561_hmk4_soln

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 02/27/2010.

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online