Principles of Communications -...

Meixia Tao @ SJTU

Principles of Communications

Meixia Tao

Dept. of Electronic EngineeringShanghai Jiao Tong University

Chapter 3: Analog ModulationSelected from Ch 3, Ch 4.1-4.4, Ch 6.1-6.2 of of Fundamentals of Communications Systems, Pearson Prentice Hall 2005, by

Proakis & Salehi

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Topics to be Covered

AM/FM radio FM radio

Source Modulator Channel Demodulator Output

Amplitude modulation Angle modulation (phase/frequency) Effect of noise on amplitude modulation Effect of noise on frequency modulation

TV broadcast

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Modulation What is modulation?

Transform a message into another signal to facilitate transmission over a communication channel

Generate a carrier signal at the transmitter Modify some characteristics of the carrier with the information

to be transmitted Detect the modifications at the receiver

Why modulation? Frequency translation Frequency-division multiplexing Noise performance improvement

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Analog Modulation Characteristics that be modified in sin carrier

Amplitude → Amplitude modulation Frequency Phase

→ Angle modulation

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Baseband signal (modulating wave):

Carrier wave

Modulated wave

( )m t

( )0( ) ( ) ( ) ( ) cosc cs t c t m t A m t tω θ= = +

Amplitude ModulationDouble-sideband suppressed-carrier AM (DSB-SC)

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Spectrum of DSB-SC Signals

[ ])()(21)( ccc ffMffMAfS ++−=

0

S(f)

ffc+Wfcfc-W-fc+W-fc-W -fc

(1/2)AcM(0)Spectrum of DSB-SC

USBUSB LSBLSB

W-W 0

M(f)

f

M(0)Spectrum of message

Translation of the original message spectrum to

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Bandwidth and Power Efficiency

Required channel bandwidth Required transmit power

0

S(f)


(1/2)AcM(0)

2cB W=

[ ]

/2 /22 2 2 20/2 /2

2 2/2 20/2

1 1lim ( ) lim ( )cos ( )

1lim ( ) 1 cos(2 2 )2 2

T T

s c cT TT T

Tc cc mTT

P s t dt A m t t dtT T

A Am t t dt PT

ω θ

ω θ

− −→∞ →∞

−→∞

= = +

= + + =

∫ ∫

∫

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Demodulation of DSB-SC Signals Phase-coherent demodulation

If there is a phase error φ, then

)2cos( tfcπ

Product modulator

Local oscillator

Low-pass filter

s(t) v(t) vo(t)

Scaled version of message signal Unwanted

PLL (phase-locked loop)

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Demodulation of DSB-SC Signals Pilot-tone assisted demodulation

Add a pilot-tone into the transmitted signal

Filter out the pilot using a narrowband filter

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Conventional AM Carrier wave: Baseband signal (normalized):

Modulation index: Modulated wave

a

Modulating wave

Modulated wave 1a ≤

1a >

overmodulated10

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Spectrum of Conventional AM

( ) ( ) ( ) ( )( )2 2

c cc c c c

A A aS f f f f f M f f M f fδ δ= − + + + − + +

(Ac/2)δ(f-fc)

0

S(f)


(1/2)aAcM(0)(Ac/2)δ(f+fc)

W-W 0

M(f)

f

M(0)Spectrum of message signal

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Bandwidth and Power Efficiency Required channel bandwidth Required transmit power

Modulation efficiency

message powercarrier power

22

2

2 2 22

2=1

2 2

c mm

c mc m

a A P a PEA a a PA P

= =++

power in sideband

total power

2cB W=

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Example Message signal: Carrier: Modulation index: a=0.85 Determine the power in the carrier component and in the

sideband components of the modulated signal

( ) 3cos(200 ) sin(600 )m t t tπ π= +5( ) cos(2 10 )c t t−= ×

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Demodulation of AM signalsEnvelop Detector

The simplicity of envelop detector has made Conventional AM a practical choice for

