TEST BANK FOR Digital Communications 5th Edition By Proakis Salehi (Instructor Solution Manual)

Exam (elaborations) TEST BANK FOR Digital Communications 5th Edition By Proakis Salehi (Instructor Solution Manual) 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 form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 2 Problem 2.1 a. ˆx(t) = 1  Z ∞ −∞ x(a) t − a da Hence : −ˆx(−t) = −1  R ∞ −∞ x(a) −t−ada = −1  R −∞ ∞ x(−b) −t+b (−db) = −1  R ∞ −∞ x(b) −t+bdb = 1  R ∞ −∞ x(b) t−b db = ˆx(t) where we have made the change of variables : b = −a and used the relationship : x(b) = x(−b). b. In exactly the same way as in part (a) we prove : ˆx(t) = ˆx(−t) c. x(t) = cos !0t, so its Fourier transform is : X(f) = 1 2 [(f − f0) + (f + f0)] , f0 = 2!0. Exploiting the phase-shifting property (2-1-4) of the Hilbert transform : ˆX (f) = 1 2 [−j(f − f0) + j(f + f0)] = 1 2j [(f − f0) − (f + f0)] = F−1 {sin 2f0t} Hence, ˆx(t) = sin !0t. d. In a similar way to part (c) : x(t) = sin !0t ⇒ X(f) = 1 2j [(f − f0) − (f + f0)] ⇒ ˆX (f) = 1 2 [−(f − f0) − (f + f0)] ⇒ ˆX (f) = − 1 2 [(f − f0) + (f + f0)] = −F−1 {cos 2!0t} ⇒ ˆx(t) = −cos !0t e. The positive frequency content of the new signal will be : (−j)(−j)X(f) = −X(f), f > 0, while the negative frequency content will be : j · jX(f) = −X(f), f < 0.Hence, since ˆˆX (f) = −X(f), we have : ˆˆx(t) = −x(t). f. Since the magnitude response of the Hilbert transformer is characterized by : |H(f)| = 1, we have that :  ˆX (f)  = |H(f)| |X(f)| = |X(f)| . Hence : Z ∞ −∞  ˆX (f)  2 df = Z ∞ −∞ |X(f)|2 df 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 3 and using Parseval’s relationship : Z ∞ −∞ ˆx2(t)dt = Z ∞ −∞ x2(t)dt g. From parts (a) and (b) above, we note that if x(t) is even, ˆx(t) is odd and vice-versa. Therefore, x(t)ˆx(t) is always odd and hence : R ∞ −∞ x(t)ˆx(t)dt = 0. Problem 2.2 1. Using relations X(f) = 1 2 Xl(f − f0) + 1 2 Xl(−f − f0) Y (f) = 1 2 Yl(f − f0) + 1 2 Yl(−f − f0) and Parseval’s relation, we have Z ∞ −∞ x(t)y(t) dt = Z ∞ −∞ X(f)Y ∗(f) dt = Z ∞ −∞  1 2 Xl(f − f0) + 1 2 Xl(−f − f0)   1 2 Yl(f − f0) + 1 2 Yl(−f − f0)  ∗ df = 1 4 Z ∞ −∞ Xl(f − f0)Y ∗ l (f − f0) df + 1 4 Z ∞ −∞ Xl(−f − f0)Yl(−f − f0) df = 1 4 Z ∞ −∞ Xl(u)Y ∗ l (u) du + 1 4 X∗l (v)Y (v) dv = 1 2 Re Z ∞ −∞ Xl(f)Y ∗ l (f) df  = 1 2 Re Z ∞ −∞ xl(t)y∗l (t) dt  where we have used the fact that since Xl(f − f0) and Yl(−f − f0) do not overlap, Xl(f − f0)Yl(−f − f0) = 0 and similarly Xl(−f − f0)Yl(f − f0) = 0. 2. Putting y(t) = x(t) we get the desired result from the result of part 1. 