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Solution Manual for Principles of Communications , 7th Edition

Solution Manual for Principles of Communications, 7th Edition breaks down difficult textbook problems into simple solutions, making your study time more effective.

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    APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS 1 Problem A.1 All parts of the problem are solved using the relation V rm s = p 4 kT RB where k = 1 : 38  10 23 J/K B = 30 MHz = 3  10 7 Hz a. For R = 10 ; 000 ohms and T = T 0 = 290 K V rm s = p 4 (1 : 38  10 23 ) (290) (10 4 ) (3  10 7 ) = 6 : 93  10 5 V rms = 69 : 3  V rms b. V rm s is smaller than the result in part (a) by a factor of p 10 = 3 : 16 : Thus V rm s = 21 : 9  V rms c. V rm s is smaller than the result in part (a) by a factor of p 100 = 10 : Thus V rm s = 6 : 93  V rms d. Each answer becomes smaller by factors of 2 ; p 10 = 3 : 16 ; and 10 , respectively. Problem A.2 Use I = I s  exp  eV kT  1  We want I > 20 I s or exp eV kT  1 > 20 . a. At T = 290 K, e kT = 1 : 6  10 19 1 : 38  10 23  290  = 40 , so we have exp (40 V ) > 21 giving V > ln (21) 40 = 0 : 0761 volts i 2 rm s  = 2 eIB ' 2 eBI s exp  eV kT  or i 2 rm s B = 2 eI s exp  eV kT  = 2 1 : 6  10 19  1 : 5  10 5  exp (40  0 : 0761) = 1 : 0075  10 22 A 2 /Hz b. If T = 90 K, then e kT  = 129 , and for I > 20 I s , we need exp (129 V ) > 21 or V > ln (21) 129 = 2 : 36  10 2 volts

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    APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS 2 Thus i 2 rm s B = 2 1 : 6  10 19  1 : 5  10 5  exp (129  0 : 0236) = 1 : 0079  10 22 A 2 /Hz approximately as before. Problem A.3 a. Use Nyquistí s formula to get the equivalent circuit of R eq in parallel with R L , where R eq is given by R eq = R 3 ( R 1 + R 2 ) R 1 + R 2 + R 3 The noise equivalent circuit consists of a rms noise voltage, V eq , in series with R eq and a rms noise voltage, V L , in series with with R L with these series combinations being in parallel. The equivalent noise voltages are V eq = p 4 kT R eq B V L = p 4 kT R L B The rms noise voltages across the parallel combination of R eq and R L , by using superposition and voltage division, are V 01 = V e q R L R eq + R L and V 02 = V L R e q R eq + R L Adding noise powers to get V 2 0 we obtain V 2 0 = V 2 eq R 2 L ( R eq + R L ) 2 + V 2 L R 2 eq ( R eq + R L ) 2 = (4 kT B ) R L R eq R eq + R L = 4 kT B R L R 3 ( R 1 + R 2 ) R 1 R 3 + R 2 R 3 + R 1 R L + R 2 R L + R 3 R L Note that we could have considered the parallel combination of R 3 and R L as an equivalent load resistor and found the Thevenin equivalent. Let R jj = R 3 R L R 3 + R L The Thevenin equivalent resistance of the whole circuit is then R eq2 = R jj ( R 1 + R 2 ) R jj + R 1 + R 2 = R 3 R L R 3 + R L ( R 1 + R 2 ) R 3 R L R 3 + R L + R 1 + R 2 = R L R 3 ( R 1 + R 2 ) R 1 R 3 + R 2 R 3 + R 1 R L + R 2 R L + R 3 R L