AM-radio broadcasting

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Single Sideband (SSB) AM

Common problem in AM and DSBSC: Bandwidth wastage

SSB is very bandwidth efficient

ω

H(ω)

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Expression of SSB signals The baseband signal can be written as the sum of finite

sinusoid signals

Then its USB component is

After manipulation

1( ) cos(2 ) ,

n

i i i i ci

m t x f t f fπ θ=

= + ≤∑

[ ]1

( ) cos 2 ( ) )2

nc

c i c i ii

Am t x f f tπ θ=

= + +∑

1 1( ) cos(2 ) cos 2 sin(2 ) sin 2

2

( )cos 2 ( )sin 22 2

n nc

c i i i c i i i ci i

c cc c

Am t x f t f t x f t f t

A Am t f t m t f t

π θ π π θ π

π π

= =

= + − +

= −

∑ ∑

Hilbert transform of m(t)

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About Hilbert Transform

1 ( ) 1( ) ( )xx t d x tt tτ τ

π τ π∞

−∞= = ∗

−∫( ) ( )x t x t⇔

1

ω

)(ωH

ω

相移

90°

-90°

, 0( ) , 0

0, 0

j fH f j f

f

− >=

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Generation of SSB-AM Signal

• The spectral efficiency of SSB makes it suitable for voice communication over telephone lines (0.3~3.4 kHz)

• Not suitable for signals with significant low frequency components

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Vestigial Sideband: VSB VSB is a compromise between SSB and DSBSC

W-W 0

M(f)

f

0

S(f)

ffc

fv W

-fc

fvW

VSB signal bandwidth is B = W+fv

VSB is used in TV broadcasting and similar signals where low frequency components are significant

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Comparison of AM Techniques DSB-SC:

more power efficient. Seldom used Conventional AM:

simple envelop detector. AM radio broadcast

SSB: requires minimum transmitter power and bandwidth. Suitable for point-to-point and over long distances

VSB:bandwidth requirement between SSB and DSBSC. TV transmission

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Signal Multiplexing Multiplexing is a technique where a number of

independent signals are combined and transmitted in a common channel

These signal are de-multiplexed at the receiver Two common methods for signal multiplexing

TDM (time-division multiplexing) FDM (frequency-division multiplexing)

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FDM

LPF: ensure signal bandwidth limited to W

MOD (modulator): shift message frequency range to mutually exclusive high frequency bands

BPF: restrict the band of each modulated wave to its prescribed range

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FDM application in telephone comm.

Voice signal: 300～3400Hz Message is SSB modulated. In 1st-level FDM, 12 signal are stacked in frequency, with a freq.

separation of 4 kHz between adjacent carriers A composite 48 kHz channel, called a group channel, transmits 12

voice-band signals simultaneously In the next level of FDM, a number of group channel (typically 5

or 6) are stacked to form a supergroup channel Higher-order FDM is obtained by combining several supergroup

channels => FDM hierarchy in telephone comm. systems

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Quadrature-Carrier Multiplexing Transmit two messages on the same carrier as

cos() and sin() are two quadrature carriers Each message signal is modulated by DSB-SC Bandwidth-efficiency comparable to SSB-AM

Synchronous demodulation of m1(t):

( ) ( )1 2( ) ( ) cos 2 ( )sin 2c c c cs t A m t f t A m t f tπ π= +

( ) ( ) ( ) ( )

( ) ( )

21 2

1 1 2

( ) cos 2 ( )cos 2 ( )sin 2 cos 2

( ) ( ) cos 4 ( )sin 42 2 2

c c c c c c

c c cc c

s t f t A m t f t A m t f t f tA A Am t m t f t m t f t

π π π π

π π

= +

= + +

LPF24

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Application: AM Radio Broadcasting Commercial AM radio uses conventional AM Superheterodyne receiver:

from variable carrier freq of the incoming RF to fixed IF

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Angle Modulation Either phase or frequency of the carrier is varied according

to the message signal

The general form:

θ(t): the time-varying phase

instantaneous frequency of s(t):

1 ( )( )2i c

d tf t fdtθ

π= +

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Representation of FM and PM signals Phase modulation (PM)

