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<4
–∞ < 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.
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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.
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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.
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(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.
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(a)
F (cn) = cnδ(ω)
(
)
F c n e jnωo t = c n δ(ω − nωo )
(b)
cn =
=
n = −∞
T = 2π
ωo =
∞
∑ c δ(ω − nω )
n
o
2π
=1
T
1 T
1 ⎛ π
− jnt
f (t ) e − jnωo t dt =
⎜ 1⋅ e dt + 0 ⎞⎟
∫
⎠
T 0
2π ⎝ ∫0
1 ⎛ 1 jnt
⎜− e
2π ⎜⎝ jn
But e − jnπ
cn =
⎛ ∞
⎞
F ⎜ ∑ c n e jnωo t ⎟ =
⎝ n = −∞
⎠
⎞
j
⎟⎟ =
(
e − jnπ − 1)
⎠ 2πn
= cos nπ + j sin nπ = cos nπ = (−1) n
[
π
0
]
⎡
j
(− 1)n − 1 = ⎢ 0−,j ,
2nπ
⎣ nπ
n = even
n = odd , n ≠ 0
for n = 0
cn =
1 π
1
1 dt =
∫
0
2π
2
Hence
f (t) =
∞
1
j jnt
e
− ∑
2 n = −∞ nπ
n ≠0
n = odd
F(ω) =
∞
1
j
δω − ∑
δ(ω − n )
2
n = −∞ nπ
n≠0
n = odd
Chapter 18, Problem 21.
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Show that
∫
∞
−∞
2
π
⎛ sin aω ⎞
⎟ dω =
⎜
a
⎝ aω ⎠
Hint: Use the fact that
⎛ sin aω ⎞
F[u(t + a) – u(t – a )] = 2a ⎜
⎟.
⎝ aω ⎠
Chapter 18, Solution 21.
Using Parseval’s theorem,
∞
∫− ∞ f
2
( t )dt =
1 ∞
| F(ω) | 2 dω
2π ∫− ∞
If f(t) = u(t+a) – u(t+a), then
∞
∫−∞
a
f 2 ( t )dt = ∫ (1) 2 dt = 2a =
−a
2
1 ∞
⎛ sin aω ⎞
4a 2 ⎜
⎟ dω
∫
2 π −∞
⎝ aω ⎠
or
2
4πa π
⎛ sin aω ⎞
∫− ∞ ⎜⎝ aω ⎟⎠ dω = 4a 2 = a as required.
∞
Chapter 18, Problem 22.
Prove that if F( ω ) is the Fourier transform of f(t),
F[ f (t)sin ω 0 t] =
j
[F( ω + ω 0 ) – F( ω – ω o )]
2
Chapter 18, Solution 22.
F [f ( t ) sin ωo t ] = ∫ f ( t )
∞
−∞
=
=
(e
jω o t
)
− e − jωo t − jωt
e dt
2j
∞
1⎡ ∞
f ( t )e − j(ω− ωo )t dt − ∫ e − j(ω+ ωo )t dt ⎤
∫
⎥⎦
−∞
2 j ⎢⎣ − ∞
1
[F(ω − ω o ) − F(ω + ωo )]
2j
Chapter 18, Problem 23.
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If the Fourier transform of f(t) is
F( ω ) =
10
(2 + jω )(5 + jω )
determine the transforms of the following:
(a) f(–3t)
d
f (t )
(d)
dt
(b) f(2t – 1)
(e) ∫
t
−∞
(c) f(t)cos2t
f (t )dt
Chapter 18, Solution 23.
1
10
30
⋅
=
3 (2 + jω / 3)(5 + jω / 3) (6 + jω)(15 + jω)
30
F [f (− 3t )] =
(6 − jω)(15 − jω)
(a) f(3t) leads to
(b) f(2t)
1
10
20
⋅
=
2 (2 + jω / 2)(15 + jω / 2) (4 + jω)(10 + jω)
20e − jω / 2
(4 + jω)(10 + jω)
f(2t-1) = f [2(t-1/2)]
1
1
F(ω + 2) + F(ω + 2 )
2
2
(c) f(t) cos 2t
5
=
[2 + j(ω + 2)][5 + j(ω + 2)]
(d) F [f ' (t )] = jω F(ω) =
(e)
+
5
[2 + j(ω − 2 )[5 + j(ω − 2)]]
jω10
(2 + jω)(5 + jω)
F(ω)
+ πF(0 )δ(ω)
−∞
j(ω)
10
x10
=
+ πδ(ω)
jω(2 + jω)(5 + jω)
2x5
10
+ πδ(ω)
=
jω(2 + jω)(5 + jω)
∫ f (t ) dt
t
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Chapter 18, Problem 24.
Given that F[ f (t)] = ( j / ω )(e − jω – 1), find the Fourier transforms of:
(a) x(t) = f ( t) + 3
(b) y(t) = f ( t – 2)
(c) h(t) = f ’ ( t)
⎛2 ⎞
5
(d) g(t) = 4 f ⎜ t ⎟ + 10f ⎛⎜ t ⎞⎟
⎝3 ⎠
⎝3 ⎠
Chapter 18, Solution 24.
(a) X (ω) = F(ω) + F [3]
j
= 6πδ(ω) + e − jω − 1
ω
(
)
(b) y(t ) = f (t − 2 )
Y(ω) = e
− jω 2
je − j2ω − jω
F(ω) =
e −1
ω
(
(c) If h(t) = f '(t)
H(ω) = jωF(ω) = jω
)
j − jω
(e − 1) = 1 − e− jω
ω
3 ⎛3 ⎞
3 ⎛3 ⎞
⎛5 ⎞
⎛2 ⎞
(d) g(t ) = 4f ⎜ t ⎟ + 10f ⎜ t ⎟, G (ω) = 4 x F⎜ ω ⎟ + 10x F⎜ ω ⎟
2 ⎝2 ⎠
5 ⎝5 ⎠
⎝3 ⎠
⎝3 ⎠
= 6⋅
=
j
3
ω
2
(
(e
− j3ω / 2
)
−1 +
)
(
)
6 j − j3ω / 5
e
−1
3
ω
5
(
)
j4 − j3ω / 2
j10 − j3ω / 5
e
−1 +
e
−1
ω
ω
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Chapter 18, Problem 25.
