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Contemporary Communication Systems 1st

Edition by Mesiya ISBN 0073380369

9780073380360

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Chapter 7

7.1 Consider the transmission of a 5 kHz message signal on a baseband communications channel with a power attenuation of 30 dB. The channel noise is AWGN with N / 2 =10 10 Watts/Hz. Calculate the minimum transmitted power PT

to achieve a baseband SNRBB of at least 40 dB.

7.2 The message signal s(t) =10cos(1000πt)+5cos(2000πt) modulates the carrier

signal c(t) =10 3 cos(100×103πt)using DSB-SC AM scheme. White noise with power spectral density N / 2 =10 12 W/Hz is added during transmission. The received signal is coherently demodulated after ideal BP filtering.

a. Determine the receiver input CNR.

Solution:

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Solution: SNRBB = PR N o B 104 = PR 2×10 10 ×5×103 P =104 ×10×10 7 =1×10 2 W = 10 dBm PT = PR +Loss =10+30 = 40 dBm = 10 W

The receiver input CNR is given from (7.13) as

Substituting

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CNRIN α 2 A2 s 2 = c 2N o B Now 2 2 s 2 = 10 + 5 = 125 = 62.5 2 2 2 A2 =10 6 ,α =1, B =1 kHz N = 2×10 12

CNR 10 6 ×62.5 103 ×62.5 = = =15.625×103 = 41.93 dB

4×10 12 ×103 4

b. Determine the output SNR.

Solution:

SNRDSB = CNRIN = 41.93 dB

c. What degradation in postdetection SNR occurs if instead a 4th order Butterworth filter with 3-dB bandwidth of 1.5 kHz is used as a post-detection filter. (Hint: ratio of noise-equivalent bandwidth to 3-dB bandwidth order Butterworth filter).

Solution:

BN / f3dB =1.03 for a 4th

The noise power at the postdetection filter output equals P = 2N B . The roll-

off of the 4th order Butterworth filter at 1 kHz is obtained from (2.159) as 1/ 1+(1/1.5)8 = 0.981, i.e., about 2%. Therefore,

There is approximately 2 dB degradation in SNR using a 4th order Butterworth filter with 3-dB bandwidth of 1.5 kHz.

7.3 A message signal s(t)with spectral density G s ( f ) shown in Figure P7.1(a). It modulates the carrier signal c(t) =10 3 cos(2π f t)using DSB-SC AM scheme. White noise with power spectral density N / 2 =10 12 W/Hz is added during

transmission. The received signal is coherently demodulated.

a. Assuming an ideal pre-detection BPF, determine the CNR at its output.

Solution:

The ideal predetection filter has a bandwidth BT = 2B for DSB-SC AM scheme.

Therefore, the noise power, measured in the ideal predetection filter bandwidth is 2N o B . The CNR, measured at the output of the predetection filter, is given by

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n c o N

P (α A )2 s 2 10 6 ×

DSB P c 2N o BN 4×10 12 ×1.5×103

62.5

0.98

SNR = D = c = =10×103 = 40 dB

4 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. s c o Power in the input carrier signal α 2 A2 s 2 CNRPD = = c Input noise power in the predetection filter bandwidth 4N o B Figure P7.1 (a) Gs ( f ) 1 5 0 5 H PD ( f ) f (kHz) (b) (c) 7.5 1 0 H LP ( f ) 1 5 0 5 fc 7.5 7.5 10 kHz fc f (kHz) fc +7.5 f (kHz)

Figure P7.1 (a), we have ∞ s 2 = ∫ G ( f )df =5 −∞ A2 =10 6 ,α =1, B =5 kHz N = 2×10 12 Substituting CNR 10 6 ×5 103 ×5 = = =125= 21dB PD 8×10 12 ×5×103 8×5

Determine the output SNR assuming an ideal post-detection filter.

From

Solution:

SNR =

PR = 21+3= 24 dB

c. Modify your calculations (a) and (b) for a pre-detection filter with amplitude response shown in Figure P7.1(b).

Solution:

The CNR, measured at the output of the predetection filter with amplitude response shown in Figure P7.1(b), is given by

where BN is noise-equivalent bandwidth of the predetection filter

Although the spectral density of the in-phase noise n c (t)extends up to 7.5 KHz, the demodulator output noise is limited by the ideal LPF to 5 kHz. Thus, the performance of the system is not negatively impacted by the non-ideal predetection filter. Therefore, SNRDSB = PR = 24 dB N o B

d. How is the performance in (c) modified if a non-ideal post-detection filter with amplitude responses in Figure P7.1(c) is used instead.