Frequency modulation (FM)

The phase of FM is0

( ) 2 ( )t

ft k m dθ π τ τ= ∫

where kf = frequency deviation constant/frequency sensitivity

1( ) ( ) ( )2i c f

df t f k m t tdtθ

π− = =

( ) ( )pt k m tθ = where kp = phase deviation constant

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Distinguishing Features of PM and FM No perfect regularity in spacing of zero crossing

Constant envelop, i.e. amplitude of s(t) is constant

Relationship between PM and FM

Will discuss the properties of FM only

[ ]∫+ dttmktfA pcC )(2cos πintegrator

m(t)

)2cos( tfA cC π

Phasemodulator

∫ dttm )( FM wave

differentiator Frequencymodulatorm(t)

)2cos( tfA cC π

)(tmdtd

[ ])(22cos tmktfA fcC ππ +PM wave

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Example: Sinusoidal ModulationSinusoid modulating wave m(t)

FM wave

PM wave

)(tmdtd

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Example: Square ModulationSquare modulating wave m(t)

FM wave

PM wave

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FM by a Sinusoidal Signal Message

Instantaneous frequency of resulting FM wave

Frequency deviation: Carrier phase

Modulation index:

( )0

( ) 2 ( ) sin(2 )

sin(2 )

t

i c mm

m

ft f f d f tf

f t

θ π τ τ π

β π

∆= − =

=

∫

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Example

Problem: a sinusoidal modulating wave of amplitude 5V and frequency 1kHz is applied to a frequency modulator. The frequency sensitivity is 40Hz/V. The carrier frequency is 100kHz.

Calculate (a) the frequency deviation(b) the modulation index

Solution: Frequency deviation

Modulation index

HzAkf mf 200540 =×==∆

2.01000200

==∆

=mffβ

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Spectrum Analysis of Sinusoidal FM Wave

Rewrite the FM wave as

Define the complex envelop of FM wave

retains complete information about s(t)

)2sin()()()(~ tfjcQI meAtjststsπβ=+=

)(~ ts[ ]{ } [ ]tfjtftfjc cmc etseAts ππβπ 2)2sin(2 )(~ReRe)( == +

In-phase component Quadrature-phase component

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is periodic, expanded in Fourier series as

where

n-th order Bessel function of the first kind

Hence,

∑∞

−∞=

=n

tnfjn

mects π2)(~

( )1( ) exp sin2n

J j x nx dxπ

πβ β

π −= − ∫

)(βncn JAc =

)2sin()(~ tfjc meAtsπβ=

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Substituting into

FM wave in time domain

FM wave in frequency-domain

)(~ ts

( )∑∞

−∞=

=n

mnc tnfjJAts πβ 2exp)()(~

[ ])()()(2

)( mcmcn

nc

c nfffnfffJAfS +++−−= ∑∞

−∞=

δδβ

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Properties of Bessel Function

∞→→ ββ as0)(nJ

−

=− odd,)(even,)(

)(nJ

nJJ

n

nn β

ββ

Property 1: for small β ≤0.3 (Narrowband FM) Approximations

Substituting above into 1,0)(

2)(1)(

1

0

>≈≈≈

nJJJ

n βββ

β

[ ]

[ ]tffAtffAtfAts

mcc

mcc

cc

)(2cos2

)(2cos2

)2cos()(

−−

++≈

πβ

πβπ

Approximate bandwidth =Discuss the similarity between the conventional AM wave and

a narrow band FM wave37

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General Case Goal: to investigate how and affect the spectrum Fix and vary

and are varied

f∆2

5=β

cf

1.0

f∆2

1=β

cf

1.0

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Fix and vary is fixed, but is varied

f∆2

1=β

cf

1.0

f∆2

5=β

cf

1.0

General Case

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Effective Bandwidth of FW Waves

For large B is only slightly greater than

For small The spectrum is limited to

Carson’s Rule:

