Communications II

Lecture 4: Effects of Noise on AM

Professor Kin K. LeungEEE and Computing Departments

Imperial College London© Copyright reserved

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Noise in Analog Communication Systems

• How do various analog modulation schemes perform in the presence of noise?

• Which scheme performs best?

• How can we measure its performance?

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We must find a way to quantify (=to measure) the performance of a modulation scheme.

We use the signal-to-noise ratio (SNR) at the output of the receiver:

N

S

PPSNR

outputreceivertheatnoiseofpoweraverageoutputreceivertheatsignalmessageofpoweraverage

0

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Model of an analog communication system

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PT : The transmitted power

Limited by: equipment capability, cost, government restrictions,interference with other channels, etc

The higher it is, the more the received power (PS ), the higher the SNR

For a fair comparison between different modulation schemes: PT should be the same for all

We use the baseband signal to noise ration SNRbaseband to calibrate theSNR values we obtain

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A Baseband Communication System

• It does not use modulation

• It is suitable for transmission over wires

• The power it transmits is identical to the message power: PT = P

• The results carry over to band-pass systems

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Average signal(=message) power:P =the area under the triangular curve

Assume:Additive, white noise with power spectral density PSD=N0/2

Average noise power at the receiver: PN = area under the straight line = 2WN0/2 = WN0

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SNR at the receiver output:

Note: Assume no propagation loss

Improve the SNR by:(a) increasing the transmitted power (PT ↑), (b) restricting the message bandwidth (W ↓), (c) making the receiver less noisy (N0 ↓).

WNPSNR T

baseband0

ST PP

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REVISION: Amplitude Modulation

General form of an AM signal:

A: the amplitude of the carrier

fc: the carrier frequency

m(t): the message signal

)2cos()]([)( tftmAts cAM

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Modulation index:

mp: the peak amplitude of m(t), i.e., mp = max |m(t)|

Amp

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Signal recovery

pmA 11) : use an envelope detector

2) Otherwise: use synchronous detection=product demodulation=coherent detection

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Synchronous detection

• Multiply the waveform at the receiver with a local carrier of the same frequency (and phase) as the carrier used at the transmitter:

• Use a LPF to recover (m(t) + A)/2 and finally m(t)

• Problem: At the receiver you need a signal perfectly synchronised with the transmitted carrier

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REVISION Amplitude Modulation: Double-sideband suppressed carrier (DSB-SC)

Signal recovery: With synchronous detection only

)2cos()()( tftAmts cDSBS

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Noise in DSB-SC

The true signal received is:

)2sin()()2cos()]()([)2sin()()2cos()()2cos()(

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

tftntftntAmtftntftntftAm

tftntftntstntstx

cscc

csccc

cscc

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)()(~ tntAmy c

For synchronous detection:

• multiply with 2cos(2πfct):

• Use a LPF to keep

)4sin()()]4cos(1)[()]4cos(1)[(

)4sin()()2(cos2)()2(cos)(

)()2cos(2)(22

tftntftntftAm

tftntftntftAm

txtfty

cs

ccc

csccc

c

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Signal power at the receiver output:

Power of the noise signal nc(t):

PAtmEAtmAEPS22222 )}({)}({

WNdfNPW

WN 00 2

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SNR at the receiver output:

To which transmitted power does this correspond?

WNPASNR

0

2

0 2

2

)}2(cos)({2

222

PA

tftmAEP cT

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So

Comparison with

Conclusion: a DSB-SC system provides no SNR performance gain over a baseband system.

SCDSBT SNRWN

PSNR 0

0

WNPSNR T

baseband0

basebandSCDSB SNRSNR

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Noise in standard AM, Synchronous Detection

Pre-detection signal:

)2sin()()2cos()]()([)2sin()()2cos()()2cos()]([

)()2cos()]([)(

tftntftntmAtftntftntftmA

tntftmAtx

cscc

csccc

c

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Signal Recovery:

)()(~ tntmAy c

• Multiply with 2 cos(2πfct):

• LPF

)4sin()()]4cos(1)[()]4cos(1)[()]4cos(1[)(

tftntftntftmtfAty

cscc

cc

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Signal power at the receiver output:

Noise power:

PtmEPS )}({ 2

PNPN 02

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SNR at the receiver output:

Transmitted power:

SNR of a baseband signal with the same transmitted power:

AMSNRWN

PSNR 0

0 2

222

22 PAPAPT

WNPASNRbaseband

0

2

2

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Thus:

Note:

Conclusion: the performance of standard AM with synchronous recovery is worse than that of a baseband system.

basebandAM SNRPA

PSNR

2

12 PAP

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Noise in standard AM, Envelope Detection

Phasor diagram of the signals present at an AM receiver

Ei(t): receiver output= y(t)

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Equation too complicated

Must use limiting cases to put it in a form where noise and message are added

22 )()]()([

)(ofenvelope)(

tntntmA

txty

sc

(95)

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1st Approximation: (a) Small Noise Case

Then

Then

)]([)( tmAtn

)]()([)( tntmAtn cs

)]()([)( tntmAty c

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Thus

And in terms of baseband SNR:

Valid for small noise only!

envSNRWN

PSNR 0

0 2

basebandenv SNRPA

PSNR

2

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)]([)( tmAtn

2nd Approximation: (b) Large Noise Case

Isolate the small quantity in (95):

)()()())((2

)()())((1)]()([

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

)()]()([)(

22222

222

222

222

tntntntmA

tntntmAtntn

tntntmAtntmA

tntntmAty

sc

c

scsc

scc

sc

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where

: the envelope of the noise )()()( 22 tntntEn sc

)()()]([21)(

)()()()]([21)]()([)(

22

22222

tEtntmAtE

tntntntmAtntnty

n

cn

sc

csc

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From the phasor diagram: nc(t) = En(t) cosθn(t)

Then:

Use :1for2

11 xxx

)()(cos)]([21)()(

tEttmAtEty

n

nn

)(cos)]([)()(

)(cos)]([1)()(

ttmAtEtE

ttmAtEty

nn

n

nn

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Noise is multiplicative here!

No term proportional to the message!

Result: a threshold effect, as below some carrier power level (very low A), the performance of the detector deteriorates very rapidly.

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SSB modulation

Single (lower) sideband AM:

where is the Hilbert transform of m(t).is obtained by passing m(t) through a linear filter with transfer

function −jsgn(f ).and m(t) have the same power P .

The average power is A2P/4.

tftmAtftmAts ccSSB 2sin)(ˆ2

2cos)(2

)(

)(ˆ tm)(ˆ tm

)(ˆ tm

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Noise in SSB

Receiver signal x(t) = s(t) + n(t).

Apply a band-pass filter on the lower sideband. Using coherent detection:

After low-pass filtering,

)4sin()()(ˆ2

)4cos()()(2

)()(2

)2cos(2)()(

tftntmA

tftntmAtntmAtftxty

cs

ccc

c

)()(

2)( tntmAty c

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Signal power A2P/4

Noise power for nc(t) = that for band-pass noise = N0W

SNR at output

For a baseband system with the same transmitted power A2P/4

Conclusion: SSB achieves the same SNR performance as DSB-SC (and the baseband model) but only requires half the band-width.

WNPASNRSSB

0

2

4

WNPASNRbaseband

0

2

4

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Summary