Solutions Manual for Digital Communications, 5th Edition (Chapter 2) 1 Prepared by Kostas Stamatiou January 11, 2008 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 wr itten 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 π  ∞ −∞ x(a) t −a da Hence : −ˆx(−t) = − 1 π  ∞ −∞ x(a) −t−a da = − 1 π  −∞ ∞ x(−b) −t+b (−db) = − 1 π  ∞ −∞ x(b) −t+b db = 1 π  ∞ −∞ 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 ω 0 t, so its Fourier transform is : X(f) = 1 2 [δ(f − f 0 ) + δ(f + f 0 )] , f 0 = 2πω 0 . Exploiting the phase-shifting property (2-1-4) of the Hilbert transform : ˆ X(f ) = 1 2 [−jδ(f − f 0 ) + jδ(f + f 0 )] = 1 2j [δ(f − f 0 ) −δ(f + f 0 )] = F −1 {sin 2πf 0 t} Hence, ˆx(t) = sin ω 0 t. d. In a similar way to part (c) : x(t) = sin ω 0 t ⇒ X(f) = 1 2j [δ(f − f 0 ) −δ(f + f 0 )] ⇒ ˆ X(f) = 1 2 [−δ(f − f 0 ) −δ(f + f 0 )] ⇒ ˆ X(f) = − 1 2 [δ(f − f 0 ) + δ(f + f 0 )] = −F −1 {cos 2πω 0 t} ⇒ ˆx(t) = −cos ω 0 t 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 ˆ ˆ X(f) = −X(f ), we have : ˆ ˆx(t) = −x(t). f. Since the magnitude response of the Hilbert transformer is characterized by : |H(f )| = 1, we have that :    ˆ X(f)    = |H(f)||X(f)| = |X(f)|. Hence :  ∞ −∞    ˆ X(f)    2 df =  ∞ −∞ |X(f)| 2 df PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, repro duced or distributed in any form or by any means, without the prior wri tten 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 :  ∞ −∞ ˆx 2 (t)dt =  ∞ −∞ x 2 (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 :  ∞ −∞ x(t)ˆx(t)dt = 0. Problem 2.2 1. Using relations X(f) = 1 2 X l (f − f 0 ) + 1 2 X l (−f − f 0 ) Y (f ) = 1 2 Y l (f − f 0 ) + 1 2 Y l (−f − f 0 ) and Parseval’s relation, we have  ∞ −∞ x(t)y(t) dt =  ∞ −∞ X(f)Y ∗ (f) dt =  ∞ −∞  1 2 X l (f − f 0 ) + 1 2 X l (−f − f 0 )  1 2 Y l (f −f 0 ) + 1 2 Y l (−f − f 0 )  ∗ df = 1 4  ∞ −∞ X l (f − f 0 )Y ∗ l (f − f 0 ) df + 1 4  ∞ −∞ X l (−f − f 0 )Y l (−f − f 0 ) df = 1 4  ∞ −∞ X l (u)Y ∗ l (u) du + 1 4 X ∗ l (v)Y (v) dv = 1 2 Re   ∞ −∞ X l (f)Y ∗ l (f) df  = 1 2 Re   ∞ −∞ x l (t)y ∗ l (t) dt  where we have used th e fact that since X l (f − f 0 ) and Y l (−f − f 0 ) do not overlap, X l (f − f 0 )Y l (−f − f 0 ) = 0 an d similarly X l (−f − f 0 )Y l (f −f 0 ) = 0. 2. Putting y(t) = x(t) we get the d esired result fr om the result of part 1. PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, repro duced or distributed in any form or by any means, without the prior wri tten 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 E e is obtained when the error is orthogonal to each of the fun ctions in the series expansion. Hence :  ∞ −∞  s(t) − K  k=1 s k f k (t)  f ∗ n (t)dt = 0, n = 1, 2, , K (1) since the functions {f n (t)} are orthonormal, only the term with k = n will remain in the sum, so :  ∞ −∞ s(t)f ∗ n (t)dt −s n = 0, n = 1, 2, , K or: s n =  ∞ −∞ s(t)f ∗ n (t)dt n = 1, 2, , K The corresponding residual error E e is : E min =  ∞ −∞  s(t) −  K k=1 s k f k (t)  s(t) −  K n=1 s n f n (t)  ∗ dt =  ∞ −∞ |s(t)| 2 dt −  ∞ −∞  K k=1 s k f k (t)s ∗ (t)dt −  K n=1 s ∗ n  ∞ −∞  s(t) −  K k=1 s k f k (t)  f ∗ n (t)dt =  ∞ −∞ |s(t)| 2 dt −  ∞ −∞  K k=1 s k f k (t)s ∗ (t)dt = E s −  K k=1 |s k | 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 err or with respect to the coefficients {s n }. Since s n is, in