2 2 2(1 )m mB f f fβ≈ ∆ + = +

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99% bandwidth approximation The separation between the two frequencies beyond which

none of the side-frequencies is greater than 1% of the unmodulated carrier amplitude

i.e where is the max that satisfies

01.0)( >βnJ

β 0.1 0.3 0.5 1.0 2.0 5.0 10 20 30

2nmax 2 4 4 6 8 16 28 50 70


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A universal curve for evaluating the 99% bandwidth As increases, the bandwidth occupied by the significant side-

frequencies drops toward that over which the carrier frequency actually deviates, i.e. B become less affected by

20

2

0.2 2


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FM by an Arbitrary Message Consider an arbitrary with highest freq. component W

Frequency deviation:

Modulation index:

Carson’s rule applies as

Carson’s rule underestimates the FM bandwidth requirement

Universal curve yields a conservative result

max ( )ff k m t∆ =

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Example

In north America, the maximum value of frequency deviation is fixed at for commercial FM broadcasting by radio.

Take , typically the maximum audio frequency of interest in FM transmission, the modulation index is

Using Carson’s rule,

Using universal curve,

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Exercise

Assuming that , determine the transmission bandwidth of an FM modulated signal with

Solution By Carson’s rule:

( )4( ) 10sinc 10m t t=

4000fk =

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Application: FM Radio broadcasting As with standard AM radio, most FM radio receivers are

of super-heterodyne type

Limiter

discriminator

Audio amplifier with de-emphasis

Baseband low-pass filter

loudspeaker

Typical freq parameters– RF carrier range = 88~108 MHz– Midband of IF = 10.7MHz– IF bandwidth = 200kHz– Peak freq. deviation = 75KHz

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Summary Spectrum of sinusoidal FM Wave

Carson rule approximation Universal curve approximation

[ ])()()(2

)( mcmcn

nc

c nfffnfffJAfS +++−−= ∑∞

−∞=

δδβ

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Generation of FM waves Direct approach

Design an oscillator whose frequency changes with the input voltage => voltage-controlled oscillator (VCO)

Indirect approach First generate a narrowband FM signal and then

change it to a wideband signal Due to the similarity of conventional AM signals, the

generation of a narrowband FM signal is straightforward.

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Generation of Narrow-band FM

Consider a narrow band FM wave

where

Given φ1(t)

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)()()()( 1212112 tsatsatsats

nn+++=

integratorm(t)

)2sin( 11 tfA π

ProductModulator

-900 phaseshifter )2cos( 11 tfA π

Carrier wave

+-

+Narrow-bandFM wave s1(t)

Narrow-band frequency modulator

Next, pass s1(t) through a frequency multiplier

– The input-output relationship of the non-linear device is:

– The BPF is used to Pass the FM wave centred at nf1 and with deviation n∆f1 and suppress all other FM spectra

Memorylessnonlinear device

Narrow-bandFM wave

Band-passfilter

Wideband FMWave

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Example: frequency multiplier with n = 2

Problem: Consider a square-law device based frequency multiplier

with

Specify the midband freq. and bandwidth of BPF used in the freq. multiplier for the resulting freq. deviation to be twice that at the input of the nonlinear device

Solution:

)()()( 212112 tsatsats +=

+= ∫

tdmktfAts

01111)(22cos)( ττππ

+++

+=

++

+=

∫∫

∫∫tt

tt

dmktfAaAadmktfAa

dmktfAadmktfAats

011

212

212

01111

01122

12011112

)(44cos22

)(22cos

)(22cos)(22cos)(

ττππττππ

ττππττππ

Removed by BPF withfc=2f1BW > 2∆f = 4∆f1

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Integrator

Messagesignal Narrow-band

phasemodulator

WidebandFM signalFrequency

multiplier

Crystal-controlledoscillator

+= ∫

t

fcc dmktfAts 0 )(22cos)( ττππ

1

1

1

fnfnkknff

f

c

∆=∆

==

+= ∫

tdmktfAts

01111)(22cos)( ττππ

1cos(2 )cA f tπ

Generation of Wideband FM Signal

Output

)2cos( tflπ

BPF

Mixer: perform up/down conversion to shift the signal to the desired center freq.

may not be the desired carrier frequency

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Exercise: A typical FM transmitter

Problem: Given the simplified block diagram of a typical FM transmitter used to transmit audio signals containing frequencies in the range 100Hz to 15kHz.