Obtain the inverse Fourier transform of the following signals.
5
jω − 2
12
(b) H( ω ) = 2
ω +4
(a) F( ω ) =
(c) X( ω ) =
10
( jω − 1)( jω − 2)
Chapter 18, Solution 25.
(a) g (t ) = 5e 2t u (t )
(b) h(t ) = 6e −2|t|
(c ) X (ω ) =
A=
A
B
+
,
s −1 s − 2
10
= −10,
1− 2
X (ω ) =
B=
s = jω
10
= 10
2 −1
−10
10
+
jω − 1 jω − 2
x(t ) = −10et u (t ) + 10e 2t u (t )
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Chapter 18, Problem 26.
Determine the inverse Fourier transforms of the following:
e − j 2ω
1 + jω
1
(b) H( ω ) =
( jω + 4) 2
(c) G( ω ) = 2u (ω + 1) – 2u( ω − 1 )
(a) F( ω ) =
Chapter 18, Solution 26.
(a) f ( t ) = e −( t −2) u ( t )
(b) h ( t ) = te −4 t u ( t )
(c) If x ( t ) = u ( t + 1) − u ( t − 1)
⎯
⎯→
X(ω) = 2
sin ω
ω
By using duality property,
G (ω) = 2u (ω + 1) − 2u (ω − 1)
⎯
⎯→
g( t ) =
2 sin t
πt
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Chapter 18, Problem 27.
Find the inverse Fourier transforms of the following functions:
100
(a) F( ω ) =
jω ( jω + 10)
10 jω
(b) G( ω ) =
(− jω + 2)( jω + 3)
60
(c) H( ω ) =
2
− ω + j 40ω + 1300
(d) Y( ω ) =
δ (ω )
( jω + 1)( jω + 2)
Chapter 18, Solution 27.
100
A
B
= +
, s = jω
s (s + 10) s s + 10
100
100
A=
= 10, B =
= −10
10
− 10
10
10
−
F(ω) =
jω jω + 10
f(t) = 5 sgn(t ) − 10e −10 t u(t )
(a) Let F(s ) =
10s
A
B
=
+
, s = jω
(2 − s )(3 + s ) 2 − s s + 3
− 30
20
A=
= 4, B =
= −6
5
5
4
6
−
G (ω) =
= − jω + 2 jω + 3
g(t) = 4e 2 t u(− t ) − 6e −3 t u(t )
(b) G (s ) =
(c) H(ω) =
( j ω)
60
2
+ j40ω + 1300
h(t) = 2e −20 t sin( 30t ) u(t )
(d) y(t ) =
=
60
( jω + 20)2 + 900
1 ∞ δ(ω)e jωt dω
1 1 1
= π⋅ = π
∫
2π −∞ (2 + jω)( jω + 1) 2 2 4
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Chapter 18, Problem 28.
Find the inverse Fourier transforms of:
(a)
πδ (ω )
(5 + jω )(2 + jω )
(b)
10δ (ω + 2)
jω ( jω + 1)
(c)
20δ (ω − 1)
(2 + jω )(3 + jω )
(d)
5πδ (ω )
5
+
5 + jω
jω (5 + jω )
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Chapter 18, Solution 28.
(a)
(b)
πδ(ω) e jωt
1 ∞
1 ∞
jωt
f (t) =
F(ω)e dω =
dω
2π ∫−∞
2π ∫−∞ (5 + jω)(2 + jω)
1 1
1
=
=
= 0.05
2 (5)(2) 20
f (t) =
=
(c)
1 ∞ 10δ(ω + 2) jωt
10
e − j2 t
e
d
ω
=
2π ∫−∞ jω( jω + 1)
2π (− j2)(− j2 + 1)
j5 e − j2 t
( −2 + j)e − j2 t
=
2π
2π 1 − j2
f (t) =
1 ∞ 20δ(ω − 1)e jωt
20
e jt
d
ω
=
2π ∫−∞ (2 + jω)(3 + 5ω)
2π (2 + j)(3 + j)
20e jt
(1 − j)e jt
=
=
2π(5 + 5 j)
π
(d)
Let
5πδ(ω)
5
+
= F1 (ω) + F2 (ω)
(5 + jω) jω(5 + jω)
1 ∞ 5πδ(ω) jωt
5π 1
f1 ( t ) =
e dω =
⋅ = 0.5
∫
−
∞
2π
5 + jω
2π 5
F(ω) =
5
A
B
= +
, A = 1, B = −1
s(5 + s) s s + 5
1
1
−
F2 (ω) =
jω jω + 5
F2 (s) =
f 2 (t) =
1
1
sgn( t ) − e −5 t = − + u ( t ) − e 5 t
2
2
f ( t ) = f 1 ( t ) + f 2 ( t ) = u( t ) − e − 5 t
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Chapter 18, Problem 29.
* Determine the inverse Fourier transforms of:
(a) F( ω ) = 4 δ ( ω + 3) + δ ( ω ) + 4 δ ( ω − 3 )
(b) G( ω ) = 4u( ω + 2) – 4u( ω – 2)
(c) H( ω ) = 6 cos 2 ω
* An asterisk indicates a challenging problem.
Chapter 18, Solution 29.
(a)
(b)
f(t) = F -1 [δ(ω)] + F -1 [4δ(ω + 3) + 4δ(ω − 3)]
1 4 cos 3t
1
(1 + 8 cos 3t )
=
+
=
2π
π
2π
If h ( t ) = u ( t + 2) − u ( t − 2)
H(ω) =
2 sin 2ω
ω
G (ω) = 4H(ω)
g(t) =
(c)
g( t ) =
1 8 sin 2 t
⋅
2π
t
4 sin 2t
πt
Since
cos(at)
πδ(ω + a ) + πδ(ω − a )
Using the reversal property,
2π cos 2ω ↔ πδ( t + 2) + πδ( t − 2)
or F -1 [6 cos 2ω] = 3δ(t + 2) + 3δ(t − 2)
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Chapter 18, Problem 30.
For a linear system with input x(t) and output y(t) find the impulse response for the
following cases:
(a) x(t) = e − at u(t),
y(t) = u(t) – u( – t)
(b) x(t) = e − t u(t),
y(t) = e −2t u(t)
(c) x(t) = δ (t),
y(t) = e − at sin btu(t)
Chapter 18, Solution 30.
