For example, let the experiment be that of tossing a coin and A the event

'outcome of a toss is Head'. If n is the order of 100, Ann

⎛ ⎞⎜ ⎟⎝ ⎠

may not deviate from



2.6

12

by more than, say ten percent and as n becomes larger and larger, we

expect Ann

⎛ ⎞⎜ ⎟⎝ ⎠

to converge to 12

.

Def. 2.12: The classical definition: The relative frequency definition given above has empirical flavor. In the

classical approach, the probability of the event A is found without

experimentation. This is done by counting the total number N of the possible

outcomes of the experiment. If AN of those outcomes are favorable to the

occurrence of the event A , then

( ) ANP AN

= (2.2)

where it is assumed that all outcomes are equally likely!

Whatever may the definition of probability, we require the probability

measure (to the various events on the sample space) to obey the following

postulates or axioms:

P1) ( ) 0P A ≥ (2.3a)

P2) ( ) 1P =S (2.3b)

P3) ( )AB = φ , then ( ) ( ) ( )P A B P A P B+ = + (2.3c)

(Note that in Eq. 2.3(c), the symbol + is used to mean two different things;

namely, to denote the union of A and B and to denote the addition of two real

numbers). Using Eq. 2.3, it is possible for us to derive some additional

relationships:

i) If AB ≠ φ , then ( ) ( ) ( ) ( )P A B P A P B P AB+ = + − (2.4)

ii) Let 1 2, ,......, nA A A be random events such that:

a) i jA A = φ , for i j≠ and (2.5a)



2.7

b) 1 2 ...... nA A A+ + + = S . (2.5b)

Then, ( ) ( ) ( ) ( )1 2 ...... nP A P A A P A A P A A= + + + (2.6)

where A is any event on the sample space.

Note: nA A A1 2, , ,⋅ ⋅ ⋅ are said to be mutually exclusive (Eq. 2.5a) and exhaustive

(Eq. 2.5b).

iii) ( ) ( )1P A P A= − (2.7)

The derivation of Eq. 2.4, 2.6 and 2.7 is left as an exercise.

A very useful concept in probability theory is that of conditional

probability, denoted ( )|P B A ; it represents the probability of B occurring, given

that A has occurred. In a real world random experiment, it is quite likely that the

occurrence of the event B is very much influenced by the occurrence of the

event A . To give a simple example, let a bowl contain 3 resistors and 1

capacitor. The occurrence of the event 'the capacitor on the second draw' is very

much dependent on what has been drawn at the first instant. Such dependencies

between the events is brought out using the notion of conditional probability .

The conditional probability ( )|P B A can be written in terms of the joint

probability ( )P AB and the probability of the event ( )P A . This relation can be

arrived at by using either the relative frequency definition of probability or the

classical definition. Using the former, we have

( ) lim AB

n

nP AB

n→ ∞

⎛ ⎞= ⎜ ⎟

⎝ ⎠

( ) lim An

nP An→ ∞

⎛ ⎞= ⎜ ⎟⎝ ⎠



2.8

where ABn is the number of times AB occurs in n repetitions of the experiment.

As ( )|P B A refers to the probability of B occurring, given that A has occurred,

we have

Def 2.13: Conditional Probability

( )| lim AB

n A

nP B A

n→ ∞=

( )( ) ( )lim , 0

AB

n A

nP ABn P An P A

n→ ∞

⎛ ⎞⎜ ⎟

= = ≠⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

(2.8a)

or ( ) ( ) ( )|P AB P B A P A=

Interchanging the role of A and B , we have

( ) ( )( ) ( )| , 0

P ABP A B P B

P B= ≠ (2.8b)

Eq. 2.8(a) and 2.8(b) can be written as

( ) ( ) ( ) ( ) ( )| |P AB P B A P A P B P A B= = (2.9)

In view of Eq. 2.9, we can also write Eq. 2.8(a) as

( ) ( ) ( )( )

||

P B P A BP B A

P A= , ( )P A 0≠ (2.10a)

Similarly

( ) ( ) ( )( )

P A P B AP A B

P B|

| = , ( )P B 0≠ (2.10b)

Eq. 2.10(a) or 2.10(b) is one form of Bayes’ rule or Bayes’ theorem.

Eq. 2.9 expresses the probability of joint event AB in terms of conditional

probability, say ( )|P B A and the (unconditional) probability ( )P A . Similar

relation can be derived for the joint probability of a joint event involving the

intersection of three or more events. For example ( )P ABC can be written as



2.9

( ) ( ) ( )|P ABC P AB P C AB=

( ) ( ) ( )| |P A P B A P C AB= (2.11)

Another useful probabilistic concept is that of statistical independence.

Suppose the events A and B are such that

( ) ( )|P B A P B= (2.13)

That is, knowledge of occurrence of A tells no more about the probability of

occurrence B than we knew without that knowledge. Then, the events A and B

are said to be statistically independent. Alternatively, if A and B satisfy the

Eq. 2.13, then

( ) ( ) ( )P AB P A P B= (2.14)

Either Eq. 2.13 or 2.14 can be used to define the statistical independence

of two events. Note that if A and B are independent, then

( ) ( ) ( )P AB P A P B= , whereas if they are disjoint, then ( ) 0P AB = . The notion

of statistical independence can be generalized to the case of more than two

events. A set of k events 1 2, ,......, kA A A are said to be statistically independent

if and only if (iff) the probability of every intersection of k or fewer events equal

the product of the probabilities of its constituents. Thus three events , ,A B C are

independent when

Exercise 2.1

Let 1 2, ,......, nA A A be n mutually exclusive and exhaustive events

and B is another event defined on the same space. Show that

( ) ( ) ( )( ) ( )

1

||

|

j jj n

j ji

P B A P AP A B

P B A P A=

=

∑ (2.12)

Eq. 2.12 represents another form of Bayes’ theorem.



2.10

( ) ( ) ( )P AB P A P B=

( ) ( ) ( )P AC P A P C=

( ) ( ) ( )P BC P B P C=

and ( ) ( ) ( ) ( )P ABC P A P B P C=

We shall illustrate some of the concepts introduced above with the help of

two examples.

Example 2.1

Priya (P1) and Prasanna (P2), after seeing each other for some time (and

after a few tiffs) decide to get married, much against the wishes of the parents on

both the sides. They agree to meet at the office of registrar of marriages at 11:30

a.m. on the ensuing Friday (looks like they are not aware of Rahu Kalam or they

don’t care about it).

However, both are somewhat lacking in punctuality and their arrival times

are equally likely to be anywhere in the interval 11 to 12 hrs on that day. Also

arrival of one person is independent of the other. Unfortunately, both are also

very short tempered and will wait only 10 min. before leaving in a huff never to

meet again.

a) Picture the sample space

b) Let the event A stand for “P1 and P2 meet”. Mark this event on the sample

space.

c) Find the probability that the lovebirds will get married and (hopefully) will

live happily ever after.

a) The sample space is the rectangle, shown in Fig. 2.1(a).



2.11

Fig. 2.1(a): S of Example 2.1

b) The diagonal OP represents the simultaneous arrival of Priya and

Prasanna. Assuming that P1 arrives at 11: x , meeting between P1 and P2

would take place if P2 arrives within the interval a to b , as shown in the

figure. The event A , indicating the possibility of P1 and P2 meeting, is

shown in Fig. 2.1(b).

Fig. 2.1(b): The event A of Example 2.1



2.12

c) Shaded area 11Probability of marriageTotal area 36

= =

Example 2.2: Let two honest coins, marked 1 and 2, be tossed together. The four

possible outcomes are 1 2T T , 1 2T H , 1 2H T , 1 2H H . ( 1T indicates toss of coin 1

resulting in tails; similarly 2T etc.) We shall treat that all these outcomes are

equally likely; that is the probability of occurrence of any of these four outcomes

is 14

. (Treating each of these outcomes as an event, we find that these events

are mutually exclusive and exhaustive). Let the event A be 'not 1 2H H ' and B be

the event 'match'. (Match comprises the two outcomes 1 2T T , 1 2H H ). Find

( )|P B A . Are A and B independent?

We know that ( ) ( )( )

|P AB

P B AP A

= .

AB is the event 'not 1 2H H ' and 'match'; i.e., it represents the outcome 1 2T T .

Hence ( ) 14

P AB = . The event A comprises of the outcomes ‘ 1 2T T , 1 2T H and

1 2H T ’; therefore,

( ) 34

P A =

( )1

14| 3 34

P B A = =

Intuitively, the result ( ) 1|3

P B A = is satisfying because, given 'not 1 2H H ’ the

toss would have resulted in anyone of the three other outcomes which can be



2.13

treated to be equally likely, namely 13

. This implies that the outcome 1 2T T given

'not 1 2H H ', has a probability of 13

.

As ( ) 12

P B = and ( ) 1|3

P B A = , A and B are dependent events.

2.3 Random Variables Let us introduce a transformation or function, say X , whose domain is the

sample space (of a random experiment) and whose range is in the real line; that

is, to each i ∈s S , X assigns a real number, ( )iX s , as shown in Fig.2.2.

Fig. 2.2: A mapping ( )X from S to the real line.

The figure shows the transformation of a few individual sample points as

well as the transformation of the event A , which falls on the real line segment

[ ]1 2,a a .

2.3.1 Distribution function: Taking a specific case, let the random experiment be that of throwing a

die. The six faces of the die can be treated as the six sample points in S ; that is,



2.14

, 1, 2, ...... , 6i iF s i= = . Let ( )iX s i= . Once the transformation is induced,

then the events on the sample space will become transformed into appropriate

segments of the real line. Then we can enquire into the probabilities such as

( ) P s X s a1:⎡ ⎤<⎣ ⎦

( ) P s b X s b1 2:⎡ ⎤< ≤⎣ ⎦

or

( ) :P s X s c⎡ ⎤=⎣ ⎦

These and other probabilities can be arrived at, provided we know the

Distribution Function of X, denoted by ( )XF which is given by

( ) ( ) XF x P s X s x:⎡ ⎤= ≤⎣ ⎦ (2.15)

That is, ( )XF x is the probability of the event, comprising all those sample points

which are transformed by X into real numbers less than or equal to x . (Note

that, for convenience, we use x as the argument of ( )XF . But, it could be any

other symbol and we may use ( )XF α , ( )1XF a etc.) Evidently, ( )XF is a function

whose domain is the real line and whose range is the interval [ ]0, 1 ).

As an example of a distribution function (also called Cumulative

Distribution Function CDF), let S consist of four sample points, 1s to 4s , with

each with sample point representing an event with the probabilities ( )114

P s = ,

( )218

P s = , ( )318

P s = and ( )412

P s = . If ( ) 1.5, 1, 2, 3, 4iX s i i= − = , then

the distribution function ( )XF x , will be as shown in Fig. 2.3.



2.15

Fig. 2.3: An example of a distribution function

( )XF satisfies the following properties:

i) ( ) 0,XF x x≥ − ∞ < < ∞

ii) ( ) 0XF − ∞ =

iii) ( ) 1XF ∞ =

iv) If a b> , then ( ) ( ) ( ) :X XF a F b P s b X s a⎡ ⎤⎡ ⎤− = < ≤⎣ ⎦ ⎣ ⎦

v) If a b> , then ( ) ( )X XF a F b≥

The first three properties follow from the fact that ( )XF represents the

probability and ( ) 1P =S . Properties iv) and v) follow from the fact

( ) ( ) ( ) : : :s X s b s b X s a s X s a≤ ∪ < ≤ = ≤

Referring to the Fig. 2.3, note that ( ) 0XF x = for 0.5x < − whereas

( )XF 10.54

− = . In other words, there is a discontinuity of 14

at the point

x 0.5= − . In general, there is a discontinuity in XF of magnitude aP at a point

x a= , if and only if

( ) : aP s X s a P⎡ ⎤= =⎣ ⎦ (2.16)



2.16

The properties of the distribution function are summarized by saying that

( )xF is monotonically non-decreasing, is continuous to the right at each point

x TP

1PT, and has a step of size aP at point a if and if Eq. 2.16 is satisfied.

Functions such as ( )X for which distribution functions exist are called

Random Variables (RV). In other words, for any real x , ( ) :s X s x≤ should

be an event in the sample space with some assigned probability. (The term

“random variable” is somewhat misleading because an RV is a well defined

function from S into the real line.) However, every transformation from S into

the real line need not be a random variable. For example, let S consist of six

sample points, 1s to 6s . The only events that have been identified on the sample

space are: 1 2 3 4 5 6, , , , andA s s B s s s C s= = = and their probabilities

are ( ) ( ) ( )= = =P A P B P C2 1 1, and6 2 6

. We see that the probabilities for the

various unions and intersections of A , B and C can be obtained.

Let the transformation X be ( )iX s i= . Then the distribution function

fails to exist because

[ ] ( )4: 3.5 4.5P s x P s< ≤ = is not known as 4s is not an event on the sample

space.

TP

1PT Let x a= . Consider, with 0∆ > ,

( ) ( ) ( )lim lim0 0

X XP a X s a F a F a⎡ ⎤ ⎡ ⎤< ≤ + ∆ = + ∆ −⎣ ⎦ ⎣ ⎦∆ → ∆ →

We intuitively feel that as 0∆ → , the limit of the set ( ) :s a X s a< ≤ + ∆ is the null set and

can be proved to be so. Hence,

( ) ( ) 0X X

a F aF + − = , where ( )0

lima a+ ∆ →= + ∆

That is, ( )X

F x is continuous to the right.



2.17

2.3.2 Probability density function

Though the CDF is a very fundamental concept, in practice we find it more

convenient to deal with Probability Density Function (PDF). The PDF, ( )Xf x is

defined as the derivative of the CDF; that is

( ) ( )XX

d F xf x

d x= (2.17a)

or

( ) ( )x

X XF x f d− ∞

= α α∫ (2.17b)

The distribution function may fail to have a continuous derivative at a point

x a= for one of the two reasons:

i) the slope of the ( )xF x is discontinuous at x a=

ii) ( )xF x has a step discontinuity at x a=

The situation is illustrated in Fig. 2.4.

Exercise 2.2

Let S be a sample space with six sample points, 1s to 6s . The events

identified on S are the same as above, namely, ,1 2A s s= ,

3 4 5, ,B s s s= and 6C s= with ( ) ( ) ( )1 1 1, and3 2 6

P A P B P C= = = .

Let ( )Y be the transformation,

( )1, 1, 22 , 3, 4, 53 , 6

i

iY s i

i

=⎧⎪= =⎨⎪ =⎩

Show that ( )Y is a random variable by finding ( )YF y . Sketch ( )YF y .



2.18

Fig. 2.4: A CDF without a continuous derivative

As can be seen from the figure, ( )XF x has a discontinuous slope at

1x = and a step discontinuity at 2x = . In the first case, we resolve the

ambiguity by taking Xf to be a derivative on the right. (Note that ( )XF x is

continuous to the right.) The second case is taken care of by introducing the

impulse in the probability domain. That is, if there is a discontinuity in XF at

x a= of magnitude aP , we include an impulse ( )aP x aδ − in the PDF. For

example, for the CDF shown in Fig. 2.3, the PDF will be,

( ) 1 1 1 1 1 3 1 54 2 8 2 8 2 2 2Xf x x x x x⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= δ + + δ − + δ − + δ −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ (2.18)

In Eq. 2.18, ( )Xf x has an impulse of weight 18

at 12

x = as

1 12 8

P X⎡ ⎤= =⎢ ⎥⎣ ⎦. This impulse function cannot be taken as the limiting case of

an even function (such as 1 xga ⎛ ⎞⎜ ⎟ε ε⎝ ⎠

) because,

( )1 12 2

0 01 12 2

1 1 1lim lim8 2 16Xf x dx x dx

ε → ε →− ε − ε

⎛ ⎞= δ − ≠⎜ ⎟⎝ ⎠∫ ∫



2.19

However, ( )ε →

− ε

=∫12

012

1lim8Xf x dx . This ensures,

( )⎧ − ≤ <⎪⎪= ⎨⎪ ≤ <⎪⎩

2 1 1,8 2 23 1 3,8 2 2

X

xF x

x

Such an impulse is referred to as the left-sided delta function.

As XF is non-decreasing and ( ) 1XF ∞ = , we have

i) ( ) 0Xf x ≥ (2.19a)

ii) ( ) 1Xf x dx∞

− ∞

=∫ (2.19b)

Based on the behavior of CDF, a random variable can be classified as:

i) continuous (ii) discrete and (iii) mixed. If the CDF, ( )XF x , is a continuous

function of x for all x , then X TP

1PT is a continuous random variable. If ( )XF x is a

staircase, then X corresponds to a discrete variable. We say that X is a mixed

random variable if ( )XF x is discontinuous but not a staircase. Later on in this

lesson, we shall discuss some examples of these three types of variables.

We can induce more than one transformation on a given sample space. If

we induce k such transformations, then we have a set of k co-existing random

variables.

2.3.3 Joint distribution and density functions Consider the case of two random variables, say X and Y . They can be

characterized by the (two-dimensional) joint distribution function, given by

( ) ( ) ( ) , , : ,X YF x y P s X s x Y s y⎡ ⎤= ≤ ≤⎣ ⎦ (2.20)

TP

1PT As the domain of the random variable ( )X is known, it is convenient to denote the variable

simply by X .



2.20

That is, ( ), ,X YF x y is the probability associated with the set of all those

sample points such that under X , their transformed values will be less than or

equal to x and at the same time, under Y , the transformed values will be less

than or equal to y . In other words, ( ), 1 1,X YF x y is the probability associated with

the set of all sample points whose transformation does not fall outside the

shaded region in the two dimensional (Euclidean) space shown in Fig. 2.5.

Fig. 2.5: Space of ( ) ( ) ,X s Y s corresponding to ( ), 1 1,X YF x y

Looking at the sample space S , let A be the set of all those sample

points s ∈ S such that ( ) 1X s x≤ . Similarly, if B is comprised of all those

sample points s ∈ S such that ( ) 1Y s y≤ ; then ( )1 1,F x y is the probability

associated with the event AB .

Properties of the two dimensional distribution function are:

i) ( ), , 0 , ,X YF x y x y≥ − ∞ < < ∞ − ∞ < < ∞

ii) ( ) ( ), ,, , 0X Y X YF y F x− ∞ = − ∞ =

iii) ( ), , 1X YF ∞ ∞ =

iv) ( ) ( ), ,X Y YF y F y∞ =

v) ( ) ( ), ,X Y XF x F x∞ =



2.21

vi) If 2 1x x> and 2 1y y> , then

( ) ( ) ( ), 2 2 , 2 1 , 1 1, , ,X Y X Y X YF x y F x y F x y≥ ≥

We define the two dimensional joint PDF as

( ) ( )∂=

∂ ∂X Y X Yf x y F x yx y

2

, ,, , (2.21a)

or ( ) ( ), ,, ,y x

X Y X YF x y f d d− ∞ − ∞

= α β α β∫ ∫ (2.21b)

The notion of joint CDF and joint PDF can be extended to the case of k random

variables, where 3k ≥ .

Given the joint PDF of random variables X and Y , it is possible to obtain

the one dimensional PDFs, ( )Xf x and ( )Yf y . We know that,

( ) ( ), 1 1,X Y XF x F x∞ = .

That is, ( ) ( )1

1 , ,x

X X YF x f d d∞

− ∞ − ∞

= α β β α∫ ∫

( ) ( )1

1 ,1

,x

X X Ydf x f d d

d x

∞

− ∞ − ∞

⎧ ⎫⎡ ⎤⎪ ⎪= α β β α⎢ ⎥⎨ ⎬⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

∫ ∫ (2.22)

Eq. 2.22 involves the derivative of an integral. Hence,

( ) ( ) ( )1

1 , 1,XX X Y

x x

d F xf x f x d

d x

∞

− ∞=

= = β β∫

or ( ) ( ), ,X X Yf x f x y dy∞

− ∞

= ∫ (2.23a)

Similarly, ( ) ( ), ,Y X Yf y f x y dx∞

− ∞

= ∫ (2.23b)

(In the study of several random variables, the statistics of any individual variable

is called the marginal. Hence it is common to refer ( )XF x as the marginal



2.22

distribution function of X and ( )Xf x as the marginal density function. ( )YF y and

( )Yf y are similarly referred to.)

2.3.4 Conditional density Given ( ), ,X Yf x y , we know how to obtain ( )Xf x or ( )Yf y . We might also

be interested in the PDF of X given a specific value of 1y y= . This is called the

conditional PDF of X , given 1y y= , denoted ( )X Yf x y| 1| and defined as

( ) ( )( )

X YX Y

Y

f x yf x y

f y, 1

| 11

,| = (2.24)

where it is assumed that ( )1 0Yf y ≠ . Once we understand the meaning of

conditional PDF, we might as well drop the subscript on y and denote it by

( )| |X Yf x y . An analogous relation can be defined for ( )| |Y Xf y x . That is, we have

the pair of equations,

( ) ( )( )

,|

,| X Y

X YY

f x yf x y

f y= (2.25a)

and ( ) ( )( )

,|

,| X Y

Y XX

f x yf y x

f x= (2.25b)

or ( ) ( ) ( ), |, |X Y X Y Yf x y f x y f y= (2.25c)

( ) ( )| |Y X Xf y x f x= (2.25d)

The function ( )| |X Yf x y may be thought of as a function of the variable x with

variable y arbitrary, but fixed. Accordingly, it satisfies all the requirements of an

ordinary PDF; that is,

( )| | 0X Yf x y ≥ (2.26a)

and ( )| | 1X Yf x y dx∞

− ∞

=∫ (2.26b)



2.23

2.3.5 Statistical independence In the case of random variables, the definition of statistical independence

is somewhat simpler than in the case of events. We call k random variables

1 2, , ......... , kX X X statistically independent iff, the k -dimensional joint PDF

factors into the product

( )1 i

k

X ii

f x=∏ (2.27)

Hence, two random variables X and Y are statistically independent, iff,

( ) ( ) ( ), ,X Y X Yf x y f x f y= (2.28)

and three random variables X , Y , Z are independent, iff

( ) ( ) ( ) ( ), , , ,X Y Z X Y zf x y z f x f y f z= (2.29)

Statistical independence can also be expressed in terms of conditional

PDF. Let X and Y be independent. Then,

( ) ( ) ( ), ,X Y X Yf x y f x f y=

Making use of Eq. 2.25(c), we have

( ) ( ) ( ) ( )| |X Y Y X Yf x y f y f x f y=

or ( ) ( )| |X Y Xf x y f x= (2.30a)

Similarly, ( ) ( )| |Y X Yf y x f y= (2.30b)

Eq. 2.30(a) and 2.30(b) are alternate expressions for the statistical independence

between X and Y . We shall now give a few examples to illustrate the concepts

discussed in sec. 2.3.

Example 2.3

A random variable X has

( ) 2

0 , 0, 0 10

100 , 10X

xF x K x x

K x

<⎧⎪= ≤ ≤⎨⎪ >⎩

i) Find the constant K

ii) Evaluate ( )5P X ≤ and ( )5 7P X< ≤



2.24

iii) What is ( )Xf x ?

i) ( )XF K K 1100 1100

∞ = = ⇒ = .

ii) ( ) ( ) 15 5 25 0.25100XP x F ⎛ ⎞≤ = = × =⎜ ⎟⎝ ⎠

( ) ( ) ( )5 7 7 5 0.24X XP X F F< ≤ = − =

( ) ( )0 , 00.02 , 0 100 , 10

XX

xd F x

f x x xd x

x

<⎧⎪= = ≤ ≤⎨⎪ >⎩

Note: Specification of ( )Xf x or ( )XF x is complete only when the algebraic

expression as well as the range of X is given.

Example 2.4 Consider the random variable X defined by the PDF

( ) ,b xXf x ae x−= − ∞ < < ∞ where a and b are positive constants.

i) Determine the relation between a and b so that ( )Xf x is a PDF.

ii) Determine the corresponding ( )XF x

iii) Find [ ]1 2P X< ≤ .

i) As can be seen ( ) 0Xf x ≥ for x− ∞ < < ∞ . In order for ( )Xf x to

represent a legitimate PDF, we require 0

2 1b x b xae dx ae dx∞ ∞

− −

− ∞

= =∫ ∫ .

That is, 0

12

b xae dx∞

− =∫ ; hence 2b a= .

ii) the given PDF can be written as

( ) , 0, 0 .

b x

X b x

a e xf x

a e x−

⎧ <⎪= ⎨≥⎪⎩



2.25

For 0x < , we have

( ) 12 2

xb b x

XbF x e d eα

− ∞

= α =∫ .

Consider 0x > . Take a specific value of 2x =

( ) ( )2

2X XF f x d x− ∞

= ∫

( )2

1 Xf x d x∞

= − ∫

But for the problem on hand, ( ) ( )X Xf x d x f x d x2

2

−∞

− ∞

=∫ ∫

Therefore, ( ) 11 , 02

b xXF x e x−= − >

We can now write the complete expression for the CDF as

( )1 , 0211 , 02

b x

Xb x

e xF x

e x−

⎧ <⎪⎪= ⎨⎪ − ≥⎪⎩

iii) ( ) ( ) ( )212 12

b bX XF F e e− −− = −

Example 2.5

Let ( ),

1 , 0 , 0 2, 2

0 ,X Y

x y yf x y

otherwise

⎧ ≤ ≤ ≤ ≤⎪= ⎨⎪⎩

Find (a) i) ( )| | 1Y Xf y and ii) ( )| | 1.5Y Xf y

(b) Are X and Y independent?

(a) ( ) ( )( )

,|

,| X Y

Y XX

f x yf y x

f x=



2.26

( )Xf x can be obtained from ,X Yf by integrating out the unwanted variable y

over the appropriate range. The maximum value taken by y is 2 ; in

addition, for any given ,x y x≥ . Hence,

( )2 1 1

2 2Xx

xf x d y= = −∫ , 0 2x≤ ≤

Hence,

(i) ( )|

1 1 , 1 22|1 1 0 ,2Y X

yf y

otherwise≤ ≤⎧

= = ⎨⎩

(ii) ( )|

1 2 , 1.5 22|1.5 1 0 ,4Y X

yf y

otherwise≤ ≤⎧

= = ⎨⎩

b) the dependence between the random variables X and Y is evident from

the statement of the problem because given a value of 1X x= , Y should

be greater than or equal to 1x for the joint PDF to be non zero. Also we see

that ( )| |Y Xf y x depends on x whereas if X and Y were to be independent,

then ( ) ( )| |Y X Yf y x f y= .



2.27

Exercise 2.3 For the two random variables X and Y , the following density functions

have been given. (Fig. 2.6)

Fig. 2.6: PDFs for the exercise 2.3

Find

a) ( ), ,X Yf x y

b) Show that

( ), 0 10

1001 , 10 205 100

Y

y yf y

y y

⎧ ≤ ≤⎪⎪= ⎨⎪ − < ≤⎪⎩

2.4 Transformation of Variables

The process of communication involves various operations such as

modulation, detection, filtering etc. In performing these operations, we typically

generate new random variables from the given ones. We will now develop the

necessary theory for the statistical characterization of the output random

variables, given the transformation and the input random variables.



2.28

2.4.1 Functions of one random variable Assume that X is the given random variable of the continuous type with

the PDF, ( )Xf x . Let Y be the new random variable, obtained from X by the

transformation ( )Y g X= . By this we mean the number ( )1Y s associated with

any sample point 1s is

( ) ( )( )1 1Y s g X s=

Our interest is to obtain ( )Yf y . This can be obtained with the help of the following

theorem (Thm. 2.1).

Theorem 2.1

Let ( )Y g X= . To find ( )Yf y , for the specific y , solve the equation

( )y g x= . Denoting its real roots by nx , ( ) ( )1 ,......, , .......ny g x g x= = = = ,

we will show that

( ) ( )( )

( )( )

1

1

........... .........' '

X X nY

n

f x f xf y

g x g x= + + + (2.31)

where ( )'g x is the derivative of ( )g x .

Proof: Consider the transformation shown in Fig. 2.7. We see that the equation

( )y g x1 = has three roots namely, 1 2 3, andx x x .



2.29

Fig. 2.7: X Y− transformation used in the proof of theorem 2.1

We know that ( ) [ ]Yf y d y P y Y y d y= < ≤ + . Therefore, for any given 1y , we

need to find the set of values x such that ( )1 1y g x y d y< ≤ + and the

probability that X is in this set. As we see from the figure, this set consists of the

following intervals:

1 1 1 2 2 2 3 3 3, ,x x x d x x d x x x x x x d x< ≤ + + < ≤ < ≤ +

where 1 3 20 , 0 , but 0d x d x d x> > < .

From the above, it follows that

[ ] [ ] [ ][ ]

< ≤ + = < ≤ + + + < ≤

+ < ≤ +

P y Y y d y P x X x d x P x dx X x

P x X x d x1 1 1 1 1 2 2 2

3 3 3

This probability is the shaded area in Fig. 2.7.

[ ] ( ) ( )1 1 1 1 1 11

,'Xd yP x X x d x f x d x d x

g x< ≤ + = =

[ ] ( ) ( )2 2 2 2 2 22

,'Xd yP x d x x x f x d x d x

g x+ < ≤ = =

[ ] ( ) ( )3 3 3 3 3 33

,'Xd yP x X x d x f x d x d x

g x< ≤ + = =



2.30

We conclude that

( ) ( )( )

( )( )

( )( )

= + +X X XY

f x f x f xf y d y d y d y d y

g x g xg x1 2 3

1 32' '' (2.32)

and Eq. 2.31 follows, by canceling d y from the Eq. 2.32.

Note that if ( ) 1g x y= = constant for every x in the interval ( )0 1,x x , then

we have [ ] ( ) ( ) ( )1 0 1 1 0X XP Y y P x X x F x F x= = < ≤ = − ; that is ( )YF y is

discontinuous at 1y y= . Hence ( )Yf y contains an impulse, ( )1y yδ − of area

( ) ( )1 0X XF x F x− .

We shall take up a few examples.

Example 2.6

( )Y g X X a= = + , where a is a constant. Let us find ( )Yf y .

We have ( )' 1g x = and x y a= − . For a given y , there is a unique x

satisfying the above transformation. Hence,

( ) ( )( )'

XY

f xf y d y d y

g x= and as ( )' 1g x = , we have

( ) ( )Y Xf y f y a= −

Let ( )1 , 1

0 ,X

x xf x

elsewhere

⎧ − ≤⎪= ⎨⎪⎩

and 1a = −

Then, ( ) 1 1Yf y y= − +

As 1y x= − , and x ranges from the interval ( )1, 1− , we have the range of y

as ( )2, 0− .



2.31

Hence ( )1 1 , 2 0

0 ,Y

y yf y

elsewhere

⎧ − + − ≤ ≤⎪= ⎨⎪⎩

( )XF x and ( )YF y are shown in Fig. 2.8.

Fig. 2.8: ( )XF x and ( )YF y of example 2.6

As can be seen, the transformation of adding a constant to the given variable

simply results in the translation of its PDF.

Example 2.7

Let Y b X= , where b is a constant. Let us find ( )Yf y .

Solving for X , we have 1X Yb

= . Again for a given y , there is a unique

x . As ( )'g x b= , we have ( ) 1Y X

yf y fb b

⎛ ⎞= ⎜ ⎟⎝ ⎠

.

Let ( )X

x xf x

otherwise

1 , 0 22

0 ,

⎧ − ≤ ≤⎪= ⎨⎪⎩

and 2b = − , then



2.32

( )1 1 , 4 02 4

0 ,Y

y yf y

otherwise

⎧ ⎡ ⎤+ − ≤ ≤⎪ ⎢ ⎥= ⎣ ⎦⎨⎪⎩

( )Xf x and ( )Yf y are sketched in Fig. 2.9.

Fig. 2.9: ( )Xf x and ( )Yf y of example 2.7

Example 2.8

2 , 0Y a X a= > . Let us find ( )Yf y .

Exercise 2.4

Let Y a X b= + , where a and b are constants. Show that

( ) 1Y X

y bf y fa a

−⎛ ⎞= ⎜ ⎟⎝ ⎠

.

If ( )Xf x is as shown in Fig.2.8, compute and sketch ( )Yf y for 2a = − , and

1b = .



2.33

( )' 2g x a x= . If 0y < , then the equation 2y a x= has no real solution.

Hence ( ) 0Yf y = for 0y < . If 0y ≥ , then it has two solutions, 1yxa

= and

2yxa

= − , and Eq. 2.31 yields

( )1 , 0

2

0 ,

X X

Y

y yf f ya ayf y a

aotherwise

⎧ ⎡ ⎤⎛ ⎞ ⎛ ⎞+ − ≥⎪ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎪ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦= ⎨

⎪⎪⎩

Let 1a = , and ( )21 exp ,

22Xxf x x

⎛ ⎞= − − ∞ < < ∞⎜ ⎟⎜ ⎟π ⎝ ⎠

(Note that ( )exp α is the same as eα )

Then ( ) 1 exp exp2 22 2Yy yf y

y⎡ ⎤⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦π

y y

yotherwise

1 exp , 022

0 ,

⎧ ⎛ ⎞− ≥⎪ ⎜ ⎟= ⎝ ⎠⎨⎪⎩

π

Sketching of ( )Xf x and ( )Yf y is left as an exercise.

Example 2.9 Consider the half wave rectifier transformation given by

0 , 0, 0

XY

X X≤⎧

= ⎨ >⎩

a) Let us find the general expression for ( )Yf y

b) Let ( )Xx

f xotherwise

1 1 3,2 2 20 ,

⎧ − < <⎪= ⎨⎪⎩

We shall compute and sketch ( )Yf y .



2.34

a) Note that ( )g X is a constant (equal to zero) for X in the range of ( ), 0− ∞ .

Hence, there is an impulse in ( )Yf y at 0y = whose area is equal to

( )0XF . As Y is nonnegative, ( ) 0Yf y = for 0y < . As Y X= for 0x > ,

we have ( ) ( )Y Xf y f y= for 0y > . Hence

( ) ( ) ( ) ( ) ( )00 ,X X

Yf y w y F y

f yotherwise

⎧ + δ⎪= ⎨⎪⎩

where ( ) 1 , 00 ,

yw y

otherwise≥⎧

= ⎨⎩

b) Specifically, let ( )1 1 3,2 2 20 ,

Xx

f xelsewhere

⎧ − < <⎪= ⎨⎪⎩

Then, ( )

( )1 , 041 3, 02 20 ,

Y

y y

f y y

otherwise

⎧ δ =⎪⎪⎪= < ≤⎨⎪⎪⎪⎩

( )Xf x and ( )Yf y are sketched in Fig. 2.10.

Fig. 2.10: ( )Xf x and ( )Yf y for the example 2.9



2.35

Note that X , a continuous RV is transformed into a mixed RV, Y .

Example 2.10

Let X

YX

1, 01, 0

− <⎧= ⎨+ ≥⎩

a) Let us find the general expression for ( )Yf y .

b) We shall compute and sketch ( )Yf x assuming that ( )Xf x is the same as

that of Example 2.9.

a) In this case, Y assumes only two values, namely 1± . Hence the PDF of Y

has only two impulses. Let us write ( )Yf y as

( ) ( ) ( )1 11 1Yf y P y P y−= − + +δ δ where

[ ] [ ]1 10 0P P X and P X− = < ≥

b) Taking ( )Xf x of example 2.9, we have 134

P = and 114

P− = . Fig. 2.11

has the sketches ( )Xf x and ( )Yf y .

Fig. 2.11: ( )Xf x and ( )Yf y for the example 2.10

Note that this transformation has converted a continuous random variable X into

a discrete random variable Y .



2.36

Exercise 2.5

Let a random variable X with the PDF shown in Fig. 2.12(a) be the

input to a device with the input-output characteristic shown in Fig. 2.12(b).

Compute and sketch ( )Yf y .

Fig. 2.12: (a) Input PDF for the transformation of exercise 2.5

(b) Input-output transformation

Exercise 2.6 The random variable X of exercise 2.5 is applied as input to the

X Y− transformation shown in Fig. 2.13. Compute and sketch ( )Yf y .



2.37

We now assume that the random variable X is of discrete type taking on

the value kx with probability kP . In this case, the RV, ( )Y g X= is also

discrete, assuming the value ( )k kY g x= with probability kP .

If ( )ky g x= for only one kx x= , then [ ] [ ]k k kP Y y P X x P= = = = . If

however, ( )ky g x= for kx x= and mx x= , then [ ]k k mP Y y P P= = + .

Example 2.11

Let 2Y X= .

a) If ( ) ( )Xi

f x x i6

1

16 =

= −∑ δ , find ( )Yf y .

b) If ( ) ( )Xi

f x x i3

2

16 = −

= −∑ δ , find ( )Yf y .

a) If X takes the values ( )1, 2, ......., 6 with probability of 16

, then Y takes the

values 2 2 21 , 2 , ......., 6 with probability 16

. That is, ( ) ( )Yi

f y x i6

2

1

16 =

= δ −∑ .

b) If, however, X takes the values 2, 1, 0, 1, 2, 3− − with probability 16

, then

Y takes the values 0, 1, 4, 9 with probabilities 1 1 1 1, , ,6 3 3 6

respectively.

That is,

( ) ( ) ( ) ( ) ( )1 19 1 46 3Yf y y y y y⎡ ⎤ ⎡ ⎤= δ + δ − + δ − + δ −⎣ ⎦ ⎣ ⎦

2.4.2 Functions of two random variables Given two random variables X and Y (assumed to be continuous type),

two new random variables, Z and W are defined using the transformation

( ),Z g X Y= and ( ),W h X Y=



2.38

Given the above transformation and ( ), ,X Yf x y , we would like to obtain

( ), ,Z Wf z w . For this, we require the Jacobian of the transformation, denoted

,,

z wJx y

⎛ ⎞⎜ ⎟⎝ ⎠

where

,,

z zx yz wJ

x y w wx y

∂ ∂∂ ∂⎛ ⎞

=⎜ ⎟ ∂ ∂⎝ ⎠∂ ∂

z w z wx y y x

∂ ∂ ∂ ∂= −

∂ ∂ ∂ ∂

That is, the Jacobian is the determinant of the appropriate partial derivatives. We

shall now state the theorem which relates ( ), ,Z Wf z w and ( ), ,X Yf x y .

We shall assume that the transformation is one-to-one. That is, given

( ) 1,g x y z= , (2.33a)

( ) 1,h x y w= , (2.33b)

then there is a unique set of numbers, ( )1 1,x y satisfying Eq. 2.33.

Theorem 2.2: To obtain ( ), ,Z Wf z w , solve the system of equations

( ) 1,g x y z= ,

( )h x y w1, = ,

for x and y . Let ( )1 1,x y be the result of the solution. Then,

( ) ( ), 1 1,

1 1

,,

,,

X YZ W

f x yf z w

z wJx y

=⎛ ⎞⎜ ⎟⎝ ⎠

(2.34)

Proof of this theorem is given in appendix A2.1. For a more general version of

this theorem, refer [1].



2.39

Example 2.12 X and Y are two independent RVs with the PDFs ,

( ) 1 , 12Xf x x= ≤

( ) 1 , 12Yf y y= ≤

If Z X Y= + and W X Y= − , let us find (a) ( ), ,Z Wf z w and (b) ( )Zf z .

a) From the given transformations, we obtain ( )12

x z w= + and

( )12

y z w= − . We see that the mapping is one-to-one. Fig. 2.14(a)

depicts the (product) space A on which ( ), ,X Yf x y is non-zero.

Fig. 2.14: (a) The space where ,X Yf is non-zero

(b) The space where ,Z Wf is non-zero



2.40

We can obtain the space B (on which ( ), ,Z Wf z w is non-zero) as follows:

spaceA spaceB

The line 1x = ( )1 1 22

z w w z+ = ⇒ = − +

The line 1x = − ( ) ( )1 1 22

z w w z+ = − ⇒ = − +

The line 1y = ( )− = ⇒ = −z w w z1 1 22

The line 1y = − ( )1 1 22

z w w z− = − ⇒ = +

The space B is shown in Fig. 2.14(b). The Jacobian of the transformation

is

1 1, 2

, 1 1z wJx y

⎛ ⎞= = −⎜ ⎟ −⎝ ⎠

and ( ) 2J = .

