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ERG2310A-II p. II-89

Noises

Sources of Noises: random wandering of free electrons in resistors (thermal noise), random flow of charges in semiconductor junction (shot noise), etc.

Noises : random fluctuations of power in systems

Corrupt the signal waveform

Degrade the performance of communication systems

Additive noise

Zero-mean White Gaussian-distributed Noise, power spectral density Sn(f)=η /2

Uncorrelated with the signal

Performance measures: Analog Signal-to-Noise Ratio (SNR)

Digital Bit-Error-Rate (BER)

Transmitter Receiver+Channel

Noisen(t)

x(t) y(t)

(AWGN)

s(t)

no(t)


Noises: Signal-to-Noise Ratio

Assume the channel introduces no distortion and receiver is linear

outputreceiver at the noise ofpower averageoutputreceiver thesignal message ofpower average=oSNR

)()()( tntsty oo +=

where so(t) and no(t) are the signal and noise waveforms at the receiver output.

2

2

o

oo

nsSNR =

Receiver+

Noisen(t)

y(t)

s(t)Received

signal


White Noise and Filtered Noise

White noise: flat spectral density over a wide range of frequencies

Sn(f)

η/2

f

Rn(τ)

η/2

τ

Filter H(f)n(t) nF(t)

22 )(2

)()()( fHfSfHfS nnF

η==

∫∫∞

∞−

∞

∞−

== dffHdffSnFnF

22 )(2

)( η

Filtered white noise:

f

SnF(f)

B-B f

SnF(f)

-fc fc

2B2B

Low-pass filter band-pass filter

η/2 η/2


Bandpass Noise

For bandpass systems, which limit the bandwidth of the noise, some of thenoise fluctuations are restricted.

As the bandwidth of the noise becomes small compared to the center frequency, it becomes possible to approximate it with a phasor representation.

Here nc(t) is that portion of the noise which is the in-phase component andns(t) is the quadrature component in the phasor representation.

ns(t)

nc(t)

an

θn

The complete phasor representation of the narrowband random noise istj

scoetjntn ω)]()([ +

where ωo is the center frequency.

ttnttnetjntntn osoctj

scFo ωωω sin)(cos)(})]()(Re{[)( −=+=

which is known as the bandpass representation of noise.

= −

)()(tan 1

tntn

c

snθ


Bandpass Noise

LPFnF(t) e(t)

cosωot

ttnttntttnttnttn

sc

scoF

021

021

0002

2sin)(]2cos1)[(cossin)(cos)(cos)(

ωωωωωω

−+=−=

The output of the low-pass filter is)(]cos)([)( 2

10 tnttnte cLPF == ω

The power spectral density of nc is

LP

TFTF

Tn

LPTF

T

Tc

Tn

TNN

S

Tttn

Ttn

S

c

c

++−=

ℑ=

ℑ=

∞→

∞→∞→

200

20

2

)()(lim)(

}]cos)([2{lim

)}({lim)(

ωωωωω

ωω


Bandpass Noise

For random noise, the average of the cross products goes to zero, thus

LPnnn FFcSSS )]()([)( 00 ωωωωω ++−=

The case for ns follows in the same manner by considering n(t)sin ωct, yielding

LPnnn FFsSSS )]()([)( 00 ωωωωω ++−=

Hence,LPnnnn FFsc

SSSS )]()([)()( 00 ωωωωωω ++−==

Thus, their mean-square values are equal;

)()()( 222 tntntn scF ==

The mean-square value of bandpass random noise is

)()()( 2212

212 tntntn scF += ttnttntn osocF ωω sin)(cos)()( −=Q


Bandpass Noise


Hence, and the spectral densities of nc(t) and ns(t)are

Bandpass Noise

Example: White noise with power spectral density η/2 is filtered by a rectangular bandpass filter with H(f)=1, centered at fo and having a bandwidth W. Find the power spectral density of nc(t) and ns(t). Calculate the power in nc(t), ns(t) and nF(t).

