PREREQUISTE

Signals and Systems- Abdul Rauf Khan

1. FOURIER SERIES REPRESENTATION

Representation of a function over a certain interval by a linear combination of mutually

orthogonal functions is called Fourier series representation of a function. There exists,

however, a large number of set of orthogonal functions, and hence a given function may

be expressed in terms of different sets of orthogonal functions.

Some of them are mentioned here

1.2. LEGENDRE FOURIER SERIES

A set of Legendre polynomials Pn(x), (n = 0, 1, 2, …) forms a complete set of mutually

orthogonal functions over an interval (-1 < t < 1). These polynomials can be defined by

Rodrigue’s formula:

→1

We may verify the orthogonality of these functions by showing that

→2

We can express a function in terms of Legendre polynomial over an interval (-1 < t < 1)

as

→3

Where

→4

→5

1


1.3. TRIGONOMETRIC FOURIER SERIES

We can from a complete set of orthogonal trigonometric functions through sine and

cosine functions. Because we already know that Sin not and Sin mot are orthogonal

and also sin not and cos mot are also orthogonal. So we can say any function f(t) can be

represented in terms of these functions over any interval (t0, t0 + 2/o). Hence

→6

As we have

2/o = T

Then the preceding equation no. 7 will be

→7

This Equation is the trigonometric Fourier series representation of f(t) over an interval

(t0, t0+T). From Eq:16a( Lecture 1&2) we can find various constants as follows

→8a

→8b

If we put n=0 in Eq: 8a, we get

→9

We also have

2


Hence Eq:8a and 8b will become

→10a

→10b

As we can see from Eq: 10a that a0 is the average value of f(t) over the interval (t0, t0 + T).

Thus it is the d.c. Component of f(t) over this interval.

3


Example

We shall now expand a function ƒ(t) shown in figure2.1(a,b,c). Using trigonometric Fourier series over the interval (0,1). It is evident that ƒ(t) =At, (0<t<1), the interval T = 1

→11

→12

4

ƒ(t)

A

0 1 tfig 4(a)

ƒ1(t)

A

0 1 t fig 4(b)

ƒ2(t)

A

0 1 t

Figure 2.1(a,b,c)

Thus


2. EXPONENTIAL FOURIER SERIES

2.1. Orthogonality of Exponential Function

It can be shown that a set of exponential functions , (n = 0, 1, 2, …) is

orthogonal over an interval (t0, t0 + 2/0) for any value of t0. As this is a complex

function, we can show the orthogonality of this function using Eq: 22(lecture No.1&2)

5

Which can be expressed as

1

1

1

2sin11

2

1)(

2sin

2)(

2sin2

)(

n

n

nn

ntn

Atf

ntn

AA

tf

ntbA

tf

(0<t<1)


If n = m, then integral I will be

→13

If n m, then I will be

ej2(n-m) can be evaluated as follows

because n and m are integers and n – m is also an integer. Hence I will become

Hence

→14

So we have proved that , (n = 0, 1, 2 …) is orthogonal over an interval

(t0, t0 + T) where T=2/0.

2.2. FOURIER SERIES

As we have proved the orthogonality of exponential function hence we can represent an

arbitrary function f(t) by a linear combination of exponential functions over an interval

(t0, t0 + T):

It can be summarized as follows

→15

6


Eq: 14 is known as exponential Fourier Representation of f(t) over the interval (t0, t0 + T).

Fn can be evaluated as follows using Eq: 21(Lecture 1&2)

From Eq: 13 we have the result for the denominator, hence

→16

2.3. Trigonometric and Exponential Fourier Series Coefficients

Exponential and trigonometric and exponential Fourier series are in fact the same series

but with two different ways of representation. Coefficient of one series can be obtained

from those of the other. We can prove the following relations

a0 = F0

From Eq: 10a we have

→17a

Now we put n=0 in Eq: 16

→17b

As we can see right hand sides of Eq: 17a and 17b are Equal, hence

7


a0 = F0

an = Fn + F-n

From Eq: 16 we have

→17c

if we put n = -n

→17d

To prove we take the right hand side

By comparing the right hand side of Eq: 10b and above Equation it is proved that

an = Fn + F-n.

bn = j ( Fn - F-n )

Using Eq: 17c and 17d we have

By comparing the right hand side of the above Eq: with 10b, the required relation is

proved.