general, complex-valu ed s n = a n + jb n we have to differentiate with respect to both real and imaginary parts : d da n E e = d da n  ∞ −∞  s(t) −  K k=1 s k f k (t)  s(t) −  K n=1 s n f n (t)  ∗ dt = 0 ⇒ −  ∞ −∞ a n f n (t)  s(t) −  K n=1 s n f n (t)  ∗ + a ∗ n f ∗ n (t)  s(t) −  K n=1 s n f n (t)  dt = 0 ⇒ −2a n  ∞ −∞ Re  f ∗ n (t)  s(t) −  K n=1 s n f n (t)  dt = 0 ⇒  ∞ −∞ Re  f ∗ n (t)  s(t) −  K n=1 s n f n (t)  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, repro duced or distributed in any form or by any means, without the prior wri tten 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 E e with respect to b n 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 : c 12=  ∞ −∞ s 2 (t)f ∗ 1 (t)dt and, in general for the k-th fu nction : c ik =  ∞ −∞ s k (t)f ∗ i (t)dt, i = 1, 2, , k − 1 Problem 2.5 The first basis function is : g 4 (t) = s 4 (t) √ E 4 = s 4 (t) √ 3 =    −1/ √ 3, 0 ≤ t ≤ 3 0, o.w.    Then, for the second basis function : c 43 =  ∞ −∞ s 3 (t)g 4 (t)dt = −1/ √ 3 ⇒ g ′ 3 (t) = s 3 (t) − c 43 g 4 (t) =        2/3, 0 ≤ t ≤ 2 −4/3, 2 ≤ t ≤ 3 0, o.w        Hence : g 3 (t) = g ′ 3 (t) √ E 3 =        1/ √ 6, 0 ≤ t ≤ 2 −2/ √ 6, 2 ≤ t ≤ 3 0, o.w        where E 3 denotes the energy of g ′ 3 (t) : E 3 =  3 0 (g ′ 3 (t)) 2 dt = 8/3. For the th ird basis function : c 42 =  ∞ −∞ s 2 (t)g 4 (t)dt = 0 and c 32 =  ∞ −∞ s 2 (t)g 3 (t)dt = 0 PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, repro duced or distributed in any form or by any means, without the prior wri tten 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) = s 2 (t) −c 42 g 4 (t) − c 32 g 3 (t) = s 2 (t) and g 2 (t) = g ′ 2 (t) √ E 2 =        1/ √ 2, 0 ≤ t ≤ 1 −1/ √ 2, 1 ≤ t ≤ 2 0, o.w        where : E 2 =  2 0 (s 2 (t)) 2 dt = 2. Finally for the fourth basis fun ction : c 41 =  ∞ −∞ s 1 (t)g 4 (t)dt = −2/ √ 3, c 31 =  ∞ −∞ s 1 (t)g 3 (t)dt = 2/ √ 6, c 21 = 0 Hence : g ′ 1 (t) = s 1 (t) −c 41 g 4 (t) − c 31 g 3 (t) −c 21 g 2 (t) = 0 ⇒ g 1 (t) = 0 The last resu lt is expected, since the dimensionality of the vector space generated by these signals is 3. Based on the basis functions (g 2 (t), g 3 (t), g 4 (t)) the basis representation of the signals is : s 4 =  0, 0, √ 3  ⇒ E 4 = 3 s 3 =  0,  8/3, −1/ √ 3  ⇒ E 3 = 3 s 2 =  √ 2, 0, 0  ⇒ E 2 = 2 s 1 =  2/ √ 6, −2/ √ 3, 0  ⇒ E 1 = 2 Problem 2.6 Consider the set of signals  φ nl (t) = jφ nl (t), 1 ≤ n ≤ N , then by definition of lowpass equivalent signals and by Equations 2.2-49 and 2.2-54, we see that φ n (t)’s are √ 2 times the lowpass equivalents of φ nl (t)’s and  φ n (t)’s are √ 2 times the lowpass equivalents of  φ nl (t)’s. We also note that since φ n (t)’s have unit energy, φ nl (t),  φ nl (t) = φ nl (t), jφ nl (t) = −j and since the inner product is pure imaginary, we conclude that φ n (t) and  φ n (t) are orthogonal. Using the orthon ormality of the set φ nl (t), we have φ nl (t), −jφ ml (t) = jδ mn and using the result of problem 2.2 we have φ n (t),  φ m (t) = 0 for all n, m We also have φ n (t), φ m (t) = 0 f or all n = m PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, repro duced or distributed in any form or by any means, without the prior wri tten 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. 