Solution: Since the filter is rectangular with H(f)=1, the power spectral density of the output noise n(t) is:

elsewhereWffWf

fS oonF

22

02

)(+≤≤−

=η

)()( onon ffSffSFF

−=+

2for 22)()()()( WfffSffSfSfS ononnn FFsc≤=+=++−== ηηη

Powers of nc(t) and ns(t) are

WdffSnnW

Wnsc c

η∫−

===2

2

22 )(

Power of nF(t) isWWdffSdffSn

Wf

Wf

Wf

WfnnF

o

o

o

o

ηη∫ ∫+−

−−

+

−

=

=+=

2

2

2

2

2

22)()(


Assume distortionless channel:

where K and td are the amplification/attenuation and time-delay, respectively.

Noise in Baseband Comm. Systems

Baseband transmission: signal is transmitted without any modulation.

Assume both the transmitter filter and receiver filter are ideal low-pass filters with bandwidth W (=2π B)

LPF LPF+Channel

Noisen(t)

x(t) xo(t), no(t)

(AWGN)

)()( do ttKxtx −=

Average output signal power: (received power)

BdfdffSnB

B

B

Bno ηη

∫∫−−

=

==

2)(2Average output noise power:

Ro Px =2

BP

SNR Ro η

=


Noise in Baseband Comm. SystemsExample: Consider an analog baseband communication system with additive white

noise. The transmission channel is assumed to be distortionless and the power spectral density of white noise is η/2 is 10-9 watt per hertz (W/Hz). The signal to be transmitted is an audio signal with 4-kHz bandwidth. At the receiver end, an RC low pass filter with a 3-dB bandwidth of 8kHz is used to limit the noise power at the output. Calculate the output noise power.

Solution: For a RC low-pass filter (3-dB bandwidth=8kHz), the frequency response is

)(11

1)/(1

)/(1)( RCjCjR

CjH oo

=+

=+

= ωωωω

ωω where

( )[ ] 2/121

1)(o

Hωω

ω+

= 3dB bandwidth at ω=ωo = ωo=2π(8000)

( )∫∫∞

∞−

∞

∞− +=

= ω

ωωπηωωη

πddHn

oo 2

22

11

21

2)(

221Average output noise power:

( ) ooon ηωπωπ

η41

21

22 ==

Wno µπ 2.25)108)(2)(10(2(41 392 =×= −


Noise in Amplitude Modulated Systems

The bandpass filter limits the amount of noise outside signal band that reaches the demodulator (“out-of-band” noise)

(fc-B) fc (fc+B)(-fc-B) fc (-fc+B) f0

S(f)+N(f)


SF(f)+NF(f)

N(f)

ttnttntn csccF ωω sin)(cos)()( −=Narrowband filtered noise:

Bnnn Fsc η2222 ===Power spectral density:

BPF+

Noisen(t)

y(t)

s(t)

Received modulated

signalDetector LPF

Receiver

DemodulatornF(t)

sF(t)

no(t)

)(2 tsP FR =Received signal power:


Noise in DSB-SC Systems

In a DSB-SC system, the received modulated signal is:ttxAts cc ωcos)()( =

y(t)BPF+

Noisen(t)

s(t)

Received modulated

signalLPF

Receiver

DemodulatornF(t)

sF(t)×

cosωctConsider the demodulation by synchronous detection:

Input signal power to the demodulator: RcccF PtxAttxAts === )(21cos)()( 222222 ω

Input noise power to the demodulator:

ttnttntn csccF ωω sin)(cos)()( −=

BtnF η2 )(2 =

no(t)

where x(t) is the message signal which is band-limited with bandwidth B


SF(f)+NF(f)

ttxAts ccF ωcos)()( =


Noise in DSB-SC Systems

Output signal power after LPF: )(4

)( 22

2 txAty c=

Output noise after LPF:

)()()( 2412

412 tntntn Fco ==

Output signal after LPF: [ ] )(2

cos)()( txAttsty cLPcF == ω

[ ][ ]{ }

)(2sin)(]2cos1)[(

cossin)(cos)(

cos)()(

21

21

21

2

tnttnttntttnttn

ttntn

c

LPcscc

LPccscc

LPcFo

=−+=

−=

=

ωωωωω

ω

Output noise power after LPF: 222Fsc nnn ==Q

SNR before demodulator:BP

nP

ntxA

nsSNR R

F

R

F

c

F

Fi η2

)(22

2221

2

2

====

iR

F

c

oo SNR

BP

tntxA

tn

tySNR 2

22

)()(

)(

)(2

41

2241

2

2

====η

Output SNR after demodulator:

The detector (demodulator) improves the SNR in a DSB-SC system by a factor of 2. This improvement results from the fact that the coherent detector rejects the quadrature noise components in the input noise, thereby halving the mean-square noise power.

