Communication SystemsyLecture 27

D I KiDong In KimSchool of Info/Comm Engineering

Sungkyunkwan University

1

O liOutline Ch l SNR Channel SNR

Channel SNR in DSB Channel SNR in SSB Channel SNR in SSB Channel SNR in AM Channel SNR in FM SNR comparison between AM and FM

Problem solving: SNR in DSB SNR in AM SNR in other systems SNR in other systems

FM Threshold Effect

2

Ch l S R ( ) TPSNRChannel SNR Definition: Ratio of the power of the modulated signal to the

( ) T

0N WCSNR =

Definition: Ratio of the power of the modulated signal to the average power of channel noise in the message bandwidth, both measured at the receiver input.p

( )t Predetection

( )CSNR

D d

( )TSNR ( )DSNR

( )cx t +

Noise n(t)

Predetectionfilter

Demod

Channel SNR is different from baseband SNR and predetection SNR: Baseband SNR uses message power. Predetection SNR uses the modulated signal bandwidth, which depends on the

system (e.g., W in SSB, 2W in DSB, 2(D+1)W in FM). Channel SNR is independent of the modulation scheme; therefore it is

3

Channel SNR is independent of the modulation scheme; therefore it is usually used to compare the performance of different systems.

Ch l S R i DSChannel SNR in DSB

( ) ( ) ( )y t A m t n t= +

( ) ( ) ( ) ( ) ( )2 cos ( )cos ( )sinc c c c s ce t A m t t n t t n t tw q w q w q= + + + - +

( ) ( ) ( )D c cy t A m t n t= +

Predetection SNR: ( )2c MA PSNR =

P td t ti SNR ( )2c MA PSNR

Predetection SNR: ( )04T

SNRN W

Postdetection SNR: ( )02

c MD

SNRN W

=

2 / 2A PP Channel SNR: ( ) ( )

2

0 0

/ 2c MTC D

A PPSNR SNRN W N W

= = =

4( ) ( )/ =1.

D CSNR SNR

Ch l S R i SSChannel SNR in SSB

[ ]ˆ( ) ( ) 2 ( ) i 2A f f[ ]ˆ( ) ( )cos2 ( )sin 2SSB c c cx t A m t f t m t f tp p= 2P A P= : twice of DSB power Signal Power: T c MP A P= : twice of DSB power.

Noise Power: N0W (after predetection filtering)

Signal Power:

( )2

0 0

c M TT

A P PSNRN W N W

= = Predetection SNR:0 0

Post-detection SNR: ( )2c M T

D

A P PSNR = =

( ) ( )( )D c cy t A m t n t= +

Post detection SNR: ( )0 0

D N W N W

Channel SNR: ( ) ( ) ( )TC T D

PSNR SNR SNRN W

= = =5

C e S : ( ) ( ) ( )0

C T DN W( ) ( )/ =1.

D CSNR SNR

Ch l S R i AChannel SNR in AM AM signal: ( ) ( ) ( )1 cosx t A am t tw qé ù= + +ë û AM signal: ( ) ( ) ( )1 cosc c n cx t A am t tw qé ù= + +ë û Message bandwidth: W Modulated signal bandwidth: 2W

Transmitted Power?2 2 21 1

2 2 nT c c mP A A a P= +

Noise Power after predetection?

a s tted owe ? 2 2

NT = 2N0W

Predetection SNR? ( )2 2 2

04nc c m

T

A A a PSNR

N W+

=

Noise Power at message bandwidth? N0W2 2 2A A P+

6Channel SNR? ( )

2 2 2

0 02nc c mT

C

A A a PPSNRN W N W

+= =

Ch l S R i AChannel SNR in AM( ) ( ) ( )A( i h )

2 2A a P ( ) 2ma PSNR

( ) ( ) ( )D c n cy t A am t n t= +Output (High SNR):

( )02

nc mD

A a PSNR

N W=

( )( ) 21

n

n

mDff

mC

SNE

SNR a P= =

+

nmSquare wave message: P =1

( )( )( )

1max 1/ 21 1

D

C

SNRSNR

= =+( )C

The output SNR of AM is at least 3dB worse than

7DSB and SSB, given the same channel SNR.

Ch l S R i AChannel SNR in AM

( ) cos mm t tw=Single tone modulation:112nmP =

When mod lation inde a 1:

( ) 1/ 2SNR

When modulation index a = 1:

( )( )

1/ 2max 1/ 31 1/ 2

D

C

SNRSNR

= =+

8

Ch l S R iChannel SNR in FM( ) ( )cos ( )x t A t t n tw q fé ù+ + +Recei ed FM signal: ( ) ( )cos ( )r c cx t A t t n tw q fé ù= + + +ë ûReceived FM signal:

2T

1P = A

We know that under high SNR assumption:TP .

2 cA

( )22 2

1 3nD d m d TmD

K f P f PSNR Pæ ö÷ç= = ÷ç ÷ç ÷æ ö

( ) 12 20T

0

P3 N W

nmDD

W N WK W- ÷ç ÷æ ö è ø÷ç ÷ç ÷ç ÷è ø

( )0è ø( )CSNR

( ) 2

3 dDSNR f P

æ ö÷ç ÷9

( )( )

3n

dDm

C

f PSNR W

ç= ÷ç ÷ç ÷è ø

SNR Comparison between FM and AM

Consider single tone modulation:( ) cos mm t tw=( ) m

What’s the condition on the modulation indexth t FM ff b tt i fso that FM offers better noise performance

than AM, given the same channel SNR?

