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# ch07 - Chapter 7 Binary Data Communication 7.1 Problem...

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Chapter 7 Binary Data Communication 7.1 Problem Solutions Problem 7.1 The signal-to-noise ratio is z = A 2 T N 0 = A 2 N 0 R Trial and error using the asymptotic expression Q ( x ) exp ¡ x 2 / 2 ¢ / ¡ 2 π x ¢ shows that P E = Q ³ 2 z ´ =10 5 for z =9 . 58 dB . 078 Thus A 2 N 0 R . 078 or A = p 9 . 078 N 0 R = p 9 . 078 × 10 7 × 20000 =0 . 135 V Problem 7.2 The bandwidth should be equal to the data rate to pass the main lobe of the signal spectrum. By trial and error, solving the following equation for z , P E | desired = Q ³q 2 z | required ´ ,z = E b /N 0 1

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2 CHAPTER 7. BINARY DATA COMMUNICATION we f nd that z | required =6 . 78 dB (4.76 ratio) , 8 . 39 dB (6.90 ratio), 9 . 58 dB (9.08 ratio), 10 . 53 dB (11.3 ratio) for P E | desired =10 3 , 10 4 , 10 5 , 10 6 , respectively. The required signal power is P s ,requ ired = E b /T symbol = E b R = z | required N 0 R For example, for P E 3 and R = 1000 bits/second, we get P s = z | required N 0 R =4 . 76 × 10 6 × 1000 = 4 . 76 × 10 3 W Similar calculations allow the rest of the table to be f lled in, giving the following results. R , bps P E 3 P E 4 P E 5 P E 6 1 , 000 4 . 76 × 10 3 W 6 . 9 × 10 3 W 9 . 08 × 10 3 W 11 . 3 × 10 3 W 10 , 000 4 . 76 × 10 2 W 6 . 9 × 10 2 W 9 . 08 × 10 2 W 11 . 3 × 10 2 W 100 , 000 4 . 76 × 10 1 W 6 . 9 × 10 1 W 9 . 08 × 10 1 W 1 . 13 W Problem 7.3 (a) For R = B =5 , 000 bps, we have signal power = A 2 = z | required N 0 R =1 0 10 . 53 / 10 × 10 6 × 5 , 000 1 . 3 × 5 × 10 3 =0 . 0565 W where z | required . 53 dB from Problem 7.2. (b) For R = B =50 , 000 bps, the result is signal power = A 2 . 565 W (c) For R = B = 500 , 000 bps, we get signal power = A 2 . 65 W (d) For R = B , 000 , 000 bps, the result is signal power = A 2 =11 . 3 W Problem 7.4 The decision criterion is V> ε ,choose + A V< ε A
7.1. PROBLEM SOLUTIONS 3 where V = ± AT + N N is a Gaussian random variable of mean zero and variance σ 2 = N 0 T/ 2 . Because P ( A sent )= P ( A sent )=1 / 2 , it follows that P E = 1 2 P ( AT + N> ε | A sent )+ 1 2 P ( AT + N< ε | A sent 1 2 ( P 1 + P 2 ) where P 1 = Z AT + ε e η 2 /N 0 T π N 0 T d η = Q s 2 A 2 T N 0 + s 2 ε 2 N 0 T = Q ³ 2 z + 2 ²/ σ ´ , σ = N 0 T, z = A 2 T N 0 Similarly P 2 = Z AT + ε −∞ e η 2 0 T π N 0 T d η = Z AT ε e η 2 0 T π N 0 T d η = Q s 2 A 2 T N 0 s 2 ε 2 N 0 T = Q ³ 2 z 2 ²/ σ ´ Use the approximation Q ( u e u 2 / 2 u 2 π to evaluate 1 2 ( P 1 + P 2 )=10 6 by solving iteratively for z for various values of ε / σ . The results are given in the table below. ε / σ z dB for P E =10 6 ε / σ z dB for P E 6 0 10.54 0.6 11.34 0.2 10.69 0.8 11.67 0.4 11.01 1.0 11.98

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4 CHAPTER 7. BINARY DATA COMMUNICATION Problem 7.5 Use the z | required values found in the solution of Problem 7.2 along with z | required = A 2 /N 0 R to get R = A 2 /z | required N 0 . Substitute N 0 =1 0 6 V 2 /Hz, A =2 0 mV to get R = 400 | required . This results in the values given in the table below. P E z | required ,dB(rat io) R , bps 10 4 8.39 (6.9) 57.97 10 5 9.58 (9.08) 43.05 10 6 10.53 (11.3) 35.4 Problem 7.6 The integrate-and-dump detector integrates over the interval [0 ,T δ ] so that the SNR is z 0 = A 2 ( T δ ) N 0 instead of z = A 2 T/N 0 .T h u s z 0 z T δ and the degradation in z in dB is D = 10 log 10 μ 1 T δ Using δ =10 6
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ch07 - Chapter 7 Binary Data Communication 7.1 Problem...

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