Solution:

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IN N

CNR =

DSB

Power in the input carrier signal α 2 A2 s 2 CNRPD = = c Input noise power in the predetection filter bandwidth 4N o BN

∞ ∫ H PD ( f ) df B = 0 = 2⎛5000+1000× 2.5⎞ =11.667 kHz N ⎜ ⎟ H ( f ) 3 PD max ⎝ ⎠ Substituting CNR 10 6 ×5 103 ×5 = = =53.57 =17.3 dB PD 8×10 12 ×11.667×103 8×11.667

From Figure P7.1(c), the noise-equivalent bandwidth of the post-detection filter BN = 5.8335 kHz. Now

7.4 The message signal s(t) =5cos(700πt)+2cos(7000πt)modulates the carrier signal

c(t) =cos(2π fc t)using SSB-AM AM scheme. White noise with power spectral density N / 2 =10 12 W/Hz is added during transmission. The channel attenuates the transmitted signal by 60 dB. The received signal is coherently demodulated.

a. Assuming an ideal pre-detection BPF, determine the CNR at its output.

Solution:

The CNR at the ideal pre-detection BPF output is given by

b. Determine the output SNR.

SNRSSB = CNRIN = CNRPD = 27.14 dB

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(α A )2 s 2 10 6 ×5 SNR = c = = 214.28=

dB DSB 2N o BN 4×10 12 ×5.8335×103

23.3

P α 2 A2 s 2 CNR = R = c PD N B 4N B o o Now s 2 = 5 + 2 = 29 =14.5 2 2 2 A2 =1,α =10 3 , B =3.5 kHz N = 2×10 12

CNR 10 6 ×14.5 103 ×14.5 = = =5.178×102 = 27.14 dB PD 8×10 12 ×3.5×103 8×3.5

Substituting

7.5 A conventional AM system has a zero-mean Gaussian message signal whose spectrum is limited to B Hz. The peak value of the message signal is assumed to be

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in whole or

2.5σ m where σ m is the standard deviation of its amplitude variations. Determine the output SNR in terms of CNRIN . Assume m a = 0.875

Solution: SNRAM =ηCNRIN

The power in the normalized message signal s (t) = s(t) can be expressed n max s(t) in terms the average and peak power of the message signal as

where P max is power corresponding to the peak value (max s(t) ) of the message signal. For zero-mean Gaussian message signal,

Therefore, SNRAM = 0.109×CNRIN

7.6 The message signal s(t) =1.0cos(600πt)+0.5cos(6000πt) modulates the carrier signal c(t) =2cos(100×103 πt)using conventional AM scheme. The modulator operates with a modulation index m a = 0.85 We assume that the channel adds white

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may not

where m 2 s 2 η = a n (1+ m 2 s 2

)

2 s 2 = s

2 s 2 = σ m = 1 n (2.5σ )2 6.25

m 2 s 2 η = a n 2 2 a n (0.875)2 (1/6.25) = = 0.109 1+(0.875)2 (1/6.25)

max

Substituting

Gaussian noise with power spectral density N / 2 =10 8 W/Hz during transmission. The received signal is demodulated using an envelope detector.

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a. Determine the receiver input CNR.

b. Determine the output SNR.

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Solution: The receiver input CNR from (7.42) is P P α2 A2 (1+m 2 s 2 ) CNR = R = R = IN P N B 2N B Now ni o o m a = 0.85 2 2 2 s 2 = s = 1 ⎛1 + 0.5 ⎞ = 1.25 = 0.278 n P 1.52 ⎜ 2 2 ⎟ 2.25×2 max ⎝ ⎠ A2 = 4,α =1, B =3 kHz N o CNR = 2×10 8 4(1+0.852 ×0.278) = = 40.03×103 = 46 dB IN 4×10 8 ×3×103

Solution: SNRAM =ηCNRIN m 2 s 2 0.852 ×0.278 η = a n = = 0.167 (1+ m 2 s 2 ) 1 0.852 ×0.278 + a n Therefore, SNRAM =ηCNRIN = 0.167×40.03×103 = 6.685×103 =38.25 dB

c. What degradation in postdetection SNR occurs if instead a fourth-order Butterworth filter with 3-dB bandwidth of 4.5 kHz is used as a postdetection filter.

