561_hmk2_soln - ⇒ e = q 2 = 0.039 V b 128 = 2 ⇒ = 7...

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ECE 561 – Digital Communications Systems Solutions for Homework #2 1. 100 char * 8 bits/char = 800 bits bit rate = 800/2 = 400 bps in all cases a. 400bps/(1bit/sym) = 400 sym/sec b. 400bps/(2bit/sym) = 200 sym/sec c. 400bps/(3bit/sym) = 133.3 sym/sec d. 400bps/(4bit/sym) = 100 sym/sec e. 400bps/(5bit/sym) = 80 sym/sec 2. a. the maximum frequency is 2 KHz, thus T s = 1/4000 = 0.25 mS b. (4000 samp/sec) * 60 * 60 sec = 14400000 samples c. 14400000 samples * 8 bits/sample = 115200000 bits 3. a. T < 1/(2*4000) = 0.125 mS b. f s = 2*4000 = 8KHz 1 T s X ( f 8000 n ) c. τ /T s = .02/.125 = .16 .16sinc .16 n ( ) X ( f 8000 n ) f (KHz) -4 4 X(f) 10 -3 *(1/.125m) = 8 8 16 -16 - 8 f (KHz) -4 4 X(f) 8 16 -16 - 8 .16*10 -3 .16*10 -3 sinc(.16) .16*10 -3 sinc(.32)
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4. a. V pp L = q = 10 128 = 0.078125
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Unformatted text preview: ⇒ e = q 2 = 0.039 V b. 128 = 2 ⇒ = 7 bits / sample ⇒ 8000 * 7 = 56000 bps c. k=5 56000/5 = 11200 sym/sec 5. a. L ≥ 1 2 p = 1 .04 = 25 ⇒ L = 32 b. L = 32 = 2 ⇒ = 5 bits / sample R = 8000 * 5 = 40000 bps R W = 40000 4000 = 10 bits / s / Hz c. R = 40000 * 5 = 200000 bps R W = 200000 4000 = 50 bits / s / Hz 6. 7. 1 0 1 0 1 0 1 a. b. c. 1 1 1 1 1 1 1 a. b. c. 8. a. f s = 2(15KHz) = 30 KHz = 30000 samples/sec b. L = 512 word length = 9 R = 30000 samp/sec * 9 bits/samp = 270000 bits/sec 9. f s = 4.5 M * 2 *1.15 = 10.35 Msamp / sec L = 1024 = 2 ⇒ = 10 R = 10.35 M *10 = 103.5 Mbps T b = 1 103.5 Mbps = 9.66 ns...
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