8


Fn = ( an – jbn )

By comparing the right hand sides of the above Equation and Eq: 16 the required relation

is proved.

2.4. PERIODIC FUNCTION REPRESENTATION IN (-∞, ∞)

From Eq: 15 we have

This is true over the interval (t0 < t < t0 + T). The two sides need not be equal outside this

interval. Although it can be proved that the right hand side of the equation is periodic, i.e.

=

We take the right hand side of the equation

So this proves the periodicity of the right hand side.

So it is therefore obvious that if f(t) is periodic with period T, then the Equality in the

Eq: 32 holds over the entire interval (-∞,∞). Thus for a periodic function f(t),

9


where

10


Example

A rectified sine wave is given by.

ƒ(t)=Asinπt with T=1

Express in Exponential Fourier series.

11

ƒ(t) =Asinπt A

-2 -1 0 1 2 t


3. The Complex Fourier Spectrum

DEFINITION

In signal analysis a Fourier series expansion of a periodic function is Equivalent to resolving the function in terms of its components of various frequencies.

EXPLANATION

A periodic function with period T has frequency components of angular frequency o, 2o, 3o, …, no, where o=2/T. Thus the periodic function f(t) possesses its spectrum of frequencies. If we specify f(t), we can find its spectrum and vice versa. So a periodic function f(t) can be represented as

The time domain representation where f(t) is expressed as function of time The frequency domain representation where the spectrum – the amplitudes of

various frequencies components – is specified.

The spectrum has the following properties It exists only at o, 2o, …, no etc. Due to above reason the spectrum is not continuous but exists at some discrete

values of . It is a discrete spectrum, sometimes called a line spectrum. The spectrum can be represented graphically by drawing vertical lines at

o, 2o, …, no etc. with their heights proportional to the amplitude of the corresponding frequency component.

Either trigonometric or exponential series can be used to represent the spectrum.

We will use the exponential form, in which the periodic function is expressed as a sum of exponential functions of frequency 0, 0, 20, 30… etc.For a periodic function of period T, the exponential series is given by

We thus have the frequencies 0, 0, –0, 20, –20, …., n0, –n0 … etc. and the amplitudes of these components are, respectively, F0, F1, F-1, F2, F-2, …, Fn, F–n, etc.The amplitudes are usually complex and hence can be described by magnitude and phase.

Therefore, in general, we need two line spectra: the magnitude spectrum and the phase spectrum for the frequency domain representation of a periodic function. In most of the cases however, the amplitude of frequency components are either real or imaginary, and thus it is possible to describe the function by only one spectrum.

12


Example

→Expand the periodic gate function shown in figure 2.2 by the exponential Fourier series and plot the frequency spectrum.

The gate function has a width δ and repeats every T seconds. The function may be described analytically over one period as follows:

A (-δ/2<t<δ/2)ƒ(t) =

0 (δ/2<t<T-δ/2)

For convenience we shall choose as the limits of integration –δ/2 to (T-δ/2).

The function in the bracket has a form (sin x)/x. This function plays an important role in communication theory and is known as the sampling function, abbreviated by Sa(x). Sa(x) = sinx / x

The sampling function is shown in figure 2.3. Note that the function oscillates with a period 2π, with decaying amplitude in either direction of x, and has zeros at

13

2/

2/

2/

2/

2/

2/.0

11)(

1

TT tjntjn

TAe

Tdtetf

TFn oo

2/

2/

1

dtAeT

Fn tjn o

2/

2/

tjne

Tjn

Ao

o

j

ee

Tn

A oo jnjn

o 2

)(2 2/2/

2/sin2 o

o

nTn

A

2/

2/sin

o

o

n

n

T

AFn

ƒ(t) A

δ

-/2 -/2 /2 /2 T

T T

Figure 2.2


x = ±π,±2π,±3π,….,etc.

It is evident from Equation 18 that Fn is real, and hence we need only one spectrum for frequency domain representation. Also, since Sa(x) is an even function, it is obvious from Eq. 18 that Fn = F-n

Figure 2.3

The fundamental frequency ωο =2π/T. The frequency spectrum is a discrete function and exists only at ω = 0, ±2π/T, ±4π/T, ±6π/T,…..,etc. respectively. We shall consider the spectrum for some specific values of δ and T. The pulse width δ will be taken as 1/20 second, and the period T will be chosen as ¼ second, ½ second and 1 second, successively.