(a)
2
,
jω
Y(ω) 2(a + jω)
2a
H(ω) =
=
= 2+
X(ω)
jω
jω
y( t ) = sgn( t )
(b) X(ω) =
1
,
1 + jω
H(ω) =
⎯
⎯→
Y(ω) =
Y(ω) =
X(ω) =
⎯
⎯→
1
a + jω
h ( t ) = 2δ( t ) + a[u ( t ) − u (− t )]
1
2 + jω
1 + jω
1
= 1−
2 + jω
2 + jω
(c) In this case, by definition, h ( t ) =
⎯
⎯→
h ( t ) = δ( t ) − e − 2 t u ( t )
y( t ) = e −at sin bt u ( t )
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Chapter 18, Problem 31.
Given a linear system with output y(t) and impulse response h(t), find the corresponding
input x(t) for the following cases:
(a) y(t) = te − at u(t),
h(t) = e − at u(t)
h(t) = δ (t)
(b) y(t) = u(t + 1) - u(t – 1),
(c) y(t) = e − at u(t),
h(t) = sgn(t)
Chapter 18, Solution 31.
(a)
Y(ω) =
1
(a + jω) 2
X(ω) =
,
H(ω) =
1
a + jω
Y(ω)
1
=
H(ω) a + jω
⎯
⎯→
x ( t ) = e − at u ( t )
(b)
By definition, x ( t ) = y( t ) = u ( t + 1) − u ( t − 1)
(c )
Y(ω) =
X(ω) =
1
(a + jω)
,
H(ω) =
Y(ω)
jω
1
a
=
= −
H(ω) 2(a + jω) 2 2(a + jω)
2
jω
⎯
⎯→
x(t) =
1
a
δ( t ) − e − at u ( t )
2
2
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Chapter 18, Problem 32.
* Determine the functions corresponding to the following Fourier transforms:
e jw
(a) F 1 ( ω ) =
− jω + 1
1
(c) F 3 ( ω ) =
(1 + ω 2 ) 2
(b) F 2 ( ω ) = 2e
(d) F 4 ( ω ) =
ω
δ (ω )
1 + j 2ω
* An asterisk indicates a challenging problem.
Chapter 18, Solution 32.
e − jω
(a)
Since
jω + 1
and F(− ω)
e − ( t −1) u ( t − 1)
f(-t)
jω
e
f 1 (t ) = e − (− t −1) u (− t − 1)
− jω + 1
f1(t) = e (t +1 )u(− t − 1)
F1 (ω) =
(b)
From Section 17.3,
2
−ω
2πe
t +1
−ω
If F2 (ω) = 2e , then
2
f2(t) =
2
π t +1
2
(
(d)
)
By partial fractions
F3 (ω) =
1
( jω + 1) ( jω − 1)2
2
(
1
1
1
1
4
4
=
+ 4 +
− 4
( jω + 1)2 ( jω + 1) ( jω − 1)2 jω − 1
)
1 −t
te + e − t + te t − e t u (t )
4
1
1
= (t + 1)e − t u(t ) + (t − 1)e t u(t )
4
4
Hence f 3 (t ) =
(d)
f 4 (t ) =
1
1 ∞
1 ∞ δ(ω)e jωt
jωt
(
)
F
e
d
ω
ω
=
=
1
∫
∫
2π − ∞
2π − ∞ 1 + j2ω
2π
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Chapter 18, Problem 33.
* Find f(t) if:
(a) F( ω ) = 2sin πω[u (ω + 1) − u (ω − 1)]
1
j
(b) F( ω ) = (sin 2 ω – sin ω ) + (cos 2 ω – cos ω )
ω
ω
* An asterisk indicates a challenging problem.
Chapter 18, Solution 33.
(a)
Let x (t ) = 2 sin πt[u (t + 1) − u (t − 1)]
From Problem 17.9(b),
4 jπ sin ω
π 2 − ω2
Applying duality property,
X(ω) =
2 j sin (− t )
1
X(− t ) = 2 2
2π
π −t
2 j sin t
f(t) = 2
t − π2
f (t ) =
(b)
j
(cos 2ω − j sin 2ω) − j (cos ω − j sin ω)
ω
ω
− jω
j
e
e j2 ω
= e j 2 ω − e − jω =
−
ω
jω
jω
1
1
f (t ) = sgn (t − 1) − sgn (t − 2)
2
2
But sgn( t ) = 2u ( t ) − 1
1
1
f (t ) = u (t − 1) − − u (t − 2 ) +
2
2
= u(t − 1) − u(t − 2 )
F(ω) =
(
)
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Chapter 18, Problem 34.
Determine the signal f(t) whose Fourier transform is shown in Fig. 18.38. (Hint: Use the
duality property.)
Figure 18.38
For Prob. 18.34.
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Chapter 18, Solution 34.
First, we find G(ω) for g(t) shown below.
g (t ) = 10[u (t + 2 ) − u (t − 2 )] + 10[u (t + 1) − u (t − 1)]
g ' (t ) = 10[δ(t + 2 ) − δ(t − 2 )] + 10[δ(t + 1) − δ(t − 1)]
The Fourier transform of each term gives
g(t)
20
10
–2
–1
0
1
t
2
g ‘(t)
10δ(t+2)
–2
10δ(t+1)
–1
0
1
–10δ(t-1)
(
)
(
2
t
–10δ(t-2)
)
jωG (ω) = 10 e jω2 − e − jω2 + 10 e jω − e − jω
= 20 j sin 2ω + 20 j sin ω
20 sin 2ω 20 sin ω
+
= 40 sinc(2ω) + 20 sinc(ω)
G (ω) =
ω
ω
Note that G(ω) = G(-ω).
F(ω) = 2πG (− ω)
1
G (t )
f (t ) =
2π
= (20/π)sinc(2t) + (10/π)sinc(t)
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Chapter 18, Problem 35.
A signal f(t) has Fourier transform
1
2 + jω
F( ω ) =
Determine the Fourier transform of the following signals:
(a) x(t) = f(3t – 1)
(b) y(t) = f(t) cos 5t
(c) z(t) =
d
f(t)
dt
(d) h(t) = f(t) * f(t)
(e) i(t) = tf(t)
Chapter 18, Solution 35.
