Chapter 18 solutions_to_exercises(engineering circuit analysis 7th)

Engineering Circuit Analysis, 7th Edition

1.

Chapter Eightenn Solutions

10 March 2006

(a) 5, 10, 15, 20, 25 (all rad/s)
(b) 5, 10, 15, 20, 25 (all rad/s)
(c) 90, 180, 270, 360, 450 (all rad/s)

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2.

(a) ωo = 2π rad/s, f = 1 Hz,

10 March 2006

therefore T = 1 s.

(b) ωo = 5.95 rad/s = 2π f rad/s,
(c) ) ωo = 1 rad/s = 2πf rad/s,


f = 0.947 Hz,
f = 1/2π Hz,

therefore T = 1.056 s.
therefore T = 2π s.




10 March 2006

3.

v(t ) = 3 − 3cos(100π t − 40°) + 4sin(200π t − 10°) + 2.5cos 300π t V
(a)

Vav = 3 − 0 + 0 + 0 = 3.000 V

(b)

1
Veff = 32 + (32 + 42 + 2.52 ) = 4.962 V
2

(c)

T=

(d)

v(18ms ) = 3 − 3cos(−33.52°) + 4sin(2.960°) + 2.5cos(19.440°) = −2.459 V

2π

ωo

=

2π
= 0.02 s
100π




10 March 2006

4. (a)
t

t

v

0

2

0.55

-0.844

0.05

2.96

0.6

0.094

0.1

3.33

0.65

0.536

0.15

2.89

0.7

0.440

0.2

1.676

0.75

0

0.25

0

0.8

-0.440

0.3

-1.676

0.85

-0.536

0.35

-2.89

0.9

-0.094

0.4

-3.33

0.95

0.844

0.45

-2.96

1

2

0.5

(b)

v

-2

v′ = −4π sin 2π t + 7.2π cos 4π t = 0
∴ 4sin 2π t = 7.2(cos 2 2π t − sin 2 2π t )
−4 ± 16 + 414.72
= 0.5817, − 0.8595 = sin 2π t
28.8
= 3.330 (0.5593 for smaller max)

∴ 4sin 2π t = 7.2(1 − 2sin 2 2π t ) ∴ x =
∴ t = 0.09881, 0.83539 ∴ vmax
(c)

vmin = 3.330



5.


10 March 2006

(a) a0 = 0
(b) a0 = 0
(c) a0 = 5
(d) a0 = 5



6.


10 March 2006

(a) a0 = 0
(b) a0 = 0
(c) a0 = 100
(d) a0 = 100



7.


10 March 2006

(a) a0 = 3, a1 = 0, a2 = 0, b1 = 0, b2 = 0
(b) a0 = 3, a1 = 3, a2 = 0, b1 = 0, b2 = 0
(c) a0 = 0, a1 = 0, a2 = 0, b1 = 3, b2 = 3
(d) 3 cos(3t − 10o ) = 3cos 3t cos10o + 3sin 3t sin10o
a0 = 0, a1 = 3cos10o = 2.954, a2 = 0, b1 = 3sin10o = 0.521, b2 = 0




ao =

1
T

b1 =

2 T
2 2
5
10
f (t ) sin ω0tdt =
∫0
∫1 5sin π tdt = − 2π cos π t 1 = − π
T
2−0

b2 =

8.

10 March 2006

2 T
2 2
5
∫0 f (t ) sin 2ω0tdt = 2 − 0 ∫1 5sin 2π tdt = − 2π cos 2π t 1 = 0
T

∫

T

0

f (t )dt = 2.5 . a1 = a2 = 0 since function has odd symmetry
2

2




2

ao =

1
T

a1 =

9.

