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978-0072957167 Chapter 10
Solutions Manual for Digital Communications, 5th Edition (Chapter 10) 1 Prepared by Kostas Stamatiou January 15, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 11
Solutions Manual for Digital Communications, 5th Edition (Chapter 11) 1 Prepared by Kostas Stamatiou January 15, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 12
Solutions Manual for Digital Communications, 5th Edition (Chapter 12) 1 Prepared by Kostas Stamatiou January 29, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 13
Solutions Manual for Digital Communications, 5th Edition (Chapter 13) 1 Prepared by Kostas Stamatiou January 20, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 14
Solutions Manual for Digital Communications, 5th Edition (Chapter 14) 1 Prepared by Kostas Stamatiou March 9, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 15 Part 1
Solutions Manual for Digital Communications, 5th Edition (Chapter 15) 1 Prepared by Kostas Stamatiou March 12, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 15 Part 2
13 10-9 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100 0 5 10 15 20 Outage probability Average SNR (dB) Outage capacity is 2 bps/Hz (d) The following plots are of the “success probability” 1 −Pout vs. Cfor γ= 10,20dB, […]
978-0072957167 Chapter 16
Solutions Manual for Digital Communications, 5th Edition (Chapter 16) 1 Prepared by Kostas Stamatiou March 12, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 2 Part 1
Solutions Manual for Digital Communications, 5th Edition (Chapter 2) 1 Prepared by Kostas Stamatiou January 11, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 2 Part 2
Problem 2.28 1) By Chernov bound, for t > 0, P[X≥α]≤e−tαE[etX ] = e−tαΘX(t) This is true for all t > 0, hence ln P[X≥α]≤min t≥0[−tα + ln ΘX(t)] = −max t≥0[tα −ln ΘX(t)] 2) Here ln P[Sn≥α] = ln […]
978-0072957167 Chapter 2 Part 3
32 The power density spectrum is S(f) = P∞ k=−∞ R(k)e−j2πfk =P−1 k=−∞ 1 2−ke−j2πfk +P∞ k=0 1 2ke−j2πfk Problem 2.48 We will denote the discrete-time process by the subscript dand the continuous-time (analog) process by the subscript a. Also, […]
978-0072957167 Chapter 3 Part 1
Solutions Manual for Digital Communications, 5th Edition (Chapter 3) 1 Prepared by Kostas Stamatiou January 11, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 3 Part 2
17 1. Since : µa= 0, σ2 a= 1,we have : Sss(f) = 1 T|G(f)|2.But : G(f) = T 2 sin πfT/2 πfT /2e−j2πf T /4−T 2 sin πfT/2 πfT /2e−j2πf 3T /4 sin πfT/2 2. For non-independent information sequence […]
978-0072957167 Chapter 4 Part 1
Solutions Manual for Digital Communications, 5th Edition (Chapter 4) 1 Prepared by Kostas Stamatiou January 11, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 4 Part 2
21 At the sampling instant t=t0=T, the signal component at the output of the matched filter is y(T) = Z∞ −∞ Y(f)ej2πfT df =Z∞ −∞ s(τ)g(T−τ)dτ |S(f)|2 Problem 4.23 1. The number of bits per symbol is k=4800 R=4800 2400 […]
978-0072957167 Chapter 4 Part 3
41 hence, D1⇔Z1 e−(r1−2A)2+(r2−2)2 N0dA > Z1 e−(r1+2A)2+(r2+2)2 N0dA ⇔e 0 e−4A2−4r1A 0 e−4A2+4r1A ⇔4r2 N0 + ln Z1 0 e−4A2−4r1A N0dA > −4r2 N0 + ln Z1 0 e−4A2+4r1A N0dA ⇔r2>N0 8ln g(r1) g(−r1) where N0dA 4r2 N0Z1 N0dA […]
978-0072957167 Chapter 4 Part 4
53 1. For detection of m1, we treat m2as a RV taking 1 or 2 with equal probability. The detection rule is arg max s1∈{−1,1} pn1(r1−s1−α)pn2(r2−β) + pn1(r1−s1+α)pn2(r2+β) similarly for detection of m2we have (s21,s22)∈{(α,β),(−α,−β)} These relations reduce to D+1 […]
978-0072957167 Chapter 5
Solutions Manual for Digital Communications, 5th Edition (Chapter 5) 1 Prepared by Kostas Stamatiou January 11, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 6 Part 1
Solutions Manual for Digital Communications, 5th Edition (Chapter 6) 1 Prepared by Kostas Stamatiou January 11, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 6 Part 2
21 Problem 6.28 1. We have H(X) = −Pi=1 52−ilog22−i−1 32 log21 32 = 31/16 = 1.9375 and H(Y) = log26 = 2. For Xthe Huffman tree is ✉ ✉ ❅ ❅ 1/16 ❡❡ ✓ ✓ ✓ 1/8 […]
978-0072957167 Chapter 6 Part 3
PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the […]
978-0072957167 Chapter 7 Part 1
Solutions Manual for Digital Communications, 5th Edition (Chapter 7) 1 Prepared by Kostas Stamatiou January 11, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 7 Part 2
19 and G= [I P ], 1 0 0 1 1 0 2. H= 0 1 1 0 1 0 1 1 0 1 0 0 1 0 1 […]
978-0072957167 Chapter 8 Part 1
Solutions Manual for Digital Communications, 5th Edition (Chapter 8) 1 Prepared by Kostas Stamatiou January 30, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 8 Part 2
20 YZ2 YZ2 YZ2 Y2Z3 Y2Z3 Y2Z2 Y2Z2 Z 1 YZ YZ YZ YZ and we have the following equations Xb=Y Z2Xa+Y Z2Xb+Y ZXc+Y ZX2 Xc=Y Z2Xa+Y Z2Xb+Y ZXc+Y ZXd Xd=Y2Z2Xa+Y2Z2Xb+Y2Z3Xc+Y2Z3Xd Xe=Xb+ZXc+ZXd from which, after eliminating, we obtain T(Y, Z) […]
978-0072957167 Chapter 9 Part 1
Solutions Manual for Digital Communications, 5th Edition (Chapter 9) 1 Prepared by Kostas Stamatiou January 15, 2008 1PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any […]
978-0072957167 Chapter 9 Part 2
21 The output of the matched filter is sampled at t=Tand the samples are passed to the detector. The detector is a simple threshold device that decides if a binary 1 or 0 was transmitted depending on the sign of […]
978-0072957167 Chapter 9 Part 3
38 Thus, the autocorrelation function of the noise at the output of the equalizer is : where c(k) denotes the discrete time impulse response of the equalizer. Therefore, the autocorrela- tion sequence of the noise at the output of the […]