7 and   φ n (t),  φ m (t) = 0 f or all n = m Using the fact that the energy in lowpass equivalent signal is twice the energy in the bandpass signal we conclude that the energy in φ n (t)’s and  φ n (t)’s is unity an d hence the set of 2N signals {φ n (t),  φ n (t)} constitute an orthonormal set. The fact that this orthonormal set is s ufficient for expansion of bandp ass signals follows from Equation 2.2-57. Problem 2.7 Let x(t) = m(t) cos 2πf 0 t where m(t) is real and lowpass with bandwidth less than f 0 . Then F[ˆx(t)] = −j sgn(f)  1 2 M(f − f 0 ) + 1 2 M(f + f 0 )  and hence F[ˆx(t)] = − j 2 M(f −f 0 ) + j 2 M(f + f 0 ) where we have used that fact that M(f − f 0 ) = 0 for f < 0 and M(f + f 0 ) = 0 for f > 0. This shows that ˆx(t) = m(t) sin 2πf 0 t. Similarly we can show that Hilbert transform of m(t) sin 2πf 0 t is −m(t) cos 2πf 0 t. From above and Equation 2.2-54 we have H[φ n (t)] = √ 2φ ni (t) sin 2πf 0 t + √ 2φ nq (t) cos 2πf 0 t = −  φ n (t) Problem 2.8 For real-valued signals the correlation coefficients are given by : ρ km = 1 √ E k E m  ∞ −∞ s k (t)s m (t)dt and the Euclidean distances by : d (e) km =  E k + E m −2 √ E k E m ρ km  1/2 . For th e signals in this problem : E 1 = 2, E 2 = 2, E 3 = 3, E 4 = 3 ρ 12 = 0 ρ 13 = 2 √ 6 ρ 14 = − 2 √ 6 ρ 23 = 0 ρ 24 = 0 ρ 34 = − 1 3 and: d (e) 12 = 2 d (e) 13 =  2 + 3 −2 √ 6 2 √ 6 = 1 d (e) 14 =  2 + 3 + 2 √ 6 2 √ 6 = 3 d (e) 23 = √ 2 + 3 = √ 5 d (e) 24 = √ 5 d (e) 34 =  3 + 3 + 2 ∗3 1 3 = 2 √ 2 PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, repro duced or distributed in any form or by any means, without the prior wri tten 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. 8 Problem 2.9 We know from Fourier transform properties that if a signal x(t) is real-valued then its Fourier transform satisfies : X(−f ) = X ∗ (f) (Hermitian property). Hence the condition under which s l (t) is real-valued is : S l (−f) = S ∗ l (f) or going back to the bandpass signal s(t) (using 2-1-5): S + (f c − f) = S ∗ + (f c + f) The last condition s hows that in order to have a real-valued lowpass signal s l (t), the positive fre- quency content of the corresponding bandpass signal must exhibit hermitian symmetry around the center frequency f c . In general, bandpass s ignals do not satisfy this property (they h ave Hermitian symmetry around f = 0), hence, the lowpass equivalent is gen erally complex-valued. Problem 2.10 a. To show that the waveforms f n (t), n = 1, . . . , 3 are orthogonal we have to prove that:  ∞ −∞ f m (t)f n (t)dt = 0, m = n Clearly: c 12 =  ∞ −∞ f 1 (t)f 2 (t)dt =  4 0 f 1 (t)f 2 (t)dt =  2 0 f 1 (t)f 2 (t)dt +  4 2 f 1 (t)f 2 (t)dt = 1 4  2 0 dt − 1 4  4 2 dt = 1 4 × 2 − 1 4 × (4 −2) = 0 Similarly: c 13 =  ∞ −∞ f 1 (t)f 3 (t)dt =  4 0 f 1 (t)f 3 (t)dt = 1 4  1 0 dt − 1 4  2 1 dt − 1 4  3 2 dt + 1 4  4 3 dt = 0 and : c 23 =  ∞ −∞ f 2 (t)f 3 (t)dt =  4 0 f 2 (t)f 3 (t)dt = 1 4  1 0 dt − 1 4  2 1 dt + 1 4  3 2 dt − 1 4  4 3 dt = 0 PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, repro duced or distributed in any form or by any means, without the prior wri tten 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. 