Solution manual fundamentals of electric circuits 3rd edition chapter18

Chapter 18, Problem 1.

Obtain the Fourier transform of the function in Fig. 18.26.

Figure 18.26
For Prob. 18.1.

Chapter 18, Solution 1.

f ' ( t ) = δ( t + 2) − δ( t + 1) − δ( t − 1) + δ( t − 2)
jωF(ω) = e j2 ω − e jω − e − jω + e − jω2
= 2 cos 2ω − 2 cos ω
2[cos 2ω − cos ω]
F(ω) =
jω

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Chapter 18, Problem 2.

What is the Fourier transform of the triangular pulse in Fig. 18.27?

Figure 18.27
For Prob. 18.2.
Chapter 18, Solution 2.
⎡t,

f (t) = ⎢
⎣0,

0 < t <1
otherwise
f ”(t)

f ‘(t)
1

δ(t)

0

t

t

1

–δ’(t-1)
-δ(t-1)

-δ(t-1)

f"(t) = δ(t) - δ(t - 1) - δ'(t - 1)
Taking the Fourier transform gives
-ω2F(ω) = 1 - e-jω - jωe-jω
F(ω) =

(1 + jω)e jω − 1
ω2

1

or F(ω) = ∫ t e − jωt dt
0

But

ax
∫ x e dx =

F(ω) =

e − jω

(− jω)

2

eax
(ax − 1) + c
a2
(− jωt − 1) 10 =

[

]

1
(1 + jω)e − jω − 1
2
ω

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Chapter 18, Problem 3.

Calculate the Fourier transform of the signal in Fig. 18.28.

Figure 18.28
For Prob. 18.3.

Chapter 18, Solution 3.
f (t) =

1
t , − 2 < t < 2,
2

1
f ' (t) = , − 2 < t < 2
2

1 jωt
e − jωt
F(ω) = ∫ t e dt =
(− jωt − 1) 2− 2
2
−2 2
2(− jω)
1
= − 2 e − jω2 (− jω2 − 1) − e jω2 ( jω2 − 1)
2ω
1
=−
− jω2 e − jω2 + e jω2 + e jω2 − e − jω2
2
2ω
1
= − 2 (− jω4 cos 2ω + j2 sin 2ω)
2ω
j
(2ω cos 2ω − sin 2ω)
F(ω) =
ω2
2

[

]

[

(

)

]

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Chapter 18, Problem 4.

Find the Fourier transform of the waveform shown in Fig. 18.29.

Figure 18.29
For Prob. 18.4.

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Chapter 18, Solution 4.
2δ(t+1)
g’

2
–1
0

1

t

–2
–2δ(t–1)

4δ(t)
2δ’(t+1)
g”
–1
0
–2δ(t+1)

1

t

–2
–2δ(t–1)
–2δ’(t–1)

g ′′ = −2δ( t + 1) + 2δ′( t + 1) + 4δ( t ) − 2δ( t − 1) − 2δ′( t − 1)
( jω) 2 G (ω) = −2e jω + 2 jωe jω + 4 − 2e − jω − 2 jωe − jω
= −4 cos ω − 4ω sin ω + 4

G (ω) =

4
ω2

(cos ω + ω sin ω − 1)

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Chapter 18, Problem 5.

Obtain the Fourier transform of the signal shown in Fig. 18.30.

Figure 18.30
For Prob. 18.5.

Chapter 18, Solution 5.
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h’(t)

1

0

–1

t

1

–2δ(t)

h”(t)
1

δ(t+1)

1
t

0

–1
–2δ’(t)

–δ(t–1)

h ′′( t ) = δ( t + 1) − δ( t − 1) − 2δ′( t )
( jω) 2 H(ω) = e jω − e − jω − 2 jω = 2 j sin ω − 2 jω

H(ω) =

2j 2j
−
sin ω
ω ω2

Chapter 18, Problem 6.
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Find the Fourier transforms of both functions in Fig. 18.31 on the following page.

