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Unformatted text preview: UCLA Electrical Engineering Dept. EE132A: Communication Systems Final Exam Solutions 1. A source generates information bits to be digitally modulated at a bit rate of R b = 18000 bps. The bits are grouped in triplets, say b 1 b 2 b 3 . If b 1 = 0 , then the bit pair b 2 b 3 is sent to the modulator M1. If b 1 = 1 , then the bit pair b 2 b 3 is sent to the modulator M2. M1 is a QPSK modulator that generates the signals s (1) i ( t ) = r 2 E 1 T cos 2 f c t + ( i 1) 2 , i = 1 , 2 , 3 , 4 , t T, and M2 is a QPSK modulator which generates the signals s (2) j ( t ) = r 2 E 2 T cos 2 f c t + ( j 1) 2 , j = 1 , 2 , 3 , 4 , t T, with f c T 1 , E 1 = 2 E and E 2 = (2+ 2) 2 E , for a given E . The combined modulator output is then one out of the eight signals S = { s (1) 1 ( t ) , s (1) 2 ( t ) , s (1) 3 ( t ) , s (1) 4 ( t ) , s (2) 1 ( t ) , s (2) 2 ( t ) , s (2) 3 ( t ) , s (2) 4 ( t ) } . (a) What is the symbol period, T ? R s = 1 /T = R b /b = 18000 / 3 = 6000. (b) Draw a geometric representation of the constellation of signals, S , and the corre sponding decision regions. s (1) 1 = [ 2 , 0] E , s (1) 2 = [0 , 2] E , s (1) 3 = [ 2 , 0] E , s (1) 4 = [0 , 2] E , s (2) 1 = [2 + 2 , 0] E , s (2) 2 = [0 , 2 + 2] E , s (2) 3 = [ 2 2 , 0] E , s (2) 4 = [0 , 2 2] E . See Fig. 1. (c) Compute d min in terms of E . From the drawing, d min = 2 E . (d) Compute P ( e  s (1) 1 ) and P ( e  s (2) 1 ) in terms of E /N , and again in terms of E b /N , where E b is the energy per bit. We approximate the error probabilities by considering only the constellation point closest to the point under consideration: P ( e  s (1) 1 ) 3 Q (( d min / 2) / p N / 2) and P ( e  s (2) 1 ) Q (( d min / 2) / p N / 2) where Q (( d min / 2) / p N / 2) = Q ( p 2 E /N )....
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This note was uploaded on 04/12/2010 for the course EE 132A taught by Professor Walker during the Spring '08 term at UCLA.
 Spring '08
 Walker
 Electrical Engineering

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