For δ = 1/20 and T =1/4 second, Eq: 18 is given by

14

)2/( onSa

T

AFn

But

To

2

Therefore

and

T

nn o

2

Hence

T

nSa

T

AFn

and

tjn

n

oeT

nSa

T

Atf

)(

→18

tjn

n

oeFntf

)(

Therefore


The fundamental frequency ωο = 2π/T = 8π. Thus the spectrum exists at ω = 0, ±8π, ±16π,….,etc, and is shown in Fig. 2.4(a).For δ = 1/20 and T =1/2 second, Eq:18 is given by

The spectrum is shown in Figure: 2.4(b) and exists at frequencies ω = 0, ±4π, ±8π,….,etc, For δ = 1/20 and T =1 second, Eq:18 is given by

The spectrum exists at ω = 0, ±2π, ±4π,….,etc, and is shown in Fig. 1.4(c).

It is evident that as the period T becomes larger and larger, the fundamental frequency 2π/T becomes smaller and smaller, and hence there are more and more frequency components in a given range of frequency. The spectrum therefore becomes denser as the period T becomes larger. However, the amplitudes of the frequency components become smaller and smaller as t is increased. In the limit as T is made infinity, as we have a single rectangular pulse of width δ, and the fundamental frequency become zero. The spectrum become and continuous and exists at every frequency. Note, however, that the shape of the frequency spectrum does not change with the period T.

15


Figure 2.4

16


2.4. Arbitrary Function Representation over Entire Interval – The Fourier Transforms

INTRODUCTION

We can represent any arbitrary non periodic function over an entire interval (–,) by first construction a periodic function of period T so that f(t) represents the first cycle of this periodic function. In the limit we let the period T become infinity, and this periodic function then has only one cycle in the interval (–,) and is represented by f(t). As we are familiar with the concept that as the period T is made larger, the following observations can be easily made

1. The fundamental frequency becomes smaller (o = 2/T).2. The frequency spectrum becomes denser; i.e. in a given frequency range there are

more frequency components. 3. The amplitude becomes smaller.4. The shape of the spectrum remains same.

2.4.1. CONTINUOUS SPECTRUM

The spectrum for an arbitrary function is a continuous function of . To show it we take an arbitrary function as given in the figure, and we want to represent over the entire interval (–,). As mentioned earlier we shall construct a new periodic function fT(t) with period T where the function f(t) repeats itself every T seconds as shown in the figure2.5(a,b)

Figure 2.5(a,b)As this new function is a periodic function hence an exponential Fourier series can represent it. In the limit if we let T become infinite, then the pulses in the periodic function repeat after an infinite interval. Then if in the limit T , fT(t) and f(t) are identical. That is,

Hence we can say that Fourier series representation of fT(t) over the entire interval will also represent f(t) over the entire interval if we let T= in this series.Using the Eq: 15 and 16 the exponential Fourier series for fT(t) can be given as

→19

17

t

f(t)

t

FT(t)

0TT


where

→20

Here the term Fn represents the amplitude of the component of frequency no. Now if we make T larger, o becomes smaller and the spectrum becomes denser and changes take place, which we mentioned just now. Now when in the limit T=,

1. The magnitude of each component becomes infinitesimally small2. There are also an infinite number of frequency components. 3. The spectrum exists an every value of .4. It is no longer a discrete but a continuous function of .

To show the continuity of the function we take the following substitutionsi.e. Fn is a function of n

→21Now Eq:19 will become

by substituting T=2/o we get

→22from Eq: 20 and 21 we have

→23

Equation 22 shows that fT(t) can be expressed as a sum of exponential signals of frequencies o, 1, 2, …etc. The amplitude of the component of frequency n is Fn= F(n)o,/2. We therefore note that the amplitude content in fT(t) of frequency n is not F(n) but is proportional to F(n).If we represent Eq: 22 then we observe that each frequency component is separated by a

distance o. hence area of the shaded rectangle in fig: is .