Figure 18.31
For Prob. 18.6.
Chapter 18, Solution 6.
(a) The derivative of f(t) is shown below.
f’(t)
5δ(t)
0

5δ(t-1)
1

2

t
-10δ(t-2)

f '(t ) = 5δ (t ) + 5δ (t − 1) − 10δ (t − 2)

Taking the Fourier transform of each term,
jω F (ω ) = 5 + 5e − jω − 10e − j 2ω
F (ω ) =

5 + 5e − jω − 10e − j 2ω
jω

(b) The derivative of g(t) is shown below.
g’(t)
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10δ(t)
0

1

2
-5

-5δ(t-1)

The second derivative of g(t) is shown below.
g’’(t)
10δ’(t)
0

5δ(t-2)

1

2

t

-5δ’(t-1)
-5δ(t-1)
g”(t) = 10δ’(t) – 5δ’(t–1) – 5δ(t–1) + 5δ(t–2)
Take the Fourier transform of each term.
(jω)2G(jω) = 10jω – 5jωe–jω – 5e–jω + 5e–j2ω which leads to
G(jω) = (–10jω + 5jωe–jω + 5e–jω – 5e–j2ω )/ω2

Chapter 18, Problem 7.
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Find the Fourier transforms of the signals in Fig. 18.32.

Figure 18.32
For Prob. 18.7.
Chapter 18, Solution 7.
(a) Take the derivative of f1(t) and obtain f1’(t) as shown below.
2δ(t)

0

1

2

t

-δ(t-1) -δ(t-2)
f1' (t ) = 2δ (t ) − δ '(t − 1) − δ (t − 2)

Take the Fourier transform of each term,
jω F1 (ω ) = 2 − e − jω − e− j 2ω
F1 (ω ) =

(b) f2(t) = 5t
F2 (ω ) =

∞

∫

−∞

F2 (ω ) =

5e − j 2ω

ω

2

2

2 − e − jω − e − j 2ω
jω

f 2 (t )e − jω dt = ∫ 5te− jω dt =

(1 + jω 2) −

0

2
5
e− jωt (− jω − 1)
2
0
(− jω )

5

ω2

Chapter 18, Problem 8.
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Obtain the Fourier transforms of the signals shown in Fig. 18.33.

Figure 18.33
For Prob. 18.8.
Chapter 18, Solution 8.
1

(a)

F(ω) = ∫ 2e
0

=

− jωt

dt + ∫ (4 − 2 t )e − jωt dt
1

2 − jωt 1
4 − jωt 2
2 − jωt

2
e
+
e
−
e
(− jωt − 1) 1
0
1
2
− jω
− jω
−ω
F(ω) =

(b)

2

2
ω

2

+

2 − jω 2
4 − j2ω 2
e
+

−
e
−
(1 + j2ω)e − j2ω
2
jω
jω jω
ω

g(t) = 2[ u(t+2) – u(t-2) ] - [ u(t+1) – u(t-1) ]
G (ω) =

4 sin 2ω 2 sin ω
−
ω
ω

Chapter 18, Problem 9.
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Determine the Fourier transforms of the signals in Fig. 18.34.

Figure 18.34
For Prob. 18.9.
Chapter 18, Solution 9.

(a)

y(t) = u(t+2) – u(t-2) + 2[ u(t+1) – u(t-1) ]
Y(ω) =

1

(b) Z(ω) = ∫ (−2 t )e

− jωt

dt =

0

2
4
sin 2ω + sin ω
ω
ω

− 2e − jωt
− ω2

2 2e − j ω
1
(− jωt − 1) 0 =
−
(1 + jω)
2

2
ω

ω

Chapter 18, Problem 10.
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Obtain the Fourier transforms of the signals shown in Fig. 18.35.

Figure 18.35
For Prob. 18.10.
Chapter 18, Solution 10.
x(t) = e2tu(t)

(a)

X(ω) = 1/(–2 + jω)
(b)

e

−( t )

⎡e − t , t > 0

=⎢ t
⎣⎢e , t < 0
1

0

1

−1

−1

0

Y(ω) = ∫ y( t )e jωt dt = ∫ e t e jωt dt + ∫ e − t e − jωt dt
e (1− jω) t
=
1 − jω

=

0
−1

e − (1+ jω) t
+
− (1 + jω)

1
0

⎡ cos ω + jsin ω cos ω − jsin ω ⎤
2
− e −1 ⎢
+
⎥
2
1+ ω
1 − jω
1 + jω
⎣
⎦

Y(ω) =

[

2
1 − e −1 (cos ω − ω sin ω)
2
1+ ω

]

Chapter 18, Problem 11.
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Find the Fourier transform of the “sine-wave pulse” shown in Fig. 18.36.

Figure 18.36
For Prob. 18.11.