Hence ( ),

11 , ,4, 82 0 ,

Z W

z wf z w

otherwise

⎧ ∈⎪= = ⎨⎪⎩

B

b) ( ) ( ), ,Z Z Wf z f z w d w∞

− ∞

= ∫

From Fig. 2.14(b), we can see that, for a given z ( 0z ≥ ), w can take

values only in the toz z− . Hence

( ) 1 1 , 0 28 4

z

Zz

f z d w z z+

−

= = ≤ ≤∫

For z negative, we have

( ) 1 1 , 2 08 4

z

Zz

f z d w z z−

= = − − ≤ ≤∫



2.41

Hence ( ) 1 , 24Zf z z z= ≤

Example 2.13

Let the random variables R and Φ be given by, 2 2R X Y= + and

tan YarcX

⎛ ⎞Φ = ⎜ ⎟⎝ ⎠

where we assume 0R ≥ and − π < Φ < π . It is given that

( )2 2

,1, exp , ,

2 2X Yx yf x y x y

⎡ ⎤⎛ ⎞+= − − ∞ < < ∞⎢ ⎥⎜ ⎟π ⎝ ⎠⎣ ⎦

.

Let us find ( ), ,Rf rΦ ϕ .

As the given transformation is from cartesian-to-polar coordinates, we can

write cosx r= φ and siny r= φ , and the transformation is one -to -one;

cos sin1

sin cos

r rx y

Jr

r rx y

∂ ∂ϕ ϕ∂ ∂

= = =− ϕ ϕ∂ϕ ∂ϕ∂ ∂

Hence, ( )R

r r rf r

otherwise

2

,

exp , 0 ,, 2 2

0 ,Φ

⎧ ⎛ ⎞− ≤ ≤ ∞ − π < ϕ < π⎪ ⎜ ⎟ϕ = π⎨ ⎝ ⎠

⎪⎩

It is left as an exercise to find ( )Rf r , ( )fΦ ϕ and to show that R and Φ are

independent variables.

Theorem 2.2 can also be used when only one function ( ),Z g X Y= is

specified and what is required is ( )Zf z . To apply the theorem, a conveniently

chosen auxiliary or dummy variable W is introduced. Typically W X= or

W Y= ; using the theorem ( ), ,Z Wf z w is found from which ( )Zf z can be

obtained.



2.42

Let Z X Y= + and we require ( )Zf z . Let us introduce a dummy variable

W Y= . Then, X Z W= − , and Y W= .

As J 1= ,

( ) ( ), ,, ,Z W X Yf z w f z w w= −

and ( ) ( ), ,Z X Yf z f z w w d w∞

− ∞

= −∫ (2.35)

If X and Y are independent, then Eq. 2.35 becomes

( ) ( ) ( )Z X Yf z f z w f w d w∞

− ∞

= −∫ (2.36a)

That is, Z X Yf f f= ∗ (2.36b)

Example 2.14 Let X and Y be two independent random variables, with

( )1 , 1 120 ,

Xx

f xotherwise

⎧ − ≤ ≤⎪= ⎨⎪⎩

( )1 , 2 130 ,

Y

yf y

otherwise

⎧ − ≤ ≤⎪= ⎨⎪⎩

If Z X Y= + , let us find [ ]2P Z ≤ − .

From Eq. 2.36(b), ( )Zf z is the convolution of ( )Xf z and ( )Yf z . Carrying

out the convolution, we obtain ( )Zf z as shown in Fig. 2.15.



2.43

Fig. 2.15: PDF of Z X Y= + of example 2.14

[ ]2P Z ≤ − is the shaded area 112

= .

Example 2.15

Let XZY

= ; let us find an expression for ( )Zf z .

Introducing the dummy variable W Y= , we have

X Z W=

Y W=

As 1Jw

= , we have

( ) ( ), ,Z X Yf z w f zw w dw∞

− ∞

= ∫

Let ( ),

1 , 1, 1, 4

0 ,X Y

x y x yf x y

elsewhere

+⎧ ≤ ≤⎪= ⎨⎪⎩

Then ( )21

4Zzwf z w dw

∞

− ∞

+= ∫



2.44

( ), ,X Yf x y is non-zero if ( ),x y ∈ A . where A is the product space

1x ≤ and 1y ≤ (Fig.2.16a). Let ( ), ,Z Wf z w be non-zero if ( ),z w ∈ B . Under

the given transformation, B will be as shown in Fig. 2.16(b).

Fig. 2.16: (a) The space where ,X Yf is non-zero

(b) The space where ,Z Wf is non-zero

To obtain ( )Zf z from ( ), ,Z Wf z w , we have to integrate out w over the

appropriate ranges.

i) Let 1z < ; then

( )1 12 2

1 0

1 124 4Zzw zwf z w dw w dw

−

+ += =∫ ∫

1 14 2

z⎛ ⎞= +⎜ ⎟⎝ ⎠

ii) For 1z > , we have

( )1

2

2 30

1 1 1 124 4 2

z

Zzwf z w dw

z z⎛ ⎞+

= = +⎜ ⎟⎝ ⎠

∫

iii) For 1z < − , we have

( )1

2

2 30

1 1 1 124 4 2

z

Zzwf z w dw

z z

−⎛ ⎞+

= = +⎜ ⎟⎝ ⎠

∫



2.45

Hence ( )2 3

1 1 , 14 21 1 1 , 14 2

Z

z zf z

zz z

⎧ ⎛ ⎞+ ≤⎜ ⎟⎪ ⎝ ⎠⎪= ⎨⎛ ⎞⎪ + >⎜ ⎟⎪ ⎝ ⎠⎩

In transformations involving two random variables, we may encounter a

situation where one of the variables is continuous and the other is discrete; such

cases are handled better, by making use of the distribution function approach,

as illustrated below.

Example 2.16 The input to a noisy channel is a binary random variable with

[ ] [ ] 10 12

P X P X= = = = . The output of the channel is given by Z X Y= +

Exercise 2.7

Let Z X Y= and W Y= .

a) Show that ( ) ,1 ,Z X Y

zf z f w d ww w

∞

− ∞

⎛ ⎞= ⎜ ⎟⎝ ⎠∫

b) X and Y be independent with

( )2

1 , 11

Xf x xx

= ≤π +

and ( )2

2 , 00 ,

y

Yy e yf y

otherwise

−⎧⎪ ≥= ⎨⎪⎩

Show that ( )2

21 ,2

z

Zf z e z−= − ∞ < < ∞

π.



2.46

where Y is the channel noise with ( )2

21 ,2

y

Yf y e y−

= − ∞ < < ∞π

. Find

( )Zf z .

Let us first compute the distribution function of Z from which the density

function can be derived.

( ) [ ] [ ] [ ] [ ]| 0 0 | 1 1P Z z P Z z X P X P Z z X P X≤ = ≤ = = + ≤ = =

As Z X Y= + , we have

[ ] ( )| 0 YP Z z X F z≤ = =

Similarly [ ] ( ) ( )| 1 1 1YP Z z X P Y z F z⎡ ⎤≤ = = ≤ − = −⎣ ⎦

Hence ( ) ( ) ( )1 1 12 2Z Y YF z F z F z= + − . As ( ) ( )Z Z

df z F zd z

= , we have

( ) ( )22 11 1 1exp exp2 2 22 2Z

zzf z⎡ ⎤⎛ ⎞⎛ ⎞ −⎢ ⎥⎜ ⎟⎜ ⎟= − + −⎢ ⎥⎜ ⎟⎜ ⎟π π⎝ ⎠ ⎝ ⎠⎣ ⎦

The distribution function method, as illustrated by means of example 2.16, is a

very basic method and can be used in all situations. (In this method, if

( )Y g X= , we compute ( )YF y and then obtain ( ) ( )YY

d F yf y

d y= . Similarly, for

transformations involving more than one random variable. Of course computing

the CDF may prove to be quite difficult in certain situations. The method of

obtaining PDFs based on theorem 2.1 or 2.2 is called change of variable

method.) We shall illustrate the distribution function method with another

example.

Example 2.17

Let Z X Y= + . Obtain ( )Zf z , given ( ), ,X Yf x y .



2.47

[ ] [ ]P Z z P X Y z≤ = + ≤

[ ]P Y z x= ≤ −

This probability is the probability of ( ),X Y lying in the shaded area shown in Fig.

2.17.

Fig. 2.17: Shaded area is the [ ]P Z z≤

That is,

( ) ( ), ,z x

Z X YF z d x f x y d y−∞

− ∞ − ∞

⎡ ⎤= ⎢ ⎥

⎢ ⎥⎣ ⎦∫ ∫

( ) ( ), ,z x

Z X Yf z d x f x y d yz

−∞

− ∞ − ∞

⎡ ⎤⎡ ⎤∂= ⎢ ⎥⎢ ⎥

∂ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦∫ ∫

( ), ,z x

X Yd x f x y d yz

−∞

− ∞ − ∞

⎡ ⎤∂= ⎢ ⎥

∂⎢ ⎥⎣ ⎦∫ ∫

( ), ,X Yf x z x d x∞

− ∞

= −∫ (2.37a)

It is not too difficult to see the alternative form for ( )Zf z , namely,



2.48

( ) ( ), ,Z X Yf z f z y y d y∞

− ∞

= −∫ (2.37b)

If X and Y are independent, we have ( ) ( ) ( )Z X Yf z f z f z= ∗ . We note that Eq.

2.37(b) is the same as Eq. 2.35.

So far we have considered the transformations involving one or two

variables. This can be generalized to the case of functions of n variables. Details

can be found in [1, 2].

2.5 Statistical Averages The PDF of a random variable provides a complete statistical

characterization of the variable. However, we might be in a situation where the

PDF is not available but are able to estimate (with reasonable accuracy) certain

(statistical) averages of the random variable. Some of these averages do provide

a simple and fairly adequate (though incomplete) description of the random

variable. We now define a few of these averages and explore their significance.

The mean value (also called the expected value, mathematical

expectation or simply expectation) of random variable X is defined as

[ ] ( )X Xm E X X x f x d x∞

− ∞

= = = ∫ (2.38)

where E denotes the expectation operator. Note that Xm is a constant. Similarly,

the expected value of a function of X , ( )g X , is defined by

( ) ( ) ( ) ( )XE g X g X g x f x d x∞

− ∞

⎡ ⎤ = =⎣ ⎦ ∫ (2.39)

Remarks: The terminology expected value or expectation has its origin in games

of chance. This can be illustrated as follows: Three small similar discs, numbered

1,2 and 2 respectively are placed in bowl and are mixed. A player is to be



2.49

blindfolded and is to draw a disc from the bowl. If he draws the disc numbered 1,

he will receive nine dollars; if he draws either disc numbered 2, he will receive 3

dollars. It seems reasonable to assume that the player has a '1/3 claim' on the 9

dollars and '2/3 claim' on three dollars. His total claim is 9(1/3) + 3(2/3), or five

dollars. If we take X to be (discrete) random variable with the PDF

( ) ( ) ( )1 21 23 3Xf x x x= δ − + δ − and ( ) 15 6g X X= − , then

( ) ( ) ( )15 6 5XE g X x f x d x∞

− ∞

⎡ ⎤ = − =⎣ ⎦ ∫

That is, the mathematical expectation of ( )g X is precisely the player's claim or

expectation [3]. Note that ( )g x is such that ( )1 9g = and ( )2 3g = .

For the special case of ( ) ng X X= , we obtain the n-th moment of the

probability distribution of the RV, X ; that is,

( )n nXE X x f x d x

∞

− ∞

⎡ ⎤ =⎣ ⎦ ∫ (2.40)

The most widely used moments are the first moment ( 1n = , which results in the

mean value of Eq. 2.38) and the second moment ( 2n = , resulting in the mean

square value of X ).

( )2 2 2XE X X x f x d x

∞

− ∞

⎡ ⎤ = =⎣ ⎦ ∫ (2.41)

If ( ) ( )nXg X X m= − , then ( )E g X⎡ ⎤⎣ ⎦ gives n-th central moment; that is,

( ) ( ) ( )n nX X XE X m x m f x d x

∞

− ∞

⎡ ⎤− = −⎣ ⎦ ∫ (2.42)

We can extend the definition of expectation to the case of functions of

( )2k k ≥ random variables. Consider a function of two random variables,

( ),g X Y . Then,



2.50

( ) ( ) ( ),, , ,X YE g X Y g x y f x y d x d y∞ ∞

− ∞ − ∞

⎡ ⎤ =⎣ ⎦ ∫ ∫ (2.43)

An important property of the expectation operator is linearity; that is, if

( ),Z g X Y X Y= = α + β where α and β are constants, then Z X Y= α + β .

This result can be established as follows. From Eq. 2.43, we have

[ ] ( ) ( ), ,X YE Z x y f x y d x dy∞ ∞

− ∞ − ∞

= α + β∫ ∫

( ) ( ) ( ) ( ), ,, ,X Y X Yx f x y d x dy y f x y d x dy∞ ∞ ∞ ∞

− ∞ − ∞ − ∞ − ∞

= α + β∫ ∫ ∫ ∫

Integrating out the variable y in the first term and the variable x in the second

term, we have

[ ] ( ) ( )X YE Z x f x d x y f y d y∞ ∞

− ∞ − ∞

= α + β∫ ∫

X Y= α + β

2.5.1 Variance Coming back to the central moments, we have the first central moment

being always zero because,

( ) ( ) ( )X X XE X m x m f x d x∞

− ∞

⎡ ⎤− = −⎣ ⎦ ∫

0X Xm m= − =

Consider the second central moment

( )2 2 22X X XE X m E X m X m⎡ ⎤ ⎡ ⎤− = − +⎣ ⎦⎣ ⎦

From the linearity property of expectation,

[ ]2 2 2 22 2X X X XE X m X m E X m E X m⎡ ⎤ ⎡ ⎤− + = − +⎣ ⎦ ⎣ ⎦

2 2 22 X XE X m m⎡ ⎤= − +⎣ ⎦

22 2 2

XX m X X⎡ ⎤= − = − ⎣ ⎦



2.51

The second central moment of a random variable is called the variance and its

(positive) square root is called the standard deviation. The symbol 2σ is

generally used to denote the variance. (If necessary, we use a subscript on 2σ )

The variance provides a measure of the variable's spread or randomness.

Specifying the variance essentially constrains the effective width of the density

function. This can be made more precise with the help of the Chebyshev

Inequality which follows as a special case of the following theorem.

Theorem 2.3: Let ( )g X be a non-negative function of the random variable X . If

( )E g X⎡ ⎤⎣ ⎦ exists, then for every positive constant c ,

( ) ( )E g XP g X c

c⎡ ⎤⎣ ⎦⎡ ⎤≥ ≤⎣ ⎦ (2.44)

Proof: Let ( ) :A x g x c= ≥ and B denote the complement of A .

( ) ( ) ( )XE g X g x f x d x∞

− ∞

⎡ ⎤ =⎣ ⎦ ∫

( ) ( ) ( ) ( )X XA B

g x f x d x g x f x d x= +∫ ∫

Since each integral on the RHS above is non-negative, we can write

( ) ( ) ( )XA

E g X g x f x d x⎡ ⎤ ≥⎣ ⎦ ∫

If x A∈ , then ( )g x c≥ , hence

( ) ( )XA

E g X c f x d x⎡ ⎤ ≥⎣ ⎦ ∫

But ( ) [ ] ( )XA

f x d x P x A P g X c⎡ ⎤= ∈ = ≥⎣ ⎦∫

That is, ( ) ( )E g X c P g X c⎡ ⎤ ⎡ ⎤≥ ≥⎣ ⎦ ⎣ ⎦ , which is the desired result.

Note: The kind of manipulations used in proving the theorem 2.3 is useful in

establishing similar inequalities involving random variables.



2.52

To see how innocuous (or weak, perhaps) the inequality 2.44 is, let ( )g X

represent the height of a randomly chosen human being with ( ) 1.6E g X m⎡ ⎤ =⎣ ⎦ .

Then Eq. 2.44 states that the probability of choosing a person over 16 m tall is at

most 1 !10

(In a population of 1 billion, at most 100 million would be as tall as a

full grown Palmyra tree!)

Chebyshev inequality can be stated in two equivalent forms:

i) 2

1 , 0X XP X m k kk

⎡ ⎤− ≥ σ ≤ >⎣ ⎦ (2.45a)

ii) ⎡ ⎤− < σ > −⎣ ⎦X XP X m kk 2

11 (2.45b)

where Xσ is the standard deviation of X .

To establish 2.45(a), let ( ) ( )2Xg X X m= − and 2 2

Xc k= σ in theorem

2.3. We then have,

( )2 2 22

1X XP X m k

k⎡ ⎤− ≥ σ ≤⎣ ⎦

In other words,

2

1X XP X m k

k⎡ ⎤− ≥ σ ≤⎣ ⎦

which is the desired result. Naturally, we would take the positive number k to be

greater than one to have a meaningful result. Chebyshev inequality can be

interpreted as: the probability of observing any RV outside k± standard

deviations off its mean value is no larger than 2

1k

. With 2k = for example, the

probability of 2X XX m− ≥ σ does not exceed 1/4 or 25%. By the same token,

we expect X to occur within the range ( )2X Xm ± σ for more than 75% of the

observations. That is, smaller the standard deviation, smaller is the width of the

interval around Xm , where the required probability is concentrated. Chebyshev



2.53

inequality thus enables us to give a quantitative interpretation to the statement

'variance is indicative of the spread of the random variable'.

Note that it is not necessary that variance exists for every PDF. For example, if

( ) 2 2 , and 0Xf x xx

απ= − ∞ < < ∞ α >

α +, then 0X = but 2X is not finite.

(This is called Cauchy’s PDF)

2.5.2 Covariance An important joint expectation is the quantity called covariance which is

obtained by letting ( ) ( ) ( ), X Yg X Y X m Y m= − − in Eq. 2.43. We use the

symbol λ to denote the covariance. That is,

[ ] ( ) ( )X Y X YX Y E X m Y mcov , ⎡ ⎤λ = = − −⎣ ⎦ (2.46a)

Using the linearity property of expectation, we have

[ ]X Y X YE X Y m mλ = − (2.46b)

The [ ]X Ycov , , normalized with respect to the product X Yσ σ is termed as the

correlation coefficient and we denote it by ρ . That is,

[ ] X YX Y

X Y

E X Y m m−ρ =

σ σ (2.47)

The correlation coefficient is a measure of dependency between the variables.

Suppose X and Y are independent. Then,

[ ] ( ), ,X YE X Y x y f x y d x d y∞ ∞

− ∞ − ∞

= ∫ ∫

( ) ( )X Yx y f x f y d x d y∞ ∞

− ∞ − ∞

= ∫ ∫

( ) ( )X Y X Yx f x d x y f y d y m m∞ ∞

− ∞ − ∞

= =∫ ∫



2.54

Thus, we have X Yλ (and X Yρ ) being zero. Intuitively, this result is appealing.

Assume X and Y to be independent. When the joint experiment is performed

many times, and given 1X x= , then Y would occur sometimes positive with

respect to Ym , and sometimes negative with respect to Ym . In the course of

many trials of the experiments and with the outcome 1X x= , the sum of the

numbers ( )1 Yx y m− would be very small and the quantity, sumnumber of trials

,

tends to zero as the number of trials keep increasing.

On the other hand, let X and Y be dependent. Suppose for example, the

outcome y is conditioned on the outcome x in such a manner that there is a

greater possibility of ( )Yy m− being of the same sign as ( )Xx m− . Then we

expect X Yρ to be positive. Similarly if the probability of ( )Xx m− and ( )Yy m−

being of the opposite sign is quite large, then we expect X Yρ to be negative.

Taking the extreme case of this dependency, let X and Y be so conditioned

that, given 1X x= , then 1Y x= ± α , α being constant. Then 1X Yρ = ± . That

is, for X and Y be independent, we have 0X Yρ = and for the totally dependent

( )y x= ± α case, 1X Yρ = . If the variables are neither independent nor totally

dependent, then ρ will have a magnitude between 0 and 1.

Two random variables X and Y are said to be orthogonal if [ ]E X Y 0= .

If ( )or 0X Y X Yρ λ = , the random variables X and Y are said to be

uncorrelated. That is, if X and Y are uncorrelated, then XY X Y= . When the

random variables are independent, they are uncorrelated. However, the fact that

they are uncorrelated does not ensure that they are independent. As an example,

let



2.55

( )1 ,

2 20 ,

X

xf x

otherwise

α α⎧ − < <⎪= α⎨⎪⎩

,

and 2Y X= . Then,

2

3 3 3

2

1 1 0X Y X x d x x d xα

∞

α− ∞ −

= = = =α α∫ ∫

As 0 , 0X X Y= = which means 0X Y X Y X Yλ = − = . But X and Y are

not independent because, if 1X x= , then 21 !Y x=

Let Y be the linear combination of the two random variables 1X and 2X .

That is 1 1 2 2Y k X k X= + where 1k and 2k are constants. Let [ ]i iE X m= ,

2 2iX iσ = σ , 1, 2i = . Let 12ρ be the correlation coefficient between 1X and 2X .

We will now relate 2Yσ to the known quantities.

[ ] 22 2Y E Y E Y⎡ ⎤ ⎡ ⎤σ = − ⎣ ⎦⎣ ⎦

[ ] 1 1 2 2E Y k m k m= +

2 2 21 1 2 2 1 2 1 22E Y E k X k X k k X X⎡ ⎤ ⎡ ⎤= + +⎣ ⎦ ⎣ ⎦

With a little manipulation, we can show that

2 2 21 1 2 2 12 1 2 1 22Y k k k kσ = σ + σ + ρ σ σ (2.48a)

If 1X and 2X are uncorrelated, then

2 2 21 1 2 2Y k kσ = σ + σ (2.48b)

Note: Let 1X and 2X be uncorrelated, and let 1 2Z X X= + and 1 2W X X= − .

Then 2 2 2 21 2Z Wσ = σ = σ + σ . That is, the sum as well as the difference random

variables have the same variance which is larger than 21σ or 2

2σ !

The above result can be generalized to the case of linear combination of

n variables. That is, if



2.56

1

n

i ii

Y k X=

= ∑ , then

2 2 2

12

n

Y i i i j i j i ji i j

i j

k k k=

<

σ = σ + ρ σ σ∑ ∑∑ (2.49)

where the meaning of various symbols on the RHS of Eq. 2.49 is quite obvious.

We shall now give a few examples based on the theory covered so far in

this section.

Example 2.18 Let a random variable X have the CDF

( ) 2

0 , 0

, 0 28

, 2 4161 , 4

X

xx x

F xx x

x

<⎧⎪⎪ ≤ ≤⎪= ⎨⎪ ≤ ≤⎪⎪ ≤⎩

We shall find a) X and b) 2Xσ

a) The given CDF implies the PDF

( )

0 , 01 , 0 281 , 2 480 , 4

X

x

xf x

x x

x

<⎧⎪⎪ ≤ ≤⎪= ⎨⎪ ≤ ≤⎪⎪ ≤⎩

Therefore,

[ ]2 4

2

0 2

1 18 8

E X x d x x d x= +∫ ∫



2.57

1 7 314 3 12

= + =

b) [ ] 22 2X E X E X⎡ ⎤ ⎡ ⎤σ = − ⎣ ⎦⎣ ⎦

2 42 2 3

0 2

1 18 8

E X x d x x d x⎡ ⎤ = +⎣ ⎦ ∫ ∫

1 15 473 2 6

= + =

22 47 31 167

6 12 144X⎛ ⎞σ = − =⎜ ⎟⎝ ⎠

Example 2.19 Let cosY X= π , where

( )Xx

f xotherwise

1 11,2 2

0,

⎧ − < <⎪= ⎨⎪⎩

Let us find [ ]E Y and 2Yσ .

From Eq. 2.38 we have,

[ ] ( )12

12

2cos 0.0636E Y x d x−

= π = =π∫

( )12

2 2

12

1cos 0.52

E Y x d x−

⎡ ⎤ = π = =⎣ ⎦ ∫

Hence 22

1 4 0.962Yσ = − =

π

Example 2.20 Let X and Y have the joint PDF

( ),

, 0 1, 0 1,

0 ,X Y

x y x yf x y

elsewhere+ < < < <⎧

= ⎨⎩



2.58

Let us find a) 2E X Y⎡ ⎤⎣ ⎦ and b) X Yρ

a) From Eq. 2.43, we have

( )2 2, ,X YE X Y x y f x y dx dy

∞ ∞

− ∞ − ∞

⎡ ⎤ =⎣ ⎦ ∫ ∫

( )1 1

2

0 0

1772

x y x y dx dy= + =∫ ∫

b) To find X Yρ , we require [ ] [ ] [ ], ,E X Y E X E Y , Xσ and Yσ . We can easily

show that

[ ] [ ] 2 27 11,12 144X YE X E Y= = σ = σ = and [ ] 48

144E X Y =

Hence 111X Yρ = −

Another statistical average that will be found useful in the study of

communication theory is the conditional expectation. The quantity,

( ) ( ) ( )|| |X YE g X Y y g x f x y dx∞

− ∞

⎡ ⎤= =⎣ ⎦ ∫ (2.50)

is called the conditional expectation of ( )g X , given Y y= . If ( )g X X= , then

we have the conditional mean, namely, [ ]|E X Y .

[ ] [ ] ( )|| | |X YE X Y y E X Y x f x y dx∞

− ∞

= = = ∫ (2.51)

Similarly, we can define the conditional variance etc. We shall illustrate the

calculation of conditional mean with the help of an example.

Example 2.21

Let the joint PDF of the random variables X and Y be



2.59

( ),

1 , 0 1, 0,

0 ,X Y

x y xf x y x

outside

⎧ < < < <⎪≡ ⎨⎪⎩

.

Let us compute [ ]|E X Y .

To find [ ]|E X Y , we require the conditional PDF, ( ) ( )( )

,|

,| X Y

X YY

f x yf x y

f y=

( )1 1 ln , 0 1Yy

f y d x y yx

= = − < <∫

( )|

11| , 1

ln lnX Yxf x y y x

y x y= = − < <

−

Hence ( )1 1|

lny

E X Y x d xx y

⎡ ⎤= −⎢ ⎥

⎣ ⎦∫

1lny

y−

=

Note that [ ]|E X Y y= is a function of y .

2.6 Some Useful Probability Models In the concluding section of this chapter, we shall discuss certain

probability distributions which are encountered quite often in the study of

communication theory. We will begin our discussion with discrete random

variables.

2.6.1. Discrete random variables i) Binomial: Consider a random experiment in which our interest is in the occurrence or non-

occurrence of an event A . That is, we are interested in the event A or its

complement, say, B . Let the experiment be repeated n times and let p be the



2.60

probability of occurrence of A on each trial and the trials are independent. Let X

denote random variable, ‘number of occurrences of A in n trials'. X can be

equal to 0, 1, 2, ........., n . If we can compute [ ], 0, 1, .........,P X k k n= = , then

we can write ( )Xf x .

Taking a special case, let 5n = and 3k = . The sample space

(representing the outcomes of these five repetitions) has 32 sample points, say,

1 32, ...........,s s . The sample point 1s could represent the sequence ABBBB . The

sample points such as ABAAB , A A ABB etc. will map into real number 3 as

shown in the Fig. 2.18. (Each sample point is actually an element of the five

dimensional Cartesian product space).

Fig. 2.18: Binomial RV for the special case 5n =

( ) ( ) ( ) ( ) ( ) ( )P AB A AB P A P B P A P A P B= , as trails are independent.

( ) ( ) ( )22 31 1 1p p p p p p= − − = −

There are ( )510 sample points for which 3

3X s⎛ ⎞

= =⎜ ⎟⎝ ⎠

.

In other words, for 5n = , 3k = ,

[ ] ( )2353 1

3P X p p⎛ ⎞

= = −⎜ ⎟⎝ ⎠

Generalizing this to arbitrary n and k , we have the binomial density, given by



2.61

( ) ( )0

n

X ii

f x P x i=

= δ −∑ (2.52)

where ( )1 n iii

nP p p

i−⎛ ⎞

= −⎜ ⎟⎝ ⎠

As can be seen, ( ) 0Xf x ≥ and

( ) ( )0 0

1n n

n iiX i

i i

nf x d x P p p

i

∞−

= =− ∞

⎛ ⎞= = −⎜ ⎟

⎝ ⎠∑ ∑∫

( )1 1n

p p⎡ ⎤= − + =⎣ ⎦

It is left as an exercise to show that [ ]E X n p= and ( )2 1X n p pσ = − . (Though

the formulae for the mean and the variance of a binomial PDF are simple, the

algebra to derive them is laborious).

We write X is ( ),b n p to indicate X has a binomial PDF with parameters

n and p defined above.

The following example illustrates the use of binomial PDF in a

communication problem.

Example 2.22 A digital communication system transmits binary digits over a noisy

channel in blocks of 16 digits. Assume that the probability of a binary digit being

in error is 0.01 and that errors in various digit positions within a block are

statistically independent.

i) Find the expected number of errors per block

ii) Find the probability that the number of errors per block is greater than or

equal to 3.

Let X be the random variable representing the number of errors per block.

Then X is ( )16, 0.01b .

i) [ ] 16 0.01 0.16;E X n p= = × =



2.62

ii) ( ) [ ]3 1 2P X P X≥ = − ≤

( ) ( )2

16

0

161 0.1 1i i

ip

i−

=

⎛ ⎞= − −⎜ ⎟

⎝ ⎠∑

0.002=

ii) Poisson: A random variable X which takes on only integer values is Poisson

distributed, if

( ) ( )0 !

m

Xm

ef x x mm

− λ∞

=

λ= δ −∑ (2.53)

where λ is a positive constant.

Evidently ( ) 0Xf x ≥ and ( ) 1Xf x d x∞

− ∞

=∫ because 0 !

m

me

m

∞λ

=

λ=∑ .

We will now show that

[ ] 2XE X = λ = σ

Since,

0 !

m

me

m

∞λ

=

λ= ∑ , we have

( ) 1

0 1

1! !

m m

m m

d e me md m m

λ −∞ ∞λ

= =

λ λ= = =

λ λ∑ ∑

[ ]1 !

m

m

m eE X e em

− λ∞λ − λ

=

λ= = λ = λ∑

Differentiating the series again, we obtain,

Exercise 2.8

Show that for the example 2.22, Chebyshev inequality results in

3 0.0196 0.02P X⎡ ⎤≥ ≤⎣ ⎦ . Note that Chebyshev inequality is not very

tight.



2.63

2 2E X⎡ ⎤ = λ + λ⎣ ⎦ . Hence 2Xσ = λ .

2.6.2 Continuous random variables i) Uniform:

A random variable X is said to be uniformly distributed in the interval

a x b≤ ≤ if,

( )1 ,

0 ,X

a x bb af x

elsewhere

⎧ ≤ ≤⎪ −= ⎨⎪⎩

(2.54)

A plot of ( )Xf x is shown in Fig.2.19.

Fig.2.19: Uniform PDF

It is easy show that

[ ]2

a bE X += and

( )22

12X

b a−σ =

Note that the variance of the uniform PDF depends only on the width of the

interval ( )b a− . Therefore, whether X is uniform in ( )1, 1− or ( )2, 4 , it has the

same variance, namely 13

.

ii) Rayleigh: An RV X is said to be Rayleigh distributed if,



2.64

( )2

exp , 02

0 ,X

x x xf x b b

elsewhere

⎧ ⎛ ⎞− ≥⎪ ⎜ ⎟= ⎨ ⎝ ⎠

⎪⎩

(2.55)

where b is a positive constant,

A typical sketch of the Rayleigh PDF is given in Fig.2.20. ( ( )Rf r of

example 2.12 is Rayleigh PDF.)

Fig.2.20: Rayleigh PDF

Rayleigh PDF frequently arises in radar and communication problems. We will

encounter it later in the study of narrow-band noise processes.



2.65

iii) Gaussian

By far the most widely used PDF, in the context of communication theory

is the Gaussian (also called normal) density, specified by

( ) ( )2

21 exp ,

22X

XXX

x mf x x

⎡ ⎤−= ⎢− ⎥ − ∞ < < ∞

σπ σ ⎢ ⎥⎣ ⎦ (2.56)

where Xm is the mean value and 2Xσ the variance. That is, the Gaussian PDF is

completely specified by the two parameters, Xm and 2Xσ . We use the symbol

( )2,X XN m σ to denote the Gaussian density1PT. In appendix A2.3, we show that

( )Xf x as given by Eq. 2.56 is a valid PDF.

As can be seen from the Fig. 2.21, The Gaussian PDF is symmetrical with

respect to Xm .

TP

1PT In this notation, ( )0, 1N denotes the Gaussian PDF with zero mean and unit variance. Note that

if X is ( )2,X XN m σ , then X

X

X mY

−=

σ

⎛ ⎞⎜ ⎟⎝ ⎠

is ( )0, 1N .

Exercise 2.9

a) Let ( )Xf x be as given in Eq. 2.55. Show that ( )0

1Xf x d x∞

=∫ . Hint: Make

the change of variable 2x z= . Then, 2

d zx d x = .

b) Show that if X is Rayleigh distributed, then [ ]2bE X π

= and

2 2E X b⎡ ⎤ =⎣ ⎦



2.66

Fig. 2.21: Gaussian PDF

Hence ( ) ( ) 0.5Xm

X X XF m f x d x− ∞

= =∫

Consider [ ]P X a≥ . We have,

[ ] ( )2

21 exp

22X

Xa X

x mP X a d x

∞ ⎡ ⎤−≥ = ⎢− ⎥

σπ σ ⎢ ⎥⎣ ⎦∫

This integral cannot be evaluated in closed form. By making a change of variable

X

X

x mz⎛ ⎞−

= ⎜ ⎟σ⎝ ⎠, we have

[ ]2

212

X

X

z

a m

P X a e d z∞

−

−σ

≥ =π∫

X

X

a mQ⎛ ⎞−

= ⎜ ⎟σ⎝ ⎠

where ( )21 exp

22y

xQ y d x∞ ⎛ ⎞

= −⎜ ⎟π ⎝ ⎠

∫ (2.57)

Note that the integrand on the RHS of Eq. 2.57 is ( )0, 1N .

( )Q function table is available in most of the text books on

communication theory as well as in standard mathematical tables. A small list is

given in appendix A2.2 at the end of the chapter.



2.67

The importance of Gaussian density in communication theory is due to a

theorem called central limit theorem which essentially states that:

If the RV X is the weighted sum of N independent random components,

where each component makes only a small contribution to the sum, then ( )XF x

approaches Gaussian as N becomes large, regardless of the distribution of the

individual components.

For a more precise statement and a thorough discussion of this theorem,

you may refer [1-3]. The electrical noise in a communication system is often due

to the cumulative effects of a large number of randomly moving charged

particles, each particle making an independent contribution of the same amount,

to the total. Hence the instantaneous value of the noise can be fairly adequately

modeled as a Gaussian variable. We shall develop Gaussian random processes in detail in Chapter 3 and, in Chapter 7, we shall make use of this

theory in our studies on the noise performance of various modulation schemes.

Example 2.23 A random variable Y is said to have a log-normal PDF if lnX Y= has a

Gaussian (normal) PDF. Let Y have the PDF, ( )Yf y given by,

( )( )2

2

ln1 exp , 022

0 ,Y

yy

f y y

otherwise

⎧ ⎡ ⎤− α⎪ ⎢− ⎥ ≥⎪= βπ β⎨ ⎢ ⎥⎣ ⎦⎪⎪⎩

where α and β are given constants.

a) Show that Y is log-normal

b) Find ( )E Y

c) If m is such that ( ) 0.5YF m = , find m .

a) Let lnX Y= or lnx y= (Note that the transformation is one-to-one)



2.68

1 1d x Jd y y y

= → =

Also as 0 ,y x→ → − ∞ and as ,y x→ ∞ → ∞

Hence ( ) ( )2

21 exp ,

22X

xf x x

⎡ ⎤− α= ⎢− ⎥ − ∞ < < ∞

βπ β ⎢ ⎥⎣ ⎦

Note that X is ( )2N α, β

b) ( )2

2212

xX xY E e e e d x

− α∞ −β

− ∞

⎡ ⎤⎢ ⎥⎡ ⎤= =⎣ ⎦ ⎢ ⎥π β⎣ ⎦

∫

( ) 22

2222 1

2

x

e e d x

⎡ ⎤− α + β⎣ ⎦∞β −α + β

− ∞

⎡ ⎤⎢ ⎥= ⎢ ⎥π β⎢ ⎥⎣ ⎦

∫

As the bracketed quantity being the integral of a Gaussian PDF

between the limits ( ),− ∞ ∞ is 1, we have

2

2Y eβ

α +=

c) [ ] [ ]lnP Y m P X m≤ = ≤

Hence if [ ] 0.5P Y m≤ = , then [ ]ln 0.5P X m≤ =

That is, ln m = α or m eα=

iv) Bivariate Gaussian

As an example of a two dimensional density, we will consider the bivariate

Gaussian PDF, ( ), ,X Yf x y , ,x y− ∞ < < ∞ given by,

( ) ( ) ( ) ( ) ( )2 2

, 2 2

1 2

1 1, exp 2X Y X Y

X Y

X Y X Y

x m y m x m y mf x y

k k

− − − −= − + − ρ

σ σ σ σ

⎧ ⎡ ⎤⎫⎨ ⎬⎢ ⎥⎩ ⎣ ⎦⎭

(2.58)

where,

21 2 1X Yk = π σ σ − ρ



2.69

( )22 2 1k = − ρ

Correlation coefficient between andX Yρ =

The following properties of the bivariate Gaussian density can be verified:

P1) If andX Y are jointly Gaussian, then the marginal density of orX Y is

Gaussian; that is, X is ( )2,X XN m σ and Y is ( )2,Y YN m σ TP

1PT

P2) ( ) ( ) ( ), iff 0X Y X Yf x y f x f y= ρ =

That is, if the Gaussian variables are uncorrelated, then they are

independent. That is not true, in general, with respect to non-Gaussian

variables (we have already seen an example of this in Sec. 2.5.2).

P3) If Z X Y= α + β where α and β are constants and andX Y are jointly

Gaussian, then Z is Gaussian. Therefore ( )Zf z can be written after

computing Zm and 2Zσ with the help of the formulae given in section 2.5.

Figure 2.22 gives the plot of a bivariate Gaussian PDF for the case of

0ρ = and X Yσ = σ .

TP

1PT Note that the converse is not necessarily true. Let Xf and Yf be obtained from ,X Yf and let Xf

and Yf be Gaussian. This does not imply ,X Yf is jointly Gaussian, unless andX Y are

independent. We can construct examples of a joint PDF ,X Yf , which is not Gaussian but results in

Xf and Yf that are Gaussian.



2.70

Fig.2.22: Bivariate Gaussian PDF ( X Yσ = σ and 0ρ = )

For 0ρ = and X Yσ = σ , ,X Yf resembles a (temple) bell, with, of course,

the striker missing! For 0ρ ≠ , we have two cases (i) ρ , positive and (ii)

ρ , negative. If 0ρ > , imagine the bell being compressed along the

X Y= − axis so that it elongates along the X Y= axis. Similarly for

0ρ < .

Example 2.24

Let X and Y be jointly Gaussian with 2 21, 1X YX Y= − = σ = σ = and

12X Yρ = − . Let us find the probability of ( ),X Y lying in the shaded region D

shown in Fig. 2.23.



2.71

Fig. 2.23: The region D of example 2.24

Let A be the shaded region shown in Fig. 2.24(a) and B be the shaded

region in Fig. 2.24(b).

Fig. 2.24: (a) Region A and (b) Region B used to obtain region D

The required probability = ( ) ( ), ,P x y P x y⎡ ⎤ ⎡ ⎤∈ − ∈⎣ ⎦ ⎣ ⎦A B

For the region A , we have 1 12

y x≥ − + and for the region B , we have

1 22

y x≥ − + . Hence the required probability is,



2.72

X XP Y P Y1 22 2

⎡ ⎤ ⎡ ⎤+ ≥ − + ≥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Let XZ Y2

= +

Then Z is Gaussian with the parameters,

1 12 2

Z Y X= + = −

2 2 21 12 .4 2Z X Y X Yσ = σ + σ + ρ

1 1 1 31 2 . .4 2 2 4

= + − =

That is, Z is 1 3,2 4

N ⎛ ⎞−⎜ ⎟⎝ ⎠

. Then

12

34

ZW

+= is ( )0, 1N

[ ]1 3P Z P W⎡ ⎤≥ = ≥⎣ ⎦

[ ] 523

P Z P W⎡ ⎤≥ = ≥⎢ ⎥

⎣ ⎦

Hence the required probability ( ) ( )53 0.04 0.0013

Q Q ⎛ ⎞= − −⎜ ⎟

⎝ ⎠0.039=



2.73

Exercise 2.10 X and Y are independent, identically distributed (iid) random variables,

each being ( )0, 1N . Find the probability of ,X Y lying in the region A shown in

Fig. 2.25.

