Noise in Linear Modulations - German University in Cairo...Dr. Ahmed El-Mahdy COMM 601: Modulation I...

Dr. Ahmed El-Mahdy COMM 601: Modulation I

COMM 601: Modulation I

Lecture 8

Noise in Linear Modulations


White Noise• A white noise process has a flat PSD over a wide range

of frequencies:

is called the height of the power spectral

density of the noise . 2

oN

2)( o

w

NfS


Ex: White Noise With zero mean

Time domain representation


Narrow band white noise• Narrow band white noise : is a white noise passed through a

band pass filter which is generally used in the receiver.

• The power spectral density of Narrow band noise at the output of an ideal BPF with center frequency and bandwidth 2B is:

•

cf


Time domain representation of narrow band noise

(1) In-phase and quadrature phase representation:

The in-phase and quadrature phase are low pass

Signals.

(2) Amplitude and phase representation:

)2sin()()2cos()()( tftntftntn cQcI

)(tnI )(tnQ

)(

)(tan)(,)()(

where

))(2cos()()(

122

tn

tnttntnr(t)

ttftrtn

I

Q

QI

c


Power spectral density of the in-phase and quadrature phase components

where:

• Proof:

--------------------------

• The PSD of the in-phase and quadrature phase components is:

)()()()( cNcNNN ffSffSfSfSQI

)2sin()()2cos()()( tftntftntn cQcI


Noise in AM System Envelope Demodulation:

• The transmitted signal in an AM system has the form:

LPF

tdBPF

)(tnts i )(ty )(tVEnvelope Detector

)(ts

)2cos()](1[)( tftmkAts cac

)2sin()()2cos()]()](1[[

)()(

tftntftntmka

tntKsy(t)

cQcIac


Noise in AM System (cont.)

))(2cos()()( ttftRty ycy

)()]())(1([)(:where 22 tntntmkatR QIacy

)())(1(

)(tan)( 1

tntmka

tnt

Iac

Q

y


Noise in AM System (Cont.)

• The output of the envelope detector will be proportional to :

• The expression of can be simplified if we assume that the signal power is very large with respect to the noise power.

• We consider the strong signal case. The phasor diagram in this case is shown in the following figure. We expect that

to be much larger than most of the time. Hence we can approximate as

)(tRy

)(tRy

)(tnI

)(tnQ

)(tRy

y n

)(tRn

)(and)( tntn QI

)(tRy

a b c

d

)](1[ tmka ac

)()](1[

)(

tntmka

acoflengthadoflengthtR

Iac

y

)](1[ tmka ac

)()]())(1([)(:where 22 tntntmkatR QIacy


Noise in AM System (Cont.)• Hence the demodulator output is given by:

• The output from the LPF (assuming that the filter removes the dc component ) is given by:

• The output signal to noise ratio is given by:

)()](1[)( tntmkatd Iac

)()()( tntmkatV Iac

signal.message

theofpoweraveragetheis)()]([where 22 tmtmEPm

WN

PkaSNR

WN

tmEka

tnE

tmEkaSNR

o

maco

o

ac

I

aco

2

2

)]([

)]([

)]([

22

222

2

222

f

fHLPF

WW cf...


• SNRo in terms of average received power:

For AM signal, the received power is given by:

]1[2

]1[

:,2

]1[2

)]]([1[2

)]([

2

222

22

22

222

22

mao

mamaco

o

maco

mac

ac

r

PkWN

PkPkaSNR

asitwritecanweThenWN

PkaSNRSince

Pka

tmEka

tsKEP

ma

ma

o

ro

Pk

Pk

WN

PSNR

2

2

1

)2cos()](1[)( tftmkAts cac


Noise in AM System (Cont.)

• The output signal to noise ratio is given by:

Power

Efficiency of

AM

(less than one)

ma

ma

o

ro

Pk

Pk

WN

PSNR

2

2

1

The noise performance of AM is less than

DSB and SSB due to the wastage of

transmitted power


SNRo in AM for single tone modulating signal

The AM system must transmit 3 times as much as average power as DSB and SSB to achieve same performance.

WN

PSNRWhen

WN

PSNR

Ak

AP

o

ro

o

ro

m

am

m

3

1,1

2

,2

2

2

2


Noise in Coherent DSB System

CARRIER

BPF

f

fHLPF

WW cf...

