Solution_4_2007

Solution_4_2007 - ECE 154B Homework #4 Due: February 12,...

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Unformatted text preview: 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 sin / M = 2 Q 2 E b log 2 M N sin / M P E ,bit P E ,symbol log 2 M 2 log 2 M Q 2 E b log 2 M N sin / M 10 5 E b N 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: 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 ] 2 1 2 [ 1 2Q E s 5 N ][ 1 Q E s 5 N ] 1 4 [ 1 Q E s 5 N ] 2 = 3Q E b log 2 M 5 N 9 4 Q E b log 2 M 5 N 2 P E ,bit 1 log 2 M 3Q E b log 2 M 5 N 9 4 Q E b log 2 M 5 N 2 = 10 5 9 4 x 2 3x 10...
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This note was uploaded on 11/11/2009 for the course ECE 670377 taught by Professor Wolf during the Winter '07 term at UCSD.

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Solution_4_2007 - ECE 154B Homework #4 Due: February 12,...

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