Fig. 2.25: Region A of Exercise 2.10

Note: It would be easier to calculate this kind of probability, if the space is a

product space. From example 2.12, we feel that if we transform ( ),X Y into

( ),Z W such that Z X Y= + , W X Y= − , then the transformed space B

would be square. Find ( ), ,Z Wf z w and compute the probability ( ),Z W ∈ B .



2.74

Exercise 2.11 Two random variables X and Y are obtained by means of the

transformation given below.

( ) ( )12

1 22 log cos 2eX U U= − π (2.59a)

( ) ( )12

1 22 log sin 2eY U U= − π (2.59b)

1U and 2U are independent random variables, uniformly distributed in the

range 1 20 , 1u u< < . Show that X and Y are independent and each is

( )0, 1N .

Hint: Let 1 12 logeX U= − and 1 1Y X=

Show that 1Y is Rayleigh. Find ( ), ,X Yf x y using 1 cosX Y= Θ and

1 sinY Y= Θ , where 22 UΘ = π .

Note: The transformation given by Eq. 2.59 is called the Box-Muller

transformation and can be used to generate two Gaussian random number

sequences from two independent uniformly distributed (in the range 0 to 1)

sequences.



2.75

Appendix A2.1 Proof of Eq. 2.34 The proof of Eq. 2.34 depends on establishing a relationship between the

differential area d z d w in the z w− plane and the differential area d x d y in

the x y− plane. We know that

( ) [ ], , ,Z Wf z w d z d w P z Z z d z w W w d w= < ≤ + < ≤ +

If we can find d x d y such that

( ) ( ), ,, ,Z W X Yf z w d z d w f x y d x d y= , then ,Z Wf can be found. (Note that

the variables x and y can be replaced by their inverse transformation

quantities, namely, ( )1 ,x g z w−= and ( )1 ,y h z w−= )

Let the transformation be one-to-one. (This can be generalized to the case of

one-to-many.) Consider the mapping shown in Fig. A2.1.

Fig. A2.1: A typical transformation between the ( )x y− plane and ( )z w− plane

Infinitesimal rectangle ABC D in the z w− plane is mapped into the

parallelogram in the x y− plane. (We may assume that the vertex A transforms

to A' , B to B' etc.) We shall now find the relation between the differential area of

the rectangle and the differential area of the parallelogram.



2.76

Consider the parallelogram shown in Fig. A2.2, with vertices P P P1 2 3, ,

and P4 .

Fig. A2.2: Typical parallelogram

Let ( )x y, be the co-ordinates of P1. Then the P2 and P3 are given by

g hP x d z y d zz z

1 1

2 ,− −⎛ ⎞∂ ∂

= + +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠

x yx d z y d zz z

,⎛ ⎞∂ ∂= + +⎜ ⎟∂ ∂⎝ ⎠

⎛ ⎞∂ ∂

= + +⎜ ⎟∂ ∂⎝ ⎠

x yP x d w y d ww w3 ,

Consider the vectors 1V and 2V shown in the Fig. A2.2 where

( )P P1 2 1= −V and ( )= −P P2 3 1V .

That is,

x yd z d zz z1

∂ ∂= +

∂ ∂V i j

x yd w d ww w2∂ ∂

= +∂ ∂

V i j

where i and j are the unit vectors in the appropriate directions. Then, the area

A of the parallelogram is,

A 1 2= ×V V



2.77

As 0× =i i , 0× =j j , and ( )× = − × =j ii j k where k is the unit vector

perpendicular to both i and j , we have

x y y x d z d wz w z w1 2

∂ ∂ ∂ ∂× = −

∂ ∂ ∂ ∂V V

x yA J d z d wz w

,,

⎛ ⎞= ⎜ ⎟

⎝ ⎠

That is,

( ) ( )Z W X Yx yf z w f x y Jz w, ,

,, ,,

⎛ ⎞= ⎜ ⎟

⎝ ⎠

( )X Yf x y

z wJx y

, ,

,,

=⎛ ⎞⎜ ⎟⎝ ⎠

.



2.78

Appendix A2.2

( )Q Function Table

( )2

212

xQ e d x

∞−

α

α =π∫

It is sufficient if we know ( )Q α for 0α ≥ , because ( ) ( )1Q Q− α = − α . Note

that ( )0 0.5Q = .

y ( )Q y y ( )Q y y ( )Q y

0.05

0.10

0.15

0.20

0.25

0.4801

0.4602

0.4405

0.4207

0.4013

1.05

1.10

1.15

1.20

1.25

0.1469

0.1357

0.1251

0.1151

0.0156

2.10

2.20

2.30

2.40

2.50

0.0179

0.0139

0.0107

0.0082

0.0062

0.30

0.35

0.40

0.45

0.50

0.3821

0.3632

0.3446

0.3264

0.3085

1.30

1.35

1.40

1.45

1.50

0.0968

0.0885

0.0808

0.0735

0.0668

2.60

2.70

2.80

2.90

3.00

0.0047

0.0035

0.0026

0.0019

0.0013

0.55

0.60

0.65

0.70

0.75

0.2912

0.2743

0.2578

0.2420

0.2266

1.55

1.60

1.65

1.70

1.75

0.0606

0.0548

0.0495

0.0446

0.0401

3.10

3.20

3.30

3.40

3.50

0.0010

0.00069

0.00048

0.00034

0.00023

0.80

0.85

0.90

0.95

1.00

0.2119

0.1977

0.1841

0.1711

0.1587

1.80

1.85

1.90

1.95

2.00

0.0359

0.0322

0.0287

0.0256

0.0228

3.60

3.70

3.80

3.90

4.00

0.00016

0.00010

0.00007

0.00005

0.00003

y ( )Q y 310− 3.10 310

2

−

3.28

410− 3.70 410

2

−

3.90

510− 4.27 610− 4.78



2.79

Note that some authors use ( )erfc , the complementary error function which is

given by

( ) ( ) 221erfc erf e d∞

− β

α

α = − α = βπ ∫

and the error function, ( ) 2

0

2erf e dα

− βα = βπ ∫

Hence ( ) 12 2

Q erfc α⎛ ⎞α = ⎜ ⎟

⎝ ⎠.



2.80

Appendix A2.3

Proof that ( )σ2,X XN m is a valid PDF

We will show that ( )Xf x as given by Eq. 2.56, is a valid PDF by

establishing ( ) 1Xf x d x∞

− ∞

=∫ . (Note that ( ) 0Xf x ≥ for x− ∞ < < ∞ ).

Let 22

2 2yv

I e d v e d y∞ ∞

− −

− ∞ − ∞

= =∫ ∫ .

Then, 22

2 2 2yv

I e d v e d y∞ ∞

− −

− ∞ − ∞

⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∫ ∫

2 2

2v y

e d v d y∞ ∞ +

−

− ∞ − ∞

= ∫ ∫

Let cosv r= θ and siny r= θ . Then, 2 2r v y= + and 1tan yv

− ⎛ ⎞θ = ⎜ ⎟⎝ ⎠

,

and d x d y r d r d= θ . (Cartesian to Polar coordinate transformation).

22

2 2

0 0

r

I e r d r dπ ∞

−= θ∫ ∫

2= π or 2I = π

That is, 2

21 12

v

e d v∞

−

− ∞

=π ∫ (A2.3.1)

Let X

x

x mv −=

σ (A2.3.2)

Then, x

d xd v =σ

(A2.3.3)

Using Eq. A2.3.2 and Eq. A2.3.3 in Eq. A2.3.1, we have the required result.



2.81

References 1) Papoulis, A., ‘Probability, Random Variables and Stochastic Processes’,

McGraw Hill (3P

rdP edition), 1991.

2) Wozencraft, J. M. and Jacobs, I. J., ‘Principles of Communication

Engineering’, John Wiley, 1965.

3) Hogg, R. V., and Craig, A. T., ‘Introduction to Mathematical Statistics”, The

Macmillan Co., 2P

ndP edition, 1965.



4.1

4 CHAPTER 4

Linear Modulation

4.1 Introduction We use the word modulation to mean the systematic alteration of one

waveform, called the carrier, according to the characteristic of another waveform,

the modulating signal or the message. In Continuous Wave (CW) modulation

schemes, the carrier is a sinusoid. We use ( )c t and ( ),m t to denote the carrier

and the message waveforms respectively.

The three parameters of a sinusoidal carrier that can be varied are:

amplitude, phase and frequency. A given modulation scheme can result in the

variation of one or more of these parameters. Before we look into the details of

various linear modulation schemes, let us understand the need for modulation.

Three basic blocks in any communication system are: 1) transmitter 2) Channel

and 3) Receiver (Fig. 4.1).

Fig. 4.1: A basic communication system

The transmitter puts the information from the source (meant for the

receiver) onto the channel. The channel is the medium connecting the transmitter

and the receiver and the transmitted information travels on this channel until it

reaches the destination. Channels can be of two types: i) wired channels or ii)

wireless channels. Examples of the first type include: twisted pair telephone



4.2

channels, coaxial cables, fiber optic cable etc. Under the wireless category, we

have the following examples: earth’s atmosphere (enabling the propagation of

ground wave and sky wave), satellite channel, sea water etc.

The main disadvantage of wired channels is that they require a man-made

medium to be present between the transmitter and the receiver. Though wired

channels have been put to extensive use, wireless channels are equally (if not

more) important and have found a large number of applications.

In order to make use of the wireless channels, the information is to be

converted into a suitable form, say electromagnetic waves. This is accomplished

with the help of a transmitting antenna. The antenna at the receiver (called the

receiving antenna) converts the received electromagnetic energy to an electrical

signal which is processed by the receiver.

The question is: can we radiate the baseband1 information bearing signal

directly on to the channel?

For efficient radiation, the size of the antenna should be 10λ or more

(preferably around 4λ ), where λ is the wavelength of the signal to be radiated.

Take the case of audio, which has spectral components almost from DC upto 20

kHz. Assume that we are designing the antenna for the mid frequency; that is,10

kHz. Then the length of the antenna that is required, even for the 10λ situation

is, 8

34

3 10 3 1010 10 10

cf

×= = ×

⋅ × meters, c being the velocity of light.

1 Baseband signals have significant spectral content around DC. Some of the baseband signals

that are of interest to us are: a) Speech b) music and c) video (TV signals).

Approximate spectral widths of these signals are: Speech: 5 kHz, Audio : 20 kHz, Video : 5 MHz



4.3

Even an antenna of the size of 3 km, will not be able to take care of the entire

spectrum of the signal because for the frequency components around 1 kHz, the

length of the antenna would be 100λ . Hence, what is required from the point of

view of efficient radiation is the conversion of the baseband signal into a

narrowband, bandpass signal. Modulation process helps us to accomplish this;

besides, modulation gives rise to some other features which can be exploited for

the purpose of efficient communication. We describe below the advantages of

modulation.

1. Modulation for ease of radiation Consider again transmission of good quality audio. Assume we choose the

carrier frequency to be 1 MHz. The linear modulation schemes that would be

discussed shortly give rise to a maximum frequency spread (of the modulated

signal) of 40 kHz, the spectrum of the modulated signal extending from (1000 -

20) = 980 kHz to (1000 + 20) = 1020 kHz. If the antenna is designed for 1000

kHz, it can easily take care of the entire range of frequencies involved because

modulation process has rendered the signal into a NBBP signal.

2. Modulation for efficient transmission Quite a few wireless channels have their own appropriate passbands. For

efficient transmission, it would be necessary to shift the message spectrum into

the passband of the channel intended. Ground wave propagation (from the lower

atmosphere) is possible only up to about 2 MHz. Long distance ionospheric

propagation is possible for frequencies in the range 2 to 30 MHz. Beyond 30

MHz, the propagation is line of sight. Preferred frequencies for satellite

communication are around 3 to 6 GHz. By choosing an appropriate carrier

frequency and modulation technique, it is possible for us to translate the

baseband message spectrum into a suitable slot in the passband of the channel

intended. That is, modulation results in frequency translation.



4.4

3. Modulation for multiplexing Several message signals can be transmitted on a given channel, by assigning to

each message signal an appropriate slot in the passband of the channel. Take

the example of AM broadcast, used for voice and medium quality music

broadcast. The passband of the channel used to 550 kHz to 1650 kHz. That is,

the width of the passband of the channel that is being used is 1100 kHz. If the

required transmission bandwidth is taken as 10 kHz, then it is possible for us to

multiplex, atleast theoretically, 110 distinct message signals on the channel and

still be able to separate them individually as and when we desire because the

identity of each message is preserved in the frequency domain.

4. Modulation for frequency assignment Continuing on the broadcast situation, let us assume that each one of the

message signals is being broadcast by a different station. Each station can be

assigned a suitable carrier so that the corresponding program material can be

received by tuning to the station desired.

5. Modulation to improve the signal-to-noise ratio Certain modulation schemes (notably frequency modulation and phase

modulation) have the feature that they will permit improved signal-to-noise ratio

at the receiver output, provided we are willing to pay the price in terms of

increased transmission bandwidth (Note that the transmitted power need not be

increased). This feature can be taken advantage of when the quality of the

receiver output is very important.

Having understood the need and the potential benefits due to modulation,

let us now get into the details of various linear modulation schemes. The four

important types of linear modulation schemes are

1) Double SideBand, Suppressed Carrier (DSB-SC)

2) Double SideBand, Large Carrier (DSB-LC) (also called conventional AM or

simply AM)



4.5

3) Single SideBand (SSB)

4) Vestigial SideBand (VSB)

We shall begin our discussion with DSB-SC.

4.2 DSB-SC Modulation

4.2.1. Modulation The DSB-SC is the simplest of the four linear modulation schemes listed

above (simplest in terms of the mathematical description of modulation and

demodulation operations). Consider the scheme shown in Fig. 4.2

Fig. 4.2: DSB-SC modulation scheme

( )m t is a baseband message signal with =M f( ) 0 for ( ),f W c t> is a high

frequency carrier, usually with >>cf W .

DSB-SC modulator is basically a multiplier. Let mg denotes the amplitude

sensitivity (or gain constant) of the modulator, with the units per volt (we assume

that ( )m t and cA are in volts). Then the modulator output ( )s t is,

( ) ( ) ( )( )cosm c cs t g m t A t= ω (4.1a)

For convenience, let 1mg = . Then,

( ) ( ) ( )cosc cs t A m t t= ω (4.1b)



4.6

As DSB-SC modulation involves just the multiplication of the message

signal and the carrier, this scheme is also known as product modulation and can

be shown as in Fig. 4.3.

Fig. 4.3: Product Modulation scheme

The time domain behavior of the DSB-SC signal (with 1cA = ) is shown in

Fig. 4.4(b), for the ( )m t shown in Fig. 4.4(a).



4.7

Fig. 4.4: (a) The message signal

(b) The DSB-SC signal

Note that the carrier undergoes a 180ο phase reversal at the zero crossings of

( )m t . This is brought out more clearly in the oscillograms, shown in Fig. 4.5 and

Fig. 4.6, where ( )m t is a sinusoidal signal.

With reference to Fig. 4.5, between the points ‘a’ and ‘b’, the carrier in the

DSB-SC signal and the actual carrier (bottom picture) are in phase whereas

between the points ‘b’ and ‘c’, they are 1800 out of phase. Fig. 4.6 is an

expanded version of the central part of the waveforms in Fig. 4.5. Here, we can



4.8

very clearly observe that to the left of ‘b’, both the carriers are in phase whereas

to the right, they are 1800 out of phase.

Fig. 4.5: (top) DSB-SC signal with tone modulation

(bottom) The carrier

Fig. 4.6: Expanded versions of a part of the waveforms in Fig. 4.5

Consider waveforms shown in Fig. 4.7. We have on the top, modulating tone

signal and at the bottom, the corresponding DSB-SC. What do we observe on

the oscilloscope, if we feed the X-plates the tone signal and the Y-plates, the

DSB-SC signal? The result is shown in Fig. 4.8, which can be explained as

follows.



4.9

At the point ‘a’ in Fig. 4.7, the modulating tone is at its maximum and

hence the DSB-SC signal has the maximum value. Point ‘a’ in Fig. 4.8

corresponds to the point ‘a’ in Fig. 4.7. Between the points ‘a’ and ‘b’ in Fig. 4.7,

the tone amplitude decreases (reaching the value zero at point b); hence the

maximum value reached by the DSB-SC signal during each carrier cycle keeps

decreasing. As the X-plates are being fed with the same tone signal, this

decrease will be linear and this corresponds to segment ‘a’ to ‘b’ in Fig. 4.8.

(Note that DSB-SC signal is zero at point b). In the time interval between ‘b’ and

‘c’ of Fig. 4.7, the DSB signal increases and this increase is seen as a straight

line between the points ‘b’ and ‘c’ in Fig. 4.8. Between the points ‘c’ and ‘e’ in Fig.

4.7, the tone amplitude varies from the most negative value to the most positive

value. Correspondingly, the display on the oscilloscope will follow the trace

c d e→ → shown in Fig. 4.8.

Fig. 4.7: (top) modulating signal

(bottom) DSB-SC signal



4.10

Fig. 4.8 Display on the oscilloscope with the following inputs:

X-plates: Tone signal

Y-plates: DSB-SC signal

Taking the Fourier transform of Eq. 4.1(b), we have

( ) ( ) ( )2c

c cAS f M f f M f f⎡ ⎤= − + +⎣ ⎦ (4.2)

If we ignore the constant 2cA on the R.H.S of Eq. (4.2), we see that the

modulation process has simply shifted the message spectrum by cf± . As the

frequency translation of a given spectrum occurs quite often in the study of

modulation and demodulation operations, let us take a closer look at this.

i) Let ( )m t be a real signal with the spectrum ( )M f shown below (Fig.

4.9(a)). Let cf be 100 kHz. Assuming = 12cA , we have ( )S f as shown in

Fig. 4.9(b).



4.11

Fig. 4.9: Frequency translation (a) baseband spectrum (real signal)

(b) Shifted spectrum.

Note that ( ) ( ) ( )102 kHz2 kHz 202 kHz

fS f M M

== +

1 0= + 1=

and is the point ‘a’ in Fig. 4.9

ii) Let ( )m t be a complex signal with ( )M f as shown in Fig. 4.10(a). The

corresponding shifted spectrum (with 100cf = kHz) is shown in Fig.

4.10(b)



4.12

Fig. 4.10: Frequency translation (a) Baseband spectrum (complex signal)

(b) Shifted spectrum.

In figures 4.9(b) and 4.10(b), the part that is hatched in red is called the

Upper Sideband (USB) and the one hatched in blue is called the Lower Sideband

(LSB). Any one of these two sidebands has the complete information about the

message signal. As we shall see later, SSB modulation conserves the bandwidth

by transmitting only one sideband and recovering the ( )m t with appropriate

demodulation.

Example 4.1 Consider the scheme shown in Fig. 4.11(a). The ideal HPF has the cutoff

frequency at 10 kHz. Given that f1 10= kHz and f2 15= kHz, let us sketch

( )Y f for the ( )X f given at (b).



4.13

Fig. 4.11: (a) The scheme of example 4.1

(b) The input spectrum, ( )X f

We have ( ) ( ) ( )V f X f f X f f1 1= − + + , which is as shown in Fig. 4.12(a).

The HPF eliminates the spectral components for f 10≤ kHz. Hence ( )W f is

as shown in Fig. 4.12(b).

( ) ( ) ( )Y f W f f W f f2 2= − + + . This is shown in Fig. 4.12(c).



4.14

Fig. 4.12: Spectra at various points in the scheme of Fig. 4.11

4.2.2. Coherent demodulation The process of demodulation of a DSB-SC signal, at least theoretically, is

quite simple. Let us assume that the transmitted signal ( )s t has been received

without any kind of distortion and is one of the inputs to the demodulator as

shown in Fig. 4.13. That is, the received signal ( ) ( )r t s t= . Also, let us assume

that we are able to generate at the receiving end a replica of the transmitted

carrier (denoted ( ) ( )' cosr c cc t A t= ω in Fig. 4.13) which is the other input to the

demodulator.



4.15

Fig. 4.13: Coherent demodulation of DSB-SC

The demodulation process consists of multiplying these two inputs and lowpass

filtering the product quantity ( )v t .

From Fig. 4.13, we have

( ) ( ) ( )( ) ( )( )'cos cosg c c c cv t d A m t t A t= ω ω

where gd is the gain constant of the multiplier, called the detector gain constant,

in the context of demodulation. For convenience, let us take = 1gd

( ) ( ) ( )2' cosc c cv t A A m t t= ω

( ) ( )1 cos 2'2

cc c

tA A m t

⎡ ⎤+ ω⎣ ⎦=

Assuming that =' 2c cA A we have

( ) ( ) ( ) ( )cos 4 cv t m t m t f t= + π (4.3)

The second term on the R.H.S of Eq. 4.3 has the spectrum centered at 2 cf±

and would be eliminated by the lowpass filter following ( )v t . Hence ( )0v t , the

output of the demodulation scheme of Fig. 4.13 is the desired quantity, namely,

( )m t .

Let us illustrate the operation of the detector in the frequency domain. Let

( )m t be real with the spectrum shown in Fig. 4.14(a). Let



4.16

( ) ( ) ( ) ( )2 cos cr t s t m t t= = ω . Then ( ) ( ) ( )c cS f M f f M f f= − + + , shown in

Fig. 4.14(b).

Fig. 4.14: Spectra at various points in the demodulation scheme of Fig. 4.13

(Note that the positive frequency part of ( )S f is shown in red and the negative

frequency part in blue). Assuming ( ) ( ) ( )cos cv t s t t= ω (Fig. 4.13 with cA' 1= ),

then ( ) ( ) ( )12 c cV f S f f S f f⎡ ⎤= − + +⎣ ⎦ . ( )1

2 cS f f− and ( )12 cS f f+ are shown in



4.17

Fig. 4.14(c) and (d) respectively. ( )0V f is the sum of the outputs of the lowpass

filters shown in Fig. 4.14(c) and (d) which is the desired message spectrum,

( )M f .

From the discussion of the demodulation process so far, it appears that

demodulation of DSB-SC is quite simple. In practice, it is not. In the scheme of

Fig. 4.13, we have assumed that, we have available at the receiver, a carrier

term that is coherent (of the same frequency and phase) with the carrier used to

generate the DSB-SC signal at the transmitter. Hence this demodulation scheme

is known as coherent (or synchronous) demodulation. As the receiver and the

transmitter are, in general, not collocated, the carrier source at the receiver is

different from that used at the transmitter and it is almost impossible to

synchronize two independent sources. Fairly sophisticated circuitry has to be

used at the receiver in order to generate the coherent carrier signal, from an

( )r t that has no carrier component in it. Before we discuss at the generation of

the coherent carrier at the receiver, let us look at the degradation caused to the

demodulated message due to a local carrier that has phase and frequency

differences with the transmitted one.

Case i): Constant phase difference between ( )c t and ( )rc t

Let ( ) ( )cos 2 cc t f t= π and ( ) [ ]cos 2r cc t f t= π + ϕ (the amplitude quantities,

cA and 'cA can be treated as 1)

( ) ( ) ( ) ( )cos cosc cv t m t t t= ω ω + ϕ

( ) ( ) ( ) ( )cos cos cos sin sinc c cm t t t t⎡ ⎤= ω ω ϕ − ω ϕ⎣ ⎦

( ) ( ) ( ) ( )2cos cos sin cos sinc c cm t t t t⎡ ⎤= ω ϕ − ω ω ϕ⎣ ⎦

( ) ( ) ( ) ( )1 cos 2 sin 2cos sin

2 2c ct m t t

m t⎡ ⎤+ ω ω

= ϕ − ϕ⎢ ⎥⎣ ⎦



4.18

At the output of the LPF, we will have only the term ( ) cos2

m t ϕ. That is, the

output of the demodulator, ( )0v t , is proportional to ( ) cosm t ϕ . As long as ϕ

remains a constant, the demodulator output is a scaled version of the actual

message signal. But values of ϕ close to 2π will force the output to near about

zero. When 2ϕ = π we have zero output from the demodulator. This is called

the quadrature null effect of the coherent detector.

Case ii): Constant frequency difference between ( )c t and ( )rc t

Let ( ) ( )cos 2 cc t f t= π and ( ) ( )cos 2r cc t f f t⎡ ⎤= π + ∆⎣ ⎦ . Then,

( ) ( ) ( ) ( )cos 2 cos 2c cv t m t f t f f t⎡ ⎤= π π + ∆⎣ ⎦

By carrying out the analysis similar to case (i) above, we find that

( ) ( ) ( )0 cos 2v t m t f t⎡ ⎤α π ∆⎣ ⎦ (4.4a)

Let us look in some detail the implications of Eq. 4.4(a). For convenience,

let ( ) ( ) ( )0 cos 2v t m t f t⎡ ⎤= π ∆⎣ ⎦ (4.4b)

Assume 100f∆ = Hz and consider the spectral component at 1 kHz in ( )M f .

After demodulation, this gives rise to two spectral components, one at 900 Hz

and the other at 1100 Hz, because

( ) ( ) ( ) ( )3 1cos 2 10 cos 2 100 cos 2 1100 cos 2 9002

t t t t⎡ ⎤ ⎡ ⎤ ⎡ ⎤π × π × = π × + π ×⎣ ⎦ ⎣ ⎦⎣ ⎦

The behavior of the sum of these two components is shown in Fig. 4.15.

As can be seen from this figure, the envelope of sum signal (broken red line)

attains the peak value twice in a cycle of the beat frequency f∆ . Also, it goes

through zero twice in a cycle of the beat frequency.



4.19

Fig. 4.15: Time-domain behavior of ( ) ( )3cos 2 10 cos 2 100t tπ × π ×

Let us examine the effect of frequency offset in the frequency domain. Let ( )M f

be as shown in Fig. 4.16(a). Assume 300f∆ = Hz. Then,

( ) ( ) ( )012

V f M f f M f f⎡ ⎤= − ∆ + + ∆⎣ ⎦ will be as shown in Fig. 4.16(d), which is

one-half the sum of the spectra shown at (b) and (c). Comparing Fig. 4.16(a) and

(d), we are tempted to surmise that the output of the demodulator is a fairly

distorted version of the actual message signal. A qualitative feeling for this

distortion can be obtained by listening to the speech files that follow.

Introduction

Output 1

Output 2

Output 3

Output 4



4.20

Fig. 4.16: The effect of frequency offset in the demodulation of DSB-SC:

(a) Typical message spectrum, ( )M f

(b) ( )300M f +

(c) ( )300M f −

(d) ( ) ( )M f M f1 300 3002⎡ ⎤+ + −⎣ ⎦



4.21

Example 4.2 In this example, we will show that the DSB-SC signal can be demodulated

with the help of any periodic function ( )p t , as long as ( )p t has a spectral

component at cf , the carrier frequency. This component could be due to the

fundamental or some harmonic of the fundamental.

a) Let ( ) ( ) ( )c cs t A m t tcos= ω . Consider the product ( ) ( )ps t x t where ( )px t

is any periodic signal with the period c

Tf01

= . That is,

( ) cj n f tp n

nx t x e 2

∞π

= − ∞= ∑

where nx is the nth Fourier coefficient. We will show that if x1 0≠ , then it is

possible to extract ( )m t from the product ( ) ( )ps t x t .

b) Let ( )py t be another periodic signal with the period T NT0 0' = . We will

show that, appropriate filtering of the product ( ) ( )ps t y t , will result in ( )m t .

a) As ( ) π − π⎡ ⎤π = +⎣ ⎦2 21cos 2

2c cj f t j f t

cf t e e , we have

( ) ( ) ( ) ( ) ( )c cj n t j n tcp n n

n n

A m ts t x t x e x e1 1

2+ ω − ω⎡ ⎤

= +⎢ ⎥⎣ ⎦∑ ∑

( ) ( ) ( )c cj n t j n tcn n

n nn n

A m tx x e x x e1 1

1 1, ,

1 12

+ ω − ω−

≠ − ≠

⎡ ⎤⎢ ⎥

= + + +⎢ ⎥⎢ ⎥⎣ ⎦

∑ ∑

as [ ]x x

x1 11Re

2− +

= , the output, after lowpass filtering would be,

[ ] ( )cx A m t1Re . (We assume that the LPF will reject all the other

spectral components)

b) The product ( ) ( )ps t y t can be written as



4.22

( ) ( ) ( ) c cn nj f t j f t

c N Np n n

n n

A m ts t y t y e y e

2 1 2 1

2

⎛ ⎞ ⎛ ⎞π + π −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎧ ⎫⎪ ⎪= +⎨ ⎬⎪ ⎪⎩ ⎭∑ ∑

( ) c cn nj f t j f t

c N NN n N n

n nn N n N

A m ty y e y y e

2 1 2 1

, ,2

⎛ ⎞ ⎛ ⎞π + π −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

−

≠ − ≠

⎧ ⎫⎪ ⎪

= + + +⎨ ⎬⎪ ⎪⎩ ⎭

∑ ∑

We assume that Ny 0≠ . Then, the output of the LPF would be

[ ] ( )N cy A m tRe . (Note that ( ) ( )py t s t has spectral lobes at

c cf fN N

20, , , etc.± ± We assume that the LPF will extract the lobe at f 0=

and reject others).

Example 4.3

Consider the scheme shown in Fig. 4.17. ( )s t is the DSB-SC signal

( ) ( )cm t tcos ω with

( )f

S foutside

1 , 99 kHz 101kHz0 ,⎧ ≤ ≤⎪= ⎨⎪⎩

Let ( )g t be another bandpass signal with

( )f

G foutside

1 , 98 kHz 102 kHz0 ,⎧ ≤ ≤⎪= ⎨⎪⎩

a) We will show that the output ( ) ( )y t m tα .

b) We will show that it would not be possible to recover ( )m t from ( )v t if

( )⎧ < ≤⎪= ⎨⎪⎩

1 , 98.5 kHz 101.5 kHz0 ,

fG f

outside



4.23

Fig. 4.17: Scheme of DSB-SC demodulation (example 4.3)

a) Let ( ) ( ) ( )cg t m t f t1 cos 2= π where cf 100 kHz= and

( )f

M foutside1

2 , 2 kHz0 ,⎧ ≤⎪= ⎨⎪⎩

From ( )S f , we see that

( )f

M foutside

2 , 1kHz0 ,⎧ ≤⎪= ⎨⎪⎩

We have,

( ) ( ) ( ) ( )cv t m t m t t21 cos= ω

( ) ( ) ( )c tm t m t1

1 cos 22

⎡ ⎤+ ω= ⎢ ⎥

⎣ ⎦

( ) ( ) ( ) ( ) ( )cm t m t m t m t

t1 1 cos 22 2

= + ω

We will assume that the LPF rejects the spectrum around cf2± ,

( ) ( ) ( ) ( )m t m t M f M f1 12 2

∗←⎯→

( ) ( )M f M f1∗ will have a flat spectrum for f 1kHz≤ . By using an ILPF

with cutoff at 1 kHz, we can recover ( )m t from ( )v t .

b) For this case ( )M f1 would be

( )f

M foutside1

2 , 1.5 kHz0 ,⎧ ≤⎪= ⎨⎪⎩

( ) ( )M f M f1 ∗ will be flat only for f 0.5 kHz≤ . Hence ( )m t cannot be

recovered.



4.24

Exercise 4.1

A signal ( )m t whose spectrum is shown in Fig. 4.18(a) is generated

using the signals ( )m t1 and ( )m t2 . ( )M f1 and ( )M f2 are shown at (b) and

(c) respectively in Fig. 4.18. The signal ( ) ( ) ( )s t m t t52 cos 10= π is

transmitted on the channel.

a) Suggest a scheme to obtain ( )m t from ( )m t1 and ( )m t2 .

b) ( )m t1 and ( )m t2 are to be recovered from the received signal

( ) ( )r t s t= . A part of this receiver is shown in Fig. 4.18(d). Complete

the receiver structure by indicating the operations to be performed by the

boxes with the question mark inside.

(a)

(b) (c)

(d)

Fig. 4.18: Proposed receiver structure for the exercise 4.1



4.25

4.2.3 Carrier recovery for coherent demodulation As explained in detail in sec. 4.2.2, coherent demodulation requires a

carrier at the receiving end that is phase coherent with the transmitted carrier.

Had there been a carrier component in the transmitted signal, it would have been

possible to extract it at the receiving end and use it for demodulation. But the

DSB-SC signal has no such component and other methods have to be devised to

generate a coherent carrier at the receiver. Two methods are in common use for

the carrier recovery (and hence demodulation) from the suppressed carrier

modulation schemes, namely (a) Costas loop and (b) squaring loop.

a) Costas loop: This scheme is shown in Fig. 4.19.

Fig. 4.19: Costas loop

The VCO (Voltage Controlled Oscillator) is a source that produces a periodic

waveform1 whose frequency is controlled by the input voltage ( )ce t . The output

frequency of of the VCO when ( ) 0ce t = is called the free running frequency of

the VCO. The frequency put out by the VCO at any instant depends on the sign

and magnitude of the control voltage, ( )ce t .

1 Here, we shall assume that the VCO output is sinusoidal.



4.26

To understand the loop operation, let us assume that the frequency and

phase of the VCO output are the same as that of the incoming carrier. Then,

( ) ( ) ( )( ) ( )( )1 0cos cosc c cv t A m t t A t= ω ω

( ) ( )20 cosc cA A m t t= ω

( ) ( )0

1 2cos2

cc

tA A m t

⎡ ⎤+ ω⎣ ⎦=

The output of the LPF1 is

( ) ( )02 2

cA A m tv t = ; that is, ( ) ( )2v t m tα , the desired signal.

Similar analysis shows

( )4 0v t =

Now suppose that VCO develops a small phase offset of ϕ radians. The I-

channel output will remain essentially unchanged but a small voltage will develop

at the output of the Q-channel which will be proportional to ϕsin (If the phase

shift is − ϕ rad, then the Q channel output is proportional to sin− ϕ ). Because of

this, ( )e t is a non-zero quantity given by

( ) ( ) ( ) ( ) 22 4 0

1 cos sin4 ce t v t v t A A m t⎡ ⎤= = ϕ ϕ⎣ ⎦

( ) 20

1 sin28 cA A m t⎡ ⎤= ϕ⎣ ⎦

( )e t is input to LPF3, which has very narrow passband (Note that LPF1 and

LPF2 should have a bandwidth of at least W Hz). Hence ( ) 0 sin2ce t C= ϕ

where 0C is the DC value of ( ) 20

18 cA A m t⎡ ⎤⎣ ⎦ . This DC control voltage ensures

that the VCO output is coherent with the carrier used for modulation.

b) Squaring loop The operation of the squaring loop can be explained with the help of Fig.

4.20.



4.27

Fig. 4.20: Demodulation of DSB-SC using a squaring loop

Let ( ) ( ) ( ) ( )cosc cr t s t A m t t= = ω . Then,

( ) ( ) ( ) ( )2

2 2 1 cos 22c

cAv t r t m t t⎡ ⎤= = + ω⎣ ⎦

( )2m t will have nonzero DC value which implies its spectrum has an impulse at

= 0f . Because of this, ( )V f will have a discrete spectral component at 2 cf .

( )v t is the input to a very narrowband bandpass filter, with the centre frequency

2 cf . By making the bandwidth of LPF1 very narrow, it is possible to make the

VCO to lock on to the discrete component at 2 cf , present in ( )w t . (The dotted

box enclosing a multiplier, LPF and a VCO, connected in the feedback

configuration shown is called the Phase Locked Loop (PLL)). The VCO output

goes through a factor of two frequency divider, yielding a coherent carrier at its

output. This carrier is used to demodulate the DSB-SC signal. Note that LPF2

must have adequate bandwidth to pass the highest frequency component

present in ( )m t .

Both the Costas loop and squaring loop have one disadvantage, namely,

an 1800 phase ambiguity. Consider the Costas loop; if the input to the loop were



4.28

to be ( ) ( )cosc cA m t t− ω , output of the LPF1 would be ( )01 cos2 cA A m t− ϕ and

that of LPF2 would be ( )01 sin2 cA A m t− ϕ with the result that, ( )e t would be the

same as in the case discussed earlier. Similarly, for the squaring loop, ( )v t would

be the same whether ( ) ( ) ( )cosc cr t A m t t= ω or ( ) ( )cosc cA m t t− ω . Hence

the demodulated output could be either ( )m t or ( )m t− . However, this will not

cause any problem for audio transmission because ( )m t and ( )m t− , would

sound the same to our ears.

Though DSB-SC modulation schemes place the entire transmitted power

into the useful sidebands, the demodulation has to be coherent. The circuit

required to generate a coherent carrier increases the cost of the receiver. If only

a few receivers are to be built for a specific communication need, the cost may

not be a major factor. But in a broadcast situation, there would be a large number

of receivers tuned to a given station and in that scenario, it is better make the

receiver fairly cheap and push the cost up of the transmitter, if required. As will

seen later, the Envelope Detector(ED) is fairly cheap to implement as compared

to a coherent detector. But to make use of ED, the modulated carrier should

carry ( )m t in its envelope. Evidently, DSB-SC does not satisfy this property as

explained below.

Let ( ) ( ) ( )cosc cs t A m t t= ω

Pre-envelope of ( ) ( ) ( ) ci tcpes t s t A m t e ω⎡ ⎤= =⎣ ⎦

Complex envelope of ( ) ( ) ( )cces t s t A m t⎡ ⎤= =⎣ ⎦

Hence the envelope of DSB-SC ( ) ( )pe ces t s t= =

( )m tα



4.29

We shall now describe a modulation scheme that has ( )m t as its envelope

which can easily be extracted.

4.3 DSB-LC Modulation (or AM) By adding a large carrier component to the DSB-SC signal, we will have

DSB-LC, which, for convenience, we shall call simply as AM. By choosing the

carrier component properly, it is possible for us to generate the AM signal such

that it preserves ( )m t in its envelope. Consider the scheme shown in Fig. 4.21.

Fig. 4.21: Generation of an AM signal from a DSB-SC signal

Let ( ) ( ) ( )cosc m cv t A g m t t= ω . Then,

( ) ( ) ( )cosc cAMs t A t v t⎡ ⎤ = ω +⎣ ⎦

( ) ( )1 cosc m cA g m t t⎡ ⎤= + ω⎣ ⎦ (4.5)

In this section, unless there is confusion, we use ( )s t in place of [ ]( ) AMs t . We

shall assume that ( )m t has no DC component and ( )( ) ( )( )max minm t m t= − . Let

( )mg m t be such that ( ) 1mg m t ≤ for all t. Then ( )1 0mg m t⎡ ⎤+ ≥⎣ ⎦ and

( ) AMs t⎡ ⎤⎣ ⎦ preserves ( )m t in its envelope because

( ) ( )1 cj tc mpe

s t A g m t e ω⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦

( ) ( )1c mces t A g m t⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦



4.30

As ( )1 0mg m t⎡ ⎤+ ≥⎣ ⎦ , we have

Envelope of [ ]= = +( ) ( ) 1 ( )ce c ms t s t A g m t

The quantity after the DC block is proportional to ( )m t .

If ( )1 mg m t⎡ ⎤+⎣ ⎦ is not nonnegative for all t , then the envelope would be different

from ( )m t . This would be illustrated later with a few time domain waveforms of

the AM signal. Fig. 4.22(b) illustrates the AM waveform for the case

( )1 0mg m t⎡ ⎤+ ≥⎣ ⎦ for all t .

Fig. 4.22: (a) An arbitrary message waveform ( )m t

(b) Corresponding AM waveform



4.31

A few time instants have been marked in both the figures (a) and (b). At

the time instants when ( ) 0m t = , carrier level would be cA which we have

assumed to be 1. Maximum value of the envelope (shown in red broken line)

occurs when ( )m t has the maximum positive value. Similarly, the envelope will

reach its minimum value when ( )m t is the most negative. As can be seen from

the figure, the envelope of ( )s t follows ( )m t in a one-to-one fashion.