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Solution:

The noise-equivalent bandwidth BN of the fourth-order Butterworth filter equals its 3-dB bandwidth, i.e., 4.5 kHz. The roll-off of the filter at 3 kHz is about 2%, which is negligible. Therefore,

7.7 A conventional AM system transmits speech signal with bandwidth 3.3kHz and amplitude PDF given by

The peak value of the message signal is assumed to be 3.5

, where

is standard deviation of its amplitude variations.

The channel introduces white Gaussian noise with power spectral density N / 2 =10 10 W/Hz is added during transmission. Assume m

0.875

a. Find the carrier power required so that the output SNR exceeds 45 dB.

b. Determine the threshold value of carrier power. Solution:

The threshold in a conventional AM system occurs whenCNRPD ≈ 10 dB. That is,

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α 2 A2 m 2 s 2 4×(0.85)2 ×0.278 SNR = c a n = = 4463=36.5 dB 2N o BN 4×10 8 ×4.5×103

fs (s) = 2 5e 5 s

σ m

a =

Solution: 2 2 m 10 m a = 0.875, s n = (3.5σ = 0.0816, N = 2×10 )2 o , α =1, B =3.3 kHz α 2 A2 m 2 s 2 A2 ×(0.875)2 ×0.0816 SNR = c a n = c =104.5 AM or 4.5 2N o B 7 4×10 10 ×3.3×103 A2 = 10 ×10 ×4×3.3 = 0.668 c (0.875)2 ×0.0816 Carrier power P = A2 / 2 =

c c

0.668= 28.25 dBm

A2 (1+m 2 s 2 ) A2 ×(1+(0.875)2 ×0.0816)

13 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 10 = CNRPD c a n c = = 4N o B 8×10 10 ×3.3×103 or

7.8 Consider a message signal with bandwidth B = 5 kHz, and a normalized power s 2 = 0.25. It is required to transmit this signal via a channel with that has a power attenuation of 30 dB. The channel noise is AWGN with N / 2 =10 10 W/Hz. It is desirable to have an SNR of at least 40 dB at the receiver output.

a. Find the required carrier power if DSB-SC modulation is used.

Solution:

The output SNR for DSB-SC scheme is given by

where P = A2 / 2 is the carrier power. Since the channel produces an attenuation c c of 30 dB, the carrier power P c is

PR = 0.01 = 40 W = 46 dBm

10 3 ×0.25 10 3 ×0.25

b. Find the carrier power required if conventional AM with modulation index of m a = 0.75 is used.

Solution:

For the conventional AM scheme, the output SNR is given by

14 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. n o 7 6 A2 = 10×10 ×8×3.3 = 26.4×10 = 24.9×10 6 c 1+(0.875)2 ×0.0816 1.06 Threshold value of carrier power P = A2 / 2 =12.45×10 6 = 19.04 dBm c c

SNRDSB = PR N o B Substituting 104 = PR ⇒ P = 0.01 W =10 dBm 2×10 10 ×5×103 R Now 1 2 2 2 2 2 PR = α 2 A c s =α s P c

c. With the carrier power calculated in part (b), what is the maximum SNR that one can obtain for conventional AM (assuming of course that we do not overmodulate the message signal)?

15 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. a n α= 2 A2 a n a n a n c SNRAM =ηCNRIN =η PR , where N o B m 2 s 2 η = a n (1+ m 2 s 2 ) 0.752 ×0.25 = = 0.1233 1+0.752 ×0.25 104 = 0.1233 PR ⇒ P = 0.081 W =19.09dBm 2×10 10 ×5×103 R Now P = c (1+ m 2 s 2 )=α 2 (1+ m 2 s 2 ) P R 2 a n a n c where P = A2 / 2 is the carrier power. Since the channel produces an attenuation c c of 30 dB, the required carrier power P c by in the conventional AM case is given P = PR c α 2 (1+ m 2 s 2 ) = 0.081 = 71W = 48.54 dBm 10 3 ×(1+0.752 ×0.25)

Solution: Assuming m a =1.0 yields m 2 s 2 η = a n (1+ m 2 s 2 ) = 1×0.25 = 0.2 1+1×0.25 Therefore, P α2 (1+m 2 s 2 )P SNRAM =ηCNRIN = 0.2× R = 0.2× N o B N o B 10 3 ×(1+0.25)×710 = 0.2× =177.5×103 =52.49 dB 2×10 10 ×5×103

7.9 Consider an FM system with peak frequency deviation ∆fmax = 60 kHz, message signal bandwidth B = 10 kHz ,and a normalized power s 2 = 0.25. The channel introduces AWGN with N / 2 =10 9 W/Hz. Assuming received power level

PR = 250 mW , determine

a. Output SNR. Solution:

b. Required transmit power PT for an output SNR = 45 dB when the channel introduces attenuation of 30 dB.