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    APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS 3 and the mean-square output noise voltage is now V 2 0 = 4 kT BR eq2 which is the same result as obtained before. b. With R 1 = 2000 ; R 2 = R L = 300 ; and R 3 = 500 , we have V 2 0 B = 4 kT B R L R 3 ( R 1 + R 2 ) R 1 R 3 + R 2 R 3 + R 1 R L + R 2 R L + R 3 R L = 4 1 : 38  10 23  (290) (300) (500) (2000 + 300) 2000 (500) + 300 (500) + 2000 (300) + 300 (300) + 500 (300) = 2 : 775  10 18 V 2 /Hz Problem A.4 Find the equivalent resistance for the R 1 ; R 2 ; R 3 combination and set R L equal to this to get R L = R 3 ( R 1 + R 2 ) R 1 + R 2 + R 3 Problem A.5 Using Nyquistí s formula, we Önd the equivalent resistance looking back into the terminals with V rm s across them. It is R eq = 50 k k 20 k k (5 k + 10 k + 5 k ) = 50 k k 20 k k 20 k = 50 k k 10 k = (50 k ) (10 k ) 50 k + 10 k = 8 ; 333 Thus V 2 rm s = 4 kT R eq B = 4 1 : 38  10 23  (400) (8333) 2  10 6  = 3 : 68  10 10 V 2 which gives V rm s = 19 : 18  V rms

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    APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS 4 Problem A.6 To Önd the noise Ögure, we Örst determine the noise power due to a source at the output, then due to the source and the network, and take the ratio of the latter to the former. Initally assume unmatched conditions. The results are V 2 0 due to R S , only =  R 2 k R L R S + R 1 + R 2 k R L  2 (4 kT R S B ) V 2 0 due to R 1 and R 2 =  R 2 k R L R S + R 1 + R 2 k R L  2 (4 kT R 1 B ) +  R L k ( R 1 + R S ) R 2 + ( R 1 + R S ) k R L  2 (4 kT R 2 B ) V 2 0 due to R S ; R 1 and R 2 =  R 2 k R L R S + R 1 + R 2 k R L  2 [4 kT ( R S + R 1 ) B ] +  R L k ( R 1 + R S ) R 2 + ( R 1 + R S ) k R L  2 (4 kT R 2 B ) The noise Ögure is (after some simpliÖcation) F = 1 + R 1 R S +  R L k ( R 1 + R S ) R 2 + ( R 1 + R S ) k R L  2  R S + R 1 + R 2 k R L R 2 k R L  2 R 2 R S In the above, R a k R b = R a R b R a + R b Note that the noise due to R L has been excluded because it belongs to the next stage. Since this is a matching circuit, we want the input matched to the source and the output matched to the load. Matching at the input requires that R S = R in = R 1 + R 2 k R L = R 1 + R 2 R L R 2 + R L and matching at the output requires that R L = R out = R 2 k ( R 1 + R S ) = R 2 ( R 1 + R S ) R 1 + R 2 + R S Next, these expressions are substituted back into the expression for F . After some simpliÖcation, this gives F = 1 + R 1 R S +  2 R 2 L R S ( R 1 + R 2 + R S ) = ( R S R 1 ) R 2 2 ( R 1 + R S + R L ) + R 2 L ( R 1 + R 2 + R S )  2 R 2 R S Note that if R 1 >> R 2 we then have matched conditions of R L  = R 2 and R S  = R 1 : Then, the noise Ögure simpliÖes to F = 2 + 16 R 1 R 2

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    APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS 5 Note that the simple resistive pad matching circuit is very poor from the standpoint of noise. The equivalent noise temperature is found by using T e = T 0 ( F 1) = T 0  1 + 16 R 1 R 2  Problem A.7 a. The important relationships are F l = 1 + T e l T 0 T e l = T 0 ( F l 1) T e 0 = T e 1 + T e 2 G a 1 + T e 3 G a 1 G a 2 Completion of the table gives Ampl. No. F T e i G a i 1 not needed 300 K 10 dB = 10 2 6 dB 864.5 K 30 dB = 1000 3 11 dB 3360.9 K 30 dB = 1000 Therefore, T e 0 = 300 + 864 : 5 10 + 3360 : 9 (10) (1000) = 386 : 8 K Hence, F 0 = 1 + T e 0 T 0 = 2 : 33 = 3 : 68 dB b. With ampliÖers 1 and 2 interchanged T e 0 = 864 : 5 + 300 10 + 3360 : 9 (10) (1000) = 865 : 14 K This gives a noise Ögure of F 0 = 1 + 865 : 14 290 = 3 : 98 = 6 dB