Desired FM wave: fc = 100MHz, ∆f = 75kHz. Set β1 = 0.2 in the narrowband phase modulation to limit harmonic

distortion. Specify the two-stage frequency multiplier factors n1 and n2

Integrator

Messagesignal Narrow-band

phasemodulator

FM signalFrequencymultiplier

n1


Mixer


Frequencymultiplier

n2

0.1MHz ?

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Demodulation of FMBalanced Frequency Discriminator

Given FM wave

Differentiator + envelop detector = FM demodulator Frequency discriminator: a “freq to amplitude” transform device

FMwave

Basebandsignal

Envelopdetector

Slope circuitH1(f)

Slope circuitH2(f)

Envelopdetector

∑+

-

( )( )

+−≤≤−−−++≤≤−+−

=elsewhere ,0

2/2/,2/22/2/ ,2/2

)(1 BffBfBffajBffBfBffaj

fH cccccc

ππ

)()( 12 fHfH −=

+= ∫

t

fcc dmktfAts 0 )(22cos)( ττππ

0( ) 2 2 ( ) sin 2 2 ( )

t

c c f c fd s t A f k m t f t k m ddt

π π π π τ τ = − + + ∫Hybrid-modulated wave with AM and FM

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Circuit diagram and frequency response

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Application: FM Radio broadcasting As with standard AM radio, most FM radio receivers are

of super-heterodyne type

Limiter

discriminator

Audio amplifier with de-emphasis

Baseband low-pass filter

loudspeaker

Typical freq parameters– RF carrier range = 88~108

MHz– Midband of IF = 10.7MHz– IF bandwidth = 200kHz– Peak freq. deviation = 75KHz

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Meixia Tao @ SJTU10/4/2004 57

FM Radio Stereo Multiplexing Stereo multiplexing is a form of FDM

designed to transmit two separate signals via the same carrier.

Widely used in FM broadcasting to send two different elements of a program (e.g. vocalist and accompanist in an orchestra) so as to give a spatial dimension to its perception by a listener at the receiving end

∑

∑

Frequencydoubler

K

∑

+

++

-

ml(t)

mr(t)+

+ m(t)

)2cos( tf cπ

[ ][ ]

)2cos()4cos()()(

)()()(

tfKtftmtm

tmtmtm

c

crl

rl

ππ

+−+

+=

The sum signal is left unprocessed in its baseband form

The difference signal and a 38-kHz subcarrier produce a DSBSC wave

The 19-kHz pilot is included as a reference for coherent detection

fc = 19kHz

Meixia Tao @ SJTU10/4/2004 58

FM-Stereo Receiver

Baseband LPF

BPF centered at 2fc=38kHz

Narrow-band filter tuned to

fc=19kHz

Frequency doubler

m(t)

∑

∑

+

++

-

2ml(t)

2mr(t)

Baseband LPF

ml(t)+mr(t)

ml(t)-mr(t)

To two loudspeakers

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Think …

Compared with AM, FM requires a higher implementation complexity and a higher bandwidth occupancy. What is the advantage of FM then?

AM vs. FM

Why AM radio is mostly for news broadcasting while FM radio is mostly for music program

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Suggested Reading Chapter 3 and Chapter 4.1-4.4 of Fundamentals of

Communications Systems, Pearson Prentice Hall 2005, by Proakis & Salehi

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A Benchmark System Baseband system:

No carrier demodulation The receiver is an ideal LPF with bandwidth W Noise power at the output of the receiver

Baseband SNR is given by

+ LPFn(t)

)(tsm

0

002

W

n W

NP df N W−

= =∫

0

R

b

PSN N W

=

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Example: Find the SNR in a baseband system with a bandwidth

of 5 kHz and with W/Hz. The transmitter power is 1kW and the channel attenuation is