7.10 A PM system with transmission bandwidth BT =140 kHz is used to transmit a message signal bandwidth B = 10 kHz, and normalized power s 2 = 0.25. The channel introduces AWGN with N / 2 =10 9 W/Hz. Assuming received power level PR = 250 mW , determine

a. Output SNR. Solution:

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D = ∆fmax = 60 = 6, s 2 = 0.25, P = 250 mW , N = 2×10 9 B 10 n R o

SNR =3D2 s 2 PR =3×62 ×0.25 250×10 FM n N o B 2×10 9 ×10×103 = 27×

×102 =337.5

103

Substituting

125

=55.28 dB

Solution: 104 5 =3×62 ×0.25 PR 2×10 9 ×10×103 0.5 0.5 P = 2×10 = 2×10 = 23.42×10 3 W =13.7 dBm R 3×62 ×0.25 27

T =13.7

= 43.7 dBm

b. Required transmit power PT for output SNR = 45 dB when the channel introduces power attenuation of 30 dB.

c. Repeat (a) and (b) if the peak phase deviation ∆φmax is limited to πradians. Calculate the required transmission bandwidth B

17 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. n IN PM IN T 140 = 2(∆φmax +1)10 or ∆φmax = 7 1= 6 SNRPM =(∆φmax )2 s 2CNR where CNR P 250×10 3 = R = = 125×102 N o B 2×10 9 ×104 SNR = 62 ×0.25×125×102 =112.5×103 =50.51 dB

Solution: 104 5 = 62 ×0.25 PR 2×10 9 ×10×103 2×10 0.5 2 P = = = 70.27×10 3 W =18.47 dBm R 62 ×0.25 9 10 PT =18.47 + 30 = 48.47 dBm

T Solution: ( )2 2 2 2 3 SNRPM = ∆φmax s n CNRIN =π ×0.25×125×10 =30.84×10 = 44.9 dB 104 5 =π 2 ×0.25 PR 2×10 9 ×10×103 2×10 0.5 P = = 256.32×10 3 W = 24.087 dBm R π 2 ×0.25 PT = 24.087 + 30 = 54.087 dBm B = 2(π+1)10×103 =82.83 kHz

7.11 A message signal with bandwidth B = 15 kHz, and normalized power s 2 = 0.25 is

transmitted using an FM system Assuming the deviation ratio D = 3, find the output SNR for input CNR values (i) 11 dB, (ii) 25 (iii) 35 dB. In which case is the system operating below threshold?

Solution:

(CNRIN ) ≈ 20(3+1) =80 =19 dB

For input CNR values of 25 and 35 dB, the output SNR is given by SNR =3D2 s 2CNR FM n IN In dB form SNRFM =10log10 (3D2 s2 )+ CNR =10log10 (6.75)+ CNRIN =8.29+ CNRIN

For CNRIN = 25 dB, SNRFM =8.29+25 =33.29

For CNRIN =35 dB, SNRFM =8.29+35 = 43.29

For input CNR of 11 dB, the system is operating below threshold and the output SNR is given by 3 β 2CNR SNR = 2 FM CNRIN 1+ CNR e 2(1+β) π IN

Substituting values for β and CNRIN , we obtain SNRFM = 7.4033 dB

7.12 An RC filter with time constant 25 µsec is used for de-emphasis in an FM system. Find the output SNR improvement in an FM system broadcasting signals with bandwidth 15 kHz. How much output SNR is improved if the signal bandwidth is increased to 53 kHz.

Solution:

IN 12

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β th

With deemphasis filtering, the output noise power is given by

For B = 15 kHz,

For B = 53 kHz,

Thus the output SNR is improved by 10log

13.7 = 7.62 dBif the signal 10 2.37 bandwidth is increased to 53 kHz.

7.13 Consider a message signal with bandwidth B = 5 kHz and a normalized power s 2 = 0.15. It is required to transmit this signal via a channel that attenuates the transmitted signal by 30 dB. The channel noise is AWGN with spectral density N / 2 =10 10 W/Hz. It is desirable to have an SNR of at least 60 dB at the receiver output.