W2

DSB Coherent Demodulation

tfc2cos2

LPF td )(2cos tntftmA icc )(2cos)( tntftmaty cc )(tV

W2

ccN

cc

c

m

mccT

cc

KAafS

tn

tntftma

Wf

tmP

PAtmAP

tftmAts

and)(

psdawithnoisenarrowbandis)(where

(**))()2cos()(

isfiltertheofoutputThe

.2bandwidthandfrequencycenterahasBPFThe

signal.message

theofpoweraveragetheis)(where

2

1)(

2

1

:ispowerdtransmitteaverageThe

2cos

2

222


Noise in Coherent DSB System (Cont.)Using the in-phase and quadrature phase components of the narrow

band noise:

We have,

)2sin()()2(cos)()( tftntftntn cQcI

)4sin()()4cos()]()([)]()([)2cos()(2)(

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

tftntftntmatntmatftytd

and

tftntftntmaty

cQcIcIcc

cQcIc

])()([)(

bygivenisoutputtheandfilterthebyremovedaretermsfrequencydoubleThe

tntmatV Ic

WNtn

Patma

oI

mcc

2:ispowernoiseoutputtheand)(iscomponentnoiseoutputThe

:ispowersignaloutputtheand)(iscomponentsignaloutputThe 2


Noise in Coherent DSB System (Cont.)• Then, the signal to noise ratio at the output is given by:

• From equation (**), we have

WN

PaSNR

o

mco

2

2

inputreceivertheatpoweraveragetheiswhere,2

1 2

rmcr PPaP

WN

PSNR

o

ro


Noise in Coherent SSB system

• Assume USB

)()( tnts i

tfc2cos2

LPF td

)(2sinˆ2cos)( tntftmatftmaty cccc

)(tVBPF

tftmAtftmAts cccc 2sinˆ2cos

W

)(2

have we*),*(*eq.from then,))(ˆ())((Since.and

)(psdawithnoisenarrowbandis)(where

*)*(*)()2sin()(ˆ)2cos()(

isfiltertheofoutputThe

.bandwidthandfrequencycenterahasBPFThe

222

mmc

rmcc

N

cccc

mc

PPa

PPtmEtmEKAa

fStn

tntftmatftma

ff

SSBN fS )(

cfcf Wfc Wfc

W2

DSBN fS )(


Noise in Coherent SSB system (Cont.)• Similar to the DSB case, we have

)4sin()]()(ˆ[

)4cos()]()([)]()([)2cos()(2)(

and

)2sin(])()(ˆ[)2(cos])()([)(

tftntma

tftntmatntmatftytd

tftntmatftntmaty

cQc

cIcIcc

cQccIc

])()([)(

bygivenisoutputtheandfilterthebyremovedaretermsfrequencydoubleThe

tntmatV Ic

WNtn

Patma

oI

mcc

:ispowernoiseoutputtheand)(iscomponentnoiseoutputThe

:ispowersignaloutputtheand)(iscomponentsignaloutputThe 2


Noise in Coherent SSB System (Cont.)• Then, the signal to noise ratio at the output is given by:

• Using

WN

PaSNR

o

mco

2

inputreceivertheatpoweraveragetheiswhere,2

rmcr PPaP

WN

PSNR

o

ro

Which is similar to the Coherent DSB System


Example• Compare the average transmitted power and channel bandwidth

requirements of DSB, SSB, and AM schemes for transmitting an audio signal with a bandwidth of 10 kHz with output signal to noise ratio of 40 dB. Assume that the channel introduces a 50 dB power loss and that the noise power spectral density at the receiver input is No/2= W/Hz. Assume for AM.

Solution

The Bandwidth requirements are:

For DSB and AM, the bandwidth= 2*10=20 kHz

For SSB, the bandwidth=10 kHz

1210

mo

ro

fN

PSNR :SSB&DSBFor

5.02 ma Pk


Watts60)10()10(6henceand

][)10(6)10)(2(3103

3

1

1SNR

detection, envelopeassumingAM,theFor

Watts20)10)(10)(2(

getwe),10(50islosspowerchanneltheSince

][)10)(2()10)(10)(2(*

havewe),10)(10)(2(,40with

54

484

0

2

2

o

54

5

448

412

AMT

mor

mo

r

ma

ma

mo

r

DSBT

moDSBr

moo

P

WfNSNRP

fN

P

Pk

Pk

fN

P

P

ordB

WfNSNRoP

fNdBSNR

Noise in Linear Modulations - German University in Cairo...Dr. Ahmed El-Mahdy COMM 601: Modulation I...

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