561_hmk4_soln

561_hmk4_soln - 2 (1 + r ) ⇒ 6 K = 5 K (1 + r ) ⇒ r =...

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ECE 561 – Digital Communications Systems Solutions for Homework #4 1. a. W = 1 2 1 + r ( ) R s 100 K = 1 2 1 + 0.6 ( ) R s R s = 125 Kbps b. 32 = 2 5 f s = 125 K 5 = 25 Ks / s c. f s = 2 f m f m = 12.5 KHz d. R s = 125 Ksym / sec 8 = 2 3 R = (125 Ksym / sec)(3 bits / sym ) = 375 Kbps 375 Kbps 5 bits / samp = 75 Ksamp / sec f m = f s 2 = 37.5 KHz 2. a. W = 1 2 1 + r ( ) R s = 1 2 1 + 1 ( ) 8000 = 8 KHz b. 8 = 2 3 R = (8000 samp / sec)(3 bits / samp ) = 24000 bps W = 1 2 1 + 1 ( ) 24000 = 24 KHz c. 128 = 2 7 R = (8000 samp / sec)(7 bits / samp ) = 56000 bps W = 1 2 1 + 1 ( ) 56000 = 56 KHz 3. R = 10 Kbps W min = R 2 = 5 KHz W = R
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Unformatted text preview: 2 (1 + r ) ⇒ 6 K = 5 K (1 + r ) ⇒ r = 0.2 4. Data sequence, x k 1 0 0 1 0 1 1 0 0 1 0 Precoded seq. w k 0 1 1 1 0 0 1 0 0 0 1 1 Trans. bipolar seq, w k-1 1 1 1 -1 -1 1 -1 -1 -1 1 1 Received seq, y k 0 2 2 0 -2 0 0 -2 -2 0 2 Decoded output 1 0 0 1 0 1 1 0 0 1 0...
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This document was uploaded on 02/27/2010.

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561_hmk4_soln - 2 (1 + r ) ⇒ 6 K = 5 K (1 + r ) ⇒ r =...

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