2 ⎛ K ⎞ P = 2 FD N f 3 ⎡ B tan 1 B ⎤ n ⎜ A ⎟ o 1 ⎢ f f ⎥ ⎝ c ⎠ ⎣ 1 1 ⎦

2 ⎛ K ⎞ P = 2 FD N f 3 ⎡ 15 −tan 1 15 ⎤ n ⎜ ⎟ ⎝ A c ⎠ o 1 ⎢6.366 6.366⎥ ⎛ K ⎞ = 2 FD 2 2 N f 3 [2.356 1.17]= 2.37 ⎛ KFD ⎞ N f 3 ⎜ ⎟ o 1 ⎜ ⎟ o 1 ⎝ A c ⎠ ⎝ A c ⎠

2 ⎛ K ⎞ P = 2 FD N f 3 ⎡ 53 tan 1 53 ⎤ n ⎜ ⎟ ⎝ A c ⎠ o 1 6.366 6.366 ⎛ K ⎞ = 2 FD 2 2 N f 3 [8.325−1.45]=13.7 ⎛ KFD ⎞ N f 3 ⎜ ⎟ o 1 ⎜ ⎟ o 1 ⎝ A c ⎠ ⎝ A c ⎠

a. Find the minimum required transmit power PT 10. for a PM system with ∆φmax = Solution:

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SNRPM =(∆φmax )2 s 2 PR N o B

106 =102 ×0.15 PR 2×10 10 ×5×103

P = 1 = 1 R 1002 ×0.15 15 = 66.67×10 3 W =18.24 dBm

PT = PR +channel attenuation (dB) =18.24+30 = 48.24 dBm = 66.68 Watts

b. Find the minimum required transmit power PT for an FM system with D = 10.

Solution:

P = 1 = 1 = 22.22×10 3 W =13.467 dBm R 3×100×0.15 45

PT = PR +channel attenuation (dB) =13.467+30 = 43.467 dBm = 22.22 Watts

c. Find the minimum required transmit power PT for an FM system utilizing preemphasis and de-emphasis with D = 10 and f1 = 600 Hz. Solution:

7.14 In a certain FM satellite communication system, the output SNR is found to be 40 dB with D = 6. The modulating signal has a bandwidth B = 10 kHz and a normalized power s 2 = 0.15. The system with D = 6 is not in threshold, but the

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SNR

R FM

PR 2×10 10 ×5×103

=3D2 s

n N o

=3×10

×0.15

(SNR ) 2 = D2 ⎛ B ⎞ s 2 PR FM DE ⎜ ⎟ ⎝ f1 ⎠ N o B 106 =102 ×8.332 ×0.15 PR 2×10 10 ×5×103

= 1 = 1 = 0.96×10 3 W =

dBm

0.15

×69.44

0.177

R 100×69.44×

PT = PR +channel attenuation (dB) = 0.177+30 = 29.82 dBm = 0.96 Watts

output SNR is required to be at least 50 dB. The increase of the output SNR can be accomplished by increasing either (i) D (that is, the transmission bandwidth) and/or (ii) the transmitted power.

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a. If option (i) is selected, what are the maximum values of D and the corresponding transmission bandwidth that can be used without running into threshold? What is the corresponding output SNR?

Solution:

For the output SNR = 40 dB and D = 6, the input CNRIN is given by

104

Next we determine the maximum value of D using the threshold condition (7.103).

Solving by trial and error or using MATLAB yields the maximum value of D ≈ 19.43.

Transmission bandwidth BT corresponding to deviation ratio D ≈ 19.43

The minimum input (CNRIN ) required to operate above the threshold is (CNRIN ) = 20(D +1) = 408.6= 26.113 dB

For CNRIN = 27.9 dB, the system is not in the threshold and the output SNR is given by SNR =3D2 s 2CNR

n IN

3×19.432 ×0.15×617.3

=104.87×103 =50.2 dB

b. If option (ii) is chosen instead, calculate the minimum increase in transmitted power required to attain an output SNR of 50 dB subject to the constraint that the channel bandwidth is limited to 250 kHz? What is the corresponding value of D?

Solution:

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SNR

FM CNRIN = 3D2 s 2 = = 617.3

3×62 ×0.15

(27.9 dB)

(SNRFM =105 = 60D2 (D +1)s 2 = 9D2 (D +

= 2(D +1)B =

2(19.43+1)10

= 408.6 kHz

If the available channel bandwidth is limited to 250 kHz, the maximum value of D is given by

Increase in transmitted power required = 32.25 26.113 = 6.1375 dB

7.15 The baseband frequency spectrum for stereo FM broadcasting is depicted in Figure 5.41.

a. Calculate the noise output power without pre-emphasis.