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    APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS 6 c. See part (a) for the noise temperatures. d. For B = 50 kHz, T S = 1000 K, and an overall gain of G a = 10 7 , we have, for the conÖguration of part (a) P na; out = G a k ( T 0 + T e 0 ) B = 10 7 1 : 38  10 23  (1000 + 386 : 8) 5  10 4  = 9 : 57  10 9 watts We desire P sa; out P na; out = 10 4 = P sa; out 9 : 57  10 9 which gives P sa; out = 9 : 57  10 5 watts For part (b), we have P na; out = 10 7 1 : 38  10 23  (1000 + 865 : 14) 5  10 4  = 1 : 29  10 8 watts Once again, we desire P sa; out P na; out = 10 4 = P sa; out 1 : 29  10 8 which gives P sa; out = 1 : 29  10 4 watts and P sa; in = P sa; out G a = 1 : 29  10 11 watts

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    APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS 7 Problem A.8 a. The noise Ögure of the cascade is F overall = F 1 + F 2 1 G a 1 = L + F 1 (1 =L ) = LF b. For two identical attenuator-ampliÖer stages F overall = L + F 1 (1 =L ) + L 1 (1 =L ) L + F 1 (1 =L ) L (1 =L ) = 2 LF 1  2 LF; L >> 1 c. Generalizing, for N stages we have F overall  N F L Problem A.9 a. The data for this problem is Stage F i G i 1 (preamp) 2 dB = 1.58 G 1 2 (mixer) 8 dB = 6.31 1.5 dB = 1.41 3 (ampliÖer) 5 dB = 3.16 30 dB = 1000 The overall noise Ögure is F = F 1 + F 2 1 G 1 + F 3 1 G 1 G 2 which gives 5 dB = 3 : 16 = 1 : 58 + 6 : 31 1 G 1 + 3 : 16 1 1 : 41 G 1 or 3 : 16 1 : 58 = 6 : 31 1 G 1 + 3 : 16 1 1 : 41 G 1 or G 1 = 5 : 31 1 : 58 + 2 : 16 (1 : 41) (1 : 58) = 4 : 33 = 6 : 37 dB b. First, assume that G 1 = 15 dB = 31 : 62 : Then F = 1 : 58 + 6 : 31 1 31 : 62 + 3 : 16 1 (1 : 41) (31 : 62) = 1 : 8 = 2 : 54 dB

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    APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS 8 Then T es = T 0 ( F 1) = 290 (1 : 8 1) = 230 : 95 K and T overall = T es + T a = 230 : 95 + 300 = 530 : 95 K Now use G 1 as found in part (a): F = 3 : 16 T es = 290 (3 : 16 1) = 626 : 4 K T overall = 300 + 626 : 4 = 926 : 4 K c. For G 1 = 15 dB = 31 : 62 ; G a = (31 : 62) (1 : 41) (1000) = 4 : 46  10 4 : Thus P na; out = G a kT overall B = 4 : 46  10 4  1 : 38  10 23  (530 : 95) 10 7  = 3 : 27  10 6 watts For G 1 = 6 : 37 dB = 4 : 33 ; G a = (4 : 33) (1 : 41) (1000) = 6 : 11  10 3 : Thus P na; out = 6 : 11  10 3  1 : 38  10 23  (926 : 4) 10 7  = 7 : 81  10 7 watts Note that for the second case, we get less noise power out even wth a larger T overall . This is due to the lower gain of stage 1, which more than compensates for the larger input noise power. d. A transmission line with loss L = 2 dB connects the antenna to the preamp. We Örst Önd T S for the transmission line/preamp/mixer/amp chain: F S = F TL + F 1 1 G TL + F 2 1 G TL G 1 + F 3 1 G TL G 1 G 2 ; where G TL = 1 =L = 10 2 = 10 = 0 : 631 and F TL = L = 10 2 = 10 = 1 : 58 Assume two cases for G 1 : 15 dB and 6.37 dB. First, for G 1 = 15 dB = 31 : 6 , we have F S = 1 : 58 + 1 : 58 1 0 : 631 + 6 : 31 1 (0 : 631) (31 : 6) + 3 : 16 1 (0 : 631) (31 : 6) (1 : 41) = 2 : 84