Let ( )max1mg m t x= ≤ . Then ( )s t is said to have (100x) percentage

modulation. For the case of 100% modulation,

( ) ( )max min1m mg m t g m t⎡ ⎤ ⎡ ⎤= − =⎣ ⎦ ⎣ ⎦ . If ( )max

1mg m t > , then we have over

modulation which results in the envelope distortion. This will be illustrated in the

context of tone modulation, discussed next.

Exercise 4.2

For the waveform ( )m t shown in Fig. 4.23, sketch the AM signal with

the percentage modulation of 40. Assume cA 1= (the figure has to be shown

with reference to ( )m t )

Fig. 4.23: Baseband signal for the exercise 4.2



4.32

4.3.1. Tone Modulation

Let ( )m t to be a tone signal; that is, ( ) ( )cos 2m mm t A f t= π where

m cf f<< . Then Eq. 4.5 becomes

( ) ( ) ( )1 cos cosc m m m cs t A g A t t⎡ ⎤= + ω ω⎣ ⎦ (4.6a)

( ) ( )cos cA t t= ω (4.6b)

Let m mg A = µ . Then for tone modulation,

( ) ( ) ( )1 cos cosc m cs t A t t⎡ ⎤= + µ ω ω⎣ ⎦ (4.7)

µ is called the modulation index or modulation factor. 100µ × is the

percentage modulation. To avoid envelope distortion, we require, µ ≤ 1.

As ( ) ( )1 cosc mA t A t⎡ ⎤= + µ ω⎣ ⎦ , we have

( ) [ ]max1cA t A⎡ ⎤ = + µ⎣ ⎦

( ) [ ]min1cA t A⎡ ⎤ = − µ⎣ ⎦

( )( )

max

min

11

A t

A t

⎡ ⎤ + µ⎣ ⎦ =− µ⎡ ⎤⎣ ⎦

or ( ) ( )( ) ( )

max min

max min

A t A t

A t A t

⎡ ⎤ ⎡ ⎤−⎣ ⎦ ⎣ ⎦µ =⎡ ⎤ ⎡ ⎤+⎣ ⎦ ⎣ ⎦

Fig. 4.24 to 4.26 illustrate the experimentally generated AM waveforms for

µ = 0.5, 1 and 1.5 respectively (with 1µ > , we have overmodulation).



4.33

Fig. 4.24: AM with tone modulation ( 0.5µ = )

Fig. 4.25: AM with tone modulation ( 1µ = )



4.34

Fig. 4.26: AM with tone modulation ( 1.5µ = )

Fig. 4.27: Envelope of the AM signal of Fig. 4.26

As can be seen from 4.24 and 4.25, the envelope (shown with a red

broken line) is one-to-one related to the message sinusoid. Note that, for µ = 1,

the carrier amplitude (and hence the envelope) goes to zero, corresponding to

the time-instant when the sinusoid is going through the negative peak. However,

when µ > 1, the one-to-one relationship between the envelope of the modulated

carrier and the modulating tone is no longer maintained. This can be more clearly

seen in Fig. 4.27 which shows the output of the envelope detector when the input



4.35

is the modulated carrier of Fig. 4.26. Notice that the tone signal between ( )t t1 2,

and to the right of t3 of Fig. 4.26 is getting inverted; in other words, the output of

the ED is proportional to ( )+ µ ω1 cos mt which is not equal to ( )+ µ ω1 cos mt ,

when µ > 1.

Fig. 4.28: Oscillogram when the CRO inputs are:

X-plates: tone signal

Y-plates: AM signal with 12

µ =

Fig. 4.29: Oscillogram when the CRO inputs are:

X-plates: tone signal

Y-plates: AM signal with ( 1µ = )



4.36

Fig. 4.28 and Fig. 4.29 illustrate the oscillograms when the X-plates of the

CRO are fed with the modulating tone and the Y-plates with the AM signal with

µ = 0.5 and µ = 1 respectively. In Fig. 4.28, A represents the peak-to-peak

value of the carrier at its minimum (that is, [ ]= − µ2 1cA A ) where as B is the

peak-to-peak value of the carrier at its maximum (that is, [ ]= + µ2 1cB A ).

Hence µ can be calculated as

B AB A−

µ =+

In Fig. 4.29, as A = 0 we have µ = 1

4.3.2. Spectra of AM signals

Taking the Fourier transform of Eq. 4.5,

( ) ( ) ( ) ( ) ( )2 2c c

c c m c cAM

A AS f f f f f g M f f M f f⎡ ⎤ ⎡ ⎤ ⎡ ⎤= δ − + δ + + − + +⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(4.8)

The plot of [ ( )]AMS f is given in Fig. 4.30.

Exercise 4.3

Picture the oscillogram when the X-plates of the CRO are fed with the

modulating tone and the Y-plates with the AM signal with µ = 1.5 .



4.37

Fig. 4.30: (a) Baseband message spectrum ( )M f

(b) Spectrum of the AM signal

Based on Fig. 4.30, we make the following observations:

1) The spectrum has two sidebands, the USB [between cf to cf W+ , and

( )cf W− − to cf− , hatched in red] and the LSB ( cf W− to cf and cf− to

( )cf W− + , hatched in blue).

2) If the baseband signal has bandwidth W , then the AM signal has

bandwidth 2 W . That is, the transmission bandwidth TB , required for the

AM signal is 2 W .

3) Spectrum has discrete components at cf f= ± , indicated by impulses of

area 2cA

4) In order to avoid the overlap between the positive part and the negative part

of ( )S f , cf W> (In practice, cf W>> , so that ( )s t is a narrowband

signal)



4.38

The discrete components at cf f= ± , do not carry any information and as

such AM does not make efficient use of the transmitted power. Let us illustrate

this taking the example of tone modulation.

Example 4.4

For AM with tone modulation, let us find Total sideband powerη =Total power

, as a

function of modulation index µ .

For tone modulation, we have

( ) ( ) ( )1 cos cosc m cs t A t t⎡ ⎤= + µ ω ω⎣ ⎦

Carrier term = ( )cosc cA tω

Carrier Power = 2

2cA

USB term = ( )cos2c

c mA tµ

ω + ω

Power in USB =

2

222 8

c2c

AA

µ⎛ ⎞⎜ ⎟ µ⎝ ⎠ =

Power in LSB = Power in USB

Total sideband Power = 2

2 cc

AA22

28 4

µµ× =

Total Power=( )22 2 2

cc c cAA A A

22 2 21

2 4 2 2 4

+ µ⎛ ⎞µ µ+ = + =⎜ ⎟

⎝ ⎠

Hence, ( )

c

c

A

A

2 2

2

22 24

22

4

µµ

η = =+ µ+ µ

Calculating the value of η for a few value of µ , we have



4.39

µ η

0.25 0.03

0.50 0.11

0.75 0.22

1.0 0.33

As can be seen from the above tabulation, η increases as 1µ → ;

however, even at 1µ = , only 1/3 of the total power is in the sidebands (or side

frequencies), the remaining 2/3 being in the carrier. From this example, we see

that AM is not an efficient modulation scheme, in terms of the utilization of the

transmitted power.

The complex envelope behavior of ( )s t for tone modulation is quite

illustrative.

( ) ( ) ( )c m cs t A t t1 cos cos⎡ ⎤= + µ ω ω⎣ ⎦

( ) ( )c m c mc j j tj t tcc

AA e e eRe2

ω + ω ω − ωω µ⎧ ⎫⎡ ⎤= + +⎨ ⎬⎣ ⎦⎩ ⎭

( ) m mj t j tc ccce

A As t A e e2 2

ω − ωµ µ⎡ ⎤ = + +⎣ ⎦ (4.9)

Let us draw a ‘phasor’ diagram, using the carrier quantity as the reference.

The term mj tcA e2

ωµ can be represented as a rotating vector with a magnitude

of cA2µ , rotating counterclockwise at the rate of mf rev/sec. Similarly,

mj tcA e2

− ωµ can be shown as a vector with clockwise rotational speed of mf

rev/sec.



4.40

Fig. 4.31: Phasor diagram for AM with tone modulation

Fig. 4.31 depicts the behavior of all the quantities on the RHS of Eq. 4.9.

From Eq.4.9, we find that the complex envelope is real and is given by

( )c mA t1 cos⎡ ⎤+ µ ω⎣ ⎦ . This can also be seen from the phasor diagram, because

at any given time, the quadrature components of the sideband phasors cancel

out where as the in-phase components add up; the resultant of the sideband

components is collinear with the carrier.

The length of the in-phase component of ( ) ces t⎡ ⎤⎣ ⎦ depends on the sign of

the resultant of the sideband phasors. As can be seen from Fig. 4.31 this varies

between the limits [ ]cA 1 − µ to [ ]cA 1 + µ . If the modulation index µ is less than

1, [ ]cA 1 0− µ > and envelope of ( )s t is

( ) ( ) ( )⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + µ ω = + µ ω⎣ ⎦ ⎣ ⎦ ⎣ ⎦1 cos 1 cosc m c mces t A t A t

Phasor diagrams such as the one shown in Fig. 4.31 are helpful in the

study of unequal attenuation of the sideband components. We shall illustrate this

with an example.

Example 4.5

Let 1cA = , µ =12

and let the upper sideband be attenuated by a factor of

2. Let us find the expression for the resulting envelope, ( )A t .



4.41

The phasor diagram for this case is shown in Fig. 4.32.

Fig. 4.32: Phasor diagram for an AM signal with unequal sidebands

As can be seen from the figure, the resultant of the sidebands is no longer

collinear with the carrier.

( ) ( ) ( )( ) ( ) ( )( )m m m mces t t j t t j t1 11 cos sin cos sin

8 4⎡ ⎤ = + ω + ω + ω − ω⎣ ⎦

( ) ( )m mt j t3 11 cos sin8 8

= + ω − ω

( ) ( ) ( )m mA t t t

12 2 23 11 cos sin

8 8

⎡ ⎤⎛ ⎞ ⎛ ⎞= + ω + ω⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

Evidently, it is not possible for us to recover the message from the above ( )A t .

4.4 Generation of AM and DSB-SC signals Let ( )x t be the input to an LTI system with the impulse response ( )h t

and let ( )y t be the output. Then,

( ) ( ) ( )y t x t h t= ∗

( ) ( ) ( )Y f X f H f=

That is, an LTI system can only alter a frequency component (either boost or

attenuate), that is present in the input signal. In other words, an LTI system

cannot generate at its output frequency components that are not present in

( )X f . We have already seen that the spectrum of a DSB or AM signal is



4.42

different from that of the carrier and the baseband signal. That is, to generate a

DSB signal or an AM signal, we have to make use of nonlinear or time-varying

systems.

4.4.1 Generation of AM We shall discuss two methods of generating AM signals, one using a

nonlinear element and the other using an element with time-varying

characteristic.

a) Square law modulator Consider the scheme shown in Fig. 4.33(a).

Fig. 4.33 (a): A circuit with a nonlinear element

(b): v i− characteristic of the diode in Fig. 4.28(a)

A semiconductor diode, when properly biased has a v i− characteristic that

nonlinear, as shown in Fig. 4.33(b).



4.43

For fairly small variations of v around a suitable operating point, ( )v t2 can be

written as

( ) ( ) ( )v t v t v t22 1 1 2 1= α + α (4.10)

where 1α and 2α are constants.

Let ( ) ( ) ( )c cv t A f t m t1' cos 2= π + . Then,

( ) ( ) ( ) ( ) ( )c c c cv A m t f t m t m t A f tt 2 2 222 1 1 2 2

1

2' 1 cos 2 ( ) cos 2'α= α + π + α + α + α π

α

⎡ ⎤⎢ ⎥⎣ ⎦

(4.11)

The first term (on the RHS of Eq. 4.11) is ( )⎡ ⎤⎣ ⎦AMs t , with the carrier amplitude

c cA A1'= α and mg 2

1

2α=

α.

Now the question is: can we extract ( )⎡ ⎤⎣ ⎦AMs t from the sum of terms on

the RHS of Eq. 4.11? This can be answered by looking at the spectra of the

various terms constituting v t2( ) . Fig. 4.34 illustrates these spectra (quantities

marked A to E) for the ( )M f of Fig. 4.14(a). The time domain quantities

corresponding to A to E are listed below.

Fig. 4.34: Spectra of the components of ( )v t2 of Eq. 4.11



4.44

Component Spectrum indicated by

(i) ( )c cA f tcos 2π A

(ii) ( ) ( )c cA m t f t2'2 cos 2α π B

(iii) ( )m t1α C

(iv) ( )m t22α D

(v) ( )c cA f t2

22

' cos (2 )α π E

( )⎡ ⎤⎣ ⎦AMs t consists of the components (i) and (ii) of the above list. If it is possible

for us to filter out the components (iii), (iv) and (v), then the required AM signal

would be available at the output of the filter. This is possible by placing a BPF

with centre at cf and bandwidth 2W provided ( )cf W W2− > or cf W3> .

Usually, this is not a very stringent requirement. However, this scheme

suffers from a few disadvantages.

i) The required square-law nonlinearity of a given device would be available

only over a small part of the ( )v i− characteristic. Hence, it is possible to

generate only low levels of the desired output.

ii) If cf is of the order of 3W, then we require a BPF with very sharp cut off

characteristics. b) Switching modulator

In the first method of generation of the AM signals, we have made use of

the nonlinearity of a diode. In the second method discussed below, diode will be

used as a switching element. As such, it acts as a device with time-varying

characteristic, generating the desired AM signals when it is used in the circuit

configuration shown in Fig. 4.35.



4.45

Fig. 4.35: (a) Switching modulator

(b) Switching characteristic of the diode-load combination.

The (ideal) transfer characteristics of the diode-load combination is shown at (b)

in Fig. 4.35. This is explained as follows. We have,

( ) ( ) ( )v t c t m t1 = +

( ) ( )c cA f t m tcos 2= π +

If we assume that ( ) cm t Amax

<< , then the behavior of the diode is governed by

( )c t and can be approximated as

( )( ) ( )

( )v t c t

v tc t

12

, 0

0 , 0

⎧ >⎪⎨

≤⎪⎩

(That is, the diode offers infinite impedance when reverse biased and has zero

impedance, when forward biased. Hence, whether ( )v t1 is switched to the output

or not depends on the carrier cycle)

We can express ( )v t2 as

( ) ( ) ( )pv t v t x t2 1= (4.12)

where ( )px t is the periodic rectangular pulse train of example 1.1. That is,

( )( ) ( )( )( ) ( )( )p

c t c tx t

c t c t

if

if

1 , 0 positivehalf cycleof

0 , 0 negativehalf cycleof

⎧ >⎪= ⎨<⎪⎩

But from example 1.1 ( with =0 ),cf f



4.46

( )∞

π

= − ∞

= ∑ 2 cj n f tp n

nx t x e

where nnx c1 sin

2 2⎛ ⎞= ⎜ ⎟⎝ ⎠

(4.13a)

From Eq. 4.13(a), we find that 0nx = , for n 2, 4= ± ± etc. Combining the terms

− nx and nx , we obtain the trigonometric Fourier series, namely,

( ) ( ) ( ) ( )n

p c cn

x t f t n f tn

1

2

11 2 cos 2 cos 2 2 12 2 1

−∞

=

−⎡ ⎤= + π + π −⎣ ⎦π −∑ (4.13b)

From Eq. 4.12 and 4.13(b), we see that ( )v t2 is composed of two components,

namely,

a) The desired quantity: ( ) ( )⎡ ⎤+ π⎢ ⎥π⎣ ⎦

41 cos 22c

cc

A m t f tA

b) The undesired terms with

i) Impulses in spectra at c cf f f0, 2 , 4= ± ± etc.

ii) Spectral lobes (same in shape as M(f)) of width 2W, centered at

c cf f0, 3 , 5± ± etc.

As compared to the square law modulator, switching modulator has the following

advantages:

a) Generated AM signals can have larger power levels.

b) Filtering requirements are less stringent because we can separate the

desired AM signal if cf W2> .

However, the disadvantage of the method is that percentage modulation has to

be low in order that the switching characteristics of the diode are controlled only

by the carrier.



4.47

4.4.2. Generation of DSB-SC a) Product modulator

Generation of a DSB-SC signal involves the multiplication of ( )m t with

( )c cA tcos ω . To generate this signal in the laboratory, any of the commercially

available multipliers can be used. Some of them are:

National: LM 1496

Motorola: MC 1496

Analog Devices: AD 486, AD 632 etc

Signetics: 5596

The power levels that can be generated, the carrier frequencies that can be used

will depend on the IC used. The details can be obtained from the respective

manuals. Generally, only low power levels are possible and that too over a

limited carrier frequency range.

b) Ring modulator Consider the scheme shown in Fig. 4.36. We assume that the carrier

Fig. 4.36: Ring modulator

signal ( )c t is much larger than ( )m t . Thus ( )c t controls the behavior of diodes

which would be acting as ON-OFF devices. Consider the carrier cycle where the



4.48

terminal 1 is positive and terminal 2 is negative. T1 is an audio frequency

transformer which is essentially an open circuit at the frequencies near about the

carrier. With the polarities assumed for ( )c t , D1, D4 are forward biased, where as

D2, D3 are reverse biased. As a consequence, the voltage at point ‘a’ gets

switched to a' and voltage at point ‘b’ to b' . During the other half cycle of ( )c t ,

D2 and D3 are forward biased where as D1 and D4 are reverse biased. As a

result, the voltage at ‘a’ gets transferred to b' and that at point ‘b’ to a' . This

implies, during, say the positive half cycle of ( )c t , ( )m t is switched to the output

where as, during the negative half cycle, ( )m t− is switched. In other words,

( )v t can be taken as

( ) ( ) ( )pv t m t x t= (4.14)

where ( )px t is square wave as shown in Fig. 4.37.

Fig. 4.37: ( )px t of Eq. 4.14



4.49

Fig. 4.38: (a) A message waveform ( )m t

(b) ( )v t of the ring modulator

Fig. 4.38(b) illustrates the product quantity ( ) ( )pm t x t , for the ( )m t shown in

Fig. 4.38(a). The Fourier series expansion of ( )px t can be written as

( ) ( ) ( )n

p cn

x t n tn

12

1, 3, 5, ...

14 cos

−

=

−= ω

π ∑ .

When ( )v t is passed through a BPF tuned to cf , the output is the desired DSB-

SC signal, namely, ( ) ( ) ( )cs t m t t4 cos= ωπ

.

Example 4.6: Generation of DSB-SC Consider the scheme shown in Fig. 4.39. The non-linear device has the

input-output characteristic given by

( ) ( ) ( )y t a x t a x t30 1= +



4.50

Fig. 4.39: The scheme for the example 4.6

Let ( ) ( ) ( )x t A f t m t1cos 2= π + where ( )m t is the message signal. If the

required output ( )s t is a DSB-SC signal with a carrier frequency of 1 MHz, let us

find the value of f1, assuming that a suitable BPF is available.

( ) ( ) ( ) ( ) ( )y t a A f t m t a A f t m t 30 1 1 1cos 2 cos 2⎡ ⎤ ⎡ ⎤= π + + π +⎣ ⎦ ⎣ ⎦

1 2

( ) ( ) ( ) ( ) ( ) ( )a A f t m t A f t m t A f t m t3 3 3 2 2 21 1 11cos 2 3 cos 2 3 cos 2= π + + π + π⎡ ⎤⎣ ⎦2

In the equation for the quantity 2 above, the only term on the RHS that

can give rise to the DSB-SC signal is ( ) ( )a A m t f t2 21 13 cos 2π .

( ) ( ) ( )( ) f t

a A m t f t a A m t 12 2 21 1 1

1 cos 2 23 cos 2 3

2

⎡ ⎤+ π⎣ ⎦π =

Assume that the BPF will pass only the components centered around f12 . Then,

choosing f1 500= kHz, we will have

( ) ( ) ( )c cs t A m t f tcos 2= π

where cA a A213= and cf 1= MHz.



4.51

Exercise 4.4

Consider the circuit configuration (called Cowan modulator) shown in Fig.

4.40. Show that the circuit can produce at its output the DSB-SC signal. T1 is the

audio frequency transformer where as T2 and T3 are designed to operate around

the carrier frequency.

Fig. 4.40: Cowan modulator

Exercise 4.5: Balanced Modulator (BM)

Consider the scheme shown in Fig. 4.41. This configuration is usually

called a balanced modulator. Show that the output ( )s t is a DSB-SC signal,

thereby establishing that BM is essentially a multiplier.

Fig. 4.41: Balanced modulator (BM)



4.52

4.5 Envelope Detector As mentioned earlier, the AM signal, when not over modulated allows the

recovery of ( )m t from its envelope. A good approximation to the ideal envelope

detector can be realized with a fairly simple electronic circuit. This makes the

receiver for AM somewhat simple, there by making AM suitable for broadcast

applications. We shall briefly discuss the operation of the envelope detector,

which is to be found in almost all the AM receivers.

Consider the circuit shown in Fig. 4.42.

Fig. 4.42: The envelope detector circuit

We assume the diode D to be ideal. When it is forward biased, it acts as a short

circuit and thereby, making the capacitor C charge through the source

resistance sR . When D is reverse biased, it acts as an open circuit and C

discharges through the load resistance LR .

As the operation of the detector circuit depends on the charge and

discharge of the capacitor C , we shall explain this operation with the help of Fig.

4.43.



4.53

Fig. 4.43: Envelope detector waveforms

(a) ( )v t1 (before DC block)

(b) ( )outv t (after DC block)

If the time constants sR C and LR C are properly chosen, ( )v t1 follows the

envelope of ( )s t fairly closely. During the conduction cycle of D , C quickly

charges to the peak value of the carrier at that time instant. It will discharge a

little during the next off cycle of the diode. The time constants of the circuit will

control the ripple about the actual envelope. BC is a blocking capacitor and the

final ( )outv t will be proportional to ( )m t , as shown in Fig. 4.43(b). (Note that a

small high frequency ripple, at the carrier frequency could be present on ( )outv t .

For audio transmission, this would not cause any problem, as cf is generally

much higher than the upper limit of the audio frequency range).



4.54

How do we choose the time constants? sR , though not under our control

can be assumed to be fairly small. Values for LR and C can be assigned by us.

During the charging cycle, we want the capacitor to charge to the peak value of

the carrier in as short a time as possible. That is,

1s

cR C

f<< (4.15a)

Discharge time constant should be large enough so that C does not discharge

too much between the positive peaks of the carrier but small enough to be able

follow the maximum rate of change of m(t). This maximum rate depends on W,

the highest frequency in M(f). That is

Lc

R Cf W1 1

<< << (4.15b)

Too small a value for LR C will make ( )1V t somewhat ragged (sort of saw

tooth ripple on the top) where as, with too large value for LR C , ED fails to follow

the envelope during the periods when ( )m t is decreasing. A more accurate

analysis of the behavior of ED with ( )m t as a tone signal is given in appendix

A4.1.

Example 4.7

Consider the scheme shown in Fig. 4.44. ( )x t is a tone signal given by

( ) ( )x t t4cos 2 10⎡ ⎤= π ×⎣ ⎦ and ( ) ( )cc t f tcos 2= π with cf 10= MHz. ( )c t is the

HT of ( )c t . ( )v t , the output of the Balanced Modulator (BM), is applied as input

to an ideal HPF, with cutoff at 10 MHz. We shall find the expression for ( )y t , the

output of an ideal envelope detector.



4.55


It is not too difficult to see that

( ) ( )w t f t61 cos 2 10 102

⎡ ⎤= π + ∆⎣ ⎦ where f 0.01∆ = Hz. That is,

( ) ( ) ( ) ( ) ( )c cw t f t t f t t1 cos 2 cos sin 2 sin2⎡ ⎤⎡ ⎤ ⎡ ⎤= π ∆ ω − π ∆ ω⎣ ⎦ ⎣ ⎦⎣ ⎦

As ( ) ( ) ( )cz t w t f tsin 2⎡ ⎤= + π⎣ ⎦ , we have

( ) ( ) ( ) ( ) ( )c cz t f t t f t t1 1cos 2 cos sin 2 1 sin2 2

⎡ ⎤⎡ ⎤ ⎡ ⎤= π ∆ ω − π ∆ − ω⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

( )z t represents a narrowband signal with the in phase component

( )f t1 cos 22

⎡ ⎤π ∆⎣ ⎦ and the quadrature component ( )f t1 sin 2 12

⎧ ⎫⎡ ⎤π ∆ −⎨ ⎬⎣ ⎦⎩ ⎭. Hence,

( ) ( ) ( )y t f t f t

12 2

21 1cos 2 sin 2 14 2

⎧ ⎫⎪ ⎪⎡ ⎤⎡ ⎤ ⎡ ⎤= π ∆ + π ∆ −⎨ ⎬⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦⎪ ⎪⎩ ⎭

( ) t

1245 sin 2 10

4⎧ ⎫⎡ ⎤= − π ×⎨ ⎬⎣ ⎦⎩ ⎭

Example 4.8 Consider the scheme shown in Fig.4.45.


Let us find the output ( )y t when,



4.56

a) ( ) ( ) ( )m cx t g m t t1 cos⎡ ⎤= + ω⎣ ⎦ . ( )mg m t 1< and ( )m t is band-limited to

W Hz and the LPF has a bandwidth of W2 . Assume that cf W2>> .

b) ( )x t is a DSB-SC signal; that is ( ) ( ) ( )cx t m t tcos= ω .

a) ( ) ( ) ( ) ( )m cv t x t g m t t22 21 cos⎡ ⎤= = + ω⎣ ⎦

( ) ( )cm

tg m t 2 1 cos 2

12

⎡ ⎤+ ω⎣ ⎦⎡ ⎤= +⎣ ⎦

( ) ( ) ( )m m

cg m t g m t

t2 21 1

cos 22 2

⎡ ⎤ ⎡ ⎤+ +⎣ ⎦ ⎣ ⎦= + ω .

The second term on the RHS will be eliminated by the LPF. Hence,

( ) ( )mg m tw t

212

⎡ ⎤+⎣ ⎦= . As ( )mg m t1 0⎡ ⎤+ ≥⎣ ⎦ , we have

( ) ( )mg m ty t

12

⎡ ⎤+⎣ ⎦= .

b) When ( ) ( ) ( )cx t m t tcos= ω , we have

( ) ( ) ( ) ( ) ( )cc

tv t m t t m t2 2 2 1 cos 2

cos2

⎡ ⎤+ ω⎣ ⎦= ω =

The output of the LPF would be

( ) ( )m tw t

2

2=

As the squaring operation removes the information about the sign of the

signal, the output of ( )y t is

( ) ( )m ty t

2=



4.57

4.6 Theory of Single-Sideband Assume that from a DSB-SC signal, we have completely suppressed one

of the sidebands, say, the LSB. Let ( ) ( ) ( )2c

c cDSBAS f M f f M f f⎡ ⎤⎡ ⎤ = − + +⎣ ⎦ ⎣ ⎦

where ( )M f is as shown in Fig. 4.30(a). The resulting spectrum ( )⎡ ⎤⎣ ⎦USBS f will

be as shown in Fig. 4.47. Can we get back ( )m t from the above signal? The

answer is YES. Let

( ) ( ) ( )⎡ ⎤= × π⎣ ⎦ cos 2 cUSBv t s t f t

If ( ) ( )⎡ ⎤= ⎣ ⎦'

USBS f S f , then ( ) ( ) ( )1 ' '

2 c cV f S f f S f f⎡ ⎤= − + +⎢ ⎥⎣ ⎦

Exercise 4.6

Consider the waveform ( )m t shown in Fig. 4.46. A DSB-SC is

generated using ( )m t and a suitable high frequency carrier. Sketch the output

of an ideal envelope detector when the input to the detector is the DSB-SC

signal.

Fig. 4.46: ( )m t for the exercise 4.6



4.58

Fig 4.47: Spectrum of the upper sideband signal

By plotting the spectrum of ( )v t and extracting the spectrum for f W≤ , we see

that it is ( )12 cA M f . A similar analysis will show that it is possible to extract

( )m t from ( )⎡ ⎤⎣ ⎦LSBS f . In other words, with coherent demodulation, it is possible

for us to recover the message signal either from USB or LSB and the

transmission of both the sidebands is not a must. Hence it is possible for us to

conserve transmission bandwidth, provided we are willing to go for the

appropriate demodulation.

Let us now derive the time domain equation for an SSB signal. Let us start

with the two-sided spectrum and then eliminate the unwanted sideband. We shall

retain the upper sideband and try to eliminate the lower sideband. Consider

( ) ( ) ( )sgn2c

c c cA M f f f f M f f⎡ ⎤− + − −⎣ ⎦

But ( ) ( )( )( )

,sgn

,c c

c cc c

M f f f ff f M f f

M f f f f

⎧ − >⎪− − = ⎨− − <⎪⎩

Hence ( ) ( ) ( )⎧ − >⎪⎡ ⎤− + − = ⎨⎣ ⎦ <⎪⎩

,1 sgn

2 0 ,c c cc

c cc

A M f f f fA M f f f ff f

That is, the lower sideband has been eliminated from the positive part of the

spectrum.

( ) ( ) 2 cj f tcM f f m t e π− ←⎯→



4.59

( ) ( ) ( ) 21sgn cj f tc cf f M f f m t e

jπ− − ←⎯→ − ,

where ( )m t is the HT of ( )m t .

That is, ( ) ( ) ( ) ( ) 211 sgn2 2

cj f tc cc c

A AM f f f f m t m t ej

π⎡ ⎤⎡ ⎤− + − ←⎯→ −⎢ ⎥⎣ ⎦

⎣ ⎦ (4.16a)

Similarly, ( ) ( ) ( )⎧ + < −⎪⎡ ⎤+ − + = ⎨⎣ ⎦ > −⎪⎩

,1 1 sgn2 0 ,

c cc c

c

M f f f fM f f f f

f f

( ) ( ) ( ) ( )2 21 sgn2 2 2

c cj f t j f tc c cc c

A A AM f f f f m t e m t ej

− π − π⎡ ⎤+ − + ←⎯→ +⎣ ⎦ (4.16b)

Combining Eq. 4.16(a) and Eq. 4.16(b), we have the time domain equation for

the upper single sideband signal, namely,

( ) ( ) ( ) ( ) ( )cos sinc c cUSBs t A m t t m t t⎡ ⎤⎡ ⎤ = ω − ω⎣ ⎦ ⎣ ⎦ (4.17)

Assume that the USSB signal is obtained from the DSB-SC signal,

( ) ( )cosc cA m t tω , by filtering out the LSB part. Then,

( ) ( ) ( ) ( ) ( )cos sin2c

c cUSBAs t m t t m t t⎡ ⎤⎡ ⎤ = ω − ω⎣ ⎦ ⎣ ⎦ (4.18)

A few authors take Eq. 4.18 as representative of the SSB signal. Eq. 4.18

has the feature that the average power of the SSB signal is one-half the average

power of corresponding DSB signal. We shall make use of both Eq. 4.17 and

Eq. 4.18 in our further studies.

By a procedure similar to that outlined above, we can derive a time

domain expression for the LSB signal. The result would be

( ) ( ) ( ) ( ) ( )⎡ ⎤⎡ ⎤ = ω + ω⎣ ⎦ ⎣ ⎦cos sinc c cLSBs t A m t t m t t or (4.19a)

( ) ( ) ( ) ( ) ( )cos sin2c

c cLSBA

s t m t t m t t⎡ ⎤⎡ ⎤ = ω + ω⎣ ⎦ ⎣ ⎦ (4.19b)

An SSB signal, whether upper or lower, is also a narrowband bandpass signal.

Eq. 4.18 can be treated as the canonical representation of USB signal with ( )m t

as the in-phase component and ( )m t as the quadrature component. Similarly



4.60

Eq. 4.19 provides the canonical representation of the LSB signal where ( )m t is

the in-phase component and ( )m t− is the quadrature component.

We have already seen that a narrowband signal can also be expressed in

the envelope and phase form. Taking the USB signal, we have for the complex

envelope the quantity ( ) ( )( )2cA

m t j m t+ . Hence the envelope ( )A t of the USB

signal is

( ) ( ) ( )22

2c

USBAA t m t m t⎡ ⎤ = +⎣ ⎦ (4.20a)

Similarly for the phase ( )tϕ , we have

( ) ( )( )

arc tanm t

tm t

⎡ ⎤ϕ = ⎢ ⎥

⎢ ⎥⎣ ⎦ (4.20b)

Expressing the USB signal with the envelope-phase form, we have

( ) ( ) ( ) ( )cos cUSBs t A t t t⎡ ⎤⎡ ⎤ = ω + ϕ⎣ ⎦ ⎣ ⎦ (4.21)

where ( )A t and ( )tϕ are given by Eqs. 4.20(a) and 4.20(b) respectively. The

expression for ( ) LSBs t⎡ ⎤⎣ ⎦ is identical to Eq. 4.21 but with ( )tϕ given by

( ) ( )( )

tanm t

t arcm t

⎡ ⎤ϕ = −⎢ ⎥

⎢ ⎥⎣ ⎦ (4.22)

That is, ( ) ( ) ( )( )cos cSSBs t A t t t⎡ ⎤ = ω + ϕ⎣ ⎦ (4.23)

where ( )tϕ is given either by Eq. 4.20(b) or Eq. 4.22. Eq. 4.23 indicates that an

SSB signal has both amplitude and phase variations. (AM and DSB-SC signals

have only the amplitude of the carrier being changed by the message signal.

Note that AM or DSB-SC signals do not have quadrature components.) As such,

SSB signals belong to the category of hybrid amplitude and phase modulation.



4.61

Example 4.9: SSB with tone modulation

As a simple application of the Eqs. 4.18 and 4.19, let ( )m t be ( )cos m tω .

Let us find the SSB signals.

( ) ( ) ( ) ( )m mm t t m t tcos sin= ω ⇒ = ω . Therefore,

( ) ( ) ( ) ( ) ( )cos cos sin sin2c

m c m cUSBA

s t t t t t⎡ ⎤⎡ ⎤ = ω ω − ω ω⎣ ⎦ ⎣ ⎦

( )cos2c

c mA

t⎡ ⎤= ω + ω⎣ ⎦

( ) ( ) ( ) ( ) ( )cos cos sin sin2c

m c m cLSBA

s t t t t t⎡ ⎤⎡ ⎤ = ω ω + ω ω⎣ ⎦ ⎣ ⎦

( )cos2c

c mA

t⎡ ⎤= ω − ω⎣ ⎦

Alternatively,

( ) ( ) ( )cos cosc m cDSB SCs t A t t

−⎡ ⎤ = ω ω⎣ ⎦

( ) ( )cos cos2c

c m c mA

t t⎡ ⎤= ω + ω + ω − ω⎣ ⎦

Extracting the USB, we have

( ) ( )cos2c

c mUSBA

s t t⎡ ⎤⎡ ⎤ = ω + ω⎣ ⎦ ⎣ ⎦

If we eliminate the USB, then

( ) ( )cos2c

c mLSBA

s t t⎡ ⎤⎡ ⎤ = ω − ω⎣ ⎦ ⎣ ⎦

Example 4.10

Let ( ) ( ) ( )m t x t y t= where ( )X f and ( )Y f are as shown in Fig. 4.48.

An LSB signal is generated using ( )m t as the message signal. Let us develop

the expression for the SSB signal.



4.62

Fig. 4.48: ( )X f and ( )Y f of example 4.10

Let us take the SSB signal as

( ) ( ) ( ) ( ) ( )c cLSBs t m t t m t tcos sin⎡ ⎤ = ω + ω⎣ ⎦ (4.24)

We have

( ) ( )x t c t3 32 10 sin 2 10⎡ ⎤= × ×⎣ ⎦

( ) ( ) ( )y t c t t3 2 310 sin 500 cos 4 10⎡ ⎤= π ×⎣ ⎦

( ) ( ) ( )m t x t y t= (4.25a)

What is required is ( )m t , the HT of ( )m t . ( )m t is the product of a lowpass and

a bandpass signal. Hence ( ) ( ) ( )m t x t y t= . (See the note, after example 1.25)

But ( )y t , from the result of example 1.25, is

( ) ( ) ( )y t c t t3 2 310 sin 500 sin 4 10⎡ ⎤= π ×⎣ ⎦

That is,

( ) ( ) ( ) ( )m t c t c t t6 3 2 32 10 sin 2 10 sin 500 sin 4 10⎡ ⎤ ⎡ ⎤= × × π ×⎣ ⎦ ⎣ ⎦

(4.25b)

( ) LSBs t⎡ ⎤⎣ ⎦ is obtained by using Eq. 4.25 in Eq. 4.24.



4.63

Exercise 4.7

Let ( )M f be as shown in Fig. 4.49. An upper sideband signal is

generated using the signal with this ( )M f . Compute and sketch the spectrum

of the quadrature component of the SSB signal.

Fig. 4.49: Baseband spectrum for the Exercise 4.7

Exercise 4.8

Let ( ) ( )m t c tsin= . Develop the expression for the following:

a) USB signal, in the canonical form.

b) USB signal, in the envelope and phase form

Ans. (b): ctc f t1sin cos 22 4

⎡ ⎤⎛ ⎞ ⎛ ⎞π +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

Exercise 4.9

Let ( ) ( ) ( ) ( ) ( )c cs t m t t m t tcos sin= ω − ω

where ( )M f is as shown in Fig. 4.49. Let W 5= kHz and ( )M 0 1= .

a) Sketch the spectrum of (i) ( ) ( )cs t tcos ω and (ii) ( ) ( )cs t tsin ω

b) Show that sum of the spectra of part (a) is proportional to ( )M f

c) Sketch the spectrum of ( ) ( ) ( ) ( )c cs t t s t tcos sinω − ω . Is this related to

( )M f ? Explain.



4.64

4.7 Generation of SSB Signals We shall consider two broad categories of SSB generation, namely, (i)

frequency discrimination method and (ii) phase discrimination method. The

former is based on the frequency domain description of SSB, whereas the latter

in based on the time-domain description of an SSB signal.

4.7.1 Frequency discrimination method Conceptually, it is a very simple scheme. First generate a DSB signal and

then filter out the unwanted sideband. This method is depicted in Fig. 4.50.

Fig. 4.50: Frequency discrimination method of SSB generation

( )v t is the DSB-SC signal generated by the product modulator. The BPF is

designed to suppress the unwanted sideband in ( )V f , thereby producing the

desired SSB signal.

As we have already looked at the generation of DSB-SC signals, let us

now look at the filtering problems involved in SSB generation. BPFs with abrupt

pass and stopbands cannot be built. Hence, a practical BPF will have the

magnitude characteristic ( )H f , as shown in Fig. 4.51. As can be seen from the

figure, ( ( )H f is shown only for positive frequencies) a practical filter, besides the

PassBand (PB) and StopBand (SB), also has a TransitionBand (TB), during

which the filter transits from passband to stopband. (The edges of the PB and SB

depend on the attenuation levels used to define these bands. It is a common

practice to define the passband as the frequency interval between the 3-dB



4.65

points. Attenuation requirements for the SB depend on the application. Minimum

attenuation for the SB might be in the range 30 to 50 dB.)

Fig. 4.51: Magnitude characteristic of a practical BPF. Centre frequency 0f=

Because of TB (where the attenuation is not to the desired level), a part of

the undesired sideband may get through the filter. As a rule of the thumb, it is

possible to design a filter if the permitted transitionband is not less than 1% of

center frequency of a bandpass filter. Fortunately, quite a few signals have a

spectral null around DC and if it is possible for us to fit in the transitionband into

this gap, then the desired SSB signal could be generated. In order to accomplish

this, it might become necessary to perform the modulation in more than one

stage. We shall illustrate this with the help of an example.

Example 4.11 Telephone quality speech signal has a spectrum in the range 0.3 to 3.4

kHz. We will suggest a scheme to generate upper sideband signal with a carrier

frequency of 5 MHz. Assume that bandpass filters are available, providing an

attenuation of more than 40 dB in a TB of width 0.01 0f , where 0f is the centre

frequency of the BPF.

Let us look at the generation of the SSB signal in one stage using a carrier

of 5 MHz. When a DSB signal is generated, it will have a spectral null of 600 Hz

centered at 5 MHz. That is, the transitionband is about 0.01 percent of the carrier



4.66

and hence it would not be possible to design such a sideband filter. However, it

would be possible to generate the desired SSB signal using two stages of

modulation.