Figure 2.6Hence the Eq: 22 represents the sum of are under all such rectangles corresponding to n=– to . The sum of rectangular areas represents approximately the area under the

tjn

neF )(

otj

nneF )(Area

n-1 n n+1

o

18


dotted curve. This approximation becomes better as o becomes smaller. In the limit when T and o becomes infinitesimally small may be represented by d. So the discrete sum of Eq: 22 become the integral or the area under this curve. And the Eq: 22 and 23 become

→24

From Eq: 21 and 20 we have→25

So we have succeeded in representing a no periodic function f(t) in terms of exponential function over an entire interval (–,). The Eq: 19 represents f(t) as a continuous sum of exponential function with frequencies lying in the interval (– < < ). The amplitude of the components of any frequency is proportional to F(). Therefore F() represents the frequency spectrum of f(t) and is called the spectral density function. The frequency is continuous and exists at all values of .

4.4.2. THE FOURIER TRANSFORMS

Equation 19 and 20 are referred to as the Fourier transform pair. Equation 19 is called the direct Fourier transform of f(t) and Equation 20 is known as the inverse Fourier transform of f(). These are given as follows symbolically

and →26Thus F() is the direct Fourier transform of f(t), f(t) is the inverse Fourier transform of F(), and

So the Fourier transform is a tool that resolves a given signal into its exponential components. Hence F() is the frequency-domain representation of f(t), which specifies the relative amplitudes of the frequency components of the function, whereas time-domain representation specifies a function at each instance of time. Either representation uniquely specifies the function.

19


ExampleFind a Fourier transforms of a sequence of equidistant impulses of unit strength and separated by T seconds as shown in Fig 2.7.

Figure 2.7The impulse function is very important in sampling theory, and hence it is convenient to denote it by a special symbol δт(t). Thus

This is obviously a periodic function with period T. We shall first find the Fourier series for this function.

Consequently, Fn is a constant, 1/T. it therefore follows that the impulse train function of period T contains components of frequencies ω = 0,± ωο, ±2 ωο,……, ±n ωο,……,etc,( ωο = 2π/T) in the same amount.

20

.....)(..........)2()(

....)(........)2()()()(

nTtTtTt

nTtTtTtttT

n

nTt

n

tjnnT

oeFt )(

where

2/

2/)(

1 T

T

tjnTn dtet

TF o

2/

2/)(

1 T

T

tjnTn dtet

TF o

Function δт(t) in the interval (-T/2,T/2) is simply δ(t). Hence

TFn

1

From the sampling property of an impulse function the above equation reduces to

δ(T)

-4T -3T -2T -T T -2T -3T -4T


To find the Fourier transform of δт(t), since in this case fn = 1/t, it is evident that

The above Equation states that the Fourier transform of a unit impulse train of period T is also a train of impulses of strength ωο and separated by ωο radians (ωο = 2π/T). Therefore the impulse train function is its own transform. The sequence of impulses with periods T = ½ and T = 1 second and their respective transforms are shown in fig 2.8. It is evident that as the periods of the impulses increases, the frequency spectrum becomes denser.

Figure 2.8

5. SOME PROPERTIES OF THE FOURIER TRANSFORM

5.1. SYMMETRY PROPERTY:

21

n

oT nT

t )(1

2

n

onT

)(2

n

oo n )(

)( oo

ƒ(t)

-2 -1 1 2 3 4 t

F(ω)

-8π -4π 4π 8π ω

ƒ(t)

-4 -3 -2 -1 1 2 3 4 t

F(ω)

-4 -3 -2 -1 1 2 3 4 ω t


Proof

It can be seen easily that the Fourier transform of a gate function is a sampling function, and the Fourier transform of sampling function is gate function. The symmetry property holds for all, even ƒ(t).

5.2. LINEARITY PROPERTY:

5.3. SCALING PROPERTY:

Proof

22


5.4. FREQUENCY SHIFT PROPERTY:

ProofThe theorem states that a shift of ωo in the frequency domain is Equivalent to

multiplication by ejωot in the time domain. It is evident that multiplication by a factor e jωot

translates the whole frequency spectrum F(ω) by an amount ωo . Hence this theorem is also known as the frequency-translation theorem.Observing the identity that can easily show this,

23


24


5.5. TIME SHIFTING PROPERTY:

ProofThis theorem states that if a function is shifted in the time domain by to seconds, then its

magnitude spectrum |F(ω)| remains unchanged, but the phase spectrum is changed by an amount -ωto. We may state that a shift of to in the time domain is equivalent to multiplication by e-jωto in the frequency domain.