(a)
x(t) = f[3(t-1/3)]. Using the scaling and time shifting properties,
e − jω / 3
1
1
e − jω / 3 =
(6 + jω)
3 2 + jω / 3
X(ω) =
(b)
Using the modulation property,
⎤
1⎡
1
1
1
+
Y(ω) = [F(ω + 5) + F(ω − 5)] = ⎢
2 ⎣ 2 + j(ω + 5) 2 + j(ω − 5) ⎥⎦
2
jω
2 + jω
(c )
Z(ω) = jωF(ω) =
(d)
H(ω) = F(ω)F(ω) =
(e)
I(ω) = j
1
(2 + jω) 2
(0 − j)
1
d
=
F(ω) = j
dω
(2 + jω) 2 (2 + jω) 2
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Chapter 18, Problem 36.
The transfer function of a circuit is
H( ω ) =
2
jω + 2
If the input signal to the circuit is v s (t) = e −4t u(t) V find the output signal. Assume all
initial conditions are zero.
Chapter 18, Solution 36.
H (ω ) =
Y (ω )
X (ω )
⎯⎯
→
x(t ) = vs (t ) = e −4t u (t )
Y (ω ) =
Y (ω ) = H (ω ) X (ω )
⎯⎯
→
X (ω ) =
2
2
=
,
( jω + 2)(4 + jω ) ( s + 2)( s + 4)
1
4 + jω
s = jω
A
B
+
s+2 s+4
2
2
A=
= 1,
B=
= −1
−2 + 4
−4 + 2
1
1
Y (s) =
−
s+2 s+4
y (t ) = ( e −2t − e−4t ) u (t )
Y (s) =
Please note, the units are not known since the transfer function does not give
them. If the transfer function was a voltage gain then the units on y(t) would be
volts.
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Chapter 18, Problem 37.
Find the transfer function Io( ω )/I s ( ω ) for the circuit in Fig. 18.39.
Figure 18.39
For Prob. 18.37.
Chapter 18, Solution 37.
2 jω =
j2ω
2 + jω
By current division,
j2ω
I (ω)
j2ω
2 + jω
=
=
H(ω) = o
j2ω
j2ω + 8 + j4ω
I s (ω)
4+
2 + jω
jω
H(ω) =
4 + j3ω
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Chapter 18, Problem 38.
Suppose v s (t) = u(t) for t > 0. Determine i(t) in the circuit of Fig. 18.40, using the Fourier
transform.
Figure 18.40
For Prob. 18.38.
Chapter 18, Solution 38.
1
Vs = πδ (ω ) +
jω
Vs
1 ⎛
1 ⎞
I (ω ) =
=
⎜ πδ (ω ) +
⎟
jω ⎠
1 + jω 1 + jω ⎝
πδ (ω )
1
+
Let I (ω ) = I1 (ω ) + I 2 (ω ) =
1 + jω jω (1 + jω )
1
A
B
I 2 (ω ) =
= +
s = jω
,
jω (1 + jω ) s s + 1
1
where A = = 1,
1
B=
1
−1
+
jω jω + 1
πδ (ω )
I1 (ω ) =
1 + jω
I 2 (ω ) =
i1 (t ) =
1
2π
1
= −1
−1
1
⎯⎯
→ i2 (t ) = sgn(t ) − e − t
2
πδ (ω ) jωt
1 e jωt
1
e
d
ω
=
=
∫−∞ 1 + jω
2 1 + jω ω = 0 2
∞
Hence,
i (t ) = i1 (t ) + i2 (t ) =
1 1
+ sgn(t ) − e− t
2 2
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Chapter 18, Problem 39.
Given the circuit in Fig. 18.41, with its excitation, determine the Fourier transform of i(t).
Figure 18.41
For Prob. 18.39.
Chapter 18, Solution 39.
∞
Vs (ω) =
∫ (1 − t )e
−∞
I(ω) =
Vs (ω)
10 3 + jωx10 − 3
=
− jωt
dt =
1
1
1 − jω
+
−
e
2
jω ω
ω2
10 3
⎛ 1
1
1 − jω ⎞
⎜⎜ +
⎟⎟
e
−
10 6 + jω ⎝ jω ω2 ω 2
⎠
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Chapter 18, Problem 40.
Determine the current i(t) in the circuit of Fig. 18.42(b), given the voltage source shown
in Fig. 18.42(a).
Figure 18.42
For Prob. 18.40.
Chapter 18, Solution 40.
&v&( t ) = δ( t ) − 2δ( t − 1) + δ( t − 2)
− ω 2 V(ω) = 1 − 2e − jω + e jω2
Now
1 − 2e − jω + e − jω2
V(ω) =
− ω2
1 1 + j2ω
Z(ω) = 2 +
=
jω
jω
V(ω) 2e jω − e jω2 − 1
jω
=
⋅
2
Z(ω)
1 + j2ω
ω
1
=
0.5 + 0.5e − jω2 −e − jω
jω(0.5 + jω)
1
A
B
But
= +
A = 2, B = -2
s(s + 0.5) s s + 0.5
2
2
I(ω) =
0.5 + 0.5e jω2 − e − jω −
0.5 + 0.5e − jω2 − e − jω
jω
0.5 + jω
1
1
i(t) = sgn( t ) + sgn(t − 2) − sgn( t − 1) − e − 0.5t u(t ) − e − 0.5( t − 2 ) u(t − 2) − 2e − 0.5( t −1) u(t − 1)
2
2
I=
(
(
)
)
(
)
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Chapter 18, Problem 41.
Determine the Fourier transform of v(t) in the circuit shown in Fig. 18.43.
Figure 18.43
For Prob. 18.41.
Chapter 18, Solution 41.
2
+
−
1
2 + jω
+
1/s
0.5s
V
V−
(
1
2V
2 + jω
+ jω V +
−2=0
2
jω
)
jω
− 4ω 2 + j9ω
jω − 2ω + 4 V = j4ω +
=
2 + jω
2 + jω
2
V(ω) =
2 jω(4.5 + j2ω)
(2 + jω)(4 − 2ω 2 + jω)
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Chapter 18, Problem 42.
Obtain the current io(t) in the circuit of Fig. 18.44.
(a) Let i(t) = sgn(t) A.
(b) Let i(t) = 4[u(t) – u(t – 1)] A.
Figure 18.44
For Prob. 18.42.
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Chapter 18, Solution 42.
