TEST BANK FOR Digital Communications 5th Edition By Proakis Salehi

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
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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