6. FOURIER TRANSFORM OF SOME USEFUL FUNCTIONS

6.1. SINGLE-SIDED EXPONENTIAL SIGNAL e-atu(t):

25


Note that the itegral converges only for a>0. For a<0 the Fourier transform does not exist. This also follows from the fact that for a<0, ƒ(t) is not absolutly integrable.

6.2. DOUBLE-SIDED EXPONENTIAL SIGNAL e-a|t|:

26

1

e-atu(t)

0 T

|F(ω)| 1/a

π/2

ω

θ(ω)


Note that in this case the phase spectrum θ(ω) = 0. The magnitude spectrum 2a/(a2 + ω2) is shown in figure.

6.3. A GATE FUNCTION:

1 |t| < τ/2A gate function Gτ (t) = 0 |t| > τ/2

27

1

e-a|t|

0 T

F(ω) 2/a

ω


The Fourier transform of this function is given by

Note that F(ω) is a real function and hence can be represented graphically by a single curve.

28

AGτ(t) A

-τ/2 τ/2 t


7. FOURIER TRANSFORMS INVOLVING IMPULSE FUNCTIONS

7.1. THE FOURIER TRANSFORM OF AN IMPULSE FUNCTIONS:

ƒ(t) = δ(t)The Fourier transform of unit impulse function δ(t) is given by.

It is therefore evident that an impulse function has a uniform spectral density over the entire frequency interval. In other words, an impulse function contains all frequency components with the same relative amplitudes.

29

dtett ttj )()(

unity. isfunction stepunit a of ansformFourier tr theThus

1)(

Hence

1e and 1)(

thereforeelsewhere, 0(t) and 0at t 1(t) that Know We

0

t

dtt


7.2. THE FOURIER TRANSFORM OF A CONSTANT:

ƒ(t) = A

This function does not satisfy the condition of absolute integrability. Nevertheless, it has a Fourier transform in the limit. We shall consider the Fourier transform of a gate function of height A and width τ seconds as shown in figure. In the limit as τ ∞,the gate function A. The Fourier transform of a constant A is therfore the Fourier transform of a gate function Gτ(t) as τ ∞.

7.3. FOURIER TRANSFORM OF Sgn(t):

THE SIGNUM FUNCTION ABBREVIATED AS SGN(T) IS DEFINED BY

30

ƒ(t)

δ(t)

0 t

F(ω)

1

ω


1 t>0sgn(t) = -1 t<0

it can be seen easily that sgn(t) = 2u(t) – 1

The Fourier transform of sgn(t) can be easily obtained when we observe that

31

sgn(t)

1

t

-1


7.4. FOURIER TRANSFORM OF UNIT STEP FUNCTION u(t):

THE UNIT STEP FUNCTION U(T) = 1 WHEN T ≥ 0 AND U(T) = 0 WHEN T<0, IT MEANS THAT

Convolution32

ƒ(t)

1 u(t)

0 t

|F(ω)| 1

0 Ω


What is Filtering?Filtering is one of the most important steps in geophysical data recording and processing. Often this step is explicitly called filtering; other times we easily recognize the data manipulation as a filtering operation. Every form of Earth science data from space imaginary to airborne, surface, and borehole geophysical surveying undergoes filtering before an analysis and interpretation of the data are made. In this sense, the very act of observing the data visually or graphically is filtering by our past experience and knowledge. Behavior scientists would say that any living organism has a response to a certain stimulus. This is filtering. Engineers describe filtering as an input signal being modified or transformed into a new output signal. The engineering approach of treating filtering as a "black box" in which an input signal s(t) is modified to yield a new output y(t) as shown in Figure. The only kind of filtering that we will consider is where the "black box" performs an operation called convolution. In fact, we can consider filtering to be identical to convolution

Black box filters input signal, s(t) to yield output signal, y(t). The black box performs convolution operation

The convolution, y(t) of two signals, s(t) and h(t) is expressed by the convolution integral

Discrete ConvolutionA to D conversion produces digital signals sampled at a particular sampling interval, Dt (or Dx). Assuming both s(t) and h(t) are digital functions with a sampling interval of unity, the convolution operation is defined as

Output, y(t)Input, s(t)

33


CONVOLUTION THEOREMGiven two functions f1(t) and f2(t), the following integral can be formed

This integral defines the convolution of two functions f1(t) and f2(t). This is also expressed as follows