By current division, I o =
(a)
2
⋅ I(ω)
2 + jω
For i(t) = 5 sgn (t),
10
I(ω) =
jω
2
10
20
Io =
⋅
=
2 + jω jω jω(2 + jω)
20
A
B
Let I o =
= +
, A = 10, B = −10
s(s + 2) s s + 2
10
10
I o (ω) =
−
j ω 2 + jω
io(t) = 5 sgn( t ) − 10e −2 t u(t )A
i(t)
(b)
i’(t)
4
4δ(t)
1
1
t
t
–4δ(t–1)
i' ( t ) = 4δ( t ) − 4δ( t − 1)
jω I(ω) = 4 − 4e − jω
(
4 1 − e − jω
I(ω) =
jω
(
)
)
⎛ 1
8 1 − e − jω
1 ⎞
⎟⎟ 1 − e − jω
= 4⎜⎜
−
jω(2 + jω)
⎝ j ω 2 + jω ⎠
4
4
4e − j ω 4e − j ω
=
−
−
+
jω 2 + jω
jω
2 + jω
io(t) = 2 sgn(t ) − 2 sgn(t − 1) − 4e −2 t u(t ) + 4e −2( t −1) u(t − 1)A
Io =
(
)
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Chapter 18, Problem 43.
Find vo(t) in the circuit of Fig. 18.45, where i s = 5e − t u(t) A.
Figure 18.45
For Prob. 18.43.
Chapter 18, Solution 43.
20 mF
⎯
⎯→
Vo =
Vo =
1
1
50
,
=
=
3
−
jωC j20x10 ω jω
50
250
40
Is •
=
,
50
jω (s + 1)(s + 1.25)
40 +
jω
i s = 5e − t
⎯
⎯→
Is =
5
1 + jω
s = jω
A
B
1 ⎤
⎡ 1
+
= 1000⎢
−
s + 1 s + 1.25
⎣ s + 1 s + 1.25 ⎥⎦
v o ( t ) = 1000(e −1t − e −1.25t )u ( t )V
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Chapter 18, Problem 44.
If the rectangular pulse in Fig. 18.46(a) is applied to the circuit in Fig. 18.46(b), find vo at
t = 1 s.
Figure 18.46
For Prob. 18.44.
Chapter 18, Solution 44.
1H
jω
We transform the voltage source to a current source as shown in Fig. (a) and then
combine the two parallel 2Ω resistors, as shown in Fig. (b).
Io
Vs/2
+
Io
Vs/2
+
jω
Vo
(a)
jω
Vo
(b)
V
1
⋅ s
1 + jω 2
jω Vs
Vo = jω I o =
2(1 + jω)
&v& s ( t ) = 10δ(t ) − 10δ( t − 2)
2 2 = 1Ω, I o =
jω Vs (ω) = 10 − 10e − j2 ω
Vs (ω) =
10(1 − e − j2ω )
jω
(
)
5 1 − e − j2 ω
5
5
=
−
e − j2 ω
1 + jω
1 + jω 1 + jω
v o ( t ) = 5e − t u ( t ) − 5e − ( t − 2) u ( t − 2)
Hence Vo =
v o (1) = 5e −1 − 0 = 1.839 V
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Chapter 18, Problem 45.
Use the Fourier transform to find i(t) in the circuit of Fig. 18.47 if v s (t) = 10e −2t u(t).
Figure 18.47
For Prob. 18.45.
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Chapter 18, Solution 45.
We may convert the voltage source to a current source as shown below.
1H
vs/2
2Ω
2Ω
Combining the two 2-Ω resistors gives 1 Ω. The circuit now becomes that
shown below.
I
1H
vs/2
1Ω
1 Vs
1
5
5
,
=
=
s = jω
1 + jω 2 1 + jω 2 + jω ( s + 1)( s + 2)
A
B
=
+
s +1 s + 2
B = 5 / − 1 = −5
where A = 5 /1 = 5,
5
5
I=
−
s +1 s + 2
i (t ) = 5(e − t − e −2t )u (t ) A
I=
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Chapter 18, Problem 46.
Determine the Fourier transform of io(t) in the circuit of Fig. 18.48.
Figure 18.48
For Prob. 18.46.
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Chapter 18, Solution 46.
1
F
4
1
jω
2H
3δ( t )
1
4
=
− j4
ω
jω2
3
1
1 + jω
The circuit in the frequency domain is shown below:
e − t u(t)
2Ω
Io(ω)
–j4/ω
+
−
+
−
j2ω
At node Vo, KCL gives
1
− Vo
3 − Vo
V
1 + jω
+
= o
−
j
4
2
j2ω
ω
2
j2Vo
− 2Vo + jω3 − jωVo = −
ω
1 + jω
2
+ jω3
1 + jω
Vo =
j2
2 + jω −
ω
2 + jω3 − 3ω 2
V
1 + jω
I o (ω) = o =
j2 ⎞
j2ω
⎛
j2ω⎜ 2 + jω − ⎟
ω⎠
⎝
Io(ω) =
2 + jω 2 − 3ω 2
4 − 6ω 2 + j(8ω − 2ω 3 )
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Chapter 18, Problem 47.
Find the voltage vo(t) in the circuit of Fig. 18.49. Let i s (t) = 8e −t u(t) A.
Figure 18.49
For Prob. 18.47.
Chapter 18, Solution 47.
1
F
2
⎯⎯
→
Io =
1
2
=
jωC jω
1
I
2 s
1+
jω
2
2
2
8
jω
Io =
Vo =
Is =
2
jω
2 + jω 1 + jω
1+
jω
=
16
, s = jω
(s + 1)(s + 2)
=
A
B
+
s +1 s + 2
where A = 16/1 = 16, B = 16/(–1) = –16
Thus,
vo(t) = 16(e–t – e–2t)u(t) V.
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Chapter 18, Problem 48.
Find io(t) in the op amp circuit of Fig. 18.50.
Figure 18.50
For Prob. 18.48.
Chapter 18, Solution 48.