Chapter 18, Solution 11.
f(t) = sin π t [u(t) - u(t - 2)]
2

F(ω) = ∫ sin πt e − jωt dt =
0

(

)

1 2 j πt
e − e − j πt e − jωt dt
2 j ∫0

=

1 ⎡ 2 + j( − ω + π ) t
+ e − j( ω+ π ) t )dt ⎤
(e
∫
⎢
⎥⎦

0
2j ⎣

=

1 ⎡
1
e − j( ω+ π ) t 2 ⎤
− j ( ω− π ) t 2
e
+
⎢
0
0⎥
2 j ⎣ − j(ω − π)
− j(ω + π) ⎦

=

1 ⎛ 1 − e − j2 ω 1 − e − j2 ω ⎞
⎟
⎜
+
2 ⎜⎝ π − ω
π + ω ⎟⎠

=

1
2π + 2πe − j2 ω

2
2(π − ω )

(

2

F(ω) =

(

)

)

π
e − jω 2 − 1
2
ω −π
2

Chapter 18, Problem 12.
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Find the Fourier transform of the following signals.

(a) f 1 (t) = e −3t sin(10t)u(t)
(b) f 2 (t) = e −4t cos(10t)u(t)

Chapter 18, Solution 12.

(a) F1 (ω ) =

10
(3 + jω ) 2 + 100

(b) F2 (ω ) =

4 + jω
(4 + jω ) 2 + 100

Chapter 18, Problem 13.
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Find the Fourier transform of the following signals:
(a) f(t) = cos(at – π /3),
(b) g(t) = u(t + 1)sin π t,
(c) h(t) = (1 + A sin at) cos bt,
(d) i(t) = 1 – t,
0

–∞ < t < ∞
–∞ < t < ∞
– ∞ < t < ∞ , where A, a and b are constants

Chapter 18, Solution 13.

(a) We know that F[cos at ] = π[δ(ω − a ) + δ(ω + a )] .
Using the time shifting property,
F[cos a ( t − π / 3a )] = πe − jωπ / 3a [δ(ω − a ) + δ(ω + a )] = πe − jπ / 3δ(ω − a ) + πe jπ / 3δ(ω + a )

(b) sin π( t + 1) = sin πt cos π + cos πt sin π = − sin πt
g(t) = -u(t+1) sin (t+1)
Let x(t) = u(t)sin t, then X(ω) =

1
2

( jω) + 1

=

1
1 − ω2

Using the time shifting property,
G (ω) = −

1
1 − ω2

e jω =

e jω
ω2 − 1

(c ) Let y(t) = 1 + Asin at, then Y(ω) = 2πδ(ω) + jπA[δ(ω + a ) − δ(ω − a )]
h(t) = y(t) cos bt
Using the modulation property,
1
H(ω) = [Y(ω + b) + Y(ω − b)]
2
H(ω) = π[δ(ω + b) + δ(ω − b)] +

4

(d) I(ω) = ∫ (1 − t )e − jωt dt =
0

jπA
[δ(ω + a + b) − δ(ω − a + b) + δ(ω + a − b) − δ(ω − a − b)]
2

e − jωt e − jωt
1
e − j4ω e − j4ω
4
−
(− jωt − 1) 0 =
−
−

( j4ω + 1)
− jω − ω 2
jω
ω2
ω2

Chapter 18, Problem 14.
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Find the Fourier transforms of these functions:
(a) f(t) = e −t cos(3t + π )u(t)
(b) g(t) = sin π t[u(t + 1) – u(t – 1)]
(c) h(t) = e −2t cos π tu(t – 1)
(d) p(t) = e −2t sin 4tu(–t)
(e) q(t) = 4 sgn(t – 2) + 3 δ (t) – 2u(t – 2)

Chapter 18, Solution 14.
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(a)

cos(3t + π) = cos 3t cos π − sin 3t sin π = cos 3t (−1) − sin 3t (0) = − cos(3t )
f ( t ) = −e − t cos 3t u ( t )
− (1 + jω)
F(ω) =
(1 + jω)2 + 9

(b)
g(t)
1
-1
1

t

-1

g’(t)

π
-1

1
t

-π
g ' ( t ) = π cos πt[u ( t − 1) − u ( t − 1)]

g" ( t ) = −π 2 g( t ) − πδ( t + 1) + πδ( t − 1)
− ω 2 G (ω) = − π 2 G (ω) − πe jω + πe − jω