We have two theorems Time Convolution Theorem Frequency Convolution Theorem

The Fourier transform of the convolution operation remarkably turns out to be a single complex multiplication of the respective Fourier transforms in the frequency domain. The quantities multiplied in the frequency domain are the complex spectra of the convolved signals. This simple relationship provides wonderful insight into many, many filtering operations and mathematical expressions as you will see. Explicitly we have

Next Figure graphically illustrates this Fourier transform pair using two rectangle functions each with amplitude A

34


There is a reciprocity relation between convolution in the time domain and its counterpart in the frequency domain. That is, convolution in the frequency domain becomes a multiplication in the time (or space) domain. This is sometimes called the "frequency domain convolution theorem." The Fourier transform pair representing this result is given below

35


The above figure illustrates this result using cosine and rectangle functions in the time domain

The above figure illustrates this result using cosine and rectangle functions in the time domain.Time Convolution Theorem

and

then

i.e.

proof

As we have the time-shifting property36


Using this property we observe the integral inside the bracket will be

Hence

Frequency Convolution And

Then

i.e

proof.

As we have frequency-shifting property

i.e.

By applying this to the expression in the bracket we get

Again we can observe that the integral in the inverse Fourier transform of f1(t)

ResultsThe results that we get are that

Convolution of two functions in the time domain is Equivalent to multiplication of their spectra in the frequency domain.

Multiplication of two functions in the time domain is Equivalent to convolution of their spectra in the frequency domain.

37


Convolution RelationshipThe convolution is a special kind of multiplication; it is possible though to write the laws of convolution algebra along lines that are similar to those for ordinary multiplication

Graphical Interpretation of Convolution Theorem To illustrate the convolution graphically let us consider to function f1(t) and f2(t) as shown in the figure. By definition we have

The independent variable in the convolution is . The function f1() and f2(–) are shown in the figure b. f2(–) is obtained by folding f2() about the vertical axis passing through the origin. The term f2(t–) represents the function f2(–) shifted by t seconds along the positive axis.Figure c shows f2(t–). The value of convolution integral at t = t1 is given by the integral in the above Equation evaluated at t = t1. This is clearly the area under the product curve of f1() and f1(t–). This area is shown shaded in fig: d. The value of f1(t)*f2(t) at t = t1 is Equal to this shaded area. We choose different values of t, shift the function f2(–) accordingly, and find the area under the new product curve. These areas represent the value of the convolution function f1(t)*f2(t) at the respective values of t. the plot of the are under the product curve as a function of t represent the desired convolution function f1(t)*f2(t) shown in the next fig.

MechanismThe graphical mechanism of convolution can be appreciated by visualizing the function f2(–) as a rigid frame which is being progress along the axis by t, seconds. The function represented by this frame is multiplied by f1(), and the area under the product

38

1

f1(t)

t

f2(t)

½

1 t

1

a1

1

f1( )

1

½

f2(–)

b

1

1

f1( )

1

½

f2(t1–)

c

1

f1( )

d

1

f2( t1–)f1()f2( t1–)

f2( t2–)f1( )

f1()f2( t2–)11


curve is the value of the convolution function at t = t1. Therefore, to find the value of f1(t)*f2(t) at any time, say t = t0, we displace the rigid frame representing f2(–) by t0 seconds along the axis and multiply this function with f1(). The area under the product curve is the desired values of f1(t)*f2(t), we progress the frame successively by different amounts and find the area as a function of displacement of the frame represents the required convolution function f1(t)*f2(t).Now let us take our functions and start convolving them.

We see the effect of drifting the f2(t) with respect to t as follows

39

1

f1() 1

–1-¼

-½

-¾

½

¼ ½ ¾10

f1() 1

–1

-¼

-½

-¾

½

f1()¼ ¾

10

1

-¼

-¾

½

10

f1() 1

–1-¼

-½

-¾

½

¼ ½ ¾¼ ½ ¾

10

f1() 1

–1-¼

-½

-¾

½

10

f1() 1

–1

-¼

-½

-¾

½

¼ ½ ¾¼ ½ ¾

¼ ½ ¾ 10

f1() 1

–1-¼

-½

-¾

½

t¼ ½ ¾ 1 5

47

43

2

⅛

¼

⅜

½ f1(t)*f2(t)


The convolution is shown in the figure above.

.