0.2F
1
j5
=−
jωC
ω
As an integrator,
RC = 20 x 10 3 x 20 x 10 −6 = 0.4
1 t
v i dt
RC ∫o
⎤
1 ⎡ Vi
Vo = −
+ πVi (0)δ(ω)⎥
⎢
RC ⎣ jω
⎦
vo = −
=−
Io =
⎤
1 ⎡
2
+ πδ (ω)⎥
⎢ (
0 .4 ⎣ j ω 2 + j ω )
⎦
⎡
⎤
Vo
2
mA = −0.125 ⎢
+ π δ (ω)⎥
20
⎣ jω (2 + jω)
⎦
0.125 0.125
+
− 0.125πδ (ω)
jω
2 + jω
0.125
i o ( t ) = −0.125 sgn( t ) + 0.125e − 2 t u (t ) −
πδ (ω)e jωt dt
∫
2π
0.125
= 0.125 + 0.25u ( t ) + 0.125e − 2 t u ( t ) −
2
−2 t
io(t) = 0.625 − 0.25u(t ) + 0.125e u(t ) mA
=−
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Chapter 18, Problem 49.
Use the Fourier transform method to obtain vo(t) in the circuit of Fig. 18.51.
Figure 18.51
For Prob. 18.49.
Chapter 18, Solution 49.
Consider the circuit shown below:
jω
j2ω
+
−
i1
jω
i2
+
vo
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Vs = π[δ (ω + 1) + δ (ω − 2)]
For mesh 1, − Vs + (2 + j2ω)I1 − 2I 2 − jωI 2 = 0
Vs = 2 (1 + jω) I1 − (2 + jω)I 2
For mesh 2, 0 = (3 + jω)I 2 − 2I1 − jωI1
(3 + ω)I 2
I1 =
(2 + ω)
(1)
(2)
Substituting (2) into (1) gives
2 (1 + jω)(3 + jω)I 2
− (2 + jω)I 2
2 + jω
Vs (2 + ω) = 2 3 + j4ω − ω 2 − 4 + j4ω − ω 2 I 2
Vs = 2
[(
= I 2 (2 + j4ω − ω )
(s + 2)Vs
I2 = 2
, s = jω
s + 4s + 2
) (
)]
2
Vo = I 2 =
( jω + 2) π [δ (ω + 1) + δ (ω − 1)]
( jω)2 + jω4 + 2
1 ∞
v o (ω)e jωt dω
∫
−
∞
2π
1
1
jωt
(
)
(
)
( jω + 2)e jωt δ(ω − 1)dω
j
ω
+
2
e
δ
ω
+
1
d
ω
∞
=∫ 2
+2
2
−∞
( jω) + jω4 + 2
( jω)2 + jω4 + 2
1
(− j + 2)e jt 1 ( j + 2)e jt
+ 2
= 2
− 1 − j4 + 2 − 1 + j4 + 2
1
1
(2 − j)(1 + j4)
(2 − j)(1 − j4)e jt
v o (t) = 2
e jt + 2
17
17
1
(6 + j7 )e jt + 1 (6 − j7 )e jt
=
34
34
− j ( t −13.64° )
= 0.271e
+ 0.271e j ( t −13.64° )
vo(t) = 0.542 cos(t − 13.64°)V
v o (t) =
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Chapter 18, Problem 50.
Determine vo(t) in the transformer circuit of Fig. 18.52.
Figure 18.52
For Prob. 18.50.
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Chapter 18, Solution 50.
Consider the circuit shown below:
j0.5ω
1Ω
+
−
i1
i2
jω
+
jω
vo
For loop 1,
For loop 2,
− 2 + (1 + jω)I1 + j0.5ωI 2 = 0
(1)
(1 + jω)I 2 + j0.5ωI1 = 0
(2)
From (2),
I1 =
(1 + jω)I 2
= −2
(1 + jω)I 2
− j0.5ω
jω
Substituting this into (1),
− 2(1 + jω)I 2 jω
2=
+
I2
jω
2
3 ⎞
⎛
2 jω = −⎜ 4 + j4ω − ω 2 ⎟I 2
2 ⎠
⎝
2 jω
I2 =
4 + j4ω − 1.5ω 2
− 2 jω
Vo = I 2 =
2
4 + j4ω + 1.5( jω)
4
jω
3
Vo =
8ω
8
2
+j
+ ( jω )
3
3
⎛4
⎞
− 4⎜ + jω ⎟
⎝3
⎠
=
+
2
⎛4
⎞ ⎛⎜ 8 ⎞⎟
⎜ + jω ⎟ + ⎜
⎝3
⎠ ⎝ 3 ⎟⎠
2
16
3
2
⎛4
⎞ ⎛⎜ 8 ⎞⎟
⎜ + jω ⎟ + ⎜
⎝3
⎠ ⎝ 3 ⎟⎠
2
⎛ 8 ⎞
⎛ 8 ⎞
Vo ( t ) = − 4e − 4t / 3 cos⎜⎜
t ⎟⎟ u(t ) + 5.657e − 4t / 3 sin⎜⎜
t ⎟⎟u(t ) V
⎝ 3 ⎠
⎝ 3 ⎠
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Chapter 18, Problem 51.
Find the energy dissipated by the resistor in the circuit of Fig. 18.53.
Figure 18.53
For Prob. 18.51.
Chapter 18, Solution 51.
In the frequency domain, the voltage across the 2-Ω resistor is
2
2
10
20
,
V (ω ) =
Vs =
=
s = jω
2 + jω
2 + jω 1 + jω ( s + 1)( s + 2)
A
B
V ( s) =
+
s +1 s + 2
A = 20 /1 = 20,
B = 20 / − 1 = −20
V (ω ) =
20
20
−
jω + 1 jω + 2
v(t ) = ( 20e − t − 20e −2t ) u (t )
W=
(
)
1 ∞ 2
v ( t )dt = 0.5∫ 400 e − 2 t + e − 4 t − 3e − 3t dt
2 ∫0
⎛ e − 2 t e − 4 t 2e − 3 t
= 200⎜
+
−
⎜ −2
−
−3
4
⎝
∞
⎞
⎟ = 16.667 J.
⎟
⎠0
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Chapter 18, Problem 52.
For F( ω ) =
1
, find J =
3 + jω
∫
∞
−∞
f 2 (t )dt .
Chapter 18, Solution 52.
∞
J = 2 ∫ f 2 ( t ) dt =
0
1 ∞
2
F(ω) dω
∫
π 0
∞
1 ∞
1
1
1 π
dω =
tan −1 (ω / 3) =
= (1/6)
=
2
2
∫
0
3π
3π 2
π 9 +ω
0
Chapter 18, Problem 53.