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 4 Problem 2.3 A well-known result in estimation theory based on the minimum mean-squared-error criterion states that the minimum of Ee is obtained when the error is orthogonal to each of the functions in the series expansion. Hence : Z ∞ −∞ " s(t) − XK k=1 skfk(t) # f∗n(t)dt = 0, n = 1, 2, ...,K (1) since the functions {fn(t)} are orthonormal, only the term with k = n will remain in the sum, so : Z ∞ −∞ s(t)f∗n(t)dt − sn = 0, n = 1, 2, ...,K or: sn = Z ∞ −∞ s(t)f∗n(t)dt n = 1, 2, ...,K The corresponding residual error Ee is : Emin = R ∞ −∞ h s(t) − PK k=1 skfk(t) i h s(t) − PK n=1 snfn(t) i ∗ dt = R ∞ −∞ |s(t)|2 dt − R ∞ −∞ PK k=1 skfk(t)s∗(t)dt − PK n=1 s∗n R ∞ −∞ h s(t) − PK k=1 skfk(t) i f∗n(t)dt = R ∞ −∞ |s(t)|2 dt − R ∞ −∞ PK k=1 skfk(t)s∗(t)dt = Es − PK k=1 |sk|2 where we have exploited relationship (1) to go from the second to the third step in the above calculation. Note : Relationship (1) can also be obtained by simple differentiation of the residual error with respect to the coefficients {sn} . Since sn is, in general, complex-valued sn = an + jbn we have to differentiate with respect to both real and imaginary parts : d dan Ee = d dan R ∞ −∞ h s(t) − PK k=1 skfk(t) i h s(t) − PK n=1 snfn(t) i ∗ dt = 0 ⇒ − R ∞ −∞ anfn(t) h s(t) − PK n=1 snfn(t) i ∗ + a∗nf∗n(t) h s(t) − PK n=1 snfn(t) i dt = 0 ⇒ −2an R ∞ −∞ Re n f∗n(t) h s(t) − PK n=1 snfn(t) io dt = 0 ⇒ R ∞ −∞ Re n f∗n(t) h s(t) − PK n=1 snfn(t) io dt = 0, n = 1, 2, ...,K 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 5 where we have exploited the identity : (x + x∗) = 2Re{x}. Differentiation of Ee with respect to bn will give the corresponding relationship for the imaginary part; combining the two we get (1). Problem 2.4 The procedure is very similar to the one for the real-valued signals described in the book (pages 33-37). The only difference is that the projections should conform to the complex-valued vector space : c12= Z ∞ −∞ s2(t)f∗ 1 (t)dt and, in general for the k-th function : cik = Z ∞ −∞ sk(t)f∗ i (t)dt, i = 1, 2, ..., k − 1 Problem 2.5 The first basis function is : g4(t) = s4(t) √E4 = s4(t) √3 =   −1/√3, 0 ≤ t ≤ 3 0, o.w.   Then, for the second basis function : c43 = Z ∞ −∞ s3(t)g4(t)dt = −1/√3 ⇒ g′3 (t) = s3(t) − c43g4(t) =   2/3, 0 ≤ t ≤ 2 −4/3, 2 ≤ t ≤ 3 0, o.w   Hence : g3(t) = g′3 (t) √E3 =   1/√6, 0 ≤ t ≤ 2 −2/√6, 2 ≤ t ≤ 3 0, o.w   where E3 denotes the energy of g′3(t) : E3 = R 3 0 (g′3(t))2 dt = 8/3. For the third basis function : c42 = Z ∞ −∞ s2(t)g4(t)dt = 0 and c32 = Z ∞ −∞ s2(t)g3(t)dt = 0 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 limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 6 Hence : g′2 (t) = s2(t) − c42g4(t) − c32g3(t) = s2(t) and g2(t) = g′2 (t) √E2 =   1/√2, 0 ≤ t ≤ 1 −1/√2, 1 ≤ t ≤ 2 0, o.w   where : E2 = R 2 0 (s2(t))2 dt = 2. Finally for the fourth basis function : c41 = Z ∞ −∞ s1(t)g4(t)dt = −2/√3, c31 = Z ∞ −∞ s1(t)g3(t)dt = 2/√6, c21 = 0 Hence : g′1 (t) = s1(t) − c41g4(t) − c31g3(t) − c21g2(t) = 0 ⇒ g1(t) = 0 The last result is expected, since the dimensionality of the vector space generated by these signals is 3. Based on the basis functions (g2(t), g3(t), g4(t)) the basis representation of the signals is : s4 =