( ) cos mm t tw= 1P =

Solution:

Minimum postdetection lowpass filter bandwidth is

( ) m 2nmP

10mW f=

SNR Comparison between FM and AM

( ) 22 æ öæ ö( )( )

2221 33 3

2 2n

d dDm

mC

SNR f fPSNR W f

bæ öæ ö ÷÷ çç= = =÷÷ çç ÷÷ çç ÷ ÷è ø è ø

FM:( ) mC è ø è ø

So system with larger modulation index has better noise performance,

at the price of more bandwidth: 2(+ 1) fm.

( )( )

2

21nmD

a PSNRSNR P

=+

AM: Max value is 1/3.

at the price of more bandwidth: 2( 1) fm.

( ) 21nmC

SNR a P+ FM has better performance than AM if

23 1 2 or =0.47 2 3 3b b> >

11FM with 0.5 is considered as wideband FM.b>

O liOutlineCh l SNR Channel SNR Channel SNR in DSB Channel SNR in SSB Channel SNR in SSB Channel SNR in AM Channel SNR in FM

SNR i b AM d FM SNR comparison between AM and FM Problem solving:

SNR in DSB SNR in DSB SNR in AM SNR in other systems

FM Threshold Effect

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bl 6 3 S R i DSProblem 6.3: SNR in DSB( ) ( ) ( )r cx t x t w t= +

d id h

A DSB system has predetection bandwidth BZiemer

Bandwidth BT

Bandwidth BD

A DSB system has predetection bandwidth BTand postdetection bandwidth BD.

Find the detection gain of the system.

SNRSNRDetection Gain = SNR

D

T

13

bl 6 3 S R i DSProblem 6.3: SNR in DSB( ) ( ) ( )( ) ( ) ( )r cx t x t w t= +

BT

BDfc

14

bl 6 3 S R i DSProblem 6.3: SNR in DSB

N

Sc(f)

BT/2

N0

f

BT/2

BT/2Postdet filter

115

BD

1

f

bl 6 3 S R i DSProblem 6.3: SNR in DSB

N0

Sc(f)

BT/2d fil

f

Postdet filter

1

16BDf

O liOutlineCh l SNR Channel SNR Channel SNR in DSB Channel SNR in SSB Channel SNR in SSB Channel SNR in AM Channel SNR in FM

SNR i b AM d FM SNR comparison between AM and FM Problem solving:

SNR in DSB SNR in DSB SNR in AM SNR in other systems

FM Threshold Effect

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l S R i AExample: SNR in AMConsider an AM system with AWGN noise:

120 10 / .2

N Watts Hz1. The noise psd:2

p

2. Message m(t): Bandwidth 4 kHz, 1 .6nmP =

3. Modulation index: a = 0.9.

4 Demodulation: Envelope detector4. Demodulation: Envelope detector

Find the minimum carrier amplitude Ac that yields

(SNR)D 40 dB.

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l S R i AExample: SNR in AM

2 2 2 2 1(0 9)A a P A

Solution:

6 212

0

(0 9)( ) 8437500

2 2(4000) 2 10nc m c

D c

A a P ASNR A

WN

104

10

10 log ( ) 40dB

log ( ) 4 ( ) 10D

D D

SNR

SNR SNR

42 10 volts

8437500cA 84375000 034 volts

c

cA

19

O liOutlineCh l SNR Channel SNR Channel SNR in DSB Channel SNR in SSB Channel SNR in SSB Channel SNR in AM Channel SNR in FM

SNR i b AM d FM SNR comparison between AM and FM Problem solving:

SNR in DSB SNR in DSB SNR in AM SNR in other systems

FM Threshold Effect

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Problem 6.23: SNR of Other Systems

( ) cos 2 cx t A f t

Noise n(t): AWGN with psd N0/2.Lowpass filter: unit gain, bandwidth W, centered around fc.

Find the SNR of y(t)Find the SNR of y(t).

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Problem 6.23: SNR of Other Systems

22

Th h ld ffFM Threshold Effecté ù

Consider unmodulated carrier:( ) ( ) ( ) ( )1 cos ( )cos ( )sinc c c c s ce t A t t n t t n t tw q f w q w qé ù= + + + + - +ë û

( ) 0tf =Consider unmodulated carrier: ( ) 0tf

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

1 cos ( )cos ( )sin

( ) ( )c c c c s ce t A t n t t n t t

A

w q w q w q

q q f

= + + + - +

[ ] ( ) = cos ( )cos ( )c c n c nA t r t t tw q w q f+ + + +

( ) [ ]1 can also be written as: ( )cos ( )c ne t R t t tw q y+ +

23

Th h ld ffFM Threshold Effect

Fig D.2 in the textbook (pp. 606) (random noise)

(t) here is the (t) we defined beforeq yTrajectory of R(t) /dt

( ) ( )y

(t)y (t)/dtdy

(t) <n cr A (t)y

(t)y (t)/dtdy

(t) >n cr A

(t)/dtdy(t)y

24

Th h ld ffFM Threshold Effect( )y(t) y

= 4dBSNR t=-4dBTSNR

( )/ddy Crackling noise in o tp t(t)/dtdy Crackling noise in output

=-10dBTSNR

0dB

T

0dB

2510dB

Th h ld ffFM Threshold Effect

( )DSNR Figure 6.18

Output SNR drops quickly

if PT/N0W is below a threshold.

PT

0

PN W dB

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