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Solution_4_2007

# Solution_4_2007 - ECE 154B Homework#4 Due 1 Problem 8.1...

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ECE 154B Homework #4 Due: February 12, 2007 1. Problem 8.1 from Ziemer and Tranter R b = log 2 M R s bps M = 4, R b = 2 × 2000 = 4000 bps M = 8, R b = 3 × 2000 = 6000 bps M = 64, R b = 6 × 2000 = 12000 bps 2. Problem 8.12 from Ziemer and Tranter The signal points lie on a circle of radius E s centered at the origin equally spaced at angular intervals of 22.5 degrees (360/16). The decision regions are pie wedges centered over each signal point. 3. Problem 8.13 from Ziemer and Tranter M-ary PSK: If the upper bound is lower than 10 -5 , then the probability itself is lower as well, thus achieving the desired error level. P E ,symbol 2 Q 2 E s N 0 sin  / M  = 2 Q 2 E b log 2 M N 0 sin  / M  P E ,bit P E , symbol log 2 M 2 log 2 M Q 2 E b log 2 M N 0 sin  / M  10 5 E b N 0 1 2log 2 M Q 1 log 2 M 2 10 5 1 sin  / M 2 Matlab code: M=8; ebno8=(1/(2*log2(M)))*(sqrt(2)*erfinv(1-2*((log2(M)*... 10^(-5))/2))*(1/sin(pi/M)))^2 ebno8db=10*log10(ebno8) M=16; ebno16=(1/(2*log2(M)))*(sqrt(2)*erfinv(1-2*((log2(M)... *10^(-5))/2))*(1/sin(pi/M)))^2 ebno16db=10*log10(ebno16) M=32; ebno32=(1/(2*log2(M)))*(sqrt(2)*erfinv(1-2*((log2(M)... *10^(-5))/2))*(1/sin(pi/M)))^2 ebno32db=10*log10(ebno32) M=64; ebno64=(1/(2*log2(M)))*(sqrt(2)*erfinv(1-2*((log2(M)... *10^(-5))/2))*(1/sin(pi/M)))^2 ebno64db=10*log10(ebno64) For 16-QAM:

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P E , symbol = 1 − 1 4 P C I  1 2 P C II  1 4 P C III  = 1 − 1 4 [ 1 2Q E s 5 N 0 ] 2 1 2 [ 1 2Q E s 5 N 0 ][ 1 Q E s 5 N 0 ] 1 4 [ 1 Q E s 5 N 0 ] 2 = 3Q E b log 2 M 5 N 0 − 9 4 Q E b log 2 M 5 N 0 2 P E , bit 1 log 2 M 3Q E b log 2 M 5 N 0 − 9 4 Q E b log 2 M 5 N 0 2 = 10 5 9 4 x 2 3x 10 5 log 2 M = 0 x = 3 ± 9 9log 2 M × 10 5 9 2 x = 1.333, 1.333 × 10 5 Q E b log 2 M 5 N 0 = 1.333(Not possible, Q(x) is only between 0 and 1) Q E b log 2 M 5 N 0 = 1.333 × 10
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Solution_4_2007 - ECE 154B Homework#4 Due 1 Problem 8.1...

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