Consider the scheme shown in Fig. 4.52.

Fig. 4.52: Two stage generation of SSB signal

( )1v t is a DSB-SC signal with the USB occupying the (positive frequency) range

( )1 300cf + Hz to ( )1 3400cf + Hz. The frequency range of the LSB is

( )1 3400cf − Hz to ( )1 300cf − Hz. Let us extract the upper sideband from ( )1v t

with the help of BPF1. Then the centre frequency of BPF1, ( )0 1f , is

( ) ( ) ( )1 10 1

300 34002

c cf ff

+ + +=

( )1 1850cf= +

width of the spectral null around 1 600cf = Hz.

Hence 1 1850 600100

cf +≤

or 1 60,000 1850cf ≤ −

58.1≤ kHz

Let us take 1cf as 50 kHz. Let ( )M f be as shown in Fig. 4.53(a). Then ( )2V f will

be as shown in Fig. 4.53(b).



4.67

Fig. 4.53: (a) Message spectrum

(b) The spectrum of the USB signal with 1 50cf = kHz

( )3v t is a DSB-SC signal with the carrier 2cf , ( )2v t being the modulating signal.

Then ( )3V f will be as shown in Fig. 4.54. In this figure,

Fig. 4.54: Spectrum of ( )3v t of Fig. 4.52

( )1 2 53,400cf f= − Hz

( )2 2 50,300cf f= − Hz

( )3 2 50,300cf f= + Hz

( )4 2 53,400cf f= + Hz



4.68

Hence the transition band available to design the sideband filter is

( )3 2 100.6f f− = kHz. With this TB, we can choose centre frequency of BPF2

less than or equal to 10.06 MHz. If we choose 2cf as 4.95 MHz then we will have

the upper sideband occupying frequency range (4.95 + 0.0503) = 5.0003 MHz to

(4.95 + 0.0534) = 5.0034 MHz. This is exactly what would have happened if the

modulation scheme was attempted in one step with 5 MHz as the carrier.

Note: As the spectral occupancy of the USB signal is from 5.0003 MHz to 5.0034

MHz, theoretical centre frequency of the BPF2 is 5.00185. With respect to this

frequency, we have

TB width 100.6 100 2.01 percentcentre freq. 5001.85

= × =

which is about twice the permitted ratio. Hence, it is possible to choose for 1cf a

value lower than 50 kHz.

4.7.2 Phase discrimination method This method implements Eq. 4.17 or Eq. 4.19(a), to generate the SSB

signal. Consider the scheme shown in Fig. 4.55. This scheme requires two

product modulators, two 2π phase shifters and an adder. One of the phase shifter

is actually a Hilbert transformer (HT); it should provide a 2π phase shift for all the

components in ( )M f . This is not such an easy circuit to realize. Assuming it is

possible to build the HT, the SSB can be generated for any cf , provided the

product modulators (multipliers) can work at these frequencies.



4.69

Fig. 4.55: SSB generation: Phase discrimination method

Instead of a single wide band phase shifter acting as the HT, it is possible

to have an SSB generator with two Phase Shifting Networks, (PSN), one in each

branch as shown in Fig. 4.56.

Fig. 4.56: An alternate configuration for the phase discrimination scheme

( )1H f and ( )2H f are the phase shifting networks. Let ( ) ( )11

j fH f e θ= and

( ) ( )22

j fH f e θ= . ( )1 fθ and ( )2 fθ are such that ( ) ( )1 2 2f f π

⎡ ⎤θ − θ =⎣ ⎦ for the

frequency range of interest. That is, PSN1 and PSN2 maintain a constant

difference of 2π .



4.70

Let us explain the operation of the scheme shown in Fig. 4.56 by taking

( )m t to be a tone signal; that is, ( ) ( )cosm mm t A t= ω . Let

( )1 1mf f

f=

θ = θ and ( )2 2mf f

f=

θ = θ

where 2 1 2π

θ = θ + . Then,

( ) ( )1 1cosm mv t A t= ω + θ and ( ) ( )2 2cosm mv t A t= ω + θ .

( ) ( ) ( )3 1cos cosm c m cv t A A t t= ω + θ ω

( ) ( ) ( )4 2cos sinm c m cv t A A t t= ω + θ ω

( )1cos sin2m c m cA A t tπ⎛ ⎞= ω + θ + ω⎜ ⎟

⎝ ⎠

( ) ( )1sin sinm c m cA A t t= − ω + θ ω

( ) ( ) ( )3 4 1cosm c c mv t v t A A t⎡ ⎤+ = ω + ω + θ⎣ ⎦

After coherent demodulation, we will have ( )1cos m tω + θ . We shall assume that

the additional phase shift 1θ which is actually frequency dependent will not cause

any problem after demodulation.

As it is not too difficult to design a Hilbert transformer using digital filter

design techniques, phase shift method of SSB generation is better suited for

digital implementation. For a brief discussion on SSB generation using digital

signal processing, the reader is referred to [1].



4.71

4.8 Demodulation of SSB SSB signals can be demodulated using coherent demodulation as shown

in Fig. 4.58.

Exercise 4.10 There is a third method of generating the SSB signal, known as Weaver’s method. This scheme is shown in Fig. 4.57.

Fig. 4.57: Weaver’s method of SSB generation

Let ( )M f be non-zero only in the interval l uf f f≤ ≤ ; and let 1 2l uf f

f+

= .

The lowpass filters in the I and Q channels are identical and have a cutoff

frequency of 0 2u l

cf f

f−

= . Assume that 2 0cf f>> . By sketching the spectra at

various points in the above scheme, show that ( )s t is an SSB signal. What is

the actual carrier frequency with respect to which, ( )s t would be an SSB

signal?



4.72

Fig. 4.58: Coherent demodulation of SSB

The received SSB signal is multiplied by the local carrier which is of the same

frequency and phase as the carrier used at the transmitter. From Fig, 4.58, we

have

( ) ( ) ( )' cosc cSSBv t s t A t⎡ ⎤= × ω⎣ ⎦

( ) ( ) ( ) ( ) ( )'cos sin cos2c

c c c cA

m t t m t t A t⎡ ⎤= ω ± ω × ω⎣ ⎦

( ) ( ) ( ) ( ) ( )2'

cos cos sin2

c cc c c

A A m t t m t t t⎡ ⎤= ω ± ω ω⎣ ⎦ (4.26)

The second term on the RHS of Eq. 4.26 has the spectrum centered at 2 cf±

which will be eliminated by the LPF following ( )v t . The first term of Eq. 4.26 can

be written as, ( ) ( )' 1 cos2

2 2cc c tA A m t

⎡ ⎤+ ω⎢ ⎥⎣ ⎦

.

As ( ) ( )cm t tcos 2ω has the spectrum centered at 2 cf∓ , even this will be

eliminated by the LPF. Hence ( ) ( )0

'

4c cA Av t m t= . That is,

( ) ( )0v t m tα

The difficulty in demodulation is to have a coherent carrier at the receiver. Can

we use a squaring loop or Costas loop to recover the carrier from ( ) SSBs t⎡ ⎤⎣ ⎦ ?

The answer is NO. Let us look at the squaring loop. From Eq. 4.23,

( ) ( ) ( ) ( )cos cSSBs t s t A t t t⎡ ⎤ ⎡ ⎤= = ω + ϕ⎣ ⎦ ⎣ ⎦

After squaring, we obtain,



4.73

( ) ( ) ( )2 2 2cos cs t A t t t⎡ ⎤= ω + ϕ⎣ ⎦

( ) ( )( ) 2 21 cos 2 cA t t t⎡ ⎤= + ω + ϕ⎣ ⎦

As ( )tϕ is a function of time, we do not have a discrete component at 2 cf f=

and hence, carrier acquisition is not possible. It is left as an exercise to show that

Costas loop will not be able to put out ( )m t when the input to the loop is the SSB

signal. Hence, when SSB is used in a communication system, highly stable

crystal oscillators are used both at the transmitter and receiver. If this scheme

does not work (especially, at very high frequencies) a small pilot carrier can be

sent along with the SSB signal. This pilot carrier is recovered at the receiver and

is used for demodulation.

Let us now look at the effects of frequency and phase offset in the carrier

used for demodulation. Let the carrier term at the receiver be

( )' cos 2c cA f f t⎡ ⎤π + ∆⎣ ⎦ . Let the received input to the demodulator (Fig. 4.58) be

( ) ( ) ( ) ( )1 cos sin2 c c cA m t t m t t⎡ ⎤ω − ω⎣ ⎦

Then, ( ) ( ) ( ) ( ) ( ) ( )⎡ ⎤= ω − ω ω + ∆ω⎣ ⎦1 ' cos sin cos2 c c c c cv t A A m t t m t t t

( )0v t , the output of the LPF would be

( ) ( ) ( ) ( ) ( )01 ' cos 2 sin 24 c cv t A A m t f t m t f t⎡ ⎤α π∆ + π∆⎣ ⎦ (4.27)

Assume that

( ) ( ) ( ) ( ) ( )0 cos 2 sin 2v t m t f t m t f t= π∆ + π∆ (4.28)

Consider a special case, namely, a frequency component at 1f = kHz in ( )M f

and 100f∆ = Hz. With these values, Eq. 4.28 becomes

( ) ( ) ( ) ( ) ( )3 30 cos 2 10 cos 2 100 sin 2 10 sin 2 100v t t t t t= π × π × + π × π ×

( )2 900 2 900

cos 2 9002

j t j te etπ × − π ×+

= π × = (4.29)



4.74

As this is true of every frequency component in ( )M f , we have the result that

when f∆ is positive and the input is a USB signal, there is an inward shift of

( )M f by f∆ . (We see from Eq. 4.28, 2 1000j te π × is converted to 2 900j te π × and

2 1000j te− π × is converted to 2 900j te− π × . That is, both the spectral components

have been shifted inward by 100 Hz.) By a similar analysis, we can easily see

that if f∆ is negative and the input is a USB signal, then, after demodulation, the

spectral components in ( )M f would undergo an outward shift by f∆ . In all, we

have four cases to be taken into account and the effects of non-zero f∆ on the

resulting output after demodulation are summarized below.

Case i) 0f∆ > and the input signal is USB: Spectral components in ( )M f will

undergo an inward shift by f∆

Case ii) 0f∆ > and the input signal is LSB: Spectral components in ( )M f will

undergo an outward shift by f∆

Case iii) 0f∆ < and the input signal is USB: Spectral components in ( )M f will

undergo an outward shift by f∆

Case iv) 0f∆ < and the input signal is LSB: Spectral components in ( )M f will

undergo an inward shift by f∆

Let ( )M f be as shown in Fig. 4.59(a). Let 300f∆ = Hz. Then, if the input

is a USB signal, the spectrum of the demodulated output, ( )0V f , would be as

shown in Fig. 4.59(b).



4.75

Fig. 4.59: (a) Baseband message spectrum

(b) Inward spectral shift (after demodulation) of a

1 USB signal. Frequency offset, 300f∆ = Hz



4.76

Example 4.12

Let ( ) ( ) ( ) ( )m t A t A t A t1 2 3cos 200 cos 1100 cos 4000= π + π + π . An

upper sideband signal is generated using this ( )m t . The carrier used for

demodulation had a positive offset of 150 Hz. Let us find the frequencies of the

spectral components after demodulation.

As the received signal is USB and f 0∆ > , there would be an inward shift

of the frequency components by 150 Hz. Spectral components in ( )M f are

shown in Fig. 4.61(a).

Exercise 4.11: Effect of phase error on SSB Consider the scheme shown in Fig. 4.60. Let

( ) ( ) ( ) ( ) ( )cos sinc cs t m t t m t t⎡ ⎤= ω + ω⎣ ⎦ . The carrier used to demodulate has

a phase difference of θ with respect to the carrier used for modulation. Show

that ( )0v t has phase distortion (when compared to ( )m t ) by establishing

( )( )( )

0, 0

, 0

j

j

M f e fV f

M f e f

θ

− θ

⎧ >⎪= ⎨<⎪⎩

Fig. 4.60: SSB demodulation with carrier phase difference

Hint: Show that ( ) ( ) ( )v t m t m t0 cos sin⎡ ⎤α θ + θ⎣ ⎦



4.77

Fig. 4.61: (a) message spectrum

(b) Spectrum after demodulation with frequency error

Note that the negative frequency components have been shown with a broken

line. Let ( )m t be the demodulated output. After demodulation, there would be an

inward shift by 150 Hz and this is shown in Fig. 4.61(b). From this spectrum, we

see that ( )m t' is consisting of components at 50 Hz, 350 Hz and 1850 Hz.

The speech files that follow provide a qualitative measure of the distortion

caused by the frequency offset of the local carrier in the demodulation of SSB

signals. Input to the demodulator is a USB signal.

Introduction

Output 1

Output 2a

Output 2b

Output 2c



4.78

Output 3a

Output 3b

Output 3c

Output 4a

Output 4b

Output 4c

After listening to these speech outputs, one gets the feeling that, for a

given frequency offset, SSB performs better than the DSB. Even in SSB, outward

frequency shift of the original message spectrum ( f∆ negative for USB) has

better clarity than the corresponding inward shift. Of course, voice tends to

become somewhat shrill, which, of course, is expected.



4.79

Exercise 4.12: Quadrature Carrier Multiplexing (QCM) It is possible to transmit two DSB-SC signals with in a bandwidth of 2 W ,

by using a scheme called QCM, as shown in Fig. 4.62. (QCM is also referred to

as Quadrature Amplitude Modulation (QAM) or simply quadrature modulation.

Using QAM, we are able to achieve the BW efficiency of SSB modulation.)

Fig. 4.62: Quadrature carrier multiplexing scheme

Two message signals ( )1m t and ( )2m t are used to generate two DSB-SC

signals, ( )1v t and ( )2v t respectively. The carriers used in generating ( )1v t and

( )2v t are in phase quadrature. The transmitted signal ( ) ( ) ( )1 2s t v t v t= + . At

the receiver, coherent demodulation is used to recover the two messages ( )1m t

and ( )2m t .

a) Show that ( )1m t and ( )2m t will be recovered by the receiver shown.

b) Let the local carrier have some frequency and phase offset; that is, instead

of ( )2 cos 2 cf tπ , let it be ( )2 cos 2 cf f t⎡ ⎤π + ∆ + ϕ⎣ ⎦ . Then show that the

output of the upper branch is

( ) ( ) ( ) ( ) 1 2cos 2 sin 2cA m t f t m t f t⎡ ⎤ ⎡ ⎤π ∆ + ϕ − π ∆ + ϕ⎣ ⎦ ⎣ ⎦

where as the output of the lower branch is

( ) ( ) ( ) ( ) 2 1cos 2 sin 2cA m t f t m t f t⎡ ⎤ ⎡ ⎤π ∆ + ϕ + π ∆ + ϕ⎣ ⎦ ⎣ ⎦

Note: We see from the above result that the carrier phase and frequency have to

be fairly accurate to have proper demodulation; otherwise ( )1m t will interfere with

( )2m t and vice versa. This is called cochannel interference. QAM is used in

color TV for multiplexing the chrominance signals.



4.80

4.9 Vestigial SideBand (VSB) Modulation One of the widespread applications of VSB has been in the transmission

of picture signals (video signals) in TV broadcast. The video signal has the

characteristic that it has a fairly wide bandwidth (about 5 MHz) with almost no

spectral hole around DC. DSB modulation, though somewhat easy to generate,

requires too much bandwidth (about 10 MHz) where SSB, though bandwidth

efficient, is extremely difficult to generate, as explained below.

With analog circuitry it is very difficult to build the 2π phase shifter over a 5

MHz bandwidth; as such phase shift discrimination method is not feasible. To

make use of the frequency discrimination method, we require very sharp cutoff

filters. Such filters have a highly non-linear phase characteristic at the band

edges and spectral components around the cut-off frequencies suffer from phase

distortion (also called group delay distortion). The human eye (unlike the ear)

being fairly sensitive to phase distortion, the quality of the picture would not be

acceptable.

VSB refers to a modulation scheme where in the wanted sideband (either

USB or LSB) is retained almost completely; in addition, a vestige (or a trace) of

the unwanted sideband is added to the wanted sideband. This composite signal

is used for transmitting the information. This vestige of the wanted sideband

makes it possible to come up with a sideband filter that can be implemented in

practice.

4.9.1 Frequency domain description of VSB Figure 4.63 depicts the scheme of VSB generation. In this figure, ( )v t is a

DSC-SC signal, which is applied as input to a Sideband Filter (SBF), ( )vH f , that

shapes ( )V f so that ( )s t is a VSB signal.



4.81

Fig. 4.63: Generation of VSB using the filtering method

Now the questions that arise are: what is the shape of the SBF and how

do we demodulate such a signal. As coherent demodulation is fairly general, let

us try to demodulate the VSB signal as well, using this method. In the process of

demodulation, we shift the modulated carrier spectrum (bandpass spectrum) up

and down by cf and then extract the relevant baseband. Because of the vestige

of the unwanted sideband, we expect some overlap (in the baseband) of the

shifted spectra. In such a situation, overlap should be such that, ( )M f is

undistorted for f W≤ . In other words, ( ) ( )v c v cH f f H f f− + + should result in

a filter with a rectangular passband within the frequency range ( )toW W− .

With a little intuition, it is not too difficult to think of one such ( )vH f .

Assume that we are retaining the USB almost completely and permitting

the vestige of the LSB. Consider the ( ) ( ) ( )vj fv vH f H f e θ= shown in Fig. 4.64.

We assume the phase characteristic ( )v fθ to be linear over the frequency range

l cf f f W≤ ≤ + with ( ) ( )v c v cf f m2θ = − θ − = − π , where m is an integer.

If a DSB signal is given as input to the above ( )vH f , it will partially

suppress USB (in the frequency range c uf f f≤ ≤ ) and allow the vestige of the

LSB (from l cf f f≤ ≤ ). The demodulation scheme is the same as shown in Fig.

4.13, with the input to the detector being the VSB signal. ( )0V f , the FT of the

output of the detector, is



4.82

( ) ( ) ( ) ( )v c v cV f K M f H f f H f f0 1 ⎡ ⎤= − + +⎣ ⎦ , for f W≤ .

where K1 is the constant of proportionality.

Fig. 4.64: An example of a SBF generating VSB

If ( ) ( ) dj f tv c v cH f f H f f K e 2

2− π⎡ ⎤− + + =⎣ ⎦ , f W≤ ,

where dt determines the slope of the phase characteristic and K2 is a constant,

then ( ) ( )0 dv t K m t t= − , where K K K1 2= .

Let ( ) ( ) ( ) ( )vj fv v v c vH f H f e H f f W f f1,

1, 1, ,θ= = − − ≤ ≤

where v c lf f f= − , is the width of the vestige in LSB.

For the ( )vH f shown in Fig. 4.64, ( )1, vH f is as shown in Fig. 4.65.



4.83

Fig. 4.65: ( )v cH f f− for f W≤

(a) magnitude characteristics

(b) phase characteristics

Let ( ) ( ) ( ) ( )2,2, 2,

vj fv v v cH f H f e H f fθ= = + for vf f W− ≤ ≤ . ( )2, vH f is

shown in figure 4.66.



4.84

Fig. 4.66: ( )v cH f f+ for f W≤

(a) magnitude characteristics

(b) phase characteristics

As ( )vH f1, 0= for vf f W≤ ≤ and j me 2 1± π = , we can write

( ) ( ) 21, ,dj f t

v c vH f f H f e f W− π− = ≤ (4.30a)

By a similar reasoning,

( ) ( ) 22, ,dj f t

v c vH f f H f e f W− π+ = ≤ (4.30b)

Therefore,

( ) ( ) ( ) ( ) 21, 2, ,dj f t

v c v c v vH f f H f f H f H f e f W− π⎡ ⎤− + + = + ≤⎣ ⎦ (4.30c)

Summing ( )vH f1, and ( )vH f2, , we have the ideal LPF with unity gain for

f W≤ .

Let us take a closer look at the sideband filter for VSB. The magnitude

characteristic shown in Fig. 4.64(a) can be taken as the sum of ( ) ( ) ( ) ( )1 2v vH f H f+



4.85

where ( ) ( )1vH f and ( ) ( )2

vH f are shown in Fig. 4.67(a) and (b) respectively.

Fig. 4.67: Decomposition of the ( )vH f of Fig. 4.64

Note that with ( ) ( )vH f1 alone for the sideband filter, we would have generated the

SSB signal. It is the addition of ( ) ( )vH f2 to ( ) ( )vH f1 that gives rise to VSB output.

Consider f 0> ; ( ) ( )vH f2 consists of two straight line segments, one between the

points ( )a b, and the other between the points ( )c d, . Similar is the case for

f 0< . Now the question is: should ( ) ( )vH f2 consist of only straight line

segments? The answer is NO. What is required is that ( ) ( )vH f2 exhibit odd

symmetry about cf± . Two more such characteristics have been shown in fig.

4.68, for f 0> .



4.86

Fig. 4.68: Two more examples of ( ) ( )vH f2

As such, we have a great deal of flexibility in designing ( )vH f . (Subject to some

upper limit, there could be choice in fixing vf , the width of the vestige.) These

features facilitate the design and implementation of a practical sideband filter.

4.9.2 Time domain description of VSB Let ( )vh t denote the impulse response of the sideband filter, ( )vH f . From

Fig. 4.63, we have

( ) ( ) ( ) ( )vVSBs t s t h v t d

∞

− ∞

⎡ ⎤ = = τ − τ τ⎣ ⎦ ∫

But ( ) ( ) ( )c cv t A m t tcos= ω . For convenience, let cA 1= .

Then, ( ) ( ) ( ) ( )v cs t h m t t dcos∞

− ∞

⎡ ⎤= τ − τ ω − τ τ⎣ ⎦∫



4.87

( ) ( ) ( ) ( ) ( ) ( )v c c c ch m t t t dcos cos sin sin∞

− ∞

⎡ ⎤= τ − τ ω ω τ + ω ω τ τ⎣ ⎦∫

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

v c c

v c c

h m t d t

h m t d t

cos cos

sin sin 4.31

∞

− ∞

∞

− ∞

⎡ ⎤⎢ ⎥= τ − τ ω τ τ ω⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥− − τ − τ ω τ τ ω⎢ ⎥⎣ ⎦

∫

∫

Eq. 4.31 is in the canonical form for the representation of narrowband

signal. Let ( )cm t denote the in-phase component of ( )s t and ( )sm t , the

quadrature component. Then,

( ) ( ) ( ) ( )c v cm t h m t dcos∞

− ∞

= τ − τ ω τ τ∫ (4.32a)

and ( ) ( ) ( ) ( )s v cm t h m t dsin∞

− ∞

⎡ ⎤⎢ ⎥= − τ − τ ω τ τ⎢ ⎥⎣ ⎦

∫ (4.32b)

Then, ( ) ( ) ( ) ( )c c s cVSBs t m t t m tcos sin⎡ ⎤ = ω − ω⎣ ⎦ (4.33)

Eq. 4.33 is the canonical representation of the VSB signal.

Let ( ) ( ) ( )i v ch t h t tcos= ω and ( ) ( ) ( )q v ch t h t tsin= − ω . Then,

( ) ( ) ( )c im t m t h t= ∗ (4.34a)

( ) ( ) ( )s qm t m t h t= ∗ (4.34b)

Taking the FT of Eq. 4.34, we have

( ) ( ) ( )c iM f M f H f= and ( ) ( ) ( )s qM f M f H f=

If ( )M f 0= for f W> , then ( )cM f and ( )sM f are bandlimited to atmost W .

This implies that ( )cm t and ( )sm t are lowpass signals, as required.

As ( ) ( ) ( )i v ch t h t tcos= ω we have,

( ) ( ) ( )v c v ci

H f f H f fH f

2− + +

= (4.35a)



4.88

Similarly, ( ) ( ) ( )v c v cq

H f f H f fH f

j2+ − −

= (4.35b)

But ( ) ( )v c v cH f f H f f 1⎡ ⎤− + + =⎣ ⎦ , for f W≤ . Hence, ( ) ( )c

m tm t

2= .

Let us look at ( )qH f .

From Eq. 4.35(b), ( ) ( ) ( )q

v c v cH f

H f f H f fj

2= − − +

Let ( )vH f be as shown in Fig. 4.69.

Fig. 4.69: ( )vH f with vestige in LSB

Fig. 4.70: ( )qH f

j2

(solid line) for the ( )vH f of Fig. 4.69

Then, ( )qH f

j2

for f W≤ will be as shown in Fig. 4.70.



4.89

Based on Eq. 4.33, we have another implementation for the generation of

the VSB signal called the phase discrimination method of generating VSB. This

scheme is shown in Fig. 4.71. Plus sign at the last summer will result in a VSB

signal with a vestige in LSB whereas the minus sign will result in a VSB signal

with the vestige in USB for the ( )qH f shown in Fig. 4.70.

Fig. 4.71: Phase discrimination method of VSB generation

Comparing the scheme shown in Fig. 4.71 with that shown in Fig. 4.55 for the

generation of SSB, we find a close resemblance between them. When ( )qh t is a

HT (with the magnitude response equal to 1/2), we have ( ) ( )VSB SSBs t s t⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ .

If ( )qh t 0= , we have ( ) ( )VSB DSBs t s t⎡ ⎤ ⎡ ⎤=⎣ ⎦ ⎣ ⎦ . In other terms, DSB and SSB can

be treated as special cases of VSB. Transmission bandwidth of VSB is,

[ ]T vVSBB W f= +

With vf 0= , we have the SSB situation where as, vf W= leads to DSB.

Example 4.13

A VSB signal is generated from the DSB-SC signal, ( ) ( )cm t t2 cos ω .

( )M f is as shown in Fig. 4.72(a). If the vestige is as shown in Fig. 4.72(b), let’s



4.90

find the values of all spectral components in the VSB signal for cf f> , assuming

that demodulation in done by multiplying the received VSB signal with

( )c t2cos ω . ( )M f is to be restored to its original values.

Fig. 4.72: (a) Baseband message spectrum used to generate the VSB

(b) Spectral components in the vestige in LSB

Let iα be the magnitude of the spectral component at c if f i, 1, 2, ..., 5+ = in

the VSB spectrum. When the VSB spectrum is shifted to the left by cf , we have

the spectrum shown in Fig. 4.73(a). Similarly, when the VSB spectrum is shifted

to the right by cf , we have the spectrum shown in Fig. 4.73(b).



4.91

Fig. 4.73: Shifted VSB spectra: (a) left shift (b) right shift

From the spectral plots in Fig. 4.73(a) and (b), we have

1 0.1 0.3α + = ; That is 1 0.2α =

2 0.2 0.8α + = ; That is 2 0.6α =

3 4 51, 0.65 and 0.4α = α = α =

Hence the VSB spectrum for f 0> is shown in Fig. 4.74.



4.92

Fig. 4.74: VSB spectrum of example 4.13

4.10 Envelope Detection of VSB+C As mentioned earlier, one important application of VSB has been in TV

broadcast. As a large number of receivers are involved, it is preferable to make

the detector circuit fairly simple and inexpensive. The envelope of a VSB signal is

not one-to-one related to the message ( )m t . Hence direct envelope detection of

a VSB signal would be of no use. However, if there is a carrier component along

with the VSB signal, that is, VSB+C, then ED might work. We shall now look into

this.

Exercise 4.13

Let ( ) ( ) ( ) ( )m t t t t2 4 cos 200 6 cos 300 5 cos 400= + π + π + π .

Specify the frequency response of a VSB filter that passes the LSB almost

completely and leaves a vestige of the first frequency component in USB. It is

given that the magnitude of the filter at the vestige frequency is 18

. Sketch the

spectrum of the VSB signal.



4.93

Let the input to the ED be

( ) ( ) ( ) ( ) ( )c c c c s cA t A m t t m t tcos cos sin⎡ ⎤ω + β ω ± β ω⎣ ⎦

where β is an adjustable scale factor. Then ( )A t , the output of the ED is

( ) ( ) ( )( ) c sA t A m t m t1

2 2 21⎡ ⎤= + β + β⎣ ⎦

( ) ( )( )

sc

m tA m t

m t

12 2

1 11

⎧ ⎫⎡ ⎤β⎪ ⎪⎡ ⎤= + β + ⎢ ⎥⎨ ⎬⎣ ⎦ + β⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭ (4.36)

If the distortion component ( )( )

sm tm t

11β

<<+ β

for all t , then the output of the ED

will be ( )cA m t1⎡ ⎤+ β⎣ ⎦ , which, after the DC block, will provide the message

output. This level of the distortion component can be reduced by

i) increasing vf . Note that as vf W→ , we will have the DSB signal; that is,

( )sm t 0= .

ii) decreasing the value of β .

In commercial TV broadcast, vf is about 75 kHz which is about 1/6 of the

width of a full sideband which is about 5 MHz. It has been found that with a

vestige of about 75 kHz, the distortion component is not too bothersome.

It would be instructive to compare the envelope detection of SSB+C with

that of VSB+C. Let the input to the ED be SSC+C, namely,

( ) ( ) ( ) ( ) ( )c c c cA t m t t m t tcos cos sin⎡ ⎤ω + ω ± ω⎣ ⎦

then, ( ) ( ) ( )cA t A m t m t

122 2⎧ ⎫⎡ ⎤⎡ ⎤= + +⎨ ⎬⎣ ⎦ ⎣ ⎦⎩ ⎭



4.94

( ) ( ) ( )c

c cc

m t m t m tA

A AA

12 22

22

1⎧ ⎫⎡ ⎤⎡ ⎤⎪ ⎪= + + + ⎢ ⎥⎢ ⎥⎨ ⎬

⎢ ⎥⎢ ⎥⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭

(4.37)

If ( )c

m tA

and ( )c

m tA

are very much less than unity, then ( )c

m tA

2⎡ ⎤⎢ ⎥⎣ ⎦

and ( )c

m tA

2⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

can

be dropped from Eq. 4.37, giving us

( ) ( )c

c

m tA t A

A

122

1⎡ ⎤+⎢ ⎥

⎣ ⎦

Retaining only first order term of the binomial expansion,

( ) ( ) ( )c cc

m tA t A A m t

A1⎡ ⎤+ = +⎢ ⎥

⎣ ⎦

Of course, if we ensure ( ) ( ) cm t m t Amax min= << , then ( )c

m tA

can be

neglected in comparison with unity. This may be adequate to make ( )c

m tA

much

smaller than unity, most of the time. In any case, it is obvious that ED of SSB+C

results in excessive wastage of the transmitted power. In contrast, ED of AM is

reasonably efficient as the requirement to avoid envelope distortion is

( ) cm t Amax

≤ .

Example 4.14 Consider the SSB+C signal,

( ) ( ) ( ) ( ) ( ) ( )c c c cs t A t m t t m t tcos cos sin= ω + ω + ω

where ( )m tt2

11

=+

. We will show that, if cA 1>> , then the output of the ED

can be taken as ( )+cA m t .



4.95

As ( )m tt2

11

=+

, we have ( ) tm tt21

=+

. Hence ( )A t , the envelope of

( )s t is,

( ) ctA t A

t t

12 2 2

2 21

1 1

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥+ +⎝ ⎠ ⎝ ⎠⎣ ⎦

( ) ( )

cc

A tAtt t

12

22

2 2 22 2

2111 1

⎡ ⎤⎢ ⎥= + + +⎢ ⎥++ +⎢ ⎥⎣ ⎦

cc

AAt t

122

2 221

1 1

⎡ ⎤= + +⎢ ⎥

+ +⎣ ⎦

( ) ( )c

c c

AA t A t

12

2 2 22 11

1 1

⎧ ⎫⎪ ⎪= + +⎨ ⎬

+ +⎪ ⎪⎩ ⎭

Neglecting the term ( )+2 211cA t

, we have

( ) ( )cc

A t AA t

12

221

1

⎡ ⎤⎢ ⎥+⎢ ⎥+⎣ ⎦

Using the binomial expansion upto the second term, we have

( ) ( )cc

A t AA t2

1 212 1

⎡ ⎤⎢ ⎥+⎢ ⎥+⎣ ⎦

cAt2

11

++

which is the desired result.



4.96

A note on the linearity of AM, DSB-SC, SSB and VSB: Having discussed

these modulation schemes, let us look at the linearity aspect of these schemes.

By linearity, we imply that the schemes obey the superposition property. This can

be easily verified in the case of DSB-SC, SSB and VSB. Consider DSB-SC.

When message signals ( )m t1 and ( )m t2 are applied separately, the resulting

modulated waveforms are ( ) ( )c cA m t t1 cos ω and ( ) ( )c cA m t t2 cos ω . Let

( ) ( ) ( )m t m t m t1 1 2 2= α + α

where 1α and 2α are constants. Then the modulated carrier is

( ) ( ) ( ) ( ) ( )c c c cA m t t A m t m t t1 1 2 2cos cos⎡ ⎤ω = α + α ω⎣ ⎦ ,

Exercise 4.14 Consider SSB+C with tone modulation. Let

( ) ( ) ( )c c c ms t A t tcos cos⎡ ⎤= ω + β ω +ω⎣ ⎦

Assume that c mf f>> and 1β < .

a) Construct the phasor diagram and develop the expression for the

envelope ( )A t .

b) Let β be such that terms involving m m, 3β ≥ can be neglected. Show

that

( ) ( ) ( )c m mA t A t t2 2

1 cos cos 24 4

⎡ ⎤β β+ + β ω − ω⎢ ⎥

⎢ ⎥⎣ ⎦

c) Find the value of β so that second harmonic envelope distortion, that is,

the ratio, ( )( )A t

A tAmplitude of the second harmonic in

Amplitude of the fundamental in is less than five

percent.



4.97

which establishes the linearity property. Similarly, SSB and VSB can be shown to

be linear. AM is not linear in its strict sense, because if ( )m t is applied as input

to an AM modulator, the output is

( ) ( ) ( ) ( ) ( ) c m c c m cA g m t t A g m t m t t1 1 2 21 cos 1 cos⎡ ⎤ ⎡ ⎤+ ω = + α + α ω⎣ ⎦⎣ ⎦

and this is not equal to

( ) ( ) ( ) ( )c m c c m cA g m t t A g m t t1 1 2 21 cos 1 cos⎡ ⎤ ⎡ ⎤+ α ω + + α ω⎣ ⎦ ⎣ ⎦

That is, superposition does not apply to the carrier component. As this is only a

minor deviation, all the four modulation types are put under the category of linear

modulation.

4.11 Superheterodyne Receiver The important function of a receiver is demodulation; that is, to recover the

message signal from the received modulated waveform. The other functions

which become necessary for the proper reception of a signal are: amplification

and tuning or selective filtering. Amplification becomes necessary because, most

often, the received signal is quite weak and without sufficient amplification, it may

not even be able to drive the receiver circuitry. Tuning becomes important

especially in a broadcast situation because there is more than one station

broadcasting at the same time and the receiver must pick the required station

and reject the inputs from the other (unwanted) stations; tuning also ensures that

out of band noise components do not affect the receiver's performance. Besides

these operations, most of the receivers also incorporate certain other features

such as frequency conversion, automatic gain control etc.

Two of the demodulation methods which we have already discussed are

coherent or synchronous detection and envelope detection. Coherent detection

can be used to demodulate any linear modulation scheme: DSB-SC, DSB-LC,

SSB or VSB. In practice, it is used to demodulate only suppressed carrier



4.98

signals. Envelope detection is mainly used in the demodulation of DSB-LC and

VSB+C signals. We shall now describe the receiver used in AM broadcast.

The receiver for the broadcast AM is of the superheterodyne (or superhet)

variety1. This is shown schematically in the Fig. 4.75.

Fig. 4.75: Superheterodyne Receiver

The wanted signal ( )s t , along with other signals and noise, is input to the Radio

Frequency (RF) stage of the receiver. The RF section is tuned to cf , the carrier

frequency of the desired signal ( )s t . The bandwidth of the RF stage, RFB , is

relatively broad; hence along with ( )s t , a few adjacent signals are also passed

by it. The next stage in the receiver is the frequency conversion stage consisting

of a mixer and a local oscillator. The local oscillator frequency, LOf tracks the

carrier frequency cf , (with the help of a ganged capacitor) and is usually

( )c IFf f+ , where IFf denotes the Intermediate Frequency (IF). The mixer output

consists of, among others, the frequency components at c IFf f2 + and IFf . The

following stage, called the IF stage, is a tuned amplifier, which rejects all the

other components and produces an output that is centered at IFf . The bandwidth

of the IF stage, IFB , is approximately equal to the transmission bandwidth, TB , of

1 Some of the other applications of superhet are: reception of FM and TV broadcast signals and RADAR



4.99

the modulation scheme under consideration. For example, if the input signal is of

the double sideband variety, then IFB W2≈

The IF stage constitutes a very important stage in a superheterodyne

receiver. It is a fixed frequency amplifier (it could consist of one or more stages of

amplification) and provides most of the gain of the superhet. Also, as IF TB B= ,

it also rejects the adjacent channels (carrier frequency spacing ensures this).

Next to IF, we have the detector or demodulation stage which removes the IF

carrier and produces the baseband message signal at its output. Finally, the

demodulator output goes through a baseband amplification stage (audio or video,

depending upon the type of the signal) before being applied to the final

transducer (speaker, picture-tube etc.)

The spectral drawings shown in Fig. 4.76 and 4.77 help clarify the action

of a superhet receiver. We shall assume the input to the receiver is a signal with

symmetric sidebands.

Fig. 4.76: Typical spectrum at the input to the RF stage of a superhet

Fig. 4.77: Spectrum at the input of the IF stage of a stage of a superhet



4.100

Fig. 4.76 shows the spectrum of the input signal to the RF stage. As can be

seen, it has the desired signal, plus a few adjacent channels and possibly a

signal with the carrier frequency c c IF LO IF cf f f f f f' '2= + = + ⋅ is called the image

frequency corresponding to cf . If cf' is allowed to pass through the RF stage,

then it will also give rise to an output at the IF stage which would interfere with

the wanted signal, namely the signal whose carrier frequency is cf . Hence the

main task of the RF section is to pass the frequency components in the range

c Tf B2

± while rejecting the signal with the spectrum centered at cf' (image

frequency rejection). If the 3-dB bandwidth, RFB is such that

T RF FB B f12< <

then, ( )RFH f , should be able to provide sufficient attenuation at the image

frequency.

Fig. 4.77 depicts the frequency response characteristic of the IF stage. As

seen from this figure, the IF filter takes care of adjacent channel rejection.

The superheterodyne1 structure results in several practical benefits:

i) Tuning takes place entirely in the front end (RF and mixer stage) so that the

rest of the circuitry (IF stage, detector and the final power amplifier stage)

requires no adjustments to changes in cf .

ii) Separation between cf and IFf eliminates potential instability due to stray

feedback from the amplified output to the receiver's input.

iii) Most of the gain and selectivity is concentrated in the fixed IF stage. IFf is

so selected so that IF IFB f results in a reasonable fractional bandwidth (for

1 The word superheterodyne refers to the operation of the receiver, namely, the incoming signal at the carrier frequency is heterodyned or mixed with the LO signal whose frequency is higher than ( )c LO c IFf f f f= + .



4.101

AM broadcast various frequency parameters are given in table 4.1). It has

been possible to build superhets with about 70 dB gain at the IF stage itself.

Table 4.1 Parameters of AM radio

Carrier frequency range 540-1600 kHz

Carrier spacing 10 kHz

Intermediate frequency 455 kHz

IF bandwidth 6-10 kHz

Audio bandwidth 3-5 kHz

An IF of 455 kHz has been arrived at by taking the following points into

consideration.

1) IF must not fall within the tuning range of the receiver. Assume that there is

a station broadcasting with the carrier frequency equal to IFf . This signal,

could directly be picked off by the IF stage (every piece of wire can act as

an antenna). Interference would then result between the desired station and

the station broadcasting at c IFf f= .

2) Too high an IF would result in poor selectivity which implies poor adjacent

channel rejection. Assume that IF was selected to be 2 MHz. With the

required bandwidth of less than 10 kHz, we require very sharp cutoff filters,

which would push up the cost of the receiver.

3) As IF is lowered, image frequency rejection would become poorer. Also,

selectivity of the IF stage may increase; thereby a part of the sidebands

could be lost.