BPSNR R

o η= for DSB-SC


Noise in DSB-SC SystemsExample: A DSB-SC system with additive white noise is demodulated by a

synchronous detector with a phase error φ, i.e. cos(ωct + φ). The original message signal is x(t). Show that where γ is the output SNR when the synchronous detector has no phase error.

Output signal power after LPF: φ222

2 cos)(4

)( txA

ty c=

Output noise after LPF:

( ))(

sincos)(

sin)(cos)()(

241

22241

224122

412

tn

tn

tntntn

F

F

sco

=

+=

+=

φφ

φφ

Output signal after LPF: ( )[ ] φφω cos)(2

cos)()( txA

ttsty cLPcF =+=

( )[ ][ ]φφ

φωsin)(cos)(

cos)()(

21

21 tntn

ttntn

sc

LPcFo

+=+=

Output noise power after LPF:

222Fsc nnn ==Q

φγφφ 222

22

241

22241

2

2

coscos)(

)()(

cos)()(

)(====

tntxA

tntxA

tn

tySNR

F

c

F

c

o

o


ttnttntn csccF ωω sin)(cos)()( −=ttxAts ccF ωcos)()( =

φγ 2cos=oSNR

)(

)(2

22

tn

txA

F

c=γ

Solution:


Noise in SSB Systems

In SSB system, the received modulated signal is:[ ]ttxttxAts ccc ωω sin)(ˆcos)()( +=

y(t)BPF+

Noisen(t)

s(t)

Received modulated

signalLPF

Receiver

DemodulatornF(t)

sF(t)×

cosωctConsider the demodulation by synchronous detection:


no(t)



SF(f)+NF(f)[ ]ttxttxAts cccF ωω sin)(ˆcos)()( +=

Bnnn Fsc η=== 222



Input noise power to the demodulator: BtnF η= )(2

Input signal power to the demodulator:

[ ][ ][ ]

Rc

c

ccc

cccccF

PtxA

txtxA

ttxttxA

tttxtxttxttxAts

==

+=

+=

++=

)(

)()(

sin)(ˆcos)(

cossin)(ˆ)(2sin)(ˆcos)()(

22

212

2122

22222

222222

ωω

ωωωω

ttnttntn csccF ωω sin)(cos)()( −=[ ]ttxttxAts cccF ωω sin)(ˆcos)()( +=

Output signal to the demodulator: [ ] )(2

cos)()( txAttsty cLPcF == ω

Input signal to the demodulator:Input noise to the demodulator:

Output noise to the demodulator: [ ] )(cos)()( 21 tnttntn cLPcFo == ω

Output signal power to the demodulator:

Output noise power to the demodulator:

)(4

)( 22

2 txAty c=

)()()( 2412

412 tntntn Fco ==


SNR before demodulator: BP

ntxA

nsSNR R

F

c

F

Fi η

===2

22

2

2 )(

iR

F

c

oo SNR

BP

tntxA

tn

tySNR ====

η)()(

)(

)(2

41

2241

2

2



The detector (demodulator) does not improve the SNR in a SSB-SC system.