Solution:

140 10N

−=1210−

12 3 910 10 10 WattsRP− −= × =

9

140

10 2010 5000

R

b

PSN N W

−

−

= = = ×

1010 log 20 13dB= =

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Effect of Noise on DSBSC

Modulated signal Input to the demodulator

+ BPF demodn(t)

( )s t

)(tni

)(0 tm

)(0 tn( ) ( ) cosc cs t A m t tω=

( ) ( ) ( )( ) cos ( ) cos ( )sin

i

c c c c s c

r t s t n tA m t w t n t w t n t w t

= += + −

Here ni(t) is a Gaussian narrow-band noise

0 / 2( )0 otherwisei

cn

N f f WS f

− ≤=

( )s t

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In the demodulator, the received signal is first multiplied by a locally generated sinusoid signal

Assume coherent detector, we have

( ) cos( ) ( ) cos cos( )( ) cos cos( ) ( )sin cos( )

c c c c

c c c s c c

r t w t A m t w t w tn t w t w t n t w t w t

φ φφ φ

+ = +

+ + − +

[ ]

[ ]

1 1( )cos ( ) cos(2 )2 21 ( )cos ( )sin21 ( )cos(2 ) ( )sin(2 )2

c c c

c s

c c s c

A m t A m t w t

n t n t

n t w t n t w t

φ φ

φ φ

φ φ

= + +

+ +

+ + − +

0φ =65

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Then the signal is passed through a LPF with bandwidth W

The output SNR can thus be defined as

Since the received power of DSBSC in baseband is The output SNR can be rewritten as

[ ]1( ) ( ) ( )2 c c

y t A m t n t= +

22

0

141 24

oc

c mo c m

o nn

A PP A PSN P WNP

= = =

( ) ( ) ( )c i in n c n c

S f S f f S f f

for f W

= − + +

≤where

2

2c m

RA PP =

0

R

oDSB b

PS SN WN N

= =

DBSSC does not provide any SNR improvement over a simple baseband systems

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Effect of Noise on SSB Modulated signal: Input to the demodulator

Output of LPF:

Therefore, the output SNR is

( ) ( ) cos ( )sinc c c cs t A m t w t A m t w t= ±

[ ] [ ]( ) ( ) cos ( ) ( ) sinc c c c s cA m t n t w t A m t n t w t= + + ± −

1( ) ( ) ( )2 2

cc

Ay t m t n t= +

0 / 2 / 2( )0 otherwisei

cn

N f f WS f

− ≤=

( ) ( ) ( )ir t s t n t= + where

22

0

1414

oc

c mo c m

o nn

A PP A PSN P WNP

= = =

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But in this case,

Thus,

2R c mP A P=

0

R

oSSB b

PS SN WN N

= =

SNR in an SSB system is equivalent to that of a DSBSC system

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Effect of Noise on Conventional AM Modulated signal

Input to the demodulator

With coherent detector, after mixing and LPF:

Removing DC component

[ ]( ) 1 ( ) cosc cs t A am t w t= +

[ ]( ) ( ) ( )

( ) ( ) cos ( )sini

c c c c s c

r t s t n tA A am t n t w t n t w t

= +

= + + −

[ ]11( ) ( ) ( )2 c c c

y t A A am t n t= + +

[ ]1( ) ( ) ( )2 c c

y t A am t n t= +

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Exercise: please show that the output SNR is

Hint: the received signal power is

Modulation efficiency is

2 21 12R c m

P A a P = +

SNR in conventional AM is always smaller than that in baseband.