Solution:

We refer to block diagram of the FM stereo receiver in Figure 5.42.

L+R channel

The L+R channel occupies the frequency band [0 15] kHz. From (7.88), the power spectral density of the noise n1(t)at the FM demodulator output is

The noise power at the FM demodulator output for the L+R channel is

24 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. n n n ⎩ IN o

250×103 = 2(D +1)×104 or 250×103 D = 2×104 1=11.5 CNR SNR 105 = FM = =1.68×103 (32.25 dB) 3D2 s 2 3×11.52 ×0.15

⎧⎛ K ⎞2 ⎪⎜ FD ⎟ f 2N , 0≤ f ≤15 kHz G = G 1 L+R =⎨⎝α A c ⎠ ⎪ 0, otherwise

P = 2⎛ KFD ⎞ 2 2 N B3 = ⎛ KFD ⎞ N ×2.25×1012 n1 3⎜ α A ⎟ o ⎜ α A ⎟ o ⎝ c ⎠ ⎝ c ⎠ L-R channel

The L R channel occupies the frequency band [23 53] kHz. From (7.88), the power spectral density of the noise at the FM demodulator output is

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The power spectral density of the noise n2 (t)at the DSB-AM demodulator

=19 kHzis pilot tone frequency in stereo FM signal. The noise power

the output of the DSB-AM demodulator (L R channel path) is given by

b. Calculate the same for monophonic transmission (the baseband spectral range for mono FM signal is 15 kHz).

Solution:

The noise output power at the FM demodulator output for a mono FM system is identical to that of the L+R channel. That is,

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distributed,

on a website, in whole or part. n ⎩ n n ) n = ⎩ o o ⎜ ⎟ ⎧⎛ K ⎞2 ⎪⎜ FD ⎟ f 2N , 23≤ f ≤53 kHz G L R =⎨⎝α A c ⎠ ⎪ 0, otherwise

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G = G 2 L R ( f 2 fP + G L R ( f +

fP ) ⎧ 2 2 ⎛ KFD ⎞ ( f 2 f )2 +( f +2 f )2 N ⎛ K ⎞ = FD ( f 2 +4 f 2 )2N , B ≤ f ≤ B ⎪⎜ ⎟ P P o ⎜ ⎟ P o ⎨⎝α A c ⎠ ⎝α A c ⎠ ⎪ 0, otherwise

P = 2N 2 ⎛ KFD ⎞ B ( f 2 +4 f 2 )df n2 o ⎜ α A ⎟ ∫ P ⎝ c ⎠ B 2 B B = 2N ⎛ KFD ⎞ ⎡ f 2df +4 ⎤ f 2df o ⎜ ⎟ ⎢∫ ∫ P ⎥ ⎝α A c ⎠ ⎣ B 2 3 B ⎦ = 2N ⎛ KFD ⎞ ⎡ 2B +8 f 2B ⎤ o ⎜ α A ⎟ ⎢ 3 P ⎥ ⎝ c ⎠ ⎣ ⎦ 2 = N ⎛ KFD ⎞ ⎝α A c ⎠ ×91.14×1012

output is

where

P = 2⎛ KFD ⎞ 2 2 N B3 = ⎛ KFD ⎞ N ×2.25×1012

c. How much noisier is stereo FM versus mono FM in dBs?

27 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. nFM mono 3⎜ α A ⎟ o ⎜ α A ⎟ o ⎝ c ⎠ ⎝ c ⎠

Solution:

The output of the upper branch after matrixing is given by

Similarly, the output of the lower branch after matrixing is given by

Therefore, the noise output power for both L+R and L R channels after matrixing is given by

We conclude that stereo audio is 16.18 dB noisier than mono audio without deemphasis filtering.

d. Repeat (a) through (c) taking 75 µsec pre-emphasis and de-emphasis into consideration. Show that stereo FM is 22 dB noisier than mono FM.