BPSNR R

o η= for SSB


Noise in DSB-LC (AM) Systems

In DSB-LC (AM) system, the received modulated signal is:[ ] ttxmAts cac ωcos)(1)( +=

Consider the demodulation by synchronous detection:



y(t)BPF+

Noisen(t)

s(t)

Received modulated

signalLPF

Receiver

DemodulatornF(t)

sF(t)×

cosωct

no(t)

[ ] ttxmAts cacF ωcos)(1)( +=


SF(f)+NF(f)



Input noise power to the demodulator: BtnF η2 )(2 =



Output signal to the demodulator: [ ] [ ])(1cos)()( 21 txmAttsty acLPcF +== ω

Input signal to the demodulator:Input noise to the demodulator:

Output noise to the demodulator: [ ] )(cos)()( 21 tnttntn cLPcFo == ω


[ ][ ][ ]

R

ac

aac

cacF

PtxtxmA

txmtxmA

ttxmAts

==+=

++=

+=

signal) messagemean (zero 0)( assume )(1

)(2)(1

cos)(1)(

22221

22221

2222 ω

By removing the DC term, gives )()( 21 txmAty ac=



Output signal power to the demodulator:

Output noise power to the demodulator:

)(4

)( 222

2 txmAty ac=

)()()( 2412

412 tntntn Fco ==

SNR before demodulator: BP

nsSNR R

F

Fi η22

2

==

[ ]

BP

txmtxm

txmtxmA

Btxm

txmtxm

tntxmA

tn

tySNR

R

a

a

a

aca

a

a

F

ac

oo

η

η

⋅+

=

++⋅=

++⋅==

)(1)(

)(1)(1

)2()(

)(1)(1

)()(

)(

)(

22

22

22

22221

41

2221

22

22

241

22241

2

2


For DSB-LCBP

txmtxmSNR R

a

ao η

⋅+

=)(1

)(22

22

As to avoid distortion (over-modulation) 1)(22 ≤txma

BPSNR R

o η21 ≤⇒ SNRo of DSB-LC is least 3-dB worse than that in DSB-SC

and SSB systems


Noise of DSB-LC (AM) Systems

In DSB-LC (AM) system, the received modulated signal is:[ ] ttxmAts cac ωcos)(1)( +=

Consider the demodulation by envelope detection:


r(t)BPF+

Noisen(t)

s(t)

Received modulated

signalEnvelope detector

ReceiverDemodulator

nF(t)

sF(t)


Bnnn Fsc η2222 ===


Input to the envelope detector:

[ ] [ ]ttnttnttxmAtntstf cscccacFF ωωω sin)(cos)(cos)(1)()()( −++=+=

f(t)


Noise of DSB-LC (AM) Systems

For (large input SNR):[ ] )( ),( )(1 tntntxmA scac >>+

[ ] [ ][ ]{ } ttnttntxmA

ttnttnttxmAtf

csccac

cscccac

ωωωωω

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

−++=−++=

Output of the envelope detector:

[ ]{ } )()()(1)( 22 tntntxmAtr scac +++= [ ] )()(1)(tan)( 1

tntxmAtnt

cac

s

++= −φ

)()()( tjetrtr φ=where

Input of the envelope detector:

;

[ ] )()(1)( tntxmAtr cac ++≈⇒

Removing the DC component, gives )()()( tntxmAtr cac +≈

which is basically the same as the output signal, y(t)+no(t), for synchronous detection of DSB-LC signals without the scaling factor ½.

Thus, for high SNR at receiver input, the performance of synchronous detector and envelope detector is the same.



For (small input SNR):[ ] )( ),( )(1 tntntxmA scac <<+

[ ]{ }[ ] [ ]

[ ] [ ]

[ ]

[ ]

[ ]

[ ])(1)()()(

smallfor 2

11 using )(1)()(1)(

)(1)(

)(21)(

)()()(let )(1)(

)(21)(

)(1 smallfor )(1)(2)()(

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

)()()(1)(

2

2

2222

2

22

2222

22

txmtvtnAtv

txmtvtnAtv

txmtvtnAtv

tntntvtxmtvtnAtv

txmAtxmtnAtntn

txmtnAtntntxmA

tntntxmAtr

an

ccn

an

ccn

an

ccn

scnan

ccn

acaccsc

accscac

scac

++=

+≈+

++≈

++=

+=

++=

++++≈

+++++=

+++=

ααα

Thus, for small SNR at receiver input, the signal and noise are no longer additive. The signal multiplied by noise cannot be distinguishable. No meaningful output SNR can be defined.