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Performance of Envelope-Detector Input to the envelope-detector

Envelope of r(t)

If signal component is much stronger than noise

After removing DC component, we obtain

[ ]( ) ( ) ( ) cos ( )sinc c c c s cr t A A am t n t w t n t w t= + + −

[ ]2 2( ) ( ) ( ) ( )r c c c sV t A A am t n t n t= + + +

( ) ( ) ( )r c c cV t A A am t n t≈ + +

( ) ( ) ( )c cy t A am t n t= +

At high SNR, performance of coherent detector and envelop detector is the same

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Performance of Envelope-Detector If noise power is much stronger than the signal power

( ) ( )22 2 2( ) 1 ( ) ( ) ( ) 2 ( ) 1 ( )r c c s c cV t A am t n t n t A n t am t= + + + + +

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

c cc s

c s

A n tn t n t am tn t n t

≈ + + + +

Ignore 1st term

( )

( )

2

( )( ) 1 1 ( )( )

( )( ) 1 ( )( )

c cn

n

c cn

n

A n tV t am tV t

A n tV t am tV t

≈ + +

= + +

1 12εε+ ≈ +

2 2( ) ( ) ( )n c sV t n t n t= +

The system is operating below the threshold, no meaningful SNR can be defined.

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Exercise Consider that the message is a WSS r.p M(t) with

autocorrelation function . It is given that .

We want to transmit this message to a destination via a channel with a 50dB attenuation and additive white noise with PSD . . We also want to achieve an SNR at the modulator output of at least 50dB.

What is the required transmitted power and the channel bandwidth if we employ the following modulation schemes? DSB-SC SSB AM with modulation index = 0.8

2( ) 16sinc (10000 )MR τ τ=max

( ) 6m t =

120( ) / 2 10 W/HznS f N

−= =

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Effect of Noise in FM The effect of additive noise is described by the

changes of frequency, or the changes in the zero-crossings of the modulated FM single.

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Effect of Noise on Angle Modulation

Block diagram of an angle demodulator

Input to the demodulator is

BPFBW=Bc Demod

LPFBW=W

( ) ( )ws t n t+( ) ( ) ( )r t s t n t= + ( )y t

o

SN

[ ]cos ( ) ( )cos ( )sinc c c c s cA t t n t t n t tω φ ω ω= + + −( )r t =

( ) cos( ( ))n c nR t w t tθ= +

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Assume that the signal is much larger than the noise

( )cn t

( )nR t

( )sn t

( )tφ ( )y tθ

( )e tθ () ( )

nt tθ

φ−

( )nR t

( )n tθ

( )yR

t

2 2 ( )cA E n t

cA

( )( )( )

1

( ) ( ) cos ( ) ( )

( )sin ( ) ( )cos ( )

( ) cos ( ) ( )

c n n

n nc

c n n

r t A R t t t

R t t tw t t tg

A R t t t

θ φ

θ φφ

θ φ−

≈ + − ⋅ −

+ + + −

( )( )( ) ( ) sin ( ) ( )nr nc

R tt t t tA

θ φ θ φ= + −

The phase term can be further approximated as

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Therefore, the output of the demodulator is

The noise component is inversely proportional to the signal amplitude Ac. (This is not the case for AM system)

( )( )( ) ( ) ( ) sin ( ) ( )2 2

( ) ( )2

nr f n

c

f n

R td dy t t k m t t tdt dt A

dk m t Y tdt

θ θ φπ π

π

= = + −

= +

Desired signal Noise

( ) [ ]( ) 1( ) sin ( ) ( ) ( ) cos ( ) ( )sin ( )nn n s cc c

R tY t t t n t t n t tA A

θ φ φ φ= − = −

[ ]1 ( )cos ( )sins cc

n t n tA

φ φ= − (Since is slowly varying)( )tφ

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The power spectral density of is 1 ( )2 n

d Y tdtπ

2 22 22

2

22

022

4 cos sin( ) ( ) ( )4

2( )

0 otherwise

n s c

c

Y n nc c

Cn c

c

f S f f S f S fA A

f N f Bf S f AA

π φ φπ

= +

≤= =

xf− xf2TB− 2TB f

( )nfS f( )noS f

W-W

At the output of LPF, the noise is limited to the freq. range [-W, W]

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Output SNR in FM Now we can determine the output SNR in FM First, the output signal power is