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    APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS 9 This gives T S = 290 (2 : 84 1) = 534 K and T overall = 534 + 300 = 834 K Now, for G 1 = 6 : 37 dB = 4 : 33 , we have F S = 1 : 58 + 1 : 58 1 0 : 631 + 6 : 31 1 (0 : 631) (4 : 33) + 3 : 16 1 (0 : 631) (4 : 33) (1 : 41) = 5 : 00 This gives T S = 290 (5 : 00 1) = 1160 K and T overall = 1160 + 300 = 1460 K Problem A.10 a. (a) Using P na; out = G a kT S B = 10 8  1 : 38  10 23  (1800) 3  10 6  with the given values yields P na; out = 7 : 45  10 5 watts b. We want P sa; out P na; out = 10 5 or P sa; out = 10 5  7 : 45  10 5  = 7 : 45 watts This gives P sa; in = P sa; out G a = 7 : 45 10 8 = 7 : 45  10 8 watts = 71 : 28 dBW = 41 : 28 dBm

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    APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS 10 Problem A.11 a. For  A = 1 dB, Y = 1 : 259 and the e§ective noise temperature is T e = 600 (1 : 259) (300) 1 : 259 1 = 858 : 3 K For  A = 1 : 5 dB, Y = 1 : 413 and the e§ective noise temperature is T e = 600 (1 : 413) (300) 1 : 413 1 = 426 : 4 K For  A = 2 dB, Y = 1 : 585 and the e§ective noise temperature is T e = 600 (1 : 585) (300) 1 : 585 1 = 212 : 8 K b. These values can be converted to noise Ögure using F = 1 + T e T 0 With T 0 = 290 K, we get the following values: (1) For  A = 1 dB ; F = 5 : 98 dB; (2) For  A = 1 : 5 dB ; F = 3 : 938 dB; (3) For  A = 2 dB ; F = 2 : 39 dB. Problem A.12 a. Using the data given, we can determine the following:  = 0 : 039 m   4 d  2 = 202 : 4 dB G T = 39 : 2 dB P T G T = 74 : 2 dBW This gives P S = 202 : 4 + 74 : 2 + 6 5 = 127 : 2 dBW b. Using P n = kT e B for the noise power, we get P n = 10 log 10  kT 0  T e T 0  B  = 10 log 10 [ kT 0 ] + 10 log 10  T e T 0  + 10 log 10 ( B ) = 174 + 10 log 10  1000 290  + 10 log 10 10 6  = 108 : 6 dBm = 138 : 6 dBW

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    APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS 11 c.  P S P n  dB = 127 : 2 ( 138 : 6) = 11 : 4 dB = 10 1 : 14 = 13 : 8 ratio d. Assuming the SNR = z = E b =N 0 = 13 : 8 ; we get the results for various digital signaling techniques given in the table below: Modulation type Error probability BPSK Q p 2 z  = 7 : 4  10 8 DPSK 1 2 e z = 5 : 06  10 7 Noncoh. FSK 1 2 e z= 2 = 5 : 03  10 4 QPSK Same as BPSK