We had mentioned earlier, that LO c IFf f f= + . If we have to obtain the IFf

component after mixing, this is possible even if LO c IFf f f= − . But this causes

the following practical difficulty. Consider the AM situation. If we select

LO c IFf f f= − , then the required range of variation of LOf so as to cover the



4.102

entire AM band of 540-1600 kHz is (540 - 455) = 85 kHz to (1600 - 455) = 1145

kHz. Hence the tuning ratio required is 85 :1145 1:13 . If LO c IFf f f= + , then

the tuning ratio required is 995 : 2055 1: 2 . This is much easier to obtain than

the ratio 1:13. With the exception of tuning coils, capacitors and potentiometers,

all the circuitry required for proper reception of AM signals is available in IC chips

(for example, BEL 700).

Example 4.15: Double conversion superhet receiver In receivers operating in the VHF (30 - 300 MHz) range and meant for

receiving fairly narrowband signals (say telemetry signals), to achieve good

image frequency rejection and selectivity using a single mixer stage is quite

difficult. Hence, receivers have been developed with more than one frequency

conversion stage and more than one IF. We shall now look at the image

frequency problem of a receiver with two mixer stages, usually called double or

dual conversion receiver.

Consider the scheme shown in Fig. 4.78.

Fig. 4.78: Block diagram of a double conversion receiver

In the scheme shown, the first mixer stage has a tunable LO, its output

being 30 MHz below the incoming signal frequency. The second mixer stage has

an LO producing a fixed frequency output at 40 MHz. If the RF stage is tuned to

200 MHz, let us find the possible image frequencies of the incoming signal. We

assume that none of the filters in the cascade are ideal.



4.103

First local oscillator frequency = (200 - 30) = 170 MHz. Hence (170 - 30) =

140 MHz will be an image frequency for the incoming signal w.r.t the first IF. As

the second IF is 10 MHz, a component at 50 MHz would be an image of 30 MHz.

Hence if a frequency component at ( )170 50 220± = or 120 MHz get through

the RF stage, it would interfere with the reception of the wanted carrier at 200

MHz. In other words, the image frequencies are at 220 MHz, 140 MHz and 120

MHz.



4.104

Exercise 4.15

a) For the tuned circuit shown in Fig. 4.79, ( ) ( )( )

V fH f

I f0= is,

( )H fj f C

R f L

11 12

2

=⎛ ⎞⎛ ⎞ + π −⎜ ⎟ ⎜ ⎟π⎝ ⎠ ⎝ ⎠

( )R

j R LCL

2 11 1=

+ ω −ω

Let f0 be the resonant frequency; then fLC0

12

=π

and Q-factor of

filter is CQ RL

=

Fig. 4.79: A tuned circuit with L, C and R

Show that

( ) ( )( ) ( )

V f R RH fI f fff f j Qj Q f ff f

02 2

000

0

11

= = =⎡ ⎤ ⎛ ⎞− + −⎜ ⎟⎢ ⎥+ ⎝ ⎠⎢ ⎥⎣ ⎦

and hence

( )( )

RH fQ f2 21

=+ β

where ( ) ffff f

0

0

⎛ ⎞β = −⎜ ⎟

⎝ ⎠



4.105

b) Let the RF stage in a superhet consist of a simple tuned circuit (TC1)

whose output is input to the mixer stage as shown in Fig. 4.80.

Fig. 4.80: Superhet without RF amplifier

Let λ denote the image frequency rejection, where ( )

( )c

c

H f

H f 'λ = , cf

' being

the image frequency of cf .

That is, ( )cQ f2 2 '1λ = + β . Calculate the value of λ when the receiver

is tuned to (i) 1.0 MHz and (ii) 20.00 MHz. Assume Q of the resonant

circuit TC1 to be 100.

(Ans.: (i) 138.6λ = (ii) 70.6λ = )

c) It is required to have the value of λ at 20 MHz the same as at 1.00λ =

MHz. For this purpose, the RF stage has been modified to include a

tuned amplifier stage as shown in Fig. 4.81. Calculate the required Q of

the tuned circuit, TC2

Answer: Q of TC2 5.17=

Fig. 4.81: Superhet with RF amplifier



4.106

Appendix A4.1: Analysis of ED with Tone Modulation

Consider ED circuit shown in Fig. 4.42 which is redrawn in Fig. A4.1.

Fig. A4.1: Envelope detector

Let ( ) ( ) ( )m cAMs t f t f t1 cos 2 cos 2⎡ ⎤ ⎡ ⎤= + µ π π⎣ ⎦ ⎣ ⎦

We assume 1µ < so that ( )mf t1 cos 2⎡ ⎤+ µ π⎣ ⎦ is positive for all t . Then, the

envelope quantity is, ( ) ( )mA t t1 cos⎡ ⎤= + µ ω⎣ ⎦ . We shall derive on upper bound

on µ in terms of LR C and mω so that the detector is able to follow the envelope

for all t .

Consider the waveform ( )v t1 shown in Fig. A4.2. The value of the

envelope ( )A t at t t0= ,

( ) ( )mA t t0 01 cos⎡ ⎤= + µ ω⎣ ⎦



4.107

Fig. A4.2: Waveforms of ED of Fig. A4.1

Let ( ) ( )v t A t1 0 0= . Note that ( )v t1 0 is the voltage across the capacitor at

t t0= . Assuming that the capacitor discharges until the next positive peak in the

carrier cycle, we have,

( ) L cR C f

cv t v t e

f

1 1

1 0 1 01 −⎛ ⎞

+ =⎜ ⎟⎝ ⎠

From Eq. 4.15(b), we have L cR C f 1>> . Hence

( )( )mc L c

v t tf R C f1 0 01 11 cos 1

⎛ ⎞ ⎡ ⎤+ + µ ω −⎜ ⎟ ⎢ ⎥

⎝ ⎠ ⎣ ⎦

mc c

A t tf f0 01 11 cos

⎧ ⎫⎡ ⎤⎛ ⎞ ⎛ ⎞⎪ ⎪≤ + = + µ ω +⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭

( ) ( )m mm m

c ct t

f f0 01 cos cos sin sin⎡ ⎤⎛ ⎞ ⎛ ⎞ω ω

≤ + µ ω − ω⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

Assuming m cf f<< , we approximate

m

c

ff

cos 2 1⎡ ⎤π⎢ ⎥

⎣ ⎦ and m m

c c

f ff f

2sin 2⎡ ⎤ ππ⎢ ⎥

⎣ ⎦

Hence,

( )( ) ( ) ( )mm m m

L c ct t t

R C f f0 0 011 cos 1 1 cos sin

⎡ ⎤ µω+ µ ω − ≤ + µ ω − ω⎢ ⎥

⎣ ⎦

That is,



4.108

( )( )

mL

m m

tR C

t0

0

1 cossin

+ µ ω≤

µω ω (A4.1a)

or ( )( )

µω ω≥

+ µ ω0

0

sin11 cos

m m

L m

tR C t

(A4.1b)

Rearranging Eq. A4.1(b), we have

( ) ( )m m mL L

t tR C R C0 0

1sin cos⎛ ⎞µµω ω − ω ≤⎜ ⎟⎝ ⎠

(A4.2)

Let m LR C

1 1tan−ϕ =ω

as shown

Then, m

mLR C

22

cos1

ωϕ =

⎛ ⎞ω + ⎜ ⎟

⎝ ⎠

and, L

mL

R C

R C

22

1

sin1

ϕ =⎛ ⎞

ω + ⎜ ⎟⎝ ⎠

The inequality A4.2 can be written as

( ) ( )m mL

D t D tR C0 0

1cos sin sin cosµ ϕ ω − µ ϕ ω ≤

where

mL

D

R C

22

1

1=

⎛ ⎞ω + ⎜ ⎟

⎝ ⎠

. That is,

( )L mR C D t0sin 1µ ω − ϕ ≤ (A4.3)

The inequality should hold even when ( )m t0sin 1ω − ϕ = . That is,

LR C D 1µ ≤

( )L m L

R C D R C 2

1 1

1µ ≤ =

+ ω

or ( )Lm

R C2

211 − µ

≤ω µ



4.109

Evidently, we will not be able to demodulate, using the ED circuit, a tone

modulated AM signal with 1µ = .

Fig. A4.3 displays the experimentally generated demodulator output (top

waveform) when the time constant LR C is of proper value so as to follow the

envelope for all t (modulating tone is shown at the bottom). Fig. A4.4 is the ED

output (top waveform) when the time constant is too large. As can be seen from

the figure, ED is not able to follow the negative half cycle of the tone fully,

resulting in the clipping of a part of their cycle. (These waveforms are from

Shreya’s experimentor kit.)

Fig. A4.3



4.110

Fig. A4.4

Exercise A4.1 A 1 kHz square wave, switching between the levels 1± V, amplitude

modulates a carrier to a depth of 50 %. The parameters of the carrier are:

cA 1= volt and cf 1= MHz.

a) Sketch the resulting AM signal.

b) Let the signal of (a) be the input of the ED of Fig. A4.1. Sketch the

voltage across the capacitor C for the following cases:

i) LR C 25 sec= µ

ii) LR C m2 sec=



4.111

References 1) Charles Schuler and Mahesh Chugani, Digital Signal Processing: a hands

on approach, Tata McGraw Hill, 2005

Some suggested references 1) A. Bruce Carlson, Paul B. Crilly and Janet C. Rutledge, Communication

systems (4th ed.), Mc Graw Hill international edition, 2002

2) B. P. Lathi, Modern Digital and Analog Communication Systems, (3rd ed.)

Oxford University press, 1998

3) John G. Proakis and Masoud Salehi, Communication Systems Engineering,

Prentice Hall international edition, 1994



5.1

CHAPTER 5

CHAPTER 5

Angle Modulation

5.1 Introduction Consider a sinusoid, ( )c cA f t 0cos 2π + φ , where cA is the (constant)

amplitude, cf is the (constant) frequency in Hz and 0φ is the initial phase angle.

Let the sinusoid be written as ( )cA tcos ⎡ ⎤θ⎣ ⎦ where ( ) ct f t 02θ = π + φ . In chapter

4, we have seen that relaxing the condition that cA be a constant and making it a

function of the message signal ( )m t , gives rise to amplitude modulation. We

shall now examine the case where cA is a constant but ( )tθ , instead of being

equal to cf t 02π + φ , is a function of ( )m t . This leads to what is known as the

angle modulated signal. Two important cases of angle modulation are Frequency

Modulation (FM) and Phase modulation (PM). Our objective in this chapter is to

make a detailed study of FM and PM.

An important feature of FM and PM is that they can provide much better

protection to the message against the channel noise as compared to the linear

(amplitude) modulation schemes. Also, because of their constant amplitude

nature, they can withstand nonlinear distortion and amplitude fading. The price

paid to achieve these benefits is the increased bandwidth requirement; that is,

the transmission bandwidth of the FM or PM signal with constant amplitude and

which can provide noise immunity is much larger than W2 , where W is the

highest frequency component present in the message spectrum.



5.2

Now let us define PM and FM. Consider a signal ( )s t given by

( ) ( )c is t A tcos ⎡ ⎤= θ⎣ ⎦ where ( )i tθ , the instantaneous angle quantity, is a

function of ( )m t . We define the instantaneous frequency of the angle modulated

wave ( )s t , as

( ) ( )ii

d tf t

d t1

2θ

=π

(5.1)

(The subscript i in ( )i tθ or ( )if t is indicative of our interest in the instantaneous

behavior of these quantities). If ( )i ct f t 02θ = π + φ , then ( )if t reduces to the

constant cf , which is in perfect agreement with our established notion of

frequency of a sinusoid. This is illustrated in Fig. 5.1.

Fig. 5.1: Illustration of instantaneous phase and frequency

Curve 1 in Fig. 5.1 depicts the phase behavior of a constant frequency sinusoid

with 0 0φ = . Hence, its phase, as a function of time is a straight line; that is

( )i ct f t2θ = π . Slope of this line is a constant and is equal to the frequency of

the sinusoid. Curve 2 depicts an arbitrary phase behavior; its slope changes

with time. The instantaneous frequency (in radians per second) of this signal at

t t1= is given by the slope of the tangent (green line) at that time.



5.3

a) Phase modulation

For PM, ( )i tθ is given by

( ) ( )i c pt f t k m t2θ = π + (5.2)

The term cf t2π is the angle of the unmodulated carrier and the constant pk is

the phase sensitivity of the modulator with the units, radians per volt. (For

convenience, the initial phase angle of the unmodulated carrier is assumed to be

zero). Using Eq. 5.2, the phase modulated wave ( )s t can be written as

( ) ( )c c pPMs t A f t k m tcos 2⎡ ⎤⎡ ⎤ = π +⎣ ⎦ ⎣ ⎦ (5.3)

From Eq. 5.2 and 5.3, it is evident that for PM, the phase deviation of ( )s t from

that of the unmodulated carrier phase is a linear function of the base-band

message signal, ( )m t . The instantaneous frequency of a phase modulated

signal depends on ( ) ( )d m tm t

d t'= because ( ) ( )pi

ckd t

f m td t

1 '2 2

θ= +

π π.

b) Frequency Modulation

Let us now consider the case where ( )if t is a function of ( )m t ; that is,

( ) ( )i c ff t f k m t= + (5.4)

or ( ) ( )t

i it f d2− ∞

θ = π τ τ∫ (5.5)

( )t

c ff t k m d2 2− ∞

= π + π τ τ∫ (5.6)

fk is a constant, which we will identify shortly. A frequency modulated signal

( )s t is described in the time domain by

( ) ( )t

c c fFMs t A f t k m dcos 2 2

− ∞

⎡ ⎤⎢ ⎥⎡ ⎤ = π + π τ τ⎣ ⎦ ⎢ ⎥⎣ ⎦

∫ (5.7)

fk is termed as the frequency sensitivity of the modulator with the units Hz/volt.

From Eq. 5.4 we infer that for an FM signal, the instantaneous frequency



5.4

deviation of ( )s t from the (unmodulated) carrier frequency cf is a linear function

of ( )m t . Fig. 5.2 to 5.5 illustrate the experimentally generated AM, FM and PM

waveforms for three different base-band signals. From these illustrations, we

observe the following:

i) Unlike AM, the zero crossings of PM and FM waves are not uniform (zero

crossings refer to the time instants at which a waveform changes from

negative to positive and vice versa).

ii) Unlike AM, the envelope of PM or FM wave is a constant.

iii) From Fig. 5.2(b) and 5.3(b), we find that the minimum instantaneous

frequency of the FM occurs (as expected) at those instants when ( )m t is

most negative (such as t t1= ) and maximum instantaneous frequency

occurs at those time instants when ( )m t attains its positive peak value, pm

(such as t t2= ). When ( )m t is a square wave (Fig. 5.4), it can assume

only two possible values. Correspondingly, the instantaneous frequency

has only two possibilities. This is quite evident in Fig. 5.4(b).



5.5

Fig 5.2: AM and FM with tone modulation



5.6

Fig. 5.3: AM and FM with the triangular wave shown as ( )m t



5.7

Fig. 5.4: AM and FM with square wave shown as ( )m t

iv) A triangular wave has only two possibilities for its slope. In Fig. 5.5(b), it has

a constant positive slope between t1 and t2 , and constant negative slope to

the right of t2 for the remaining part of the cycle shown. Correspondingly,

the PM wave has only two values for its ( )if t , which is evident from the

figure.

v) The modulating waveform of Fig. 5.5(c) is a square wave. Except at time

instants such as t t1= , it has zero slope and ( )t t

d m td t

1=

is an impulse.



5.8

Therefore, the modulated carrier is simply a sinusoid of frequency cf ,

except at the time instants when ( )m t changes its polarity. At t t1= , ( )if t

has to be infinity. This is justified by the fact that at t t1= , ( )i tθ undergoes

sudden phase change (as can be seen in the modulated waveform) which

implies ( )id td tθ

tends to become an impulse.

Eq. 5.3 and 5.7 reveal a close relationship between PM and FM. Let

( ) ( )t

Im t m d− ∞

= τ τ∫ . If ( )Im t phase modulates a carrier with modulator

sensitivity p fk k2= π , then the resulting signal is actually an FM signal as given

by Eq. 5.7. Similarly a PM signal can be obtained using frequency modulator by

differentiating ( )m t before applying it to the frequency modulator. (Because of

differentiation, ( )m t should not have any discontinuities.)

As both PM and FM have constant amplitude cA , the average power of a

PM or FM signal is,

cav

AP2

2= ,

regardless of the value of pk or fk .



5.9

Fig 5.5: PM with ( )m t (a) a sine wave

(b) a triangular wave

(c) a square wave



5.10

Example 5.1 An angle modulated signal is given by

( ) ( ) ( )( )s t t t t6cos 2 2 10 30sin 150 40cos 150⎡ ⎤= π × + +⎣ ⎦

Let us find the maximum phase and frequency derivations.

The terms ( ) ( )t t30sin 150 40cos 150⎡ ⎤+⎣ ⎦ can be expressed in the form

( ) ( )t tcos sin 150 sin cos 150⎡ ⎤α + α⎣ ⎦ .

As 2 230 40 50+ = , we have

( ) ( ) ( ) ( )t t t t3 430sin 150 40cos 150 50 sin 150 cos 1505 5

⎛ ⎞+ = +⎜ ⎟⎝ ⎠

( )t50sin 150= + ϕ where 1 4tan3

−ϕ =

Hence ( ) ( )s t t t6cos 4 10 100 sin 150⎡ ⎤= π × + π + ϕ⎣ ⎦

Evidently, the maximum phase deviation is ( )100π .

Let ( ) ( )t t100 sin 150ψ = π + ϕ

( ) ( )d tt

d t1 50cos 150 150

2ψ

= + ϕ ⋅π

( )t7500 cos 150= + ϕ

Hence maximum frequency deviation 7500= Hz.

Example 5.2

Let ( )m t be a periodic triangular wave with period 310− sec. with

( ) ( )m t m tmax min 1= − = volt. We shall find the maximum and minimum values

of the instantaneous frequency for

a) FM with fk 410= Hz/volt

b) PM with pk = π rad/volt



5.11

Assume the carrier frequency to be 100 kHz.

a) For FM, ( ) ( )i c ff t f k m t= +

( )( )if t 3 4min

100 10 10 90= × − = kHz

( )( )if t 3 4max

100 10 10 110= × + = kHz

b) For PM, ( ) ( )pi c

kf t f m t'

2= +

π

Note that ( )m t' is a square wave with maximum and minimum values as

4,000± .

Hence ( )( )if t 3min

1100 10 4,0002

= × − ×

98= kHz

Similarly, ( )( )if tmax

102= kHz

Example 5.3

Let ( )s t be a general angle modulated signal given by

( ) ( ) ( )c i c cs t A t A t tcos cos⎡ ⎤ ⎡ ⎤= θ = ω + ϕ⎣ ⎦ ⎣ ⎦

It is given that when ( ) mm t tcos= ω , ( )s t has the instantaneous frequency

given by ( ) ( )i c m mf t f k f t22 cos= + π ω , where k is a suitable constant. Let us

find the expression for ( )i tθ . If ( )m t is different from m tcos ω , what could be

the expression for ( )i tθ and ( )s t .

( ) ( ) ( )ii c m m

d tf t f k f t

d t21 2 cos

2θ

= = + π ωπ

( ) ( )ic m m

d tf k t

d t22 cos

θ= π + ω ω



5.12

or ( ) ( )i c m mt f t k t2 sinθ = π + ω ω

( )c mdf t k td t

2 cos⎡ ⎤= π − ω⎣ ⎦

Generalizing,

( ) ( )i cdt f t k m td t

2 ⎡ ⎤θ = π − ⎣ ⎦

and ( ) ( )c c

d m ts t A f t k

d tcos 2

⎡ ⎤= π −⎢ ⎥

⎣ ⎦

Exercise 5.1

A periodic signal ( )m t angle modulates a very high frequency carrier.

Let the modulated signal be given by, ( ) ( )c ps t A t k m t8cos 2 10⎡ ⎤= π × +⎣ ⎦ ,

where ( )m t is as shown in Fig. 5.6.

Fig. 5.6: Modulating signal for the exercise 5.1

If ( ) ( )i if fmax min− is to be 100 kHz show that pk 10= π rad/volt.



5.13

5.2 Bandwidth of FM In this section, we shall make a detailed analysis of the bandwidth

requirements of FM. (PM is considered in section 5.4.) Let f fc k2= π . For

convenience, let ( )s t denote the FM signal. With ( )−∞

= τ τ∫t

Im t m d( ) , the pre-

envelope of ( )s t is

( ) ( )( )c c f Ipes t A j t c m texp ⎡ ⎤⎡ ⎤ = ω +⎣ ⎦ ⎣ ⎦

( ) ( )( ) ( )( ) cn n j tnf f

c f I I Ic cA j c m t m t j m t e

n

2 212! !

ω⎡ ⎤= + − + ⋅ ⋅ ⋅ + + ⋅ ⋅ ⋅⎢ ⎥

⎢ ⎥⎣ ⎦

(5.8)

As ( ) ( ) pes t s tRe ⎡ ⎤= ⎣ ⎦ , we have

( ) ( ) ( ) ( ) ( )( ) ( )fc c f I c I c

cs t A t c m t t m t t2 2cos sin cos

2!⎡ ⎤

= ω − ω − ω + ⋅ ⋅ ⋅⎢ ⎥⎢ ⎥⎣ ⎦

(5.9)

If ( )m t is band-limited to W , then so is ( )Im t , as ( ) ( )( )IM f

M fj f2

=π

. But ( )s t

consists of terms ( )( ) ( ) ( )( ) ( )I c I cm t t m t t2 3cos , sinω ω etc. The spectrum of

( )( )Im t 2 is of width W2 , and that of ( )( )Im t 3 is W3 etc. The spectrum of

( )( ) ( )I cm t t2 cos ω occupies the interval cf W2± and that of ( )( ) ( )nI cm t tcos ω

occupies the interval cf nW± . Clearly, the spectrum of the FM signal is not

band-limited. It appears that, at least theoretically, it has infinite bandwidth and

( )S f is not simply related to ( )M f . A similar situation prevails even in the case

of a PM signal. (Recall that the bandwidth of any linear modulated signal is less

than or equal to W2 , where W is the message band-width). Let us take a closer

look at this issue of FM bandwidth.



5.14

5.2.1 NarrowBand FM (NBFM)

Assume that ( )f Ic m tmax

1<< . Then ( )s t of Eq. 5.9 can be well

approximated by the first two terms; that is,

( ) ( ) ( ) ( )c c f I cs t A t c m t tcos sin⎡ ⎤ω − ω⎣ ⎦ (5.10)

Eq. 5.10 is similar to that of an AM wave. Now ( )S f is band-limited to W2 as in

the case of an AM signal1. For this reason, the case ( )f Ic m tmax

1<< is called

narrowband FM (NBFM). Similarly, the narrowband PM (NBPM) signal can be

written as

( ) ( ) ( ) ( )c c p cs t A t k m t tcos sin⎡ ⎤ω − ω⎣ ⎦ (5.11)

Though the expressions for NBFM or NBPM resemble fairly closely that of an AM

signal, there is one important difference; namely, the sideband spectrum of PM

or FM has a phase shift of 2π with respect to the carrier. This makes the time

domain signal of PM or FM very much different from that of an AM signal. A

NBFM signal has phase variations with very little amplitude variations whereas

the AM signal has amplitude variations with no phase variations.

Note that from Eq. 5.10, we have the envelope of the NBFM signal given by

( ) ( )c f IA t A c m t2 21= +

If ( )f Ic m t 1<< , then ( ) cA t A .

Example 5.4

1 In general, FM is nonlinear as

( ) ( )[ ] ( ) ( ) c c f I I c c f I c c f IA t c m t m t A t c m t A t c m t1 2 1 2cos cos cosω + + ≠ ω + + ω +



5.15

A NBFM signal, ( )s t , is generated by using ( ) ( )m t c t3 2 310 sin 10= . We

shall find ( )S f .

From Eq. 5.10, we have

( ) ( ) ( ) ( )c c f I cs t A t c m t tcos sin⎡ ⎤= ω − ω⎣ ⎦

But ( ) ( )I

M fm t

j f2←⎯→

π and f fc k2= π .

Hence

( ) ( ) ( ) ( )( )

( )( )

c cc c fc c

c c

M f f M f fA A KS f f f f ff f f f2 2

⎡ ⎤− +⎡ ⎤= δ − + δ + + −⎢ ⎥⎣ ⎦ − +⎢ ⎥⎣ ⎦

where ( ) fM f tri 310⎛ ⎞= ⎜ ⎟⎝ ⎠

5.2.2 WideBand FM (WBFM)

If the condition ( )f Ic m tmax

1<< is not satisfied, then we have the

wideband FM, which, as mentioned earlier has, at least theoretically, infinite

bandwidth. However, as will be seen a little later, most of the power of the FM

signal resides in a finite bandwidth, called the transmission bandwidth. In order

to estimate this bandwidth, we observe that

( ) ( )i c ff t f k m t= +

Let ( ) ( )pm m t m tmax min= = . Then the instantaneous frequency varies in the

Exercise 5.2

Let ( ) fM f ga3 31

10 10⎛ ⎞= ⎜ ⎟⎝ ⎠

and cf610= Hz. It is given that fk 250=

Hz/volt, and cA 4= V. Sketch ( )S f for a NBFM signal.



5.16

range ( )c f pf k m− to ( )c f pf k m+ . As the total range of frequency deviation (with

centre at cf ) is f pk m2 , can we assume that, for all practical purposes, the

transmission bandwidth of an FM signal, ( )T FMB is

( )T f pFMB k m?

2=

Let f∆ denote the maximum deviation of the carrier frequency cf . That is,

f pf k m∆ = . Then, can we assume that, to a very good approximation,

( )T FMB f2= ∆ ?

The above expression for the transmission bandwidth is valid only when

f W∆ >> . In the case f W∆ << , ( )T FMB f2≠ ∆ but W2 , as will be shown

later. This is indeed the fallacy that gave birth to FM in the first place. In the

1920s radio engineers, thinking that the actual bandwidth of an FM signal is

f2∆ , felt that bandwidth requirement of FM can be made less than that of AM

(that is, less than W2 ) by choosing f∆ appropriately! The fallacy here lies in

equating the instantaneous frequency to the spectral frequency. Although ( )if t is

measured in Hz, it should not be equated with spectral frequency. Spectral

frequency f is the independent variable of the frequency domain where as ( )if t

is a time dependent quantity indicating the time behavior of a signal with angle

modulation. When we refer to the spectrum ( )X f of a signal ( )x t , we imply that

( )x t is composed of the complex exponentials with the magnitude and phase

specified by ( )X f , and each one of these exponentials exists for all t with the

given frequency. ( )if t , on the other hand represents the frequency of a cosine

signal that can be treated to have a constant frequency for a very short duration,

maybe only for a few cycles. We shall now give some justification that bandwidth

of an FM signal is never less than W2 .



5.17

Let the base-band signal ( )m t be approximated by a staircase signal, as

shown in Fig. 5.7(a). From the sampling theorem (to be discussed in chapter 6),

this staircase approximation is justified as long as the width of each rectangle

(separation between two adjacent samples) is less than or equal to W1

2. For

convenience, we shall treat the adjacent sample separation to be equal to W1

2.

The FM wave for the staircase-approximated signal will consist of a

sequence of sinusoidal pulses, each of a constant frequency and duration of

W1

2 sec. A typical pulse and its spectrum are shown in Fig. 5.7(b) and (c)

respectively.

The frequency of the RF pulse in the interval k kt tW1,

2⎛ ⎞

+⎜ ⎟⎝ ⎠

is

( ) ( )i k c f kf t f k m t= + . Hence, its spectral range (between the first nulls) as

shown in Fig. 5.7(c), is from ( )c f kf k m t W2⎡ ⎤+ −⎣ ⎦ to ( )c f kf k m t W2⎡ ⎤+ +⎣ ⎦ .

Clearly the significant part of the spectrum of the FM signal will lie in the interval

( )c f pf k m W2− − to ( )c f pf k m W2+ + . (Note that about 92% of the energy of

a rectangular pulse of duration T sec, lies in the frequency range fT1

≤ ).

Hence, we can take as one possible value of the transmission bandwidth of an

FM signal, as

( ) ( )T f pFMB k m W f W f W2 4 2 4 2 2= + = ∆ + = ∆ + (5.12)

For the wideband FM case, where f W∆ >> , ( )T FMB can be well approximated

by f2∆ .

In the literature, other rules of thumb for FM bandwidths are to be found.

The most commonly used rule, called the Carson’s rule, is

( ) ( )T FMB f W2= ∆ + (5.13)



5.18

Fig. 5.7: Estimation of FM spectral width

Carson’s rule gives a better bandwidth estimate than does Eq. 5.12 for the NBFM

( )f W∆ << case. This is in agreement with our result on NBFM, namely, its

bandwidth is approximately W2 . In other cases, where f W∆ << is not satisfied

(wideband and intermediate cases) Eq. 5.12 gives a better estimate than does

Eq. 5.13.



5.19

We now define the deviation ratio D , as

fDW∆

= (5.14)

Then Eq. 5.12 and 5.13 can be combined into

( ) ( )T FMB W D k2= + (5.15)

where k varies between 1 and 2. As can be seen, Eq. 5.12 and Eq. 5.13 are

special cases of Eq. 5.15.

5.3 Tone Modulation Let ( ) ( )m mm t A f tcos 2= π

Then ( ) ( )mI m

m

Am t tsin= ωω

, assuming ( )Im 0− ∞ =

( ) ( )mpe c c f m

m

As t A j t c texp sin⎡ ⎤⎛ ⎞

= ω + ω⎢ ⎥⎜ ⎟ω⎝ ⎠⎣ ⎦

( )f mc c m

m

k AA j t tf

exp sin⎡ ⎤⎛ ⎞

= ω + ω⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

For tone modulation, f mf k A∆ = and mW f= . In this case, the deviation ratio is

referred to as the modulation index and is usually denoted by the symbol β .

That is, for tone modulation, f m

m

k Af

β = and

( ) ( )mc j tj tpe cs t A e e sinβ ωω ⎡ ⎤= ⎢ ⎥⎣ ⎦

(5.16a)

( ) ( ) ( )( )pe c c ms t s t A t tRe cos sin⎡ ⎤= = ω + β ω⎣ ⎦ (5.16b)

5.3.1 NBFM

( ) ( )c c ms t A t tcos sin⎡ ⎤= ω + β ω⎣ ⎦

( ) ( )( ) ( ) ( )( ) c c m c mA t t t tcos cos sin sin sin sin= ω β ω − ω β ω



5.20

NBFM: small β .

Let β be small enough so that we can use the approximations

( )( )m tcos sin 1β ω , and

( )( ) ( )m mt tsin sin sinβ ω β ω .

Then, ( ) ( ) ( ) ( )c c m cs t A t t tcos sin sin⎡ ⎤= ω − β ω ω⎣ ⎦ (5.17a)

( ) ( ) ( )c c c m c mA t t tcos cos cos2β⎧ ⎫⎡ ⎤= ω + ω + ω − ω − ω⎨ ⎬⎣ ⎦⎩ ⎭

(5.17b)

Corresponding expression for AM (Eq 4.7) is

( ) ( ) ( ) ( )c c c m c mAMs t A t t tcos cos cos

2µ⎧ ⎫⎡ ⎤⎡ ⎤ = ω + ω + ω + ω − ω⎨ ⎬⎣ ⎦ ⎣ ⎦⎩ ⎭

Eq. 5.17(b) can be written as

( ) ( ) ( )c m c mc j t j tj t cc

As t A e e eRe2

ω + ω − ω − ωω β⎧ ⎫⎡ ⎤= + −⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭ (5.17c)

Fig. 5.8: Phasor diagram for NBFM with tone modulation

Using Eq. 5.17(c), we construct the phasor diagram for NBFM, shown in Fig. 5.8.

(We have taken the carrier phasor as the reference.) Comparing the phasor

diagram for the NBFM with that of the AM signal (Fig. 4.26), we make the

following observation.

In the case of AM, the resultant of the side-band phasors is collinear with

the carrier phasor whereas, it is perpendicular to the carrier phasor in NBFM.



5.21

It is this quadrature relationship between the two phasors of NBFM that

produces angle variations resulting in corresponding changes in ( )if t .

5.3.2 WBFM The exponential term in the brackets in Eq. 5.16(a) is a periodic signal

with period mf1 . Expressing this term in Fourier series, we have

( )m mj t j n tn

ne x esin

∞β ω ω

= − ∞= ∑

where ( )mm m

m

fj t j n t

n m

f

x f e e d t

12

sin

12

β ω − ω

−

= ∫

Let m tθ = ω . Then,

( )j nnx e dsin1

2

πβ θ − θ

− π

= θπ ∫ (5.18)

The integral on the RHS of Eq. 5.18 is recognized as the nth order Bessel

function of the first kind and argument β and is commonly denoted by the symbol

( )nJ β ; that is,

( ) ( )j nn nx J e dsin1

2

πβ θ − θ

− π

= β = θπ ∫ (5.19a)

That is,

( ) ( )m mj t j n tn

ne J esin

∞β ω ω

= − ∞= β∑ (5.19b)

Hence,

( ) ( ) ( )mj tn m

ne J f n fsin

∞β ω

= − ∞←⎯→ β δ −∑ (5.19c)

Using Eq. 5.19, we can express ( )pes t as



5.22

( ) ( ) ( )c mj f n f tpe c n

ns t A J e 2∞

⎡ ⎤π +⎣ ⎦

= − ∞= β∑ (5.20)

As ( ) ( )pes t s tRe ⎡ ⎤= ⎣ ⎦ , we obtain

( ) ( ) ( )c n c mn

s t A J f n f tcos 2∞

= − ∞⎡ ⎤= β π +⎣ ⎦∑ (5.21)

Taking the Fourier transform of ( )s t , we have

( ) ( ) ( ) ( )cn c m c m

n

AS f J f f n f f f n f2

∞

= − ∞⎡ ⎤= β δ − − + δ + +⎣ ⎦∑ (5.22)

Properties of ( )nJ β : The following properties of ( )nJ β can be established.

1) ( )nJ β is always real (For all n and β )

2) ( ) ( ) ( )nn nJ J1 −β = − β

3) For small values of β , ( ) ( )nnJ

n2!

ββ

Exercise 5.3

Eq. 5.19(c) gives the Fourier transform of ( )mj te sinβ ω . Show that

a) ( ) ( ) ( ) ( )m nj tn m

ne j J f n fcos

∞β ω

= − ∞←⎯→ β δ +∑

b) ( ) ( ) ( )c c m c n c mn

A t t A J n t ncos cos cos2

∞

= − ∞

π⎡ ⎤⎡ ⎤ω + β ω = β ω + ω +⎣ ⎦ ⎢ ⎥⎣ ⎦∑

Hint: ( ) mmm

j tj te e

sin2cos

⎡ ⎤⎛ ⎞πβ ω −⎢ ⎥⎜ ⎟ωβ ω ⎢ ⎥⎝ ⎠⎣ ⎦=

If ( ) ( )mj tx t e sinβ ω −= , then ( )mj t

me x tcos

2β ω ⎛ ⎞π

= −⎜ ⎟ω⎝ ⎠.

Now use the Fourier transform properties for time reversal and time shift.



5.23

Hence, for small values of β ,

( )J0 1β

( )J1 2β

β

( )nJ n0, 1β >

4) ( )nn

J2 1∞

= − ∞β =∑

Fig 5.9 depicts the behavior of ( )nJ β . We now make the following observations

regarding the spectrum ( )f 0> of a tone modulated FM signal.

1) The spectrum of a tone modulated FM wave contains carrier component

and an infinite set of side-frequencies located symmetrically on either side

of the carrier at frequency separations of mnf n, 1, 2,= ⋅ ⋅ ⋅

2) For small β , only ( )J0 β and ( )J1 β are significant. Then the FM spectrum

has only three components: at c c mf f f, ± . This situation corresponds to the

special case of NBFM.

3) The amplitude of the carrier component varies with β according to ( )J0 β .

Thus, in contrast to AM, this amplitude ‘contains’ part of the message

information. Moreover, ( )J0 0β = for some values ( )2.4, 5.5, etc.β β = .

4) From Fig. 5.9(b), we find that ( )nJ β decays monotonically for n 1>β

and

that ( )nJ 1β << for n 1>>β

.

Fig, 5.10 gives the spectral plots obtained from a spectrum analyzer with

2.4β . As can be seen from the plot, the carrier component is very nearly zero

(approximately 50 dB below the maximum value).



5.24

Fig 5.11(a) gives the theoretical plot of various spectral components for

4β . Fig. 5.11(b) is the spectral plot as observed on a spectrum analyzer. We

see from this plot that the third spectral component has the largest magnitude

which is an agreement with the theory. Let us normalize the dB plot with respect

to the largest magnitude; that is, we treat the largest magnitude as 0 dB and

compare the rest respect to with this value. For n 4= , Fig. 5.11(a), we have

( )J4 0.281β = . Hence 100.28120 log 3.70.43

= − and value indicated by the

spectrum analyzer agrees with this. Similarly, for n 5= , we expect

100.13220 log 10.250.430

= − ; the value as observed on the plot of Fig. 5.11(b) is in

close agreement with this. For n 1= , theoretical value is 0.06620 log 17.10.43

= −

dB. Spectrum analyzer display is in close agreement with this. The values of the

remaining spectral components can be similarly be verified.

From Eq. 5.21, the average power avP of the FM signal with tone

modulation is ( )cav n

n

AP J2

22

∞

= − ∞= β∑ .



5.25

Fig. 5.9: Bessel Functions:

(a) ( )nJ β vs. β for fixed n

(b) ( )nJ β vs. nβ

for fixed β



5.26

Fig. 5.10: Spectrum analyzer output with 2.4β

But from property 4 of ( )nJ β , we have ( )nn

J2 1∞

= − ∞β =∑ , which implies that the

avP of the tone modulated FM signal is cA2

2. This result is true in general

(whether ( )m t is a tone or not) because an FM signal is essentially a constant

amplitude cosine signal with time varying frequency. As the RMS value of a

sinusoid is independent of the frequency, we state that

( ) cav FM

AP2

2= (5.23)

For large β , if we assume that we can neglect ( )nJ β for n 2> β + , then

( ) ( )T m mFMB nf f2 2 2= = β + (5.24a)

This in agreement with Eq. 5.15, with k 2= . If we neglect the values of ( )nJ β

for n 1> β + , then,

( ) ( )T mFMB f2 1= β + (5.24b)

which corresponds to Eq. 5.15 with k 1= .



5.27

Fig. 5.11(a): ( )nJ β for 4β =

(b) Spectrum analyzer output for 4β

Consider 5β = . From Appendix A5.1, we have ( )J7 5 0.053= and

( )J8 5 0.018= . P1, the average power of these two is,



5.28

( ) ( )c cA AP2 2

1 0.0028 0.0003 0.00312 2

= + = .

Hence, ( )av FM

PP

1 0.0031 0.31%= = . As ( )nJ 5 for n 9≥ are much smaller than

( )J8 5 , we see that practically the entire power of the FM signal is confined to the

frequency range [ ]±c mf f6 . As such, we find that even Eq. 5.24(b) is a fairly

good measure of ( )T FMB for tone modulation.

With tone modulation, it is possible to estimate the value of TB to the

accuracy desired. In this case, the magnitude of a spectral component depends

only on ( )nJ β and these Bessel coefficients have been tabulated extensively.

For a given β , let us define transmission bandwidth such that it includes all those

spectral components whose magnitude is greater than one percent of the

unmodulated carrier amplitude. If we take cA 1= , then

T sig mB n f2= (5.24c)

where sign is such that ( )nJ 0.01β ≥ for sign n≤ . For example, if 20β = ,then

sign would be 25. (For 10, 5,β = etc., sign can be found from the table in

Appendix A5.1.) Let mf f0= when 20β = . Let us tabulate TB for a few other

values of β . We will keep f f020∆ = , but reduce the value of β by increasing

mf . These are listed in Table 5.1. Also listed in the table are TB ,1 and TB ,2 where

( )T mB f,1 2 1= β + (Carson’s rule), and

( )T mB f,2 2 2= β +



5.29

mf β 2 sign TB ,1TB ,2TB

f0 20 50 50 f0 42 f0 44 f0

2 f0 10 28 56 f0 44 f0 48 f0

4 f0 5 16 64 f0 48 f0 56 f0

10 f0 2 8 80 f0 60 f0 80 f0

20 f0 1 6 120 f0 80 f0 120 f0

40 f0 0.5 4 160 f0 120 f0 200 f0

60 f0 0.33 4 240 f0 160 f0 280 f0

200 f0 0.1 2 400 f0 440 f0 840 f0

Table 5.1: TB of Eq. 5.24c for various values of β with ∆ =f f020

From the above table, we see that

i) For small values of β (less than or equal to 0.5), f∆ becomes less and less

significant in the calculation of TB . For 0.1β = , we find that

T

ffB f

0

0

402 1400 10

∆= =

ii) For small values of β , bandwidth is essentially decided by the highest

frequency component in the input spectrum. As such, as 0β → , TB does

not go zero. In fact, in absolute terms, it increases. (From the table, with

20β = , we require a TB of f050 where as with 0.1β = , TB required is

f0400 .)

iii) TB ,1, which is based on Carson’s rule is in close agreement with TB only

for very small values of β . Otherwise, TB ,2 is better.

