Solutions Manual Digital Communications 5th Edition By Kostas Stamatiou, Proakis Salehi

Solutions Manual
Digital Communications 5th Edition
By
Kostas Stamatiou,
Proakis Salehi

( All Chapters Included - 100% Verified Solutions )

, 2

Problem 2.1


a. Z ∞
1 x(a)
x̂(t) = da
π −∞ t−a
Hence : R∞
x(a)
−x̂(−t) = − π1 −∞ −t−a da
R −∞
= − π1 ∞ x(−b)
−t+b (−db)
R
1 ∞ x(b)
= − π −∞ −t+b db
R
1 ∞ x(b)
= π −∞ t−b db = x̂(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 :

x̂(t) = x̂(−t)



c. x(t) = cos ω0 t, so its Fourier transform is : X(f ) = 21 [δ(f − f0 ) + δ(f + f0 )] , f0 = 2πω0 .
Exploiting the phase-shifting property (2-1-4) of the Hilbert transform :

1 1
X̂(f ) = [−jδ(f − f0 ) + jδ(f + f0 )] = [δ(f − f0 ) − δ(f + f0 )] = F −1 {sin 2πf0 t}
2 2j

Hence, x̂(t) = sin ω0 t.

d. In a similar way to part (c) :

1 1
x(t) = sin ω0 t ⇒ X(f ) = [δ(f − f0 ) − δ(f + f0 )] ⇒ X̂(f ) = [−δ(f − f0 ) − δ(f + f0 )]
2j 2
1
⇒ X̂(f ) = − [δ(f − f0 ) + δ(f + f0 )] = −F −1 {cos 2πω0 t} ⇒ x̂(t) = − cos ω0 t
2


e. The positive frequency content of the new signal will be : (−j)(−j)X(f ) = −X(f ), f > 0, while
ˆ
the negative frequency content will be : j · jX(f ) = −X(f ), f < 0. Hence, since X̂(f ) = −X(f ),
ˆ = −x(t).
we have : x̂(t)

f. Since the magnitude response of the Hilbert transformer is characterized by : |H(f )| = 1, we
have that : X̂(f ) = |H(f )| |X(f )| = |X(f )| . Hence :
Z ∞ 2
Z ∞
X̂(f ) df = |X(f )|2 df
−∞ −∞

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1

, 3


and using Parseval’s relationship :
Z ∞ Z ∞
2
x̂ (t)dt = x2 (t)dt
−∞ −∞




g. From parts (a) and (b) above, weR note that if x(t) is even, x̂(t) is odd and vice-versa. Therefore,
∞
x(t)x̂(t) is always odd and hence : −∞ x(t)x̂(t)dt = 0.




Problem 2.2




1. Using relations

1 1
X(f ) = Xl (f − f0 ) + Xl (−f − f0 )
2 2
1 1
Y (f ) = Yl (f − f0 ) + Yl (−f − f0 )
2 2
and Parseval’s relation, we have
Z ∞ Z ∞
x(t)y(t) dt = X(f )Y ∗ (f ) dt
−∞ −∞
Z ∞  ∗
1 1 1 1
= Xl (f − f0 ) + Xl (−f − f0 ) Yl (f − f0 ) + Yl (−f − f0 ) df
−∞ 2 2 2 2
Z ∞ Z ∞
1 1
= Xl (f − f0 )Yl∗ (f − f0 ) df + Xl (−f − f0 )Yl (−f − f0 ) df
4 −∞ 4 −∞
Z
1 ∞ 1
= Xl (u)Yl∗ (u) du + Xl∗ (v)Y (v) dv
4 −∞ 4
Z ∞ 
1 ∗
= Re Xl (f )Yl (f ) df
2 −∞
Z ∞ 
1 ∗
= Re xl (t)yl (t) dt
2 −∞

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.

2

, 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 ∞" XK
#
s(t) − sk fk (t) fn∗ (t)dt = 0, n = 1, 2, ..., K (1)
−∞ k=1

since the functions {fn (t)} are orthonormal, only the term with k = n will remain in the sum, so :
Z ∞
s(t)fn∗ (t)dt − sn = 0, n = 1, 2, ..., K
−∞

or: Z ∞
sn = s(t)fn∗ (t)dt n = 1, 2, ..., K
−∞

The corresponding residual error Ee is :
R∞ h P ih PK i∗
Emin = −∞ s(t) − K s
k=1 k kf (t) s(t) − s f
n=1 n n (t) dt

R∞ 2 R ∞ PK ∗
PK ∗
R∞ h PK i
= −∞ |s(t)| dt − −∞ k=1 sk fk (t)s (t)dt − n=1 sn −∞ s(t) − k=1 s k f k (t) fn∗ (t)dt

R∞ 2 R ∞ PK ∗ (t)dt
= −∞ |s(t)| dt − −∞ k=1 sk fk (t)s


PK 2
= Es − k=1 |sk |

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 d
R∞ h PK ih PK i∗
E
dan e = dan −∞ s(t) − s f
k=1 k k (t) s(t) − s f
n=1 n n (t) dt = 0

R∞ h PK i∗ h PK i
⇒− a f (t) s(t) − s f (t) + a∗ f ∗ (t) s(t) − s f (t) dt = 0
−∞ n n n=1 n n n n n=1 n n


R∞ n h PK io
⇒ −2an Re f ∗ (t) s(t) − s f (t) dt = 0
−∞ n n=1 n n


R∞ n h PK io
⇒ Re f ∗ (t) s(t) − s f (t) dt = 0, n = 1, 2, ..., K
−∞ n n=1 n n


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