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    APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS 1 Problem D.1 a. Consider z ( t ) = A cos ( ! 0 + ! d ) t + n ( t ) ; ! 0 = 2 f 0 ; ! d = 2 f d = A cos ( ! 0 + ! d ) t + n c ( t ) cos ! 0 t n s ( t ) sin ! 0 t with the PSD of n ( t ) given by S n ( t ) = N 0 = 2 for j f f 0 j  B= 2 and 0 otherwise. Note also that the PSD of the quadrature noise components is S n c ( f ) = S n s ( f ) =  N 0 ; j f j  B= 2 0 ; otherwise Also, the quadrature noise components are uncorrelated. It follows that n 0 c ( t ) cos ( ! 0 + ! d ) t n 0 s ( t ) sin ( ! 0 + ! d ) t = n 0 c ( t ) [cos ! 0 t cos ! d t sin ! 0 t sin ! d t ] n 0 s ( t ) [sin ! 0 t cos ! d t + cos ! 0 t sin ! d t ] =  n 0 c ( t ) cos ! d t n 0 s ( t ) sin ! d t  cos ! 0 t  n 0 c ( t ) sin ! d t + n 0 s ( t ) cos ! d t  sin ! 0 t  n c ( t ) cos ! 0 t n s ( t ) sin ! 0 t Therefore n c ( t ) = n 0 c ( t ) cos ! d t n 0 s ( t ) sin ! d t n s ( t ) = n 0 c ( t ) sin ! d t + n 0 s ( t ) cos ! d t These two equations can be solved for n 0 c ( t ) and n 0 s ( t ) . The results are n 0 c ( t ) = n c ( t ) cos ! d t + n s ( t ) sin ! d t n 0 s ( t ) = n c ( t ) sin ! d t + n s ( t ) cos ! d t To Önd the PSDs of n 0 c ( t ) and n 0 s ( t ) we Örst Önd their autocorrelation functions. By deÖnition, R n 0 c (  ) = E  [ n c ( t ) cos ! d t + n s ( t ) sin ! d t ]  [ n c ( t +  ) cos ! d ( t +  ) + n s ( t +  ) sin ! d ( t +  )]  = R c (  ) cos ! d t cos ! d ( t +  ) + R s (  ) sin ! d t sin ! d ( t +  ) Noting that R c (  ) = R s (  ) and using a trig identity, we Önd that R n 0 c (  ) = R c (  ) cos ! d 

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    APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS 2 Similarly, it can be shown that R n 0 s (  ) = R s (  ) cos ! d  The PSDs follow by applying the modulation theorem: S n 0 c ( f ) = S n 0 s ( f ) = 1 2 S n c ( f + f d ) + 1 2 S n c ( f f d ) = 8 < : N 0 ; j f j  B= 2 f d N 0 = 2 ; B= 2 f d < j f j  B= 2 + f d 0 ; otherwise b. To Önd the cross-spectral density of n 0 c ( t ) and n 0 s ( t ) , we Örst Önd their cross-correlation function: R n 0 c n 0 s (  ) = E  n 0 c ( t ) n 0 s ( t +  ) = E  [ n c ( t ) cos ! d t + n s ( t ) sin ! d t ]  [ n c ( t +  ) sin ! d ( t +  ) + n s ( t +  ) cos ! d ( t +  )]  = R c (  ) cos ! d t sin ! d ( t +  ) + R s (  ) sin ! d t cos ! d ( t +  ) Again noting that R c (  ) = R s (  ) and using a trig identity, we Önd that R n 0 c n 0 s (  ) = R s (  ) sin ! d  = R s (  )  exp ( j! d  ) exp ( j! d  ) 2 j  Application of the frequency translation theorem results in the following for the cross- spectral density: S n 0 c n 0 s ( f ) = j 2 [ S n s ( f f d ) S n s ( f + f d )] = 8 < : jN 0 = 2 ; B= 2 f d  f  B= 2 + f d jN 0 = 2 ; ( B= 2 + f d )  f  ( B= 2 f d ) 0 ; otherwise For f d 6 = 0 this cross-PSD is not identically zero as a sketch will show. Thus n 0 c ( t ) and n 0 s ( t ) are not uncorrelated. However, they are when sampled at the same instant because R n 0 c n 0 s (0) = 0 . Therefore, they are statistically independent when sampled at the same instant since they are jointly Gaussian. Problem D.2 The derivation here follows S. O. Rice, "Noise in FM Receivers," Proc. of the Symp. of Time Series Analysis , M. Rosenblatt, ed., New York: Wiley (1963), pp. 395-422.