THE SAMPLING THEOREM

Statement

A band limited signal, which has no spectral components above a frequency fm Hz is uniquely determined by its values at uniform intervals less than 1/2fm seconds apart.

Explanation

This statement of the theorem implies that if the Fourier transform of f(t) is zero beyond a certain frequency m = 2fm, then the complete information about f(t) is contained in its samples spaced uniformly at a distance less than 1/2fm seconds. The function f(t) is sampled once every T seconds (T≤1/2fm) or at a rate greater than or equal to 2fm samples per second. It follows from the sampling theorem that these samples contain the information about f(t) at every values of t. the sampling rate, however, must be at least twice the highest frequency fm present in the spectrum of f(t). To say it another way, the signal must be sampled at least twice during each period or cycle of its highest frequency component.

1. Graphical Proof

The sampling theorem can be easily proved with the help of the frequency convolution theorem. Consider a band limited signal f(t) which has no spectral components above fm cycles per second. This means that F(),the Fourier transform of f(t), is zero for ││>m (m = 2fm). Suppose we multiply the function f(t) by a periodic impulse function T(t). The product function is a sequence of impulse located at regular intervals of T seconds and having strengths equal to the values of f(t) at the corresponding instants. The product f(t)T(t) indeed represents the function f(t) sampled at a uniform interval of T seconds. Let us represent the sampled function by fs(t)

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The frequency spectrum of f(t) is F(). As the Fourier transform of a uniform train of impulse function T(t) is also a uniform train of impulse function . The impulses are separated by a uniform interval 0=2/T.

The Fourier transform of f(t)T(t) will, according to the frequency convolution theorem, be given by the convolution of F() with .

Substituting 0=2/T, we get

→ 1

The above equation shows that the spectrum of the sampled signal fS(t) is given by the convolution of F() with a train of impulses. Functions F() and can be convolved graphically. In order to perform this operation, we fold back the function about the vertical axis = 0. Since is an even function of , the folded function is the same as the original function . To perform the operation of convolution, we now progress the whole train of impulsesin a positive direction. As each impulse passes across F(), it reproduces F() itself. Since the impulses are spaced at a distance 0 = 2/T, the operation of convolution yields F() repeating itself every o radians per second. The spectral density function (the Fourier transform) of fS(t) is therefore the same as F() but repeating itself periodically, every 0 radians per second. The function will be designated as FS(). Note that F() will repeat periodically without overlap as long as 0 ≥ 2m, or

Therefore, as long as we sample f(t) at regular intervals less than 1/2fm seconds apart. FS(), the spectral density function of f(t), will be a periodic replica of F() and therefore contains all the information of f(t). We can easily recover F() from FS() by allowing the sampled signal to pass through a low-pass filter which will only allow frequency components below fm and attenuate all the higher frequency components. Thus it is evident that the sampled function fS(t) contains all of the information of f(t). To recover f(t) from fS(t), we allow the sampled function fS(t) to pass through a low-pass filter which permits the transmission of all of the components of frequencies below fm and attenuates all of the components of frequencies above fm. The ideal filter characteristic to achieve this is shown dotted in Fig 8.1

Note that if the sampling interval T becomes larger than 1/2fm, then the convolution of F() with yields F() periodically. But now there is an overlap between successive cycles, and F() cannot be recovered from FS(). Therefore, if the sampling interval T is made too large, the information is partly lost, and the signal f(t) cannot be recovered from the sampled signal fS(t). This conclusion is quite logical since it is reasonable to expect that the information will be lost if the sampling is too slow. The maximum interval of sampling T = 1/2fm is also called the Nyquist interval.

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8.2. ANALYTICAL PROOF

F() * can be readily derived by analytical procedure. We have

From Eq:1 we it follow that

As we have

By applying this to our equation we get

The right hand side of the equation represents function F() repeating itself every 0 radians per second. This is exactly the same result obtained by graphical convolution.

9. DATA ENCODING

Before the data is transmitted over some medium it is encoded into some proper format this is called data encoding. Both digital and analog information (Data) can be encoded either analog or digital signals. The encoding that has been chosen depends upon specific requirements to be met and the media and communication facilities available.

Figure 8.2

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There are four possible combinations when data or information is transmitted and encoded.

Analog data, analog signal. Digital data, digital signal Analog data, digital signal Digital data, analog signal

Then there come the techniques involved in each of these four combinations.

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