Digital Communications 5th Edition Instructor Solution Manual PDF

Solutions
Manual for
Digital Communications, 5th Edition
(Chapter 2) 1

, 2
Problem 2.1


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

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 ω0t, 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 — f ) + jδ(f + f )] = [δ(f — f ) — δ(f + f )] = F − 1 {sin 2πf t}
0 0 0 0 0
2 2j

Hence, x̂ (t) = sin ω0t.

d. In a similar way to part (c) :
1 1
x(t) = sin ω 0t ⇒ X(f ) = [δ(f — f )0 — δ(f + f )]0 ⇒ X̂ ( f ) = [—δ(f — f 0) — δ(f + f 0)]
2j 2
1
⇒ X̂ (f ) = — [δ(f — f0 ) + δ(f + f )]
0 = —F
−1 {cos 2πω t} ⇒ x̂ (t) = — cos ω t
0 0
2


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 X̂ ( f ) = —X(f ),
we have : x̂ (t ) = —x(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 :
∫ ∞ ∫∞
2 2
ˆ
X(f ) df = |X(f )| df
−∞ −∞


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, 3


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




g. From parts (a) and (b) above, we 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
∫ ∞ ∫ ∞
x(t)y(t) dt =∫ X(f )Y ∗(f ) dt
−∞ ∞
−∞ ∗
1 1 1 1
= X (f — —f — f ) Y (f — —f — f ) df
l f0) + Xl( 0
2 l f0) + Yl( 0
∫ ∞2
−∞ 2 ∫ ∞ 2
1 1
= X (f — f )Y 0 l (f — f 0) df + 4 −∞ X (—f
∗ — f 0)Y l(—f — f 0) df
4 ∫ −∞
l l
1 ∞ 1 ∗
∗
=
Xl(u)Yl (u) du + Xl (v)Y (v) dv
4 −∞ 4
1 ∫ ∞
= Re Xl(f )Yl∗(f ) df
2 −∞
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.




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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
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, 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 :
∫∞" #
Σ
K
s(t) — skfk(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 :
∫∞
s(t)f n∗(t)dt — sn = 0, n = 1, 2, ..., K
−∞

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

The corresponding residual error Ee is :
∫∞ h ΣK ih ΣK i∗
Emin = −∞
s(t) — s f
k=1 k k (t) s(t) — s f
n=1 n n (t) dt

∫∞ ∫ ∞ ΣK ΣK h ΣK i
= −∞
|s(t)|2 dt — −∞ k=1 skfk(t)s (t)dt —
∗
n=1
s∗ ∫−∞
∞
s(t) — k=1 s k fk (t) fn∗(t)dt
n


∫∞ 2
∫ ∞ ΣK
= −∞|s(t)| dt — −∞ k=1 skfk(t)s∗(t)dt

ΣK
= Es — 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 :
∫ h Σ ih Σ i∗
d
Ee = d ∞ s(t) — K skfk(t) s(t) — K snfn(t) dt = 0
dan dan −∞ k=1 n=1


∫∞ h Σ i∗ h Σ i
⇒— −∞ anfn(t) s(t) — K n=1 snfn(t) + a n∗ fn∗(t) s(t) — Kn=1 snfn(t) dt = 0

∫∞ n h ΣK i,
∗
⇒ —2an −∞
Re fn (t) s(t) — s f
n=1 n n (t) dt = 0

∫∞ n h ΣK i,
∗
⇒ −∞
Re fn (t) s(t) — n=1 sn 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,
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.