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    APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS 3 a. We need the additional clicks due to modulation. Thus, consider z ( t ) = A cos ( ! 0 + ! d ) t + n c ( t ) cos ! 0 t n s ( t ) sin ! 0 t = A cos ( ! 0 + ! d ) t + n 0 c ( t ) cos ( ! 0 + ! d ) t n 0 s ( t ) sin ( ! 0 + ! d ) t =  A + n 0 c ( t )  cos ( ! 0 + ! d ) t n 0 s ( t ) sin ( ! 0 + ! d ) t , R ( t ) cos [( ! 0 + ! d ) t +  ( t )] where R ( t ) = q [ A + n 0 c ( t )] 2 + [ n 0 s ( t )] 2 and tan  ( t ) = n 0 s ( t ) A + n 0 c ( t ) It follows from the cross-correlation function found in Problem D.1 that R n 0 c n 0 s (0) = E  n 0 c ( t ) n 0 s ( t ) = 0 From an appropriate sketch, it follows that the probability of a counter-clockwise origin encirclement in a small time interval  is P cc  = Pr  A + n 0 c ( t ) < 0 and n 0 s ( t ) undergoes a + to zero crossing in [0 ; ] The condition for n 0 s ( t ) to undergo a + to zero crossing in [0 ; ] , from an appropriate sketch, is that n 0 s (0) > 0 and that n 0 s (0) + dn 0 s ( t ) dt t =0  < 0 for all dn 0 s ( t ) dt t =0 < 0 . Therefore P cc  = Pr  A + n 0 c (0) < 0 ; n 0 s (0) > 0 ; n 0 s (0) + dn 0 s ( t ) dt t =0  < 0 ; dn 0 s ( t ) dt t =0 < 0  Thus we need the joint pdf of X , n 0 c (0) ; Y , n 0 s (0) ; and Z , dn 0 s ( t ) dt t =0 These are jointly Gaussian random variables (the derivative is a linear operation). The

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    APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS 4 following expectations hold: E [ X ] = E [ Y ] = E [ Z ] = 0 E  X 2  = E  Y 2  = Z 1 1 S n 0 c ( f ) df = Z 1 1 S n 0 s ( f ) df = N 0 B , a E  Z 2  = Z 1 1 j j 2 f j 2 S n 0 s ( f ) df = 1 3  2 N 0 B 3  + 4  2 f 2 d N 0 B , b + a! 2 d E [ XY ] = E [ Y Z ] = 0 E [ XZ ] = dR n 0 c n 0 s (  ) d  =0 = d ( R s (  ) sin ! d  ) d  =0 = a! d E [ Z j X ] = E  dn 0 s ( t ) dt t =0 j n 0 c (0)  = E  ! d n c (0) + dn s ( t ) dt t =0 j n c (0)  = ! d E [ n c (0) j n c (0)] = ! d X Therefore, the joint pdf of X; Y; Z is f XY Z ( x; y; z ) = f Z j XY ( z j x; y ) f X ( x ) f Y ( y ) = f Z j X ( z j x ) f X ( x ) f Y ( y ) = 1 (2  ) 3 = 2 a p b exp ( " x 2 + y 2 2 a + ( z + a! d ) 2 2 b #)

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