If f(t) = e
−2 t
, find J =
∫
∞
−∞
2
F (ω ) dω .
Chapter 18, Solution 53.
If f(t) = e-2|t|, find J = ∫
∞
−∞
J =
∞
∫− ∞ F(ω)
2
2
F (ω ) d ω .
dω = 2π∫
∞
−∞
f(t) =
t<0
e 2t ,
e
−2 t
f 2 ( t ) dt
,
t>0
⎡ 4t 0
−4 t ∞ ⎤
∞ −4 t ⎤
0 4t
e
e
⎡
⎥ = 2π[(1/4) + (1/4)] = π
J = 2π ⎢ ∫ e dt + ∫ e dt ⎥ = 2π⎢
+
0
⎢ 4
−4 ⎥
⎣ −∞
⎦
−∞
0 ⎥⎦
⎣⎢
Chapter 18, Problem 54.
Given the signal f(t) = 4e −t u(t) what is the total energy in f(t)?
Chapter 18, Solution 54.
W1Ω =
∫
∞
−∞
∞
∞
0
0
f 2 ( t ) dt = 16 ∫ e − 2 t dt = − 8e − 2 t
= 8J
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Chapter 18, Problem 55.
Let f(t) = 5e − (t − 2) u(t) and use it to find the total energy in f(t).
Chapter 18, Solution 55.
f(t) = 5e2e–tu(t)
F(ω) = 5e2/(1 + jω), |F(ω)|2 = 25e4/(1 + ω2)
W1Ω
1 ∞
25e 4
2
=
F
(
ω
)
d
ω
=
π ∫0
π
∫
∞
0
∞
1
25e 4
d
ω
=
tan −1 (ω)
π
1 + ω2
0
= 12.5e4 = 682.5 J
W1Ω =
or
∫
∞
−∞
∞
f 2 ( t ) dt = 25e 4 ∫ e − 2 t dt = 12.5e4 = 682.5 J
0
Chapter 18, Problem 56.
The voltage across a 1- Ω resistor is v(t) = te −2t u(t) V. (a) What is the total energy
absorbed by the resistor? (b) What fraction of this energy absorbed is in the frequency
band –2 ≤ ω ≤ 2?
Chapter 18, Solution 56.
∞ 2
e −4t
2
(16
8
2)
t
t
(a) W = ∫ V (t )dt = ∫ t e dt =
+
+
=
= 0.0313 J
0 64
(−4)3
−∞
0
(b) In the frequency domain,
1
V (ω ) =
(2 + jω ) 2
∞
∞
2
2 −4 t
1
(4 + jω ) 2
2
2
1
2
1
2
|
(
)
|
Wo =
V
ω
d
ω
=
dω
∫
∫
2π −2
2π 0 (4 + ω 2 ) 2
| V (ω ) |2 = V (ω )V * (ω ) =
⎞
1 1 ⎛ ω
⎜⎜
+ 0.5 tan −1 (0.5ω)⎟⎟
=
π 2x 4 ⎝ ω2 + 4
⎠
Fraction =
2
=
0
1
1
+
= 0.0256
32π 64
Wo 0.0256
=
= 81.79%
W 0.0313
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Chapter 18, Problem 57.
Let i(t) = 2e t u(–t)A. Find the total energy carried by i(t) and the percentage of the 1- Ω
energy in the frequency range of –5 < ω < 5 rad/s.
Chapter 18, Solution 57.
W1Ω =
or
∫
∞
−∞
0
0
−∞
−∞
i 2 ( t ) dt = ∫ 4e 2 t dt = 2e 2 t
= 2J
I(ω) = 2/(1 – jω), |I(ω)|2 = 4/(1 + ω2)
∞
W1Ω
1 ∞
4 ∞
1
4
4π
2
I(ω) dω =
dω = tan −1 (ω) =
=
= 2J
2
∫
∫
2π − ∞
2π −∞ (1 + ω )
π
π2
0
In the frequency range, –5 < ω < 5,
5
4
4
4
W =
tan −1 ω = tan −1 (5) = (1.373) = 1.7487
π
π
π
0
W/ W1Ω = 1.7487/2 = 0.8743 or 87.43%
Chapter 18, Problem 58.
An AM signal is specified by
f(t) = 10(1 + 4 cos 200 π t)cos π × 10 4 t
Determine the following:
(a) the carrier frequency,
(b) the lower sideband frequency,
(c) the upper sideband frequency.
Chapter 18, Solution 58.
ωm = 200π = 2πfm which leads to fm = 100 Hz
(a)
ωc = πx104 = 2πfc which leads to fc = 104/2 = 5 kHz
(b)
lsb = fc – fm = 5,000 – 100 = 4,900 Hz
(c)
usb = fc + fm = 5,000 + 100 = 5,100 Hz
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Chapter 18, Problem 59.
For the linear system in Fig. 18.54, when the input voltage is v i (t) = 2 δ (t) V, the output
is v 0 (t) = 10e −2t – 6e −4t V. Find the output when the input is v i (t) = 4e − t u(t) V.
Figure 18.54
For Prob. 18.9.
Chapter 18, Solution 59.
10
6
−
V (ω) 2 + jω 4 + jω
5
3
H(ω) = o
=
=
−
Vi (ω)
2
2 + jω 4 + jω
⎛ 5
3 ⎞ 4
⎟⎟
Vo (ω) = H(ω)Vi (ω) = ⎜⎜
−
⎝ 2 + jω 4 + jω ⎠ 1 + jω
20
12
=
−
, s = jω
(s + 1)(s + 2) (s + 1)(s + 4)
Using partial fraction,
Vo (ω) =
A
B
C
D
16
20
4
+
+
+
=
−
+
s + 1 s + 2 s + 1 s + 4 1 + jω 2 + jω 4 + jω
Thus,
(
)
v o ( t ) = 16e − t − 20e −2 t + 4e −4 t u ( t ) V
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Chapter 18, Problem 60.
A band-limited signal has the following Fourier series representation:
i s (t) = 10 + 8 cos(2 π t + 30º) + 5 cos(4 π t – 150º)mA
If the signal is applied to the circuit in Fig. 18.55, find v(t).
Figure 18.55
For Prob. 18.60.