(π 2 − ω2 )G(ω) = −π(e jω − e − jω ) = −2 jπ sin ω
2 jπ sin ω
G(ω) =
ω2 − π 2
Alternatively, we compare this with Prob. 17.7
f(t) = g(t - 1)
F(ω) = G(ω)e-jω
π
(e − jω − e jω )
G (ω) = F(ω)e jω = 2
ω − π2
− j2π sin ω
=
ω2 − π 2
2 jπ sin ω
G(ω) =
π 2 − ω2
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cos π( t − 1) = cos πt cos π + sin πt sin π = cos πt (−1) + sin πt (0) = − cos πt
Let x ( t ) = e −2( t −1) cos π( t − 1)u ( t − 1) = −e 2 h ( t )
y( t ) = e −2 t cos(πt )u ( t )
2 + jω
Y(ω) =
(2 + jω) 2 + π 2

y( t ) = x ( t − 1)
Y(ω) = X(ω)e − jω

(c)
and

X(ω) =

(2 + jω)e jω
(2 + jω)2 + π 2

X(ω) = −e 2 H(ω)
H(ω) = −e −2 X(ω)
=

− (2 + jω)e jω− 2

(2 + jω)2 + π 2

Let x ( t ) = e −2 t sin( −4t )u (− t ) = y(− t )
p( t ) = − x ( t )
where y( t ) = e 2 t sin 4t u ( t )
2 + jω
Y(ω) =
(2 + jω)2 + 4 2
2 − jω
X(ω) = Y(−ω) =
(2 − jω)2 + 16
jω − 2
p(ω) = −X(ω) =

(jω − 2 )2 + 16
(d)

(e)

⎛
8 − jω 2
1 ⎞
e
+ 3 − 2⎜⎜ πδ(ω) + ⎟⎟e − jω2
jω
jω ⎠
⎝
6 jω 2
e + 3 − 2πδ(ω)e − jω 2
Q(ω) =
jω
Q(ω) =

Chapter 18, Problem 15.
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Find the Fourier transforms of the following functions:
(a) f(t) = δ (t +3) – δ (t – 3)
(b) f(t) =

∫

∞

−∞

2δ (t − 1) dt

(c) f(t) = δ (3t) – δ '(2t)

Chapter 18, Solution 15.

(a)

F(ω) = e j3ω − e − jω3 = 2 j sin 3ω

(b)

Let g( t ) = 2δ( t − 1), G (ω) = 2e − jω
t
F(ω) = F ⎛⎜ ∫ g ( t ) dt ⎞⎟
⎝ −∞
⎠
G (ω)
+ πF(0)δ(ω)
=
jω

(c)

=

2e − j ω
+ 2πδ(−1)δ(ω)
jω

=

2e − jω
jω

1
⋅1
2
1 jω
1
1
F(ω) = ⋅ 1 − jω = −
3 2
3
2

F [δ(2t )] =

Chapter 18, Problem 16.
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* Determine the Fourier transforms of these functions:
(a) f(t) = 4/t 2

(b) g(t) = 8/(4 + t 2 )

* An asterisk indicates a challenging problem.
Chapter 18, Solution 16.

(a) Using duality properly
t →

or

−2
ω2

−2
→ 2π ω
t2
4
→ − 4π ω
t2
⎛4⎞
F(ω) = F ⎜ 2 ⎟ = − 4π ω
⎝t ⎠

(b)

e

−at

2a
a + ω2
2

2a
a + t2

2π e

−a ω

8
a + t2

4π e

−2 ω

2

2

⎛ 8 ⎞
−2 ω
= 4π e

G(ω) = F ⎜
2 ⎟
⎝4+t ⎠

Chapter 18, Problem 17.
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Find the Fourier transforms of:
(a) cos 2tu(t)

(b) sin 10tu(t)

Chapter 18, Solution 17.

1
[F(ω + ω0 ) + F(ω − ω0 )]
2
1
where F(ω) = F [u (t )] = πδ(ω) + , ω0 = 2
jω

(a) Since H(ω) = F (cos ω0 t f ( t ) ) =

H(ω) =

1⎡
1
1 ⎤
+ πδ(ω − 2 ) +
⎥
⎢πδ (ω + 2 ) + (
j ω + 2)
j (ω − 2 ) ⎦
2⎣

π
[δ(ω + 2) + δ(ω − 2)] − j ⎡⎢ ω + 2 + ω − 2 ⎤⎥
2
2 ⎣ (ω + 2)(ω − 2) ⎦
jω
π
H(ω) = [δ(ω + 2 ) + δ(ω − 2 )] − 2
2
ω −4
=