Solution:

Mono audio with deemphasis

For mono audio case, the output noise power of the after deemphasis filtering is given by

28 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. (P )

L(t)+ R(t)+ L(t) R(t)+ n1(t)+ n2(t) ≈ 2L(t)+ n2(t)

L(t)+ R(t) L(t)+ R(t)+ n1(t) n2(t) ≈ 2L(t)+ n2(t)

(P ) = P + P 2 ⎛ K ⎞ = FD N ×93.39×1012 no stereo n1 n2 ⎜ ⎟ o ⎝ A c ⎠ Thus, (P n ) ⎛93.39×1012 ⎞ 10log o stereo =10log ⎜ ⎟=16.18 dB 10 no mono 10 ⎝ 2.25×1012 ⎠

1 106 f = = = 2.122×103 1 2πτ 2π×75

Stereo audio with deemphasis

For stereo audio case, the output noise power at the output of the L+R channel branch after deemphasis filtering is identical to that in the mono case. That is,

For the L R channel, the power spectral density of the noise n2 (t)at the DSBAM modulator output after deemphasis filtering is given by

The output noise power at the output of the L R channel branch after deemphasis filtering is now obtained as

29 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. = 2 p o ⎩ p o 1 2 p o 1 1 1 ⎨ 2 B B (P ) ⎛ K ⎞ = 2 FD 2 2 N f 3 ⎡ B −tan 1 B ⎤ = 2 ⎛ KFD ⎞ N f 3 ⎡ 15 tan 1 15 ⎤ no mono ⎜ A ⎟ o 1 ⎢ f f ⎥ ⎜ A ⎟ o 1 2.122 2.122 ⎝ c ⎠ ⎣ 1 1 ⎦ ⎝ c ⎠ ⎛ K ⎞ =11.28 FD 2 2 N f 3 =11.28 ⎛ KFD ⎞ N (2.122×103 )3 ⎜ ⎟ o 1 ⎜ ⎟ o ⎝ A c ⎠ ⎝ A c ⎠

2 ⎛ K ⎞ P =11.28 FD N (2.122×103 )3 n ⎜ ⎟ o ⎝ A c ⎠

⎧ ⎛ ⎞2 G ⎪ H DE ( f ) KFD ⎜ α A ⎟ ( f 2 +4 f 2 )2N , B ≤ f ≤ B kHz n2 ⎝ c ⎠ ⎪ 0, otherwise

2 ⎛ KFD ⎞ B ( f 2 +4 f 2 ) P n = 2N ⎜ ⎟ ∫ df ⎝α A c ⎠ 1+( f / f )2 ⎛ K ⎞ ⎡ B f 2 B f 2 ⎤ = 2N ⎜ FD ⎟ ⎢∫ df +4 ∫ df ⎥ ⎝α A c ⎠ ⎢ ⎣ B 2 1+( f / f )2 1+( f / f )2 ⎥ = 4N ⎛ KFD ⎞ ⎡ ⎛ B B ⎞

30 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. o f 3 tan 1 +4 f 2 f tan 1 B ⎤ o ⎜ ⎟ ⎢ 1 ⎜ ⎟ p 1 ⎥ ⎝α A c ⎠ ⎣ 2 ⎝ f1 f1 ⎠ f1 ⎦ = 4N ⎛ KFD ⎞ ⎡(2.122)3 ⎛ 15 tan 1 15 ⎞ +(38)2 ×2.122×tan 1 15 ⎤ ×109 o ⎜ A ⎟ ⎢ ⎜ 2.122 2.122⎟ 2.122⎥ ⎝ c ⎠ ⎣ ⎝ ⎠ ⎦ 2 ⎛ K ⎞ =⎜ FD ⎟ N ×17.55×1012 ⎝ A c ⎠

We observe that P

. Therefore, the noise output power for both L+R and

channels after matrixing is given by

We conclude that stereo audio is 22.1dB noisier than mono audio when deemphasis filtering is applied in both cases.

7.16 In FDM telephony system, 12 voice channels are frequency-division-multiplexed to form a group that occupies a frequency band from 60 to 108 kHz. Each 3.3-kHz voice signal is SSB-AM modulated and assigned a 4 kHz frequency slot to provide 0.7 kHz guardband between channels. The frequency-division-multiplexed group signal is then frequency modulated with a peak deviation∆fmax of 400 kHz.

a. Determine the transmission bandwidth of the FM signal.

b. Determine the degradation in output SNR of the 12th channel compared with the first Solution: The 1st channel occupies the frequency band [60 64] kHz. From (7.88), the power spectral density of the noise at the FM demodulator output is

31 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. (P ⎝ ⎠ 2 ) T o T

2 1 L R

(P ) = P + P 2 ⎛ K ⎞ ≈ P = FD N ×17.55×1012 no stereo n1 n2 n ⎜ ⎟ o ⎝ A c ⎠

(P n ) ⎛ 17.55×1012 ⎞ 10log o stereo =10log ⎜ ⎟= 22.1dB 10 no mono 10 ⎜11.28×(2.122×103 )3 ⎟

Thus,

Solution: ∆f 400×

D = max =

B 108

Transmission bandwidth B = 2(D +1)B = 2(3.7+1)×108×103 =1.0152 MHz .