9 Thus, the signals f n (t) are orthogonal. It is also straightforward to prove that the signals have unit energy :  ∞ −∞ |f i (t)| 2 dt = 1, i = 1, 2, 3 Hence, th ey are orth on ormal. b. We first determine the weighting coefficients x n =  ∞ −∞ x(t)f n (t)dt, n = 1, 2, 3 x 1 =  4 0 x(t)f 1 (t)dt = − 1 2  1 0 dt + 1 2  2 1 dt − 1 2  3 2 dt + 1 2  4 3 dt = 0 x 2 =  4 0 x(t)f 2 (t)dt = 1 2  4 0 x(t)dt = 0 x 3 =  4 0 x(t)f 3 (t)dt = − 1 2  1 0 dt − 1 2  2 1 dt + 1 2  3 2 dt + 1 2  4 3 dt = 0 As it is observed, x(t) is orthogonal to the signal wavaforms f n (t), n = 1, 2, 3 and thus it can not represented as a linear combination of these functions. Problem 2.11 a. As an orthonorm al set of basis fu nctions we consider the set f 1 (t) =    1 0 ≤ t < 1 0 o.w f 2 (t) =    1 1 ≤ t < 2 0 o.w f 3 (t) =    1 2 ≤ t < 3 0 o.w f 4 (t) =    1 3 ≤ t < 4 0 o.w In matrix notation, the four waveforms can be represented as        s 1 (t) s 2 (t) s 3 (t) s 4 (t)        =        2 −1 −1 −1 −2 1 1 0 1 −1 1 −1 1 −2 −2 2               f 1 (t) f 2 (t) f 3 (t) f 4 (t)        Note that the rank of the transformation matrix is 4 and therefore, the dimensionality of the waveforms is 4 PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, repro duced or distributed in any form or by any means, without the prior wri tten 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. 10 b. The representation vectors are s 1 =  2 −1 −1 −1  s 2 =  −2 1 1 0  s 3 =  1 −1 1 −1  s 4 =  1 −2 −2 2  c. T he distance between th e first and the second vector is: d 1,2 =  |s 1 − s 2 | 2 =      4 −2 −2 −1     2 = √ 25 Similarly we find that : d 1,3 =  |s 1 − s 3 | 2 =      1 0 −2 0     2 = √ 5 d 1,4 =  |s 1 − s 4 | 2 =      1 1 1 −3     2 = √ 12 d 2,3 =  |s 2 − s 3 | 2 =      −3 2 0 1     2 = √ 14 d 2,4 =  |s 2 − s 4 | 2 =      −3 3 3 −2     2 = √ 31 d 3,4 =  |s 3 − s 4 | 2 =      0 1 3 −3     2 = √ 19 Thus, the minimum distance between any pair of vectors is d min = √ 5. Problem 2.12 As a set of orthonormal functions we consider the waveforms f 1 (t) =    1 0 ≤ t < 1 0 o.w f 2 (t) =    1 1 ≤ t < 2 0 o.w f 3 (t) =    1 2 ≤ t < 3 0 o.w The vector representation of th e signals is s 1 =  2 2 2  s 2 =  2 0 0  s 3 =  0 −2 −2  s 4 =  2 2 0  PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, repro duced or distributed in any form or by any means, without the prior wri tten 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. [...]... procedure as in example 2-1 -1 , we prove : pY (y) = 1 pX |a| y−b a 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... 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... 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... 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... 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... 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... 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... 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... 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... 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 . Solutions Manual for Digital Communications, 5th Edition (Chapter 2) 1 Prepared by Kostas Stamatiou January 11, 2008 1 PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc bandpass signal s(t) (using 2-1 -5 ): S + (f c − f) = S ∗ + (f c + f) The last condition s hows that in order to have a real-valued lowpass signal s l (t), the positive fre- quency content of the corresponding. procedure as in example 2-1 -1 , we prove : p Y (y) = 1 |a| p X  y −b a  PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, repro