(b)

j
[F(ω + ω0 ) − F(ω − ω0 )]
2
1
where F(ω) = F [u (t )] = πδ (ω) +
jω
⎤

j⎡
1
1
G (ω) = ⎢πδ(ω + 10) +
− πδ(ω − 10) −
2⎣
j(ω + 10)
j(ω − 10 ) ⎥⎦
G(ω) = F [sin ω0 t f ( t )] =

jπ
[δ(ω + 10) − δ (ω − 10)] + j ⎡⎢ j − j ⎤⎥
2
2 ⎣ ω − 10 ω + 10 ⎦
jπ
[δ(ω + 10) − δ(ω − 10 )] − 2 10
=
2
ω − 100
=

Chapter 18, Problem 18.
PROPRIETARY MATERIAL. © 2007 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.


Given that F( ω ) = F[f(t)], prove the following results, using the definition of Fourier

transform:
(a) F [ f (t − t 0 )] = e − jωt0 F( ω )
⎡ df (t ) ⎤
= j ω F( ω )
(b) F ⎢
⎣ dt ⎥⎦
(c) F[f(–t)] = F(– ω )
d
F( ω )
(d) F[tf(t)] = j
dω
Chapter 18, Solution 18.
∞

(a) F [ f (t − to )] =

∫

f (t − to )e− jωt dt

−∞

Let t − to = λ

⎯⎯
→ t = λ + to ,
∞

∫

F [ f (t − to )] =

dt = d λ

f (λ )e− jωλ e− jωto d λ =e− jωto F (ω )

−∞

(b) Given that

f (t ) = F −1[ F (ω )] =

1
2π

∫

∞

−∞

F (ω )e jωt dω

∞

jω
jωt
−1
f '(t ) =
∫ F (ω )e dt = jω F [ F (ω )]

2π −∞
or
F [ f '(t )] = jω F (ω )
(c ) This is a special case of the time scaling property when a = –1. Hence,

F [ f (−t )] =

1
F (−ω ) = F (−ω )
| −1|

(d) F (ω ) = ∫

∞

−∞

f (t )e − jωt dt

Differentiating both sides respect to ω and multiplying by t yields
∞
∞
dF (ω )
− jω t
j
= j ∫ (− jt ) f (t )e dt = ∫ tf (t )e − jωt dt
dω
−∞
−∞
Hence,

dF (ω )
= F [tf (t )]
j
dω
Chapter 18, Problem 19.
PROPRIETARY MATERIAL. © 2007 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.


Find the Fourier transform of

f(t) = cos 2 π t[u(t) – u(t – 1)]

Chapter 18, Solution 19.
∞

F(ω) = ∫ f ( t )e jωt dt =
−∞

F(ω) =

(

)

1 1 j2 πt
e + e − j2 πt e − jωt dt

∫
0
2

[

]

1 1 − j( ω + 2 π ) t
e
+ e − j(ω− 2 π )t dt
2 ∫0
1

⎤
1
1⎡
1
e − j( ω − 2 π ) t ⎥
= ⎢
e − j( ω + 2 π ) t +
2 ⎣ − j (ω + 2π )
− j(ω − 2π )
⎦0
1 ⎡ e − j( ω+ 2 π ) − 1 e − j( ω− 2 π ) − 1 ⎤
=− ⎢
+
⎥
2 ⎣ j (ω + 2π)
j(ω − 2π ) ⎦

But

e j2 π = cos 2π + j sin 2π = 1 = e − j2 π
1 ⎛ e − jω − 1 ⎞⎛ 1
1 ⎞
⎟⎟⎜
+
F(ω) = − ⎜⎜
⎟
2⎝
j ⎠⎝ ω + 2π ω − 2π ⎠
jω
= 2
e − jω − 1
2
ω − 4π

(

)

Chapter 18, Problem 20.
PROPRIETARY MATERIAL. © 2007 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.

(a) Show that a periodic signal with exponential Fourier series
∞

f(t) =

∑c e

n = −∞

jnω0t

n

has the Fourier transform

F( ω ) =

∞

∑ c δ (ω − nω )

n = −∞

n

0

where ω 0 = 2 π /T.
(b) Find the Fourier transform of the signal in Fig. 18.37.

Figure 18.37
For Prob. 18.20(b).

Chapter 18, Solution 20.
PROPRIETARY MATERIAL. © 2007 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.