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Proakis J.G., Salehi M.Fundamentals Of Communication Systems答案
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Solution Manual
Fundamentals of Communication Systems
John G. Proakis Masoud Salehi
Second Edition
2013
Proakis/Salehi/Fundamentals of Communications Systems 2E
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle
River, NJ 07458.
Chapter 2
Problem 2.1
1. Π
(
2t +5
)
= Π
2
t +
5
2
. This indicates ﬁrst we have to plot
Π(
2
t)
and then shift it to left by
5
2
. A plot is shown below:
6
−
11
4
−
9
4

t
Π
(
2t +5
)
1
2.
P
∞
n=0
Λ(t −n)
is a sum of shifted triangular pulses. Note that the sum of the left and right side
of triangular pulses that are displaced by one unit of time is equal to 1, The plot is given below
✲
✻
t
x
2
(t)
−1
1
3. It is obvious from the deﬁnition of sgn(t) that sgn(2t) = sgn(t). Therefore x
3
(t) = 0.
4. x
4
(t) is sinc(t) contracted by a factor of 10.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
3
Proakis/Salehi/Fundamentals of Communications Systems 2E
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle
River, NJ 07458.
Problem 2.2
1. x[n] = sinc(3n/9) = sinc(n/3).
−20 −15 −10 −5 0 5 10 15 20
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
2. x[n] = Π
n
4
−1
3
. If −
1
2
≤
n
4
−1
3
≤
1
2
, i.e., −2 ≤ n ≤ 10, we have x[n] = 1.
−20 −15 −10 −5 0 5 10 15 20
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
3. x[n] =
n
4
u
−1
(n/
4
) −(
n
4
−
1
)u
−1
(n/
4
−
1
)
. For
n <
0,
x[n] =
0, for 0
≤ n ≤
3,
x[n] =
n
4
and
for n ≥ 4, x[n] =
n
4
−
n
4
+1 = 1.
4
Proakis/Salehi/Fundamentals of Communications Systems 2E
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle
River, NJ 07458.
−5 0 5 10 15 20
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Problem 2.3
x
1
[n] =
1 and
x
2
[n] = cos(
2
πn) =
1, for all
n
. This shows that two signals can be diﬀerent but
their sampled versions be the same.
Problem 2.4
Let
x
1
[n]
and
x
2
[n]
be two periodic signals with periods
N
1
and
N
2
, respectively, and let
N =
LCM(N
1
, N
2
)
, and deﬁne
x[n] = x
1
[n] +x
2
[n]
. Then obviously
x
1
[n +N] = x
1
[n]
and
x
2
[n +N] =
x
2
[n], and hence x[n] = x[n + N], i.e., x[n] is periodic with period N.
For continuoustime signals
x
1
(t)
and
x
2
(t)
with periods
T
1
and
T
2
respectively, in general we
cannot ﬁnd a
T
such that
T = k
1
T
1
= k
2
T
2
for integers
k
1
and
k
2
. This is obvious for instance if
T
1
=
1 and
T
2
= π
. The necessary and suﬃcient condition for the sum to be periodic is that
T
1
T
2
be a
rational number.
Problem 2.5
Using the result of problem 2.4 we have:
1.
The frequencies are 2000 and 5500, their ratio (and therefore the ratio of the periods) is
rational, hence the sum is periodic.
2. The frequencies are 2000 and
5500
π
. Their ratio is not rational, hence the sum is not periodic.
3. The sum of two periodic discretetime signal is periodic.
4.
The ﬁst signal is periodic but
cos[
11000
n]
is not periodic, since there is no
N
such that
cos[11000(n +N)] = cos(11000n) for all n. Therefore the sum cannot be periodic.
5
Proakis/Salehi/Fundamentals of Communications Systems 2E
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle
River, NJ 07458.
Problem 2.6
1)
x
1
(t) =
e
−t
t > 0
−e
t
t < 0
0 t = 0
=⇒ x
1
(−t) =
−e
−t
t > 0
e
t
t < 0
0 t = 0
= −x
1
(t)
Thus, x
1
(t) is an odd signal
2)
x
2
(t) = cos
120πt +
π
3
is neither even nor odd. We have
cos
120πt +
π
3
= cos
π
3
cos(
120
πt)−
sin
π
3
sin(
120
πt)
. Therefore
x
2e
(t) = cos
π
3
cos(
120
πt)
and
x
2o
(t) = −sin
π
3
sin(
120
πt)
.
(Note: This part can also be considered as a special case of part 7 of this problem)
3)
x
3
(t) = e
−t
=⇒ x
3
(−t) = e
−(−t)
= e
−t
= x
3
(t)
Hence, the signal x
3
(t) is even.
4)
x
4
(t) =
t t ≥ 0
0 t < 0
=⇒ x
4
(−t) =
0 t ≥ 0
−t t < 0
The signal x
4
(t) is neither even nor odd. The even part of the signal is
x
4,e
(t) =
x
4
(t) +x
4
(−t)
2
=
t
2
t ≥ 0
−t
2
t < 0
=
t
2
The odd part is
x
4,o
(t) =
x
4
(t) −x
4
(−t)
2
=
t
2
t ≥ 0
t
2
t < 0
=
t
2
5)
x
5
(t) = x
1
(t) −x
2
(t) =⇒ x
5
(−t) = x
1
(−t) −x
2
(−t) = x
1
(t) +x
2
(t)
Clearly
x
5
(−t) ≠ x
5
(t)
since otherwise
x
2
(t) =
0
∀t
. Similarly
x
5
(−t) ≠ −x
5
(t)
since otherwise
x
1
(t) = 0 ∀t. The even and the odd parts of x
5
(t) are given by
x
5,e
(t) =
x
5
(t) +x
5
(−t)
2
= x
1
(t)
x
5,o
(t) =
x
5
(t) −x
5
(−t)
2
= −x
2
(t)
6
Proakis/Salehi/Fundamentals of Communications Systems 2E
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle
River, NJ 07458.
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