103

=3.7

×103

⎧⎛ K ⎞2 ⎪⎜ FD ⎟ f 2N , f ≤ B / 2

32 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. n ⎩ G FM ( f ) =⎨⎝α A c ⎠ ⎪ 0, otherwise

The noise power at the output of the 1st channel USB-AM demodulator is given by

The 12th channel occupies the frequency band [104 108] Hz. The noise power at the output of the 12th channel USB-AM demodulator is given by

Assuming equal signal output powers for all channels in the FDM group, the degradation in output SNR of the 12th channel compared with the 1st is

To assure equal demodulator output SNRs, the voice channels in the multiplex at centered at higher frequencies are carried at increasingly higher levels (resulting in higher frequency deviation) to compensate for the parabolic noise spectral density of FM demodulator.

7.17 A radio receiver block diagram is shown in Figure P7.2. Each block is impedance matched to the source of 50. The radio is designed to receive a signal with a bandwidth of 20MHz centered about the BP filter frequency.

a. Calculate the noise figure of the overall receiver.

Solution:

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64 64 2 P1 = 2 G ( f )df = 2 ⎛ KFD ⎞ f 2N df n ∫ nFM ∫⎜ α A ⎟ o 60 60⎝ c ⎠ 9 2 2 = 2×10 × ⎛ KFD ⎞ N ⎡643 603 ⎤= ⎛ KFD ⎞ N ×30.76×1012 3 ⎜ α A ⎟ o ⎣ ⎦ ⎜ α A ⎟ o ⎝ c ⎠ ⎝ c ⎠

108 108 2 P12 = 2 G ( f )df = 2 ⎛ KFD ⎞ f 2N df n ∫ nFM ∫ ⎜ α A ⎟ o 104 104⎝ c ⎠ 9 2 2 = 2×10 × ⎛ KFD ⎞ N ⎡1083 −1043 ⎤= ⎛ KFD ⎞ N ×89.9×1012 3 ⎜ α A ⎟ o ⎣ ⎦ ⎜ α A ⎟ o ⎝ c ⎠ ⎝ c ⎠

P12 89.9 n 10log10 1 n =10log10 30.76 = 4.657 dB

The noise factor of a lossy two-port device is equal to its attenuation or loss. For example, NF1 = L1 =3 dB. Using Frii’s

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The noise figure NFT of the receiver is 6.07 dB

b. Calculate the input signal to the receiver to achieve the SNR of at least 25 dB at the input of baseband stage. Specify the answer in dBm and in volts. Solution:

In order to achieve the SNR of at least SNR req = 25 dB at the input of baseband stage, the SNR at the receiver input should be

c. Calculate the signal and noise power at each stage of the receiver chain for input signal level determined in part (b). Solution:

Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1 TF = F + F2 1 + F3 1 + F4 1 + G1 G1G2 G1G2G3 =1.585+ 0.585 + 31.62 1 + 1.585 1 + 0.63 0.63×101 5 =1.585+0.928+1.537 = 4.05 0.63×101.5 ×105

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SNR1 = SNR req + NFT = 25+6.07 =31

noise power at the receiver input from (6.295) is Nin = kTB = 4×10 21 ×20×106 =8×10 14 = 101 dBm Input signal power = 101 + 31= 70 dBm (70.71 µV RMS) 1 BPF GLPF = 3 dB 2 LPF 3 4 5 Detector Baseband GBPF = 2 dB G1 =15 dB NF1 = 2 dB G2 =50 dB NF2 =15 dB Gdet NF2 = 4 dB =10 dB

The

The output noise power is computed using (6.324).

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d. The baseband stage can only work with signals as large as 500 mV. What is the maximum signal that can be received?

Solution:

Signal power corresponding to 500 mV maximum = (500×10 3 )2 2×50 = 4 dBm

Since the total gain of the receiver is 56 dB, the maximum signal that can be received at the receiver input is 4 56 dBm = 52 dBm.

37 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Stage Signal Power Noise Power (1) BP Filter −72 dBm −101 dBm (2) Low-noise Amplifier 57 84 (3) Amplifier 7 32 (4) LPF 10 35 (5) Detector 14 39