Example: Assuming sinusoidal modulation, show that, in an AM system with envelope detection, the output SNR is given by

where ma is the modulation index for AM.

BP

mmSNR R

a

ao η2

2

2 +=

Solution: For sinusoidal modulation, ttx mωcos)( =

2122 cos)( == ttx mω

Using for AM or DSB-LC BP

txmtxmSNR R

a

ao η

⋅+

=)(1

)(22

22

( )( ) B

Pm

mSNR R

a

ao η

⋅+

= 221

212

1

BP

mmSNR R

a

ao η2

2

2 +=

Thus,



Example: Consider an AM system with additive thermal noise having a powerspectral density η/2=10-12 W/Hz. Assume that the baseband message signal x(t) has a bandwidth of 4kHz and the amplitude distribution shown by the figure. The signal is demodulated by envelope detection and appropriate post-detection filtering. Assume modulation index ma=1.

fx(x)1

-1 0 1 x

Find the minimum value of the carrier amplitude Ac that will yield SNRo≥ 40dB.



Solution:

[ ] ttxAts cc ωcos)(1)( +=

∫ ∫∞

∞−

=+−==61)1(2)()(

1

0

222 dxxxdxxfxtx x

For ma=1, B=4kHz, η/2=10-12 W/Hz ,

RRR

a

ao PP

BP

txmtxmSNR 9

71

31261

61

22

22

108)104)(102())(1(1))(1(

)(1)(

−− ×=

××⋅

+=⋅

+=

η

[ ] [ ] 2127

612

21222

21 ))(1(1)(1 ccacR AAtxmAP =+=+=with

For

[ ]VA

A

SNR

dBSNR

c

c

o

o

1031

10108

10

40

3

42127

971

4

−

−

×≥

≥×

≥

≥

Thus the minimum value of Ac required is 31mV.


Noise in Angle Modulated Systems

For message signal x(t), angle-modulated signal: [ ])(cos)( ttAts cc φω +=

)(

)()(

0

=

∫t

f

p

dxk

txkt ττφ

for PM

for FM


The BPF (pre-detection filter) has a bandwidth of W=2(D+1)B where D is the frequency deviation ratio and B is the bandwidth of the message signal.

[ ] 221222 )(cos)( cccF AttAts =+= φω

Input noise power to the demodulator: ( ) WWtnF ηη =

= 2

2)(2

SNR before the demodulator:WA

tntsSNR c

F

Fi η2)(

)( 2

2

2

==

BPF Discriminator LPF

Noise

nF(t)s(t)

DemodulatorReceiver

sF(t)

y(t)

no(t)

n(t)

LimiterG(t)



Let G(t) the input to the demodulator such that

[ ][ ])(cos)(

sin)(cos)()(tttR

ttnttnAtG

c

csccc

θωωω

+=−+=

The limiter suppresses any amplitude variation of G(t), thus in angle modulation, SNR is derived from consideration of θ(t) only.

σσσ

θ

smallfor tan )(

)()(

)(tan)( 1

≈+

≈

+= −

QtnA

tntnA

tnt

cc

s

cc

s

{ } { }ttnttnttAtntstG csccccFF ωωφω sin)(cos)()](cos[)()()( −++=+=

[ ])(

)(tan)( ; )()()( 122

tnAtnttntnAtRcc

sscc +

=++= −θwhere

To facilitate the analysis of the noise only, we can assume φ(t)=0

Thus,



)(

)()(tnA

tntcc

s

+≈θ

For high input SNR, Ac >> |nc(t)|, Ac>>|ns(t)| , thus , )()(c

s

Atnt ≈θ

+=

c

sc A

tnttG )(cos)( ωαOutput noise from Limiter:

This operation is just like passing ns(t) through a differentiator with transfer response

)()( tndtd

Atn s

cd

α=Output noise from Discriminator (neglect DC term):

cAjH ωαω =)(

)()()( 2 ωωωsd nn SHS =

Thus, consider the relation of power spectral densities:

If the input noise is white with Sn(ω)=η/2,

[ ] ηηηωωωωω =+=++−= 2/2/)()()( LPcncnn SSSs

α=≈L

)( cAtR



With and ,

Thus,

)()()( 2 ωωωsd nn SHS =

ηωαω 2

22

)(c

n AS

d=

cAjH ωαω =)( ηω =)(

snS

for2Wf ≤

Output signal from discriminator:

2

3222

2

22

22

2

2

38

21)(

21)(

c

B

Bc

B

Bno A

BdA

dStnd

ηαπωωπ

ηαωωπ

π

π

π

π

=== ∫∫−−

Output noise power after LPF (with bandwidth B):

+=

+= )(1)(1)( tx

ktx

kAty

c

fc

c

fcLc ω

αωω

ω

Neglecting the DC term, gives )()()()( 2222 txktytxkty ff αα =⇒=

Output SNR:

( ) ( ) 32

222

2322

222

2

2

8)(3

38)(

)()(

BtxkA

ABtxk

tntySNR fc

c

f

oo ηπηαπ

α===



32

222

8)(3

BtxkA

SNR fco ηπ

=

If the message signal is sinusoidal (single-tone), i.e.

and produces a frequency deviation of ∆ω,

tatx mm ωcos)( =

)sincos()(

)(43421t

mm

cc ttAts

φ

ωω

ωω ∆+=FM signal:

Differentiating φ(t), gives))(cos()(

)(

0 43421

Q

t

t

fcc dxktAts

φ

ττω ∫+=ttxk mf ωωcos)( ∆=

Thus, ( ) ( ) ( )2212

212222 2cos)( fttxk mf ∆=∆=∆= πωωω

BP

BA

BfSNR Rc

o ηβ

η2

22

232

23 =

∆=⇒

BPSNR R

o ηβ 2

23= for FM

Note: When β=5, output SNR of FM system is 37.5 times that of a baseband system, but the bandwidth of FM system is about 12 times that of the baseband system.


SNR in AM & FM Systems

Under the most favorable conditions in AM, the modulation index is 100%,thus

For FM:

It shows that the output signal-to-noise ratio can be made much higher in FM than in AM by increasing the modulation index β.

An increase in β also increases the bandwidth so that FM systems providean improvement in signal-to-noise at the expense of an increase in bandwidth.

( )BP

BP

mmSNR RR

a

aAMo ηη 3

12 2

2

=+

=

( )BPSNR R

FMo ηβ 2

23=

To realize any signal-to-noise improvement in FM over AM we must have

0.47 31

23 2 >⇒> ββ

This condition is approximately the transition point between narrowband andwideband FM.

Narrowband FM provides no signal-to-noise improvement over AM.


Pre-emphasis and De-emphasis in FM

ηωαω 2

22

)(c

n AS

d= for

2Wf ≤

Recall: the power spectral density at the FM discriminator output

-W/2 0 W/2 f

)(ωdn

S

Noise power is larger at higher frequencies within the message bandwidth.

Pre-emphasis: the high-frequency components in the input message signal are emphasized at the transmitter before the noise is introduced.

De-emphasis: at the output of FM demodulator, the inverse operation is performed to de-emphasize the high frequency components


Pre-emphasis and De-emphasis in FM

Pre-emphasis filter

De-emphasis filter

FM transmitter

FM receiver

AWGN n(t)

x(t) output

HPE(f) HDE(f)

WfWfH

fHDE

PE <<−= , )(1)(

f f

/1

1)( ; 1)(o

DEo

PE fjfKfH

fjfKfH

+=

+=Example:

input outputR

Cinput output

R

rC

ωωπ

ηαωωπ

π

π

π

π

dA

dStnB

Bc

B

Bno d ∫∫

−−

==2

2

22

22

2

2

21)(

21)(Output noise power without de-emphasis filter:

Output noise power with de-emphasis filter:

ωω

ωπ

ηαωωωπ

π

π

π

π

dHA

dSH

tnB

B PEc

B

Bn

PEo d ∫∫

−−

==2

2

22

2

22

2

22

)(1

21)(

)(1

21)(

noises - webpages.eng.wayne.eduwebpages.eng.wayne.edu/ece4700/lecture notes/snr.pdf · noises...

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