It is interesting to note that for β = 5, 10 and 20, TB ,2 (which is generally

considered to overestimate the bandwidth requirement) is less than TB as given



5.30

by Eq. 5.24(c). This is because TB of Eq. 5.24(c) neglects only those frequency

components whose magnitude is less than or equal to cA0.01 . In other words,

the ratio

T

av

BP

Power of any spectral component not included in

is less than ( )c

c

AA

24

20.01 2

102

−=

With 5β = , TB ,2 takes into account only upto ( )J7 5 and

( )( )( )

−= ×c

c

J A

A

28 4

2

5 23.38 10

2. That is, TB as given by Eq. 5.24(c) includes few

more spectral components in the bandwidth than that taken into account by TB ,2

resulting in a larger value for the bandwidth.

It is instructive to analyze the FM signal when ( )m t is a sum of sinusoids.

This has been done in the Appendix A5.3. Another measure of bandwidth that is

useful in the study of frequency modulation is the rms bandwidth. This has been

dealt with in Appendix A5.4.

5.4 Phase Modulation All the results derived for FM can be directly applied to PM. We know that,

for phase modulation,

( ) ( ) ( )pii c

kd tf t f m t

d t1 '

2 2θ

= = +π π

(5.25a)

where ( )m t' is the derivative of ( )m t . Therefore,

pp

kf m'

2∆ =

π, where ( )pm m t

max

' '= (5.25b)



5.31

( ) [ ]T PMB f kW k2 , 1 2= ∆ + < <

p pk mkW

'2

2

⎡ ⎤⎢ ⎥= +⎢ ⎥π⎣ ⎦

(5.26)

With respect to f∆ , there is one important difference between PM and

FM. For FM, as f pf k m∆ = , it depends only on the peak value of the modulating

signal. Hence the bandwidth of WBFM is essentially independent of the spectrum

of ( )m t (It has a weak dependence on W ). On the other hand, for PM, f∆ is a

function of pm' and this value is strongly dependent on ( )M f . Strong high

frequency components in ( )M f would result in large values of pm' . Similarly,

predominance of lower-frequency components will result in a lower value of pm' .

As a result, for signals with strong high frequency components, ( )T PMB will be

much larger than for the case of the signals with strong spectral components at

lower frequencies.

Example 5.5 To determine the frequency sensitivity of an FM source, the following

method has been used. The FM generator is fed a tone modulating signal

( )mA t3cos 4 10⎡ ⎤π ×⎣ ⎦

. Starting near about zero, mA is gradually increased and

when mA 2= V, it has been found that the carrier component goes to zero for

the first time. What is the frequency sensitivity of the source? Keeping mA at 2 V,

frequency mf is decreased until the carrier component goes zero for the second

time. What is the value of mf for this to happen?



5.32

With tone modulation, we know that carrier goes to zero for the first time

when 2.4β = . (That is, the smallest value of β for which ( )J0 0β = is

2.4β = .)

But m

ff

2.4∆β = = . That is,

m f pf f k m2.4 4.8 kHz∆ = = =

As pm is 2 V, we have fk 4.8 2.42

= = kHz/V

The carrier component is zero for the second time when 5.5β = . Hence,

mf

34.8 105.5 ×=

or mf34.8 10 872

5.5×

= = Hz

Example 5.6 A 1.0 kHz tone is used to generate both an AM and an FM signal.

Unmodulated carrier amplitude is the same for both AM and FM. The modulation

index β of FM is 8. If the frequency components at ( )cf 1000± Hz have the

same magnitude in AM as well as FM, find the modulation index of AM.

For AM, the magnitude of the spectral components at ( )cf 1000± Hz is

cA2µ . For FM, the magnitude of the spectral components at ( )cf 1000± Hz is

( )cA J1 8 .

( ) ( )cc

A A J J1 18 2 82µ

= ⇒ µ =

0.47=



5.33

Example 5.7 In this example we will show that the bandwidth of the PM signal is

strongly dependent on ( )M f whereas this dependence is weak in the case of an

FM signal.

Let ( ) ( ) ( )m t f t f t1 2cos 2 0.85 cos 2= π + π be used to generate an FM as

well as a PM signal. Modulator constants are fk 5= kHz/V and pk 10= rad/V.

Let us compute (a) ( )T FMB and (b) ( )T PM

B for the following combinations

of f1 and f2 .

i) f1 500= Hz and f2 700= Hz

ii) f1 1000= Hz and f2 1400= Hz

We will assume that ( )nJ β can be neglected if n 2> β + .

a) ( )T FMB

Case i) f1 500= Hz and f2 700= Hz

From Eq. A5.3.1 (Appendix 5.3), we have

( ) ( ) ( ) ( )c m n cm n

s t A J J m n t1 2 1 2cos ⎡ ⎤= β β ω + ω + ω⎣ ⎦∑∑

3

15 10 10

500×

β = =

3

25 10 0.85 6

700× ×

β =

We shall take into account ( )mJ 1β upto ( )m 112 2= = β + and

( )nJ 2β upto ( )n 28 2= = β + . Then,

( ) ( )T FMB 2 12 500 8 700= × + ×

323.2 10= × Hz



5.34

In the equation for ( )s t above, we see that the magnitude of

any spectral component depends on the product. ( ) ( )m nJ J1 2β β . Let

us calculate the ratio ( ) ( )( ) ( )m n

J JJ J

13 8

max

10 610 6⎡ ⎤⎣ ⎦

so as to get an idea of the

magnitude of the spectral components that have been neglected in

calculating ( )T FMB .

Let ( ) ( )A J J13 910 6 0.03 0.02= = ×

0.0006=

From the tables (Appendix A5.1), we find that the maximum value of

the product ( ) ( )m nJ J10 6 occurs for m 8= and n 5= . Let,

( ) ( )B J J8 510 6 0.318 0.362= = ×

0.115= ,

Then, AB

0.0052= .

Case ii) f1 1000= Hz and f2 1400= Hz

Now, we have 1 5β = and 2 3β .

If we account for ( )mJ 1β and ( )nJ 2β upto m 7= and n 5= ,

then

( ) ( )T FMB 32 7 10 5 1400= × + ×

= × 328 10 Hz

Of course the maximum value of ( )mJ 1β and ( )nJ 2β occurs for

m 4= and n 2= , and this product is

( ) ( )C J J4 25 3 0.391 0.486 0.19= = × =

However, ( ) ( )D J J8 65 3 0.018 0.0114= = ×

0.0002=



5.35

Hence DC

0.0002 0.0010.19

=

This is much less than ( )A B which implies that we have taken into

account more number of spectral components than case (i) above. If

we restrict ourselves to the number of spectral components upto

m 1= β + , then,

( ) ( )( ) ( )

J JJ J

7 6

4 2

5 30.003

5 3= which is closer to ( )A B than 0.001.

Then,

( ) ( )T FMB 32 6000 5 1400 26 10= + × = × Hz

which is fairly close to the previous value of 323.2 10× Hz.

b) ( )T PMB

( ) ( ) ( ) ( )m t f f t f f t1 1 2 2' 2 sin 2 0.85 2 sin 2⎡ ⎤= − π π + π π⎣ ⎦

Now ( )m t' frequency modulates the carrier

pp

k f kf

11

1

2 102

πβ = ⋅ = =

π

Similarly,

pk2 0.85 8.5β = =

Case i) f1 500= Hz and f2 700= Hz

( ) ( )T PMB 32 12 500 10 700 26 10= × + × = × Hz

Case ii) f1 1000= Hz and f2 1400= Hz

As 1β and 2β remains unchanged, we have

( ) ( )T PMB 32 12 10 10 1400= × + ×

352 10= × Hz



5.36

This is twice ( )T PMB of case (i). This example shows that the

bandwidth of the PM signals is strongly dependent on ( )M f .

5.5 Generation of FM We had earlier identified two different categories of FM, namely, NBFM

and WBFM. We shall now present the schemes for their generation.

Exercise 5.4 Consider the scheme shown in Fig. 5.12.

Fig. 5.12: Scheme for the Exercise 5.4

a) Let cf be the frequency of the modulator when ( )m t 0= . If ( )s t can be

expressed in the form

( ) ( )⎡ ⎤= ω − β ω⎢ ⎥⎣ ⎦cos ' sin 2c c ms t A t t ,

what are the values of cf' and β ?

b) What are the frequency components in the spectrum of ( )s t ?

Ans: f mc c

k Af f2'

2⎛ ⎞

= +⎜ ⎟⎜ ⎟⎝ ⎠

f m

m

k Af

2

2β =



5.37

5.5.1 Narrowband FM One of the principal applications of NBFM is in the (indirect) generation of

WBFM as explained later on in this section. From the approximation (5.10) we

have,

( ) ( ) ( ) ( )c c f I cs t A t c m t tcos sin⎡ ⎤ω − ω⎣ ⎦

The system shown in Fig. 5.13 can be used to generate the NBFM signal.

Applying ( )m t directly to the balanced modulator, results in NBPM with a

suitable value for pk .

Fig. 5.13: Generation of NBFM signal

5.5.2 WBFM: Indirect and direct methods There are two distinct methods of generating WBFM signals: a) Direct FM

b) Indirect FM. Details on their generation are as follows.

a) Indirect FM (Armstrong’s method) In this method - attributed to Armstrong - first a narrowband FM signal is

generated. This is then converted to WBFM by using frequency multiplication.

This is shown schematically in Fig. 5.14.



5.38

Fig. 5.14: Generation of WBFM (Armstrong method)

The generation of NBFM has already been described. A frequency

multiplier is a nonlinear device followed by a BPF. A nonlinearity of order n can

give rise to frequency multiplication by a factor of n . For simplicity, consider a

square law device with output ( ) ( )y t x t2= where ( )x t is the input. Let ( )x t be

the FM signal given by,

( ) ( )x t tcos ⎡ ⎤= θ⎣ ⎦ , where ( ) ( )t

c ft t k m d2− ∞

θ = ω + π α α∫

Note that we have dropped the subscript i from ( )i tθ . Then,

( ) ( )y t t2cos ⎡ ⎤= θ⎣ ⎦

( ) t1 1 cos 22

⎡ ⎤= + θ⎣ ⎦

( )t

c ft k m d1 1cos 2 42 2

− ∞

⎡ ⎤⎢ ⎥= + ω + π α α⎢ ⎥⎣ ⎦

∫ (5.27)

The DC term in Eq. 5.27 can be filtered out to give an FM output with the

carrier frequency cf2 and frequency deviation twice that of the input FM signal.

An input-output relation of the type ( ) ( ) ( ) ( )nny t a x t a x t a x t2

1 2 ....= + + will

give rise to FM output components at the frequencies c c cf f n f, 2 , ...., with the

corresponding frequency deviations f f n f, 2 , ....,∆ ∆ ∆ , where f∆ is the

frequency deviation of the input NBFM signal. The required WBFM signal can be

obtained by a suitable BPF. If necessary, frequency multiplication can be

resorted to in more than one stage. The multiplier scheme used in a commercial

FM transmitter is indicated in Fig. 5.15.



5.39

Fig. 5.15: Multiplier chain used in typical commercial FM transmitter

The carrier frequency of the NBFM signal cf 1, is 200 kHz with the corresponding

f1 25∆ = Hz. Desired FM output is to have the frequency deviation f4 75∆

kHz and a carrier ( )cf 4 of 91.2 MHz.

To obtain f4 75∆ = kHz starting from f1 25∆ = Hz, we require a total

frequency multiplication of ( )375 10

300025

×= . In the scheme of Fig. 5.11, this

has been accomplished in two stages, namely, multiplication by 64 followed by

multiplication by 48, giving a total multiplication by the factor 64 48 3072× = .

(Actually each stage of multiplication is implemented by a cascade of frequency

doublers or triplers. Thus multiplication by 64 is obtained by 6 doublers in

cascade where as multiplication by 48 is implemented by a cascade of a

frequency tripler and 4 doublers.) Multiplication of cf 1 200= kHz by 3072 gives a

carrier frequency cf 4 614.4= MHz. As the final required carrier frequency is

91.2 MHz, a frequency conversion stage is used to down convert cf 2 (12.8 MHz)

to cf 3 (1.9 MHz). In this process of down conversion, frequency deviation is

unaffected ( )f f2 3 1.6 kHz∆ = ∆ = . The possible drawbacks of this scheme are

the introduction of noise in the process of multiplication and distortion in the



5.40

generation of NBFM signal especially for low modulating frequencies as m

ff∆

could become excessive.

Example 5.8 Armstrong’s method is to be used to generate a WBFM signal. The NBFM

signal has the carrier frequency cf 1 20= kHz. The WBFM signal that is required

must have the parameters cf 6= MHz and f 10∆ = kHz. Only frequency

triplers are available. However, they have a limitation: they cannot produce

frequency components beyond 8 MHz at their output. Is frequency conversion

stage required? If so, when does it become essential? Draw the schematic block

diagram of this example.

Total frequency multiplication required 6

36 10 30020 10×

= =×

. Using only

frequency triplers, we have 53 243= and 63 729= . Hence a set of six

multipliers is required. But these six cannot be used as a single cascade

because, that would result in a carrier frequency equal to 3 620 10 3 14.58× × =

MHz and the last multiplier cannot produce this output. However, cascade of 5

triplers can be used. After this, a frequency conversion stage is needed.

Fig 5.16: Generation of WBFM from NBFM of example 5.8

Block diagram of this generation scheme is shown in Fig. 5.16. As the final

frequency deviation required is 10 kHz, the NBFM must have a



5.41

( )f3

110 10 13.71

729×

∆ = = Hz. After the frequency conversion stage, we have

one more stage multiplication by 3. If cf 3 is the carrier frequency at the mixer

output, then cf 3 3 6× = MHz cf 3 2⇒ = MHz. Assuming that Lof is greater than

incoming carrier frequency of 4.86 MHz, we require Lof 6.86= MHz so that the

difference frequency component is 2 MHz.

Exercise 5.5

In the indirect FM scheme shown in Fig. 5.17, find the values of c if , and

if∆ for i 1, 2 and 3= . What should be the centre frequency, f0 , of the BPF.

Assume that LO cf f ,2> .


Exercise 5.6 In the indirect FM scheme shown in Fig. 5.18, find the values of the

quantities with a question mark. Assume that only frequency doublers are

available. It is required that LO cf f ,2< .




5.42

b) Direct FM Conceptually, the direct generation of FM is quite simple and any such system

can be called as a Voltage Controlled Oscillator (VCO). In a VCO, the oscillation

frequency varies linearly with the control voltage. We can generate the required

FM signal from a VCO by using the message signal ( )m t in the control voltage.

We shall explain below two methods of constructing a VCO.

One can construct a VCO by using an integrator and hysteric comparator

(such as Schmitt trigger circuit). Method 1 below gives the details of this scheme.

Another way of realizing a VCO is to vary the reactive component of a tuned

circuit oscillator such as Colpitt’s oscillator or Hartley oscillator. Method 2 below

gives the details of this scheme.

Method 1: Consider the scheme shown in Fig. 5.19(a). The Schmitt trigger

output ( )y t is V0+ when the integrator output ( )x t is increasing, and V0−

when ( )x t is decreasing. Further, ( )y t changes from V0+ to V0− when ( )x t

reaches E0+ and from V0− to V0+ when ( )x t reaches E0− . The relationship

between ( )x t and ( )y t is shown in Fig. 5.19(b). The electronic switch is

designed such that it is in position 1 when ( )y t V0= and goes to position 2

when ( )y t V0= − . Let us first examine the case when the input ( )mv t is a

positive constant v0 . Consider the situation at t 0= when ( )y t has just

switched to V0 and the switch going from position 2 to position 1. At this point,

( )x t has attained the value E0− . Then, for t t10 < ≤ , we have

( )t

x t E v dRC0 0

0

1= − + τ∫

When t t1= , let ( )x t become E0 . Then ( )y t goes to V0− and the electronic

switch assumes position 2. The value of t1 can be obtained from



5.43

t

E E v dRC

1

0 0 00

1= − + τ∫

or RC Etv

01

0

2=

Fig. 5.19: Direct FM generation (method 1)

The output ( )x t now keeps decreasing until t t2= when ( )x t E0= − . It is easy

to see that ( ) RC Et tv

02 1

0

2− = . That is, ( )x t and ( )y t are periodic with period

RC Ev

0

0

4 or the fundamental frequency of these waveforms is RC Efv

10

00

4−

⎡ ⎤= ⎢ ⎥⎣ ⎦

.

Note that f0 depends on the input signal v0 . If v E0 0= then fRC01

4= . By

properly choosing the values of R and C , we can have cf fRC01

4= = , where



5.44

cf is the required carrier frequency. If the triangular wave ( )x t is the input to a

narrowband BPF tuned to cf , the output ( )s t is ( )c cA f tcos 2π . Then, by making

( ) ( )mv t v k m t0= + where k is a constant such that ( )k m t v0< , we can have

the instantaneous frequency of ( )s t as ( )c ff k m t+ , which is the desired result.

Method 2: The oscillation frequency f0 of a parallel tuned circuit with inductance

L and capacitance C is given by

fLC0

12

=π

or LC01

ω =

Let C be varied by the modulating signal ( )m t , as given by

( ) ( )C t C k m t0= −

where k is an appropriate constant.

Then ( )( )i t

k m tLC

C00

1

1

ω =⎡ ⎤−⎢ ⎥

⎣ ⎦

( )k m tCLC 00

1 12

⎡ ⎤+⎢ ⎥

⎣ ⎦ when

( )k m tC0

1<<

Let c LC0

1ω = . Then,

( ) ( )i c ft c m tω = ω + , where cf

kcC02ω

= .

One of the more recent devices for obtaining electronically variable capacitance

is the varactor (also called varicap, or voltacap). In very simple terms, the

varactor is a junction diode. Though all junction diodes have inherent junction

capacitance, varactor diodes are designed and fabricated such that the value of

the junction capacitance is significant (varactors are available with nominal

ratings from 0.1 to 2000 pF). Varactor diodes, when used as voltage-variable

capacitors are reverse biased and the capacitance of the junction varies

inversely with the applied (reverse) voltage.



5.45

Fig. 5.20: Direct FM generation (method 2)

Consider the scheme shown in Fig. 5.20. The frequency of the oscillator

output depends on L1 and eqC where eq dC C C1 ||= , dC being the capacitance

of the varactor D . BV reverse biases D such that when ( )m t 0= , eqC is of the

correct value to result in output frequency cf . When the message signal ( )m t is

on, dC can be taken as, ( )dC C k m t0'= − where C0

' is the value of dC , with

( )m t 0= . Hence ( )C t of the oscillator circuit is,

( ) ( ) ( ) ( )C t C C k m t C k m t1 0 0'= + − = − , where C C C0 1 0

'= +

The other components in Fig. 5.14 have the following functions. C2 is a

blocking capacitor to isolate the DC level of the varactor from the rest of the

modulator circuit. Note that dC C2 >> or C1 and is essentially a short at the

required FM frequencies. RFC is an RF choke that prevents the RF energy of the

oscillator circuit from feeding into the audio transformer. R1 limits the current in

the circuit in the event that the peaks of the audio signal voltage exceed the

voltage of the DC source and momentarily forward bias the diode D .

Direct FM generation can produce sufficient frequency deviation and

requires little frequency multiplication. However, it suffers from the carrier



5.46

frequency instability as the carrier frequency is not obtained from a highly stable

oscillator. Some additional circuitry is required to achieve frequency stability.

Voltage controlled oscillators are available in the form of IC chips. For

example XR-2206 of the EXAR Corporation, USA has an operating frequency

range of 0.01 Hz to 1 MHz. Another chip with an operating frequency range of

about 1 MHz is LM 566. Texas Instruments CD 4046 is an inexpensive VCO chip

with a normal operating frequency range upto 1.4 MHz. (Actually CD 4046 is a

PLL and VCO is a part of it. PLLs are discussed in sec. 5.5.4.) To make the VCO

chip functional, what is required is an external capacitor and one or two external

resistors. CD 74HC7046 is another PLL chip with VCO. Free running frequency

of the VCO is 18 MHz. Koster, Waldow and Ingo Wolf describe a VCO operating

in the frequency range 100 MHz to 4 GHz [1]. Donald Tillman describes a new

VCO design (called a quadrature trapezoid VCO) especially suited for electronic

music applications. Details are available at:

http://www.till.com/articles/QuadTrapVCO/discussion.html.



5.47

Exercise 5.7 Consider the circuit of Fig. 5.20 for the direct generation of FM. The

diode capacitance dC , is related to the reverse bias as,

dd

Cv

1001 2

=+

pF

where dv is the voltage across the varactor. Let BV 4= V and

( ) ( )m t t30.054 sin 10 10⎡ ⎤= π ×⎣ ⎦ . It is given that C1 250= pF and the circuit

resonates at 2 MHz when ( )m t 0= .

a) Show that, by using the binomial approximation, dC can be put in the

form

( )dC t10

12 310 0.2 10 sin 10 103

−− ⎡ ⎤= − × π ×

⎣ ⎦

b) Show that

( ) ( ) ( )if t t6 32 10 705.86 sin 10 10⎡ ⎤= × + π ×⎣ ⎦

(Note that C2 is a short circuit at the frequencies being generated.)

c) Let cA be the amplitude of oscillations of the VCO. Write the expression

for the generated FM signal.



5.48

Exercise 5.8 A simple variation of the circuit of Fig. 5.20 is shown in Fig. 5.21.

Fig. 5.21: A variation of the VCO circuit of Fig. 5.14

The varactor diodes D1 and D2 are connected back-to-back. This

arrangement helps to mitigate the effect of the RF signal of the tuned circuit

(also called tank circuit) driving a single diode into conduction on its peaks

which will change the bias voltage (and thereby the frequency that is

generated). Of course, as far as the tuned circuit is concerned, the two diodes

are in series; this means the capacitance of the combination is one-half that of

a single diode.

Let BV 5= V, ( ) ( )m t t30.5cos 2 10= π × . Calculate the carrier

frequency cf , ( )if max and ( )if min from the above scheme, given that C1 50=

pF, L1 50= µH and the capacitance of the varactor diode dC , follows the

relation

dd

Cv

50= pF

where dv is the voltage across the diode. Do not use any approximations.

What are the values of ( )if max and ( )if min , if you use binomial approximation.

Ans: cf 91 kHz, ( )if max 91.4 kHz and ( )if min 90.6 kHz, without any

approximation.



5.49

5.6 Demodulation of FM A variety of techniques and circuits have been developed for

demodulating FM signals. We shall consider a few of these techniques falling

under the following categories:

1) FM-to-AM conversion

2) Phase shift discrimination

3) Zero crossing detection

4) Phase Locked Loop (PLL)

5.6.1 FM-to-AM conversion The instantaneous frequency of an FM signal is given by ( )i c ff f k m t= + .

Hence a frequency selective network with a transfer function of the from

( )H f f= α + β , ( f 0> , and α and β are constants) over the FM band would

yield an output proportional to the instantaneous frequency. That is, the circuit

converts the frequency deviation into a corresponding amplitude change, which

in this case is proportional to ( )m t , the message signal. It is assumed that the

time constant of the network is small enough in comparison with the variations in

the instantaneous frequency of the FM signal. We shall now indicate three ways

of implementing this demodulation scheme.

Consider the scheme shown in Fig. 5.22 where dd t

represents a band-

pass differentiator with the magnitude characteristic ( )H f f= α + β , (for f 0> ),

over the required bandwidth. BPL1 is a Band-Pass Limiter which eliminates

amplitude fluctuations from the received FM signal.

1 Band-pass limiters have been analyzed in section 5.7.



5.50

Fig. 5.22: Schematic of an FM demodulator based on FM - to - AM conversion

( )s t is the constant amplitude FM signal to be demodulated. ( )s t' , the

output of the differentiator, is given by

( ) ( ) ( )t

c c f c fs t A k m t t k m d' 2 sin 2− ∞

⎡ ⎤⎢ ⎥⎡ ⎤= − ω + π ω + π α α⎣ ⎦ ⎢ ⎥⎣ ⎦

∫ (5.28)

Eq. 5.28 represents a signal that is both amplitude and frequency modulated.

The envelope of ( ) ( )c c fs t A k m t' 2⎡ ⎤= ω + π⎣ ⎦ (we assume that f p cf k m f∆ = ≤ ;

hence, ( )c fk m t2 0⎡ ⎤ω + π ≥⎣ ⎦ ). As c cA ω represents a DC term, signal ( )m t can

be obtained from ( )s t' , after the DC-block.

The need for a BPL is as follows. Assume that the received FM signal

(with amplitude fluctuations) is applied directly as the input to the differentiator.

Let ( )cA t denote the envelope of the FM signal. Then, there would be an

additional term, ( )cd A td t

on the RHS of Eq. 5.28. Even if this term were to be

neglected, the envelope of ( )s t' would be ( ) ( )c c fA t k m t2⎡ ⎤ω + π⎣ ⎦ , which implies

that the envelope of ( )s t' does not contain a term proportional to ( )m t .

Therefore, it is essential to maintain the FM envelope at a constant level.

(Several factors such as channel noise, fading etc. cause variations in cA ).

Band-pass limiter eliminates the amplitude fluctuations, giving rise to an FM

signal with constant envelope.

We shall now indicate some schemes to implement this method of demodulation.



5.51

Scheme 1: Circuit implementation of the scheme of Fig. 5.22, can be carried out

by constructing an op-amp differentiator circuit followed by an envelope detector.

Scheme 2 (Slope detection): Another way of implementing the FM-to-AM

conversion scheme is through a simple tuned circuit followed by an envelope

detector. The transfer characteristic of a tuned circuit in a small region off

resonance is approximately linear.

Fig. 5.23: Magnitude characteristics of a tuned circuit

In Fig. 5.23, we have shown the resonance characteristic of a tuned

circuit. The parts of the characteristic shown in red have been drawn as straight

lines. (This is a good approximation to the actual characteristic.) Assuming that

( )m t 0> produces an increase in the instantaneous frequency, we can use the

straight line segment between A and B (for f 0> ; for f 0< , we have the

corresponding segment, A' to B' ) for demodulation purposes. Let us assume

that the FM signal to be demodulated has the carrier frequency cf ff 1 2

2+

= and

Tc

Bf f1 2≤ − where as T

cBf f2 2

≥ + . This is shown in Fig. 5.24.



5.52

Fig. 5.24: Segment A to B of Fig. 5.23

As can be seen from the figure, changes in the instantaneous frequency

will give rise to corresponding changes in the output amplitude. Envelope

detection of this output will produce the required message signal.

Fig. 5.25: Tuned circuit demodulator

Consider the demodulator circuit shown in Fig. 5.25. The primary of the

coupled circuit is tuned to cf1 whereas the secondary is tuned to f0 , where

cf f0 > . If, over the frequency range Tc

Bf f2

± ≤ , the output of the primary is

fairly constant, then, we can expect the ( )m t to resemble ( )m t fairly closely.

This method of demodulating an FM signal is also known as slope detection.

Though the method is fairly simple, it suffers from the following

disadvantages: the resonant circuits on the primary and secondary side are 1 In a superheterodyne receiver, the detector stage follows the IF stage. As such, cf gets

converted to IFf .



5.53

tuned to two different frequencies; the frequency range of linear amplitude

response of a tuned circuit is somewhat limited, making the demodulation of a

WBFM unsatisfactory.

Scheme 3 (Balanced slope detection): This latter problem is partially overcome

by using a balanced configuration (balanced slope detection). The scheme,

shown in Fig. 5.26(a) has three tuned circuits: two on the secondary side of the

input transformer and one on the primary. The resonant circuit on the primary is

tuned to cf whereas the two resonant circuits on the secondary side are tuned to

two different frequencies, one above cf and the other, below cf . The outputs of

the tuned circuits on the secondary are envelope detected separately; the

difference of the two envelope detected outputs would be proportional to ( )m t .

Though the balanced configuration has linearity over a wider range (as

can be seen from Fig. 5.26(b), the width of linear frequency response is about

B3 , where B2 is the width of the 3-dB bandwidth of the individual tuned circuits)

and does not require any DC bock (The two resonant frequencies of the

secondary are appropriately selected so that output of the discriminator is zero

for cf f= ), it suffers from the disadvantage that the three tuned circuits are to be

maintained at three different frequencies.



5.54

Fig. 5.26: Balanced slope detection (a) circuit schematic (b) response curve

Example 5.9 Let the FM waveform of Fig. 5.4(b) be the input to a differentiator followed

by an envelope detector. Let us find an expression for the output of the

differentiator and sketch the output of the envelope detector. We shall assume

that the differentiator will produce the appropriate step change in the output for

sudden changes in the input frequency.



5.55

Let the FM waveform between the time instants ( )t t1 2, be taken as

( )f t1cos 2π and that between time instants ( )t t2 3, be taken as ( )f t2cos 2π ,

where f f2 1> .

Then the output of the differentiator is,

t t t1 2< < : ( ) ( )f f t A f t1 1 1 12 sin 2 sin 2π π = π

t t t2 3< < : ( ) ( )f f t A f t2 2 2 22 sin 2 sin 2π π = π

with A A2 1> and A A1 2, 0> .

(Imagine Figure 5.4(a), with two different frequencies.) The output of the ED is

proportional to A1 during t t t1 2< < and proportional to A2 during t t t2 3< < .

Taking the constant of proportionality as unity, we have the output of ED as

shown in Fig 5.27

Fig. 5.27: Output of the ED

After DC block, we obtain the modulating square wave signal. Hence, dd t

followed by ED with a DC block, would act as a demodulator for the FM.

Example 5.10 a) Consider the RC network shown in Fig. 5.28. For the values of R and C

given, we will show that for frequencies around 1.0 MHz, this can act as a

differentiator.



5.56

Fig. 5.28: The RC network of Example 5.10

b) Let us find the condition on R and C such that this network can act as a

differentiator for frequencies around some frequency cf .

c) For the above network, let ( ) ( ) ( )in c f Iv t s t A t c m t6cos 2 10⎡ ⎤= = π × +⎣ ⎦

where ( )fk m t 610<< Hz. If ( )outv t is envelope detected, we will show

that, we can recover ( )m t from the ED output.

a) ( ) π=

+ πj f RCH f

j f RC2

1 2

RC 12 9100 50 10 5 10− −= × × = × and for frequencies around 1.00 MHz,

f RC 6 9 32 2 10 5 10 10 10100

− − ππ = π × × × = π × =

As 1100π

<< , we can take ( )H f as

( )H f j f RC2π , which is a differentiator.

( ) j f95 10 2−= × π

b) for frequencies around some cf , we require, π <<f RC2 1.

c) With ( )s t as the input, ( )outv t is

( ) ( )outdv t s td t

95 10− ⎡ ⎤= × ⎣ ⎦



5.57

( ) ( ) c f I fA t c m t k m t9 6 65 10 sin 2 10 2 10 2− ⎡ ⎤= × π × + π × + π⎣ ⎦

Output of the ED ( )c fA k m t9 65 10 2 10 2− ⎡ ⎤= × π × + π⎣ ⎦

( )fc

kA m t6 962 10 5 10 1

10− ⎡ ⎤= π × × × +⎢ ⎥⎣ ⎦

.

( )c fA k m t61100 10π ⎡ ⎤= +⎢ ⎥⎣ ⎦

5.6.2 Phase shift discriminator This method of FM demodulation involves converting frequency variations

into phase variations and detecting the phase changes. In other words, this

method makes use of linear phase networks instead of the linear amplitude

characteristic of the circuits used in the previous method. Under this category, we

have the Foster-Seely discriminator (and its variant the ratio detector) and the

quadrature detector. Foster-Seely discriminator and the ratio detector have

been discussed in appendix A5.2. We shall now explain the operation of the

quadrature detector.

Quadrature Detector (QD): Consider the FM signal

( ) ( )c cs t A t tcos ⎡ ⎤= ω + φ⎣ ⎦ where

( ) ( )t

ft k m d2− ∞

φ = π λ λ∫

Then ( ) ( ) ( )fd t

t k m td t

' 2φ

φ = = π

( ) ( )t t tt

1 ⎡ ⎤φ − φ − ∆⎣ ⎦∆, provided t∆ is small.

( )t tφ − ∆ can be obtained from ( )tφ with a delay line or a network with

linear phase.



5.58

Consider the scheme shown in Fig. 5.29. ( )H f has the transfer

characteristic given by

( ) ( )j A fH f e=

where ( )A f , the phase function, can be well approximated by a linear phase

function, namely,

( )( )

( )

Tc c

Tc c

Bf f t f f fA f

Bf f t f f f

2 , 0 ,2 2

2 , 0 ,2 2

π⎧− − π − ∆ > − <⎪⎪⎨

π⎪ − π + ∆ < + <⎪⎩

(5.29)

Fig. 5.29: Block diagram of a quadrature detector

With ( )H f specified as above, let us calculate the output of the network with the

input ( )( ) ( )c cj t t j t t

c c ce eA t t Acos

2

⎡ ⎤ ⎡ ⎤ω + ϕ − ω + ϕ⎣ ⎦ ⎣ ⎦+⎡ ⎤ω + φ =⎣ ⎦ . We can take

( )cj t tcA e1

2⎡ ⎤ω + ϕ⎣ ⎦ to represent the positive part of the spectrum. As ( )H f has the

linear phase term j f te 2− π ∆ , (which will contribute a delay of t∆ ), the term at the

output of the filter is

( ) ( ) ( )c c cj t t t t t j t t t

c cA e A e2 21 12 2

π π⎡ ⎤ ⎡ ⎤ω − ∆ + ϕ − ∆ − + ω ∆ ω + ϕ − ∆ −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= .



5.59

Similarly, the output corresponding to the negative part of the input spectrum

would be ( )cj t t t

cA e 212

π⎡ ⎤− ω + ϕ − ∆ −⎢ ⎥⎣ ⎦ . Combining these two terms, we have at the

output of the filter, the quantity

( ) ( )c c c cA t t t A t t tcos sin2π⎡ ⎤ ⎡ ⎤ω +ϕ − ∆ − = ω + ϕ − ∆⎣ ⎦⎢ ⎥⎣ ⎦

.

The 2π phase shift provided by ( )H f at cf f= ± gives rise to the term

quadrature detector. Multiplication of this output by ( )c cA t tcos ⎡ ⎤ω + ϕ⎣ ⎦ followed

by low pass filtering yields the output ( )y t proportional to ( ) ( )t t tsin ⎡ ⎤φ − φ − ∆⎣ ⎦ .

Assuming t∆ to be very small, ( )y t can be approximated as,

( ) ( ) ( ) y t k t t t1 φ − φ − ∆

( )k t t1'= ∆ ϕ

( )k m t2=

where k1 and k2 are constants with = ∆fk c k t2 1 .

Several tuned circuits, when properly designed, can provide the band-

pass response with the phase characteristic given by Eq. 5.29. Consider the

series RLC circuit shown in Fig. 5.30.

Fig. 5.30: A network to provide the phase of Eq. 5.29



5.60

( ) ( )( )i

V f j f CH fV f R j f L

j f C

0

12

122

π= =+ π +

π

( )j f RC f LC21

2 2 1=

π − π + (5.30)

Let fLC0

12

=π

and bRf

L2=

π. Then, Eq. 5.30 reduces to

( ) ( ) ( )b

fH ff f j f f

20

2 20

=− +

Consider f 0>

( ) bf fH ff f

12 2

0arg tan−

⎛ ⎞⎡ ⎤ = − ⎜ ⎟⎣ ⎦ ⎜ ⎟−⎝ ⎠

b

f ff f

2 21 0tan

2− ⎛ ⎞−π

= − − ⎜ ⎟⎜ ⎟⎝ ⎠

( ) ( )b

f f f ff f

1 0 0tan2

− ⎡ ⎤+ −π= − − ⎢ ⎥

⎣ ⎦

Let the circuit be operated in a small frequency interval, around f0 so that f f0 .

Then, f f f0 02+ , ( )b b

f ff f f

0 2+ and

( ) ( )b

fH f

f1 2

arg tan2

− δπ⎡ ⎤ − −⎣ ⎦

( )f Qf

1

0

2tan

2− δπ

− −

where b

fQf0= and ( )f f f0δ = −

If ( )f Qf0

21

δ<< , then,

( ) ( )Q fH f

f0

2arg

2δπ⎡ ⎤ − −⎣ ⎦ (5.31)



5.61

As ( )H f is the frequency response of a network with real impulse response, we

will have

( ) ( )⎡ ⎤ ⎡ ⎤− = −⎣ ⎦ ⎣ ⎦H f H farg arg

By choosing cf f0 = , we have ( )cf f fδ = − and c

Qtf1

∆ = ⋅π

There are other circuit configurations (other than the one given in Fig.

5.30) that can provide the required phase shift for the quadrature detector. The

circuit given in Fig. 5.31 is another possibility.

Fig. 5.31: Another phase shift circuit for the QD

Here C provides a very high reactance at the carrier frequency and the parallel

tuned circuit resonates at cf f= .

Quadrature detector is well suited to IC construction. Companies such as

Signetics, have built high quality FM receivers using QD as the FM demodulator.

Some details on these FM receivers can be found in Roddy and Coolen [2].

5.6.3 Zero-crossing detection Consider the scheme shown in Fig. 5.32(a). When the input to the hard

limiter is a sine wave of period T0 , it produces at its output a square wave of the

same period, with the transitions in the square wave occurring at the zero



5.62

crossings of the input sine wave. When the input to the hard limiter is an FM

signal, the hard limiter output appears as a square wave of varying frequency.

The hard limiter output ( )Hv t , triggers a monostable pulse generator, which

produces a short pulse of amplitude A and duration τ at each upward (or

downward) transition of ( )Hv t .

Consider a time interval T such that c

TW f1 1

>> >> , where W is the

highest frequency present in the input signal. We shall assume that during the T

sec. interval, the message signal ( )m t is essentially constant which implies that

instantaneous frequency ( )if t is also a near constant (Fig. 5.32(b)). Then the

monostable output, ( )pv t , looks like a pulse train of nearly constant period. The

number of such pulses in this interval ( )T in T f t with an average value,

( ) ( )t

I pt T

v t v dT1

−

= λ λ∫

( )T in A A f tT1

= τ τ

After the DC block, we will have ( )y t being proportional to ( )m t , which is the

desired result.



5.63

Fig 5.32: Zero-crossing detector

5.6.4 FM demodulation using PLL PLL is a versatile building block of the present day communication

systems. Besides FM demodulation, it has a large number of other applications

such as carrier tracking (in schemes with a pilot carrier and even suppressed

carrier; recall the squaring loop, Section 4.2.3), timing recovery, frequency

synthesis etc.

The basic aim of a PLL is to lock (or synchronize) the instantaneous angle

of a VCO output to the instantaneous angle of a signal that is given as input to

the PLL. In the case of demodulation of FM, the input signal to PLL is the

received FM signal.

In its simplest form, PLL consists of a phase detector and a VCO

connected as shown in Fig. 5.33(a).



5.64

Fig. 5.33: Phase lock loop (a) Basic configuration

(b) PD of (a) in functional form

(c) PLL as a negative feedback loop

PD makes the comparison of the instantaneous phase of ( )x t and ( )v t ,

and is designed such that ( )v tθ , the instantaneous phase of ( )v t locks on to

( )x tθ , the instantaneous phase of ( )x t , if necessary with some fixed phase

difference. (This will become evident later.)

