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Fundamentals Of Communication Systems答案

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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

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



t +



. This indicates ﬁrst we have to plot

Π(

and then shift it to left by

. A plot is shown below:

−

(

2t +5

)

∞

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)

−1

3. It is obvious from the deﬁnition of sgn(t) that sgn(2t) = sgn(t). Therefore x

(t) = 0.

4. x

(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

0.2

0.4

0.6

0.8

Proakis/Salehi/Fundamentals of Communications Systems 2E

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

0.2

0.4

0.6

0.8

2. x[n] = Π



−1



. If −

≤

−1

≤

, i.e., −2 ≤ n ≤ 10, we have x[n] = 1.

−20 −15 −10 −5 0 5 10 15 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

3. x[n] =

−1

(n/

) −(

−

−1

(n/

−

)

. For

n <

x[n] =

0, for 0

≤ n ≤

x[n] =

and

for n ≥ 4, x[n] =

−

+1 = 1.

Proakis/Salehi/Fundamentals of Communications Systems 2E

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

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Problem 2.3

[n] =

1 and

[n] = cos(

πn) =

1, for all

. This shows that two signals can be diﬀerent but

their sampled versions be the same.

Problem 2.4

Let

[n]

and

[n]

be two periodic signals with periods

and

, respectively, and let

N =

LCM(N

, N

)

, and deﬁne

x[n] = x

[n] +x

[n]

. Then obviously

[n +N] = x

[n]

and

[n +N] =

[n], and hence x[n] = x[n + N], i.e., x[n] is periodic with period N.

For continuous-time signals

(t)

and

(t)

with periods

and

respectively, in general we

cannot ﬁnd a

such that

T = k

= k

for integers

and

. This is obvious for instance if

1 and

= π

. The necessary and suﬃcient condition for the sum to be periodic is that

be a

rational number.

Problem 2.5

Using the result of problem 2.4 we have:

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

. Their ratio is not rational, hence the sum is not periodic.

3. The sum of two periodic discrete-time signal is periodic.

The ﬁst signal is periodic but

cos[

11000

is not periodic, since there is no

such that

cos[11000(n +N)] = cos(11000n) for all n. Therefore the sum cannot be periodic.

Proakis/Salehi/Fundamentals of Communications Systems 2E

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

(t) =











−t

t > 0

−e

t < 0

0 t = 0

=⇒ x

(−t) =











−e

−t

t > 0

t < 0

0 t = 0

= −x

(t)

Thus, x

(t) is an odd signal

(t) = cos



120πt +



is neither even nor odd. We have

cos



120πt +



= cos





cos(

120

πt)−

sin





sin(

120

πt)

. Therefore

(t) = cos





cos(

120

πt)

and

(t) = −sin





sin(

120

πt)

(Note: This part can also be considered as a special case of part 7 of this problem)

(t) = e

−|t|

=⇒ x

(−t) = e

−|(−t)|

= e

−|t|

= x

(t)

Hence, the signal x

(t) is even.

(t) =











t t ≥ 0

0 t < 0

=⇒ x

(−t) =











0 t ≥ 0

−t t < 0

The signal x

(t) is neither even nor odd. The even part of the signal is

4,e

(t) =

(t) +x

(−t)











t ≥ 0

−t

t < 0

|t|

The odd part is

4,o

(t) =

(t) −x

(−t)











t ≥ 0

t < 0

(t) = x

(t) −x

(t) =⇒ x

(−t) = x

(−t) −x

(−t) = x

(t) +x

(t)

Clearly

(−t) ≠ x

(t)

since otherwise

(t) =

∀t

. Similarly

(−t) ≠ −x

(t)

since otherwise

(t) = 0 ∀t. The even and the odd parts of x

(t) are given by

5,e

(t) =

(t) +x

(−t)

= x

(t)

5,o

(t) =

(t) −x

(−t)

= −x

(t)

Proakis/Salehi/Fundamentals of Communications Systems 2E

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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