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Homework 1 Solutions - CS 421/S03 Chapter 3(7/e 1 Problem...

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CS 421/S03 Solutions to Homework One Chapter 3 (7/e) 1. Problem 3.14 N = 10 log k + 10 log T + 10 log B = -228.6 + 10 log 10 4 + 10 log 10 7 (dBW) = -228.6 + 40 + 70 = - 118.6 (dBW) 2. Problem 3.16 Use Nyquist’s formula C = 2B log 2 M. (a) C = 9600 bps, log 2 M = 4 (bits) 9600 = 2B × 4 B = 1200 (Hz) (b) C = 9600 bps, log 2 M = 8 (bits) 9600 = 2B × 8 B = 600 (Hz) 3. Problem 3.19 Use Shannon’s formula C = B log 2 (1 + S/N). C = 20 × 10 6 , B = 3 × 10 6 20 × 10 6 = 3 × 10 6 log 2 (1 + S/N) log 2 (1 + SNR) = 6.67 SNR = 2 6.67 – 1 SNR 101. 4. Problem 3.20 (a) The output waveform is (4/ π )[Sin(2 π f t) + (1/3)sin(2 π (3 f )t)) + (1/5)sin(2 π (5 f )t)) + (1/7)sin(2 π (7 f )t))], where f = 1/T = 1000 Hz = 1 k Hz. The output power is P = (4/ π ) 2 (1 + 1/9 + 1/25 + 1/ 49) /2 = 1.62 × 0.586 (watt) P = 0.95 (watt). (b) The output noise power is N = N 0 × B = 0.1 μ Watt/Hz × 8 k Hz N = 800 × 10 -6 (watt) 1
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SNR = 0.95/0.0008 = 1187.5 (SNR) dB = 10 log 1187.5 (SNR) dB = 30.75 (dB) 5. Problem 3.21 E b / N 0 = (S/R)/ N 0 (E b / N 0 ) dB = S dB - R dB - (N 0 ) dB = -151 dBW – 10 log 2400 – 10 log k – 10 log T = -151 –33.8 + 228.6 – 31.76 = -216.6 + 228.6 = 12 (dBW) E b / N 0 = 10 1.2 = 15.85. Chapter 4 6. Problem 4.2 10 log (P out /P in ) = -20 dB. P out /P in = 10 -2 = 0.01.
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