A number of circuits are available which have been used as phase

detectors. In the context of FM demodulation, the most common PD is that of an

analog multiplier followed by a LPF (Fig. 5.33(b)). The scheme resembles closely

that of a negative feedback amplifier configuration, shown in Fig. 5.33(c). In this

figure, s is the variable of the Laplace transform, ( )G s is system function in the



5.65

forward path whereas ( )H s is the network in the feedback path. A properly

designed negative feedback system ensures that the error quantity ( )e t2 is fairly

close to zero so that ( ) ( )v t x t1 1 . This is ensured by providing sufficiently high

loop gain. Similarly, by making the amplifier gain ag sufficiently large, it is

possible to make ( )v tθ follow the changes in ( )x tθ .

Let ( ) ( )cx t t tcos ⎡ ⎤= ω + ϕ⎣ ⎦ and let the VCO output be,

( ) ( )cv t t tcos ⎡ ⎤= ω + ψ⎣ ⎦ . Then, from Fig. 5.33(b),

( ) ( ) ( ) ( ) ( )c cw t x t v t t t t tcos cos⎡ ⎤ ⎡ ⎤= = ω + ϕ ω + ψ⎣ ⎦ ⎣ ⎦

( ) ( ) ( ) ( ) c t t t t t1 cos 2 cos2

⎡ ⎤ ⎡ ⎤= ω + ϕ + ψ + ϕ − ψ⎣ ⎦ ⎣ ⎦

Only the term ( ) ( )t t1 cos2

⎡ ⎤ϕ − ψ⎣ ⎦ will appear at the output of the LPF; that is

( ) ( ) ( )e t t t11 cos2

⎡ ⎤= ϕ − ψ⎣ ⎦ (5.32)

(We are assuming that the LPF has unit gain)

As the phase detector, we want ( )e t to be zero where ( ) ( )t tϕ = ψ ; but from Eq.

5.32, ( )e t1 is maximum when ( ) ( )t tϕ = ψ . This is not the characteristic of a

proper phase detector. This anomaly can be corrected, if the loop provides a 2π

phase shift so that the output of the VCO is ( )c t tsin ⎡ ⎤ω + ψ⎣ ⎦ . That is, the loop

locks in phase quadrature. Here after, we shall assume this 2π phase shift in the

VCO output.

Now let us look at the demodulation of FM. Let

( ) ( ) ( )c cx t s t A t tcos ⎡ ⎤= = ω + ϕ⎣ ⎦



5.66

where ( ) ( )t

ft k m d2− ∞

ϕ = π τ τ∫ . The VCO is designed such that when ( )y t , the

control voltage is zero, its frequency is cf . (As mentioned earlier, in a superhet,

the demodulator follows the IF stage and cf is actually IFf , which is a known

quantity). This is called the free running frequency of the VCO. Hence, the

VCO output can be written as

( ) ( )v cv t A t tsin ⎡ ⎤= ω + ψ⎣ ⎦ (5.33a)

where ( ) ( )t

vt k y d2ψ = π τ τ∫ (5.33b)

and vk is the voltage sensitivity of the VCO, in units of Hz/volt.

Or ( ) ( )vt k y t' 2ψ = π (5.33c)

( )K y t1= , where vK k1 2= π

( )e t1 of Fig. 5.33(b) is,

( ) ( ) ( ) ( )c vA Ae t t t h t1 sin2

⎡ ⎤= ϕ − ψ ∗⎣ ⎦ (5.34a)

where ( )h t is the impulse response of the LPF,

and ( ) ( )ay t g e t1= (5.34b)

Let ( ) ( ) ( )e t t tθ = ϕ − ψ (5.35)

Then ( ) ( ) ( )c v ae

A A gy t t h tsin2

⎡ ⎤= θ ∗⎣ ⎦

( ) ( )eK t h t2 sin ⎡ ⎤= θ ∗⎣ ⎦ , where c v aA A gK2 2= (5.36)

Using Eq. 5.35, 5.36 and 5.33(c), we can draw following block diagram (Fig.

5.34) for the PLL.



5.67

Fig. 5.34: Equivalent circuit of a PLL

Fig. 5.34 brings out much more clearly the negative feedback nature of the PLL,

where in the quantities involved are instantaneous phase deviations of the input

and the VCO output.

Let the loop be in lock so that ( )θ <<e t 1 for all t . Then

( ) ( )e et tsin θ θ , ( ) ( )t tψ ϕ and ( ) ( )t t' 'ψ ϕ . As ( ) ( )t

ft k m d2− ∞

ϕ = π τ τ∫ ,

we have

( ) ( )ft k m t' 2ϕ = π

Hence ( ) ( ) ( )ft ky t m tK K1 1

' 2ψ π= . That is, ( ) ( )y t m tα .

The above has been a very elementary analysis of the operation of PLL.

Literature on PLL is very widespread. For a more detailed and rigorous analysis,

refer Taub and Shilling [3].

Summary of the Detectors: Slope detection using a single tuned circuit has been presented to show

how simple an FM demodulator could be; it is not used in practice because of its

limited range of linearity. Though balanced slope detection offers linearity over a

wider frequency range, it suffers from the problem of tuning. The Foster-Seely

discriminator and the ratio detector have been the work horses of the FM industry



5.68

until recently; they are now becoming less and less important as better circuit

configurations have been developed especially from the point of view of IC

design. The quadrature detector offers very high linearity and is commonly used

in high quality receivers. Except the phase-shifting network, the rest of the

detector, including the IF amplifier is available in a single chip (e.g. RCA CA

3089E). Commercial zero crossing detectors have better than 0.1 % linearity and

can operate from 1Hz to 10 MHz. A divide by ten counter inserted after the hard

limiter extends the range up to 100 MHz. This type of detector is best suited

when exceptional linearity over a very large frequency deviation is required. It is

less useful when the frequency deviation is a small fraction of the carrier

frequency. PLL performs better than other demodulators when the signal-to-

noise ratio at the input to the detector is somewhat low.

5.7 BandPass Limiter (BPL) We know that information in an FM wave resides in the instantaneous

frequency (or in the zero-crossings) of the signal. As such, the amplitude

changes of the carrier are irrelevant. In fact, as was pointed out in section 5.6.1,

if the envelope ( )A t is not a constant, it will give rise to distortion in the

demodulated output. In other words, from the point of view of proper

demodulation, we want ( )A t A= , a constant.

Though the FM signal that is generated at the transmitter has a constant

envelope, the received signal may not possess this property. This is due to

various impairments during propagation on the channel, namely, channel noise,

distortion introduced by the channel, etc. BPL helps us to recover the constant

envelope FM signal from the one that has envelope fluctuations.

A band-pass limiter consists of a hard limiter followed by a band-pass

filter. The input-output relationship of a hard-limiter is given by



5.69

( )( )( )

A x ty t

A x t

, 0

, 0

⎧ >⎪= ⎨− <⎪⎩

(5.37)

where ( )x t is the input, ( )y t is the output and A is a constant. Let ( )x t be as

shown in Fig. 5.35(a).

Fig. 5.35: (a) Input to the hard limiter

(b) Output of the hard limiter

Then the output with A taken as 1, would be as shown in Fig. 5.35(b). If the input

to the hard limiter is the FM signal (with or without envelope fluctuations), then

the output would be a sequence of alternate positive and negative rectangular

pulses with durations that are not uniform. Such a wave is difficult to analyze.

However, cosθ as a function of θ is always periodic with period 2π . Hence the

hard limiter output, when considered as a function of θ , will be a periodic square

wave (with period 2π ) when the input to the limiter is a cosine signal. Hence, if

( )x t tcos cos= ω = θ , then



5.70

( )y1, cos 01, cos 0

θ >⎧θ = ⎨− θ <⎩

Fig. 5.36: Hard limiter output with cosθ as input

( )y θ being a periodic signal (Fig. 5.36), we can expand it in terms of Fourier

series, namely

( )y 4 1 1cos cos3 cos53 5

⎡ ⎤θ = θ − θ + θ + ⋅ ⋅ ⋅⎢ ⎥π ⎣ ⎦ (5.38)

(Note that ( )y θ is real, and has half-wave symmetry. Hence Fourier series

consists of only cosine terms with the even harmonics missing. See Sec. 1.2.2.)

Let θ be the instantaneous angle of the FM signal; that

( ) ( )θ = θ = ω + τ τ = ω + ϕ∫t

i c f ct t c m d t t( )

Then, from Eq. 5.38, we have

( ) [ ] ( )⎡ ⎤θ = ω + ϕ − ω +ϕ + ⋅ ⋅ ⋅⎣ ⎦π πc cy t t t t4 4cos ( ) cos 3 ( )3

At the output of BPL, we have the constant envelope FM waves with carrier

frequencies cn f and frequency n f∆ respectively, where n 1, 3, 5= etc. With an

appropriate BPF, it is possible for us to obtain constant envelope FM signal with

carrier frequency cf and deviation f∆ . We will assume that BPF will pass the

required FM signal, suppressing the components with spectra centered at the



5.71

harmonics of cf . Incidentally, BPL can be used as a frequency multiplier,

multiplication factors being 3, 5, 7, etc.

A hard limiter, as defined by Eq. 5.37 can be easily be realized in practice.

One such circuit is shown in Fig. 5.37.

Fig. 5.37: Circuit realization of a hard limiter

As shown in the figure, the circuit consists of a high gain amplifier, a current

limiting resistor R and two zener diodes arranged in a back-to-back

configuration. Zeners can be chosen to have the appropriate break down

voltages.

5.8 Broadcast FM 5.8.1 Monophonic FM Reception FM stations operate in the frequency range 88.1 to 107.9 MHz with

stations being separated by 200 kHz; that is, the transmission bandwidth

allocation for each station is about 200 kHz. The receiver for the broadcast FM is

of the superheterodyne variety with the intermediate frequency of 10.7 MHz. As

the audio bandwidth is 15 kHz, these stations can broadcast high quality music.

Let us now look at the receiver for the single channel (or monophonic)

broadcast FM. Like the superhet for AM, the FM receiver (Fig. 5.38) also has the



5.72

front end tuning, RF stage, mixer stage and IF stage, operating at the allocated

frequencies. This will be followed by the limiter-discriminator combination, which

is different from the AM radio.

Fig. 5.38: FM broadcast superhet receiver

The need for a band-pass limiter has already been explained. As far as the

frequency discriminator is concerned, we have a lot of choices: Foster-Seely,

ratio detector, quadrature detector, PLL etc. This is unlike AM, where the

envelope detector is the invariable choice. Also, the output of the discriminator is

used in a feedback mode to control the frequency stability of the local oscillator.

This is very important as a change in the frequency of the LO can result in

improper demodulation. (Recall that, in the case AM, the envelope detector

output is used for AVC.) However, most of the present day receivers might be

using frequency synthesizers in place of a LO. These being fairly stable, AFC is

not a requirement.

Basic operation of the AFC block is as follows. Assume that the receiver is

properly tuned so that LO c IFf f f− = . Then the discriminator input will have equal

frequency variations with respect to IFf ; hence the discriminator output will vary

symmetrically with respect to zero output. As such, the net DC voltage is zero

and LO frequency will not be changed. Assume, however, that LO frequency is

not correct. Let LO c IFf f f '− = , where IF IFf f' < . Then the input to the S-curve of



5.73

the discriminator will be operating mostly in the negative output region. This

means that the discriminator output will have a net negative voltage. If this

voltage is applied to the varicap in the LO circuit, LOf will increase which implies

IFf ' will be increased. In other words, IFf ' tends towards IFf . Similarly, if

LO c IFf f f '− = where IF IFf f' > , then the discriminator output is positive most of

the time which implies a net positive DC value. This voltage will increase the

capacitance of the varicap which implies LOf decreases and this makes IFf ' tend

to IFf . Discriminator output goes through a base-band amplifier1 (with a

bandwidth of 15 kHz), whose output drives the speaker.

5.8.2 Two-channel (stereo) FM Two-channel (stereo) FM is fairly common these days and stereo

transmission has been made compatible with mono-aural reception. As the name

indicates, in the two-channel case, audio signal is derived as the output of two

separate microphones. These are generally called as left microphone and the

right microphone. Let the corresponding output signals be denoted by ( )Lm t

and ( )Rm t . The two- channel transmitter is shown in Fig. 5.39.

Fig. 5.39: FM stereo transmission scheme

1 Output of the discriminator goes through a de-emphasis network before being applied to a base-

band amplifier. Pre-emphasis at the transmitter and de-emphasis at the receiver are used to

improve the signal-to-noise ratio performance. Pre-emphasis and de-emphasis will be discussed

in Chapter 7.



5.74

As shown in the figure, ( ) ( ) ( )L Rv t m t m t1 = + and ( ) ( ) ( )L Rv t m t m t2 = − . The

signals ( )v t1 and ( )v t2 go through a pre-emphasis stage. The signal ( )y t2 is

applied to a balanced modulator, the other input to modulator being the carrier

with a frequency of 38 kHz. Hence ( )y t3 is a DSB-SC signal. The carrier 38 kHz

is derived from the primary source at 19 kHz and a frequency doubler. The final

base-band signal ( )Bm t consists of the sum of ( )y t1 , ( )y t3 and 19 kHz primary

carrier. Typical spectrum (for f 0> ) of the base-band signal as shown in Fig.

5.40.

( ) ( ) ( )L RV f m f m f1 = +

( ) ( ) ( )Y f K Y f f Y f f3 2 0 2 0⎡ ⎤= − + +⎣ ⎦

f0 38= kHz and K is a constant

Fig. 5.40: Spectrum of the final base-band signal

The signal ( )Bm t is used to frequency modulate the carrier allotted to the station.

The resulting signal ( )s t is transmitted on to the channel.

The block schematic of the stereo receiver upto the discriminator is the

same as shown in Fig. 5.38 (that is, monophonic case). Hence, let us look at the

operations performed by the stereo receiver after recovering ( )Bm t from ( )s t .



5.75

These operations are indicated in Fig. 5.41. The DSB-SC signal is coherently

demodulated by generating the 38 kHz carrier from the pilot carrier of 19 kHz.

(Note that if a pilot carrier of 38 kHz had been sent, it would have been difficult to

extract it at the receiver.) ( )r t1 and ( )r t2 , after de-emphasis will yield ( )v t1 and

( )v t2 . (Note that constants of proportionality are ignored and are taken as 1.)

Fig. 5.41: Scheme to recover ( )Lm t and ( )Rm t

From ( )v t1 and ( )v t2 , the individual channels signals, namely ( )Lm t and ( )Rm t

are obtained. These signals, after suitable power amplification, will drive the two

speakers, arranged such that one is on the left and the other is on the right. If the

receiver is not stereophonic, it would respond only to ( )v t1 thereby making it

stereo transmission and monophonic reception compatible.



5.76

Appendix A5.1 Table of Bessel Functions

( )nJ β

β

n 0.5 1 2 3 4 5 6 8 10

0 0.9385 0.7652 0.2239 - 0.2601 - 0.3971 - 0.1776 0.1506 0.1717 - 0.2459

1 0.2423 0.4401 0.5767 0.3391 - 0.0660 - 0.3276 - 0.2767 0.2346 0.0435

2 0.0306 0.1149 0.3528 0.4861 0.3641 0.0465 -0.2429 - 0.1130 0.2546

3 0.0026 0.0196 0.1289 0.3091 0.4302 0.3648 0.1148 - 0.2911 0.0584

4 0.0002 0.0025 0.0340 0.1320 0.2811 0.3912 0.3576 - 0.1054 - 0.2196

5 - 0.0002 0.0070 0.0430 0.1321 0.2611 0.3621 0.1858 - 0.2341

6 - 0.0012 0.0114 0.0491 0.1310 0.2458 0.3376 - 0.0145

7 0.0002 0.0025 0.0152 0.0533 0.1296 0.3206 0.2167

8 - 0.0005 0.0040 0.0184 0.0565 0.2235 0.3179

9 0.0001 0.0009 0.0055 0.0212 0.1263 0.2919

10 - 0.0002 0.0014 0.0070 0.0608 0.2075

11 - - 0.0020 0.0256 0.1231

12 0.0005 0.0096 0.0634

13 0.0001 0.0033 0.0290

14 - 0.0010 0.0120



5.77

Appendix A5.2 Phase Shift Discriminator (i) Foster-Seely Discriminator (ii) The Ratio Detector A5.2.1 Foster-Seely discriminator Fig. A5.2.1 illustrates the circuit diagram of this discriminator where all the

resonant circuits involved are tuned to the same frequency. Note the similarity

between this circuit and the circuit of Fig. 5.26. Major differences are a by-pass

capacitor C between the primary and secondary, an additional inductance L and

only a single tuned circuit on the secondary ( )L C2 2|| .

Fig. A5.2.1: Circuit schematic of Foster-Seely discriminator

In the frequency band of operation, C , C3 and C4 are essentially short circuits,

which implies that the entire input voltage inV would appear across L . The

mutually coupled double tuned circuit has high primary and secondary Q and

low mutual inductance. When evaluating the primary current, we may neglect the

primary resistance and any impedance coupled from the secondary. Hence, the

primary current



5.78

inp

VIj L1

=ω

(A5.2.1)

(Note that all the voltages and currents are phasor quantities)

The voltage induced in series with the secondary as a result of the primary

current is given by

s pV j M I= ± ω (A5.2.2)

where the sign depends on the direction of the winding. Taking the negative sign

and using Eq. A5.2.1 in Eq. A5.2.2, we have

s inMV VL1

⎛ ⎞= − ⎜ ⎟

⎝ ⎠

Assuming the diode circuit will draw very little current, we calculate the

current in the secondary because of sV as,

( )

ss

L C

VIR j X X

2 22=

+ −

where R2 is the resistance associated with the inductance L2 , LX L2 2= ω and

CXC2

2

1=

ω. Hence, the voltage across the terminals 2, 3 is given by

( )s CV I j X223 = −

( )( )

s C

L C

V j X

R j X X2

2 22

−=

+ −

in CV Xj ML R j X

2

1 2 2=

+ where ( )L CX X X

2 22 = −

The voltage applied to diode D1, V62 , is

LV V V62 2312

= +

inV V2312

= + (A5.2.3)



5.79

Similarly, the voltage applied to diode D2 , V63 , is

= −inV V V63 2312

(A5.2.4)

The final output voltage V54 is, V V V54 64 65= − which is proportional to

V V62 63− . We will consider three different cases: cf f= 1, cf f> and cf f< .

i) When the input frequency cf f= we have

in C cin

V X M Xj MV j VL R L R

2 223

1 2 1 2

⎛ ⎞= = ⎜ ⎟⎜ ⎟

⎝ ⎠ (A5.2.5a)

That is, the secondary voltage V23 leads the primary voltage by 90 .

Thus V2312

will lead inV by 90 and V2312

− will lag inV by 90 . Let us

construct a phasor diagram by taking inV as reference. This is shown in Fig.

A5.2.2(a). As the magnitude of the voltage vectors applied to the diodes D1

and D2 , V62 and V63 respectively are equal, the voltages V64 and V65 are

equal and hence the final output V54 is zero.

1 Note that in a superheterodyne receiver, the demodulator follows the IF stage. Hence, cf is

actually IFf , and the discriminator circuit is always tuned to IFf , irrespective of the incoming

carrier frequency.



5.80

Fig. A5.2.2: Phasor diagram illustrative of the operation of Foster-Seely

discriminator

ii) when the input frequency exceeds cf , L CX X X2 22 = − is positive. Let

jR j X Z e2 2 2θ+ = . Then,

( )in C in C jV X V X Mj MV eL R j X L Z

2 2 9023

1 2 2 1 2

− θ= =+

(A5.2.5b)

That is, V23 leads in inV by less than 90 and V23− lags in inV by

more than 90 . This is shown in Fig. A5.2.2(b). As the magnitude of the

vector V62 is greater than that of V63 , V V64 65> which implies the final

output V V V54 64 65= − is positive.



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Fig. A5.2.3: Response curve of the Foster-Seely discriminator

iii) Similarly, based on the phasor diagram of Fig. A5.2.2(c), we can easily

argue that the final output would be negative when cf f< . The actual value

of the final output depends on how far away the input frequency is from cf .

Fig. A5.2.3 gives the plot of the frequency response of the Foster-Seely

discriminator, which is usually termed as the S-curve of the discriminator.

Useful range of the discriminator (frequency range of linear response,

shown in red in Fig. A5.2.3) normally lies between the 3 dB points of the

tuned circuit which forms part of the discriminator circuit.

Foster-Seely discriminator responds also to input amplitude variations. Let

the input to the discriminator be ( )i cf t f= . Then, the voltages across R3 and R4

are equal and let this value be 3 V. Now, let ( )if t be such that voltage across R3

increases while that across R4 decreases. Let the voltage increase on R3 be 2

Volts. We can take the voltage decrease on R4 also as 2 V. In other words, for

the given frequency deviation, say f1∆ , we have the voltage at point 4 equal to 5

volts where as the voltage at point 5 equal to 1 V. This implies



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( )outV 5 1 4= − = . Let V64 denote voltage across R3 and V65 , the voltage

across R4 . Then VV

64

655= .

Let the input signal strength be increased such that, when ( )i cf t f= ,

V V64 65 6= = . Now let ( )if t change such that we have the deviation f1∆ as in

the previous case. Then V64 will become 10 Volts whereas V65 becomes 2 V,

with their difference being equal to 8 V. Though the ratio VV

64

65 remains at 5, outV

changes from the previous value. That is, the circuit responds not only to

frequency changes but also to changes in the incoming carrier strength. Hence,

Foster-Seely discriminator has to be the preceded by a BPL.

A5.2.2 Ratio Detector By making a few changes in the Foster-Seely discriminator, it is possible

to have a demodulator circuit which has built in capability to handle the amplitude

changes of the input FM signal, thereby obviating the need for an amplitude

limiter. The resulting circuit is called the ratio detector which has been shown in

Fig. A5.2.4.

Comparing the ratio detector circuit with that of the Foster-Seely

discriminator, we find the following differences: direction of D2 is reversed, a

parallel RC combination consisting of ( )R R5 6+ and C5 has been added and

the output outV is taken across a different pair of points. We shall now briefly

explain the operation of the circuit.



5.83

Fig. A5.2.4: Circuit schematic of a ratio detector

Reexamining Fig.A5.2.1 and the corresponding phasor diagrams, we find that by

and large, the sum V V62 63+ remains constant. Hence, any variation in the

magnitude of this sum voltage can be considered to be spurious. Suppression of

these spurious variations would result in a detector that is unaffected by input

voltage fluctuations which implies that circuit does not require a separate limiter

stage. How the sum voltage is kept constant would be explained a little later.

With the diode D2 being reversed, we find that the voltages V64 and V65

are series aiding rather than series opposing and as such, the voltage V54

represents the sum voltage. Taking R R5 6= , we find

outV V V V V64 47 64 74= + = −

V V64 5412

= −

V VV 56 6464 2

+= −

[ ]V V64 5612

= −

k V V62 63⎡ ⎤= −⎣ ⎦



5.84

Usually, C C3 4= and R R3 4= . Hence at resonance, V V64 56= which implies

that outV is zero. Above resonance, as V V64 56> , the output is positive whereas

below resonance V V56 64> , and the output is negative.

In the circuit Fig. A5.2.4, C5 is a capacitor of a rather large value. For

example, C5 is of the order of 5 Fµ whereas C3 and C4 are of the order 300

pF. If inV is constant, C5 charges to the full potential existing between the points

5 and 4, which, as indicated earlier is essentially a constant. If inV tries to

increase, C5 will tend to oppose any rise in outV . This is because as the input

voltage tries to rise, extra diode current flows trying to charge C5 . But V54

remains constant at first because C5 is a fairly large capacitance and it is not

possible for the voltage across it to change instantaneously. The situation now is

that the current in the diodes' load has risen but the voltage across the load has

not changed. This being so, the secondary of the ratio detector transformer is

more heavily damped, Q falls and so does the gain of the amplifier driving the

detector. This nearly counteracts the rise in the input voltage. Similarly, when inV

increases, the damping is reduced. The gain of the driving amplifier increases

thereby counteracting the fall in the input voltage. Thus the ratio detector

provides variable damping.

For a large number of years, the Foster-Seely discriminator and the ratio

detector have been the work horses of the FM industry. As these circuit

configurations are not very convenient from the point of view of IC fabrication, of

late, their utility has come down. Companies such as Motorola have built high

quality FM receivers using the Foster-Seely discriminator and the ratio detector.

Some details can be found in Roddy and Coolen [2].



5.85

Appendix A5.3 Multi-tone FM Let ( )m t A t A t1 1 2 2cos cos= ω + ω where f1 and f2 are arbitrary. Then,

( ) ( ) ( ) cj t j t j tpe cs t A e e e1 1 2 2sin sinβ ω β ω ω⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

where fA kf1

11

β = and fA kf2

22

β =

( ) ( ) ( ) cj tj m t j n tpe c m n

m ns t A J e J e e1 2

1 2ωω ω⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪= β β⎨ ⎬⎢ ⎥ ⎢ ⎥

⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭∑ ∑

Hence, ( ) ( ) ( ) ( )c m n cm n

s t A J J m n t1 2 1 2cos ⎡ ⎤= β β ω + ω + ω⎣ ⎦∑∑ (A5.3.1)

The (discrete) spectrum of ( )s t can be divided into 4 categories:

1) Carrier component: amplitude ( ) ( )J J0 1 0 2= β β when cf f=

2) A set of side frequency components due to f1: These components have

amplitudes ( ) ( )mJ J1 0 2β β at frequencies ( )cf mf m1 , 1, 2, 3,± = ⋅ ⋅ ⋅

3) A set of side frequency components due to f2 : These components have

amplitudes ( ) ( )nJ J0 1 2β β at frequencies ( )cf n f n2 , 1, 2, 3,± = ⋅ ⋅ ⋅

4) A set of cross modulation (or beat frequency) terms with amplitudes

( ) ( )m nJ J1 2β β at frequencies ( )cf mf n f1 2± ∓ where m 1, 2, 3,= ⋅ ⋅ ⋅ ;

n 1, 2, 3,= ⋅ ⋅ ⋅

Cross spectral terms of the type given at (4) above clearly indicate the

non-linear nature of FM. These terms are not present in linear modulation. Even

with respect to the terms of the type (2) and (3), linear modulation generates only

those components with m 1= and n 1= .



5.86

Appendix A5.4 RMS Bandwidth of WBFM We have already used a number of measures for the bandwidths of

signals and systems, such as 3-dB bandwidth, null-to-null bandwidth, noise

equivalent bandwidth, etc. In the context of WBFM, another meaningful and

useful bandwidth quantity is the r ms bandwidth. The basic idea behind the r ms

bandwidth is as follows.

Let ( )XS f denote the power spectral density of a random process ( )X t .

Then, the normalized PSD, ( ) ( )

( )

XX N

X

S fS f

S f d f, ∞

− ∞

=

∫, has the properties of a

PDF. Let fσ denote the standard deviation of ( )X NS f, . Then, f2 σ can be used

as a measure of the spectral width of the process.

The WBFM process, with cf f>> ∆ , can be treated as a band-pass

random process. We shall define the r ms bandwidth of a band-pass process as

( )rmsB f f1

2 202⎛ ⎞= −⎜ ⎟

⎝ ⎠ (A5.4.1a)

or ( )rmsB f f 2204= − (A5.4.1b)

where ( )( ) ( )

( )

X

X

f f S f d ff f

S f d f

20

2 00

0

∞

∞

−

− =∫

∫

( ) ( )X

T

f f S f d f

P

20

0

2

∞

−

=∫

(A5.4.2a)

where TP is the total power of the process and



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( )XT

f f S f d fP0

0

2 ∞

= ∫ (A5.4.2b)

Let ( )M t be a strict sense stationary message process with ( )m t as a

sample function. Let ( )Mp m denote the PDF of the process. (Note that we are

using the symbol ( )p to denote the PDF, instead of the earlier symbol ( )f ; this

has been done to avoid confusion. In this derivation, f denotes only the

frequency variable.) For the FM case, we have

i c ff f k m= +

where m is a specific value of ( )m t for some t . Then

( )i c

f

f fm

k−

= (A5.4.3)

Using quasi-static approximation, it has been shown by Peebles [4] that,

( ) c cTX M MWBFM

f f f

f f f fPS f p pk k k2

⎡ ⎤⎛ ⎞ ⎛ ⎞− +⎡ ⎤ = +⎢ ⎥⎜ ⎟ ⎜ ⎟⎣ ⎦

⎝ ⎠ ⎝ ⎠⎣ ⎦ (A5.4.4)

(Note that in quasi-static approximation, it is assumed that ( )if t remains constant

for a sufficiently long period; as such, FM wave appears to be a regular sinusoid

and if can be replaced by f .) For the WBFM process, cT T

AP S2

2= = .

In Eq. A5.4.4, cM

f

f fpk

⎛ ⎞−⎜ ⎟⎝ ⎠

is the positive part of the spectrum. We will now

show that ( ) ( )rms f MWBFMB k R

122 0⎡ ⎤= ⎣ ⎦ where ( ) ( )MR M t20 = , the mean

square value of the process. (Note that ( )MR τ is the ACF of the process.)



5.88

For f 0> , ( )XS f is symmetrical with respect to cf f= (we assume that

the input PDF is symmetric about zero) and hence cf f0 = . Using Eq. A5.4.4 in

A5.4.2(a), we have

( ) ( ) cTrms c MWBFM T f f

f fPB f f p d fP k k

22

0

82

∞ ⎛ ⎞−= − ⋅ ⎜ ⎟

⎝ ⎠∫ .

Let c

f

f fk−

= λ . Then,

( ) ( ) ( )c

f

rms f MWBFM fk

B k p d22 4∞

−

= λ λ λ∫

( ) ( )f M fk p d k M t2 2 2 24 4∞

− ∞

λ λ λ =∫

( )f Mk R24 0=

That is,

( ) ( )rms f MWBFMB k R2 0= (A5.4.5)

Example A5.4.1

Let ( )m t be a sample function of a strict sense stationary process ( )M t .

A WBFM signal is generated using ( )m t as the message signal. It is given that

( )Mm

p motherwise

1 , 120 ,

⎧ <⎪= ⎨⎪⎩

a) Find ( )X WBFMS f⎡ ⎤⎣ ⎦ .

b) What is the value of rmsB ?

As ( )Mp m is uniform, we expect the PSD of the resulting WBFM also to

be uniform over the appropriate frequency range. When m 0= , we have cf f= ,



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whereas m 1= ± results in the instantaneous frequency to be c ff k±

respectively. In other words, the frequency range over which ( )MS f exists is

c f c ff k f f k− < < + . Let c cM M

f f

f f f fp pk k

⎛ ⎞ ⎛ ⎞− ++⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ be as shown in Fig. A5.4.1.

Fig. A5.4.1: Shape of the PSD of Example A5.4.1

To find the value of α , we require that the RHS of Eq. A5.4.4, when integrated

over the entire range should be equal to cA2

2. That is,

c f

c f

f kc c

f f k

A Ad fk

2 22

4 2

+

−

⎡ ⎤⎢ ⎥α =⎢ ⎥⎣ ⎦

∫

or ( )c cf

f

A Akk

2 22

2 2⋅ α =

That is, 12

α =

Hence, ( )c

c f c fX fWBFM

A f k f f kS f k

otherwise

2,

80 ,

⎧− < < +⎪⎡ ⎤ = ⎨⎣ ⎦

⎪⎩

In a few situations ( )rmsB might be quite meaningful. For example, if ( )m t

is the sample function of a Guassian process, there is a small but finite



5.90

probability that ( )m t will assume very large values. This implies that f∆

becomes excessive and as such TB given by Carson’s rule or its variants would

be extremely high. As ( )rmsB weights the large frequency derivations with small

probabilities, ( )rmsB will not be unduly excessive. For example if ( )m t belongs

to a Guassian process and say ( )Mp m is ( )N 0, 9 , then

( )rms f fWBFMB k k2 3 6= × × =

(Note that ( )MR 0 9 3= = )

Exercise A5.4.1

We define the RMS bandwidth of any lowpass process ( )M t as,

( )( )

( )

M

rms M M

f S f d f

BR

2

2

0

∞

− ∞⎡ ⎤ =⎣ ⎦

∫ (A5.4.6)

Let ( )rms PMB denote the RMS bandwidth of the PM signal. Show that

( ) ( ) ( )⎡ ⎤=⎣ ⎦rms p M rmsPM M

B k R B2 0 (A5.4.7)



5.91

Appendix A5.5 Modulation Techniques in TV Color television is an engineering marvel. At the flick of a button on a

remote, we have access to so much information and entertainment in such a fine

detail (audio as well as video with all its hues and colors) that the angels could

envy the humans on this count. Distances are no longer a barrier; we have real

time reception almost at any point on this globe. Maybe, not too far into the

future, we may be able to watch any program of our choice in the language we

prefer to listen either in real time or near real time. We shall now take a closer

look at the modulation techniques used in (commercial) TV, based on the NTSC

standard used in North America and Japan. Our purpose here is to illustrate how

various modulation techniques have been used in a practical scheme. Hence

PAL and SECAM systems have not been discussed.

As the TV transmission and reception started with monochrome (black

and white) signals we shall begin our discussion with this scheme. (Note that

color transmission can be viewed on a monochrome receiver. Similarly, black-

and-white transmission can be viewed on a color receiver.)

The bandwidth of a monochrome video signal is about 4.2 MHz. Let

( )vm t denote this signal. The bandwidth allocated to each TV station by the

regulatory body is about 6 MHz. Hence the use of DSB is ruled out. It is very

difficult to generate the SSB of the video signal, ( )vm t . (Filtering method is ruled

out because of appreciable low frequency content in ( )vM f and because of the

fairly wide bandwidth, designing the HT with required specifications in extremely

difficult.) As such, VSB becomes the automatic choice. With a suitable carrier

component, we have seen that VSB can be envelope detected. Actually at the

transmitter, it is only partially VSB; because of the high power levels at the

transmitter, it is difficult to design a filter with an exact vestigial sideband. It is at



5.92

the IF stage of the receiver that perfect VSB shaping is achieved; subsequently it

is demodulated.

Fig. A5.5.1: Modulation stages in a monochrome TV transmitter

Fig. A5.5.1 shows the block diagram of the modulation scheme of a

monochrome TV transmitter. Details of the magnitude spectrum of the

transmitted signal is given in Fig. A5.5.2.

Fig. A5.5.2: Audio and video spectrum of a monochrome TV signal



5.93

In this figure, cvf is the video carrier and caf is the carrier for audio, which is

higher than cvf by 4.5 MHz. (The video carrier cvf is 1.25 MHz above the lower

frequency limit allotted to the station)

The USB of the video occupies the full bandwidth of 4.2 MHz. That part of

LSB spectrum between ( )cvf 0.75− to cvf is transmitted without any attenuation.

Below the frequency ( )cvf 0.75− MHz, LSB is gradually attenuated reaching

almost zero level at ( )cvf 1.25 MHz− . As this is not exactly a VSB characteristic,

it is termed as partial VSB.

The audio signal, band-limited to 10 kHz frequency modulates the audio

carrier. The maximum frequency deviation is 25 kHz. Hence, the audio

bandwidth can be taken as ( ) ( )f W2 2 25 10 70∆ + = + = kHz.

TV receiver is of the superheterodyne variety. A part of the reciever structure is

shown in Fig. A5.5.3.

Fig. A5.5.3: (Partial) Block diagram of monochrome TV receiver

The mixer output, ( )v t1 , is applied to an IF amplifier and VSB shaping

network. The IF amplifier has a pass-band of 41 to 47 MHz. The characteristic of

the VSB shaping filter is shown in Fig. A5.5.4.



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Fig. A5.5.4: VSB shaping in TV receiver

The frequency modulated audio signal is also passed by the IF stage but with

much less gain than the video part. That is, ( )v t2 consists of the video signal of

the (VSB+C) type and the FM audio signal, with a carrier at 4.5 MHz and with

ca cvA A<< where caA is the audio carrier amplitude and cvA is the amplitude of

the video carrier. Under these conditions, it can be shown that ( )v t3 , the output

of the envelope detector, has the video as well as required audio. The video

amplifier removes the audio from ( )v t3 . The output of the video amplifier is

processed further and is displayed on the picture tube. ( )v t3 is also applied as

input to an IF stage, with the IF of 4.5 MHz. The audio part of the ( )v t3 is passed

by this IF stage; the FM demodulator that follows produces the audio signal.

Color TV: The three primary colors, whose linear combination can give rise to

other colors are: Red, Blue and Green. These color signals, pertaining to the

scene that is being shot, are available at the outputs of three color cameras. Let

us denote these signals as ( )Rm t , ( )Gm t and ( )Bm t respectively. These basic

color components are linearly combined to produce (i) the video signal of the

monochrome variety (this is called the luminance signal and is denoted by

( )Lm t )) and (ii) two other independent color signals (called the in-phase

component of the color signal and the quadrature component of the color signal).



5.95

Let us denote these components as ( )Im t and ( )Qm t respectively. The

equations of the linear transformation are given below in the form of a matrix

equation.

( )( )( )

( )( )( )

L R

I G

BQ

m t m tm t m t

m tm t

0.3 0.59 0.110.6 0.28 0.320.21 0.52 0.31

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= − −⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦⎣ ⎦

M

(A5.5.1)

From Eq. A5.5.1, we obtain

( ) ( ) ( ) ( )L R G Bm t m t m t m t0.3 0.59 0.11= + + (A5.5.2)

It has been found that ( )Lm t as given by Eq. A5.5.2 closely resembles ( )vm t of

the monochrome system. ( )Lm t has the same bandwidth as ( )vm t namely, 4.2

MHz. This is required to preserve the sharp transitions in the intensity of light at

the edges in a scene. However, the eye is not as sensitive to color transitions in

a scene and is possible to reduce the bandwidth occupancy of ( )Im t and ( )Qm t .

In the NTSC system, bandwidth allocation for ( )Im t is 1.5 MHz and that of

( )Qm t is 0.5 MHz. These chrominance signals are quadrature multiplexed

(QAM) on the color subcarrier as shown in Fig. A5.5.5.

Fig. A5.5.5: Generation of composite baseband video



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The composite video signal ( )bvm t is modulated on to the carrier cvf .

This implies that color subcarrier, ccf , is 3.58 MHz above cvf and caf is 4.5 MHz

above cvf . As such, the color information gets interleaved in between the spectral

lines of the luminance signal and no additional bandwidth is required for the color

TV system. In order to facilitate coherent demodulation of the QAM signal, a few

cycles of the color subcarrier (called color burst) is sent along with the

transmitted signal. This reference carrier is tracked by a PLL in the receiver. The

VCO output of the PLL is used in the demodulation of the chrominance signal.

For more details on TV transmission and reception, refer to Carlson, Crilly and

Rutledge [5] or Leon Couch [6].

TV modulator circuit is available in an IC chip. For example, Motorola MC

1374 includes an FM audio modulator, sound carrier oscillator, RF dual input

modulator. It is designed to generate a TV signal from audio and video inputs. It

is also suited for applications such as video tape recorders, video disc players,

TV games etc. The FM system can also be used in the base station of a cordless

telephone. Circuit details and other parameters can be obtained from the manual.

Exercise A5.5.1 Indirect method can used to generate the FM signal for the audio in TV.

Required carrier is 4.5 MHz and f 25∆ = kHz. Using cf 1 200= kHz and

f1 20∆ < Hz, design the modulator such that frequency at any point in the

modulator does not exceed 100 MHz. Use the shortest possible chain of

frequency doublers and triplers.



5.97

References 1) Bettina Koster, Peter Waldow and Ingo Wolff, A unique low voltage,

source-coupled J-FET VCO, RF signal processing, April 2001, PP58-66

(website: http://www.rfdesign.com)

2) Dennis Roddy and John Coolen, Electronic communications, (4th ed.), PHI,

1995

3) Herbert Taub and Donald L. Schilling, Principles of communication

systems, (2nd ed.), McGraw Hill International ed., 1986

4) Peebles P. Z., Communication system principles, Addison-Wesley, 1976

5) A. Bruce Carlson, Paul B. Crilly and Janet C. Rutledge, Communication

systems (4th ed.), McGraw Hill International ed., 2002

6) Leon W. Couch II, Digital and Analog Communication systems (6th ed.),

Pearson Education-Asia, 2001

Principles of Communication - Iit Madras Lecture Material

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