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Chapter 18, Solution 60.
2
1/jω
jω
+
V
V = jωI s
1
jω
1
+ 2 + jω
jω
=
jωI s
1 − ω 2 + j2ω
Since the voltage appears across the inductor, there is no DC component.
V1 =
2π∠90°8
1 − 4π 2 + j4π
V2 =
=
50.27∠90°
= 1.2418∠ − 71.92°
− 38.48 + j12.566
4π∠90°5
1 − 16π 2 + j8π
=
62.83∠90°
= 0.3954∠ − 80.9°
− 156.91 + j25.13
v( t ) = 1.2418 cos(2πt − 41.92°) + 0.3954 cos(4πt + 129.1°) mV
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Chapter 18, Problem 61.
In a system, the input signal x(t) is amplitude-modulated by m(t) = 2 + cos ω 0 t. The
response y(t) = m(t)x(t). Find Y( ω ) in terms of X( ω ).
Chapter 18, Solution 61.
y (t ) = (2 + cos ωo t ) x(t )
We apply the Fourier Transform
Y(ω) = 2X(ω) + 0.5X(ω+ωo) + 0.5X(ω–ωo).
Chapter 18, Problem 62.
A voice signal occupying the frequency band of 0.4 to 3.5 kHz is used to amplitudemodulate a 10-MHz carrier. Determine the range of frequencies for the lower and upper
sidebands.
Chapter 18, Solution 62.
For the lower sideband, the frequencies range from
10,000,000 – 3,500 Hz = 9,996,500 Hz to
10,000,000 – 400 Hz = 9,999,600 Hz
For the upper sideband, the frequencies range from
10,000,000 + 400 Hz = 10,000,400 Hz to
10,000,000 + 3,500 Hz = 10,003,500 Hz
Chapter 18, Problem 63.
For a given locality, calculate the number of stations allowable in the AM broadcasting
band (540 to 1600 kHz) without interference with one another.
Chapter 18, Solution 63.
Since fn = 5 kHz, 2fn = 10 kHz
i.e. the stations must be spaced 10 kHz apart to avoid interference.
∆f = 1600 – 540 = 1060 kHz
The number of stations = ∆f /10 kHz = 106 stations
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Chapter 18, Problem 64.
Repeat the previous problem for the FM broadcasting band (88 to 108 MHz), assuming
that the carrier frequencies are spaced 200 kHz apart.
Chapter 18, Solution 64.
∆f = 108 – 88 MHz = 20 MHz
The number of stations = 20 MHz/0.2 MHz = 100 stations
Chapter 18, Problem 65.
The highest-frequency component of a voice signal is 3.4 kHz. What is the Nyquist rate
of the sampler of the voice signal?
Chapter 18, Solution 65.
ω = 3.4 kHz
fs = 2ω = 6.8 kHz
Chapter 18, Problem 66.
A TV signal is band-limited to 4.5 MHz. If samples are to be reconstructed at a distant
point, what is the maximum sampling interval allowable?
Chapter 18, Solution 66.
ω = 4.5 MHz
fc = 2ω = 9 MHz
Ts = 1/fc = 1/(9x106) = 1.11x10–7 = 111 ns
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Chapter 18, Problem 67.
* Given a signal g(t) = sinc(200 π t) find the Nyquist rate and the Nyquist interval for the
signal.
* An asterisk indicates a challenging problem.
Chapter 18, Solution 67.
We first find the Fourier transform of g(t). We use the results of Example 17.2 in
conjunction with the duality property. Let Arect(t) be a rectangular pulse of height A and
width T as shown below.
Arect(t) transforms to Atsinc(ω2/2)
F(ω)
f(t)
A
ω
t
–T/2
T/2
G(ω)
ω
–ωm/2
ωm/2
According to the duality property,
Aτsinc(τt/2)
becomes 2πArect(τ)
g(t) = sinc(200πt) becomes 2πArect(τ)
where Aτ = 1 and τ/2 = 200π or T = 400π
i.e. the upper frequency ωu = 400π = 2πfu or fu = 200 Hz
The Nyquist rate = fs = 200 Hz
The Nyquist interval = 1/fs = 1/200 = 5 ms
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Chapter 18, Problem 68.
−2 t
The voltage signal at the input of a filter is v(t) = 50e V What percentage of the total 1Ω energy content lies in the frequency range of 1 < ω < 5 rad/s?
Chapter 18, Solution 68.
The total energy is
WT =
∫
∞
−∞
v 2 ( t ) dt
Since v(t) is an even function,
WT =
∫
∞
0
2500e
−4 t
e −4 t
dt = 5000
−4
∞
= 1250 J
0
V(ω) = 50x4/(4 + ω2)
1 5
1 5 (200) 2
2
| V(ω) | dω =
dω
W =
2π ∫1
2π ∫1 (4 + ω 2 ) 2
But
∫ (a
2
1
1 ⎡ x
1
⎤
dx = 2 ⎢ 2
+ tan −1 ( x / a )⎥ + C
2 2
2
a
+x )
2a ⎣ x + a
⎦
5
2x10 4 1 ⎡ ω
1
⎤
+ tan −1 (ω / 2)⎥
W =
2
⎢
π 8 ⎣4 + ω
2
⎦1
= (2500/π)[(5/29) + 0.5tan-1(5/2) – (1/5) – 0.5tan–1(1/2) = 267.19
W/WT = 267.19/1250 = 0.2137 or 21.37%
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Chapter 18, Problem 69.
A signal with Fourier transform
F( ω ) =
20
4 + jω
is passed through a filter whose cutoff frequency is 2 rad/s (i.e., 0 < ω < 2). What fraction
of the energy in the input signal is contained in the output signal?
Chapter 18, Solution 69.
The total energy is
WT =
=
W =
1 ∞
1 ∞ 400
2
F(ω) dω =
dω
∫
2π − ∞
2π ∫−∞ 4 2 + ω 2
[
400
(1 / 4) tan −1 (ω / 4)
π
]
∞
0
=
100 π
= 50
π 2
[
1 2
400
2
F(ω) dω =
(1 / 4) tan −1 (ω / 4)
∫
0
2π
2π
]
2
0
= [100/(2π)]tan–1(2) = (50/π)(1.107) = 17.6187
W/WT = 17.6187/50 = 0.3524 or 35.24%
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