The output noise power is

Then, the output SNR is

2os f m

P k P=

320 0

2 2

23o

W

n Wc c

N N WP f dfA A−

= =∫

0

2 2

20

32

o

s f c m

o n

P k A PSN P W N W

= = ( )

2

2

3

max ( )f m

b

P SNm t

β =

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Key Observations Rewrite

Increasing increases the output SNR, in contrast to AM Increasing the bandwidth (by Carson rule )

increases the output SNR. Thus, FM provides a way to trade off bandwidth for transmitted power

Having a large B means having a large noise power. Thus, Increasing cannot continue improving the performance indefinitely due to the threshold effect

Increasing the transmitted power increases output SNR in both FM and AM systems, but the mechanisms are totally different.

β

β

Power content of the normalized message

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Threshold Effect There exists a specific SNR at the input of the

demodulator below which the signal is not distinguishable from the noise

0

FM

DSB

0

0

NS

i

i

NS

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Comparison of Analog-Modulation Bandwidth efficiency

SSB is the most bandwidth efficient, but cannot effectively transmit DC

VSB is a good compromise PM/FM are the least favorable systems

Power efficiency (reflected in performance with noise) FM provides high noise immunity Conventional AM is the least power efficient

Implementation complexity (transmitter and receiver) The simplest receiver structure is conventional AM FM receivers are also easy to implement DSB-SC and SSB-SC requires coherent detector and hence is

much more complicated.

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Think …

Compared with AM, FM requires a higher implementation complexity and a higher bandwidth occupancy. What is the advantage of FM then?

AM vs. FM

FM provides high noise immunity. This is why AM radio is mostly for news broadcasting while FM radio is mostly for music program.

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Suggested Reading Chapter 6.1-6.3 of of Fundamentals of

Communications Systems, Pearson Prentice Hall 2005, by Proakis & Salehi

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Principles of CommunicationsTopics to be CoveredModulationAnalog ModulationAmplitude ModulationSpectrum of DSB-SC SignalsBandwidth and Power EfficiencyDemodulation of DSB-SC SignalsDemodulation of DSB-SC SignalsConventional AMSpectrum of Conventional AMBandwidth and Power EfficiencyExampleDemodulation of AM signals�Envelop DetectorSingle Sideband (SSB) AM Expression of SSB signalsAbout Hilbert TransformGeneration of SSB-AM SignalVestigial Sideband: VSBComparison of AM TechniquesSignal MultiplexingFDMFDM application in telephone comm.Quadrature-Carrier MultiplexingApplication: AM Radio BroadcastingTopics to be CoveredAngle ModulationRepresentation of FM and PM signalsDistinguishing Features of PM and FMExample: Sinusoidal ModulationExample: Square ModulationFM by a Sinusoidal SignalExample幻灯片编号 34幻灯片编号 35幻灯片编号 36Properties of Bessel FunctionGeneral CaseGeneral CaseEffective Bandwidth of FW WavesEffective Bandwidth of FW WavesEffective Bandwidth of FW WavesFM by an Arbitrary MessageExampleExerciseApplication: FM Radio broadcastingSummaryGeneration of FM wavesGeneration of Narrow-band FM 幻灯片编号 50幻灯片编号 51幻灯片编号 52Exercise: A typical FM transmitterDemodulation of FM�Balanced Frequency Discriminator幻灯片编号 55Application: FM Radio broadcastingFM Radio Stereo Multiplexing幻灯片编号 58Think …Suggested ReadingTopics to be CoveredA Benchmark SystemExample:Effect of Noise on DSBSC 幻灯片编号 65幻灯片编号 66Effect of Noise on SSB幻灯片编号 68Effect of Noise on Conventional AM幻灯片编号 70Performance of Envelope-DetectorPerformance of Envelope-DetectorExercise Topics to be CoveredEffect of Noise in FM Effect of Noise on Angle Modulation幻灯片编号 77幻灯片编号 78幻灯片编号 79Output SNR in FMKey ObservationsThreshold EffectComparison of Analog-ModulationThink …Suggested Reading

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