2 T
2 2
⎛ 2π
f (t ) cos ω0tdt = ∫ 2 cos ⎜
∫0
0
T
3
⎝ 3

∫

T

0

f (t )dt =

1 2
2
∫0 2dt = 3 t 0 =
3

10 March 2006

4
.
3

4⎛ 3
⎞
t ⎟ dt = ⎜
3 ⎝ 2π
⎠

⎞ ⎛ 2π
⎟ sin ⎜
⎠ ⎝ 3

2

⎞
t ⎟ = − 0.551
⎠0
2

2 T
2 2
⎛ 4π
a2 = ∫ f (t ) cos 2ω0tdt = ∫ 2 cos ⎜
T 0
3 0
⎝ 3

⎞
t ⎟ dt =
⎠

4⎛ 3
⎜
3 ⎝ 4π

⎞ ⎛ 4π
⎟ sin ⎜
⎠ ⎝ 3

⎞
t ⎟ = 0.276
⎠0

2 T
2 2
⎛ 6π
a3 = ∫ f (t ) cos 3ω0tdt = ∫ 2 cos ⎜
T 0
3 0
⎝ 3

⎞
t ⎟ dt =
⎠

4⎛ 3
⎜
3 ⎝ 6π

⎞ ⎛ 6π
⎟ sin ⎜
⎠ ⎝ 3

⎞
t⎟ = 0
⎠0

2 T
2 2
⎛ 2π
b1 = ∫ f (t ) sin ω0tdt = ∫ 2sin ⎜
0
0
T
3
⎝ 3

2

2

⎞
⎛4⎞ 3
⎛ 2π ⎞
t ⎟ dt = − ⎜ ⎟
cos ⎜
t ⎟ = 0.955
⎠
⎝ 3 ⎠ 2π
⎝ 3 ⎠0

2 T
2 2
⎛ 4π
b 2 = ∫ f (t ) sin 2ω0tdt = ∫ 2sin ⎜
T 0
3 0
⎝ 3

⎞
⎛4⎞ 3
⎛ 4π
t ⎟ dt = − ⎜ ⎟
cos ⎜
⎠
⎝ 3 ⎠ 4π
⎝ 3

2

⎞
t ⎟ = 0.477
⎠0



10.


10 March 2006

h(t) = –3 + 8 sin πt + f(t)

Use linearity and superposition. T = 2 s.
ao = − 3 +

1 T
1
∫0 f (t )dt = −3 + 2 = − 2.5 .
T

a2 = 0
b1 = 8 +

b2 =

2 T
2 2
2
∫0 f (t ) sin ω0tdt = 8 + 2 ∫0 (1) sin π tdt = 8 − π = 7.36
T

2 T
2 2
∫0 f (t ) sin 2ω0tdt = 2 ∫1 (1) sin 2π tdt = 0
T




11.
(a)

T = 10 s, Fav = ao = 0.1(2 × 4 + 2 × 2) = 1.200

(b)

Feff =

10 March 2006

2

2

1
(4 − t ) 2 dt = 0.2 ∫ (16 − 8t + t 2 ) dt
5∫
0
0

2
2
⎡
2
1 3 ⎤
8⎞
⎛
2
= 0.2 ⎢16t −4t + t ⎥ = 0.2 ⎜ 32 − 16 + ⎟ = 1.9322
0
3 0⎥
3⎠
⎝
⎢
0
⎣
⎦

2
2π t
a3 = × 2 ∫ (4 − t ) cos 3 ×
dt = 0.4 ∫ 4 cos 0.6π t dt − 0.4 ∫ t cos 0.6π t dt
10
10
0
0
0
2

(c)

2

2

2

2

1
t
⎛ 1
⎞
= 1.6
sin 0.6π t −0.4 ⎜
cos 0.6π t +
sin 0.6π t ⎟
2
0.6π
0.6π
⎝ 0.36π
⎠0
0
=

8
10
4
sin1.2π − 2 (cos1.2π − 1) −
sin 1.2π = −0.04581
3π
9π
3π



12.
(a)


10 March 2006

T=8−2=6s
1
Hz
6

(b)

fo =

(c)

ω o = 2π f o =

(d)

1
ao = (10 × 1 + 5 × 1) = 2.5
6

(e)

b2 =

π
3

rad/s

3
4
2⎡
2π t
2π t ⎤
10sin
dt + ∫ 5sin
dt ⎥
⎢∫
6 ⎣2
3
3
3
⎦

3
1 ⎡ 30
2π t
15
2π t
cos
cos
= ⎢−
−
3 ⎢ 2π
3 2 2π
3
⎣

1 ⎡ 15 ⎛
4π
∴ b2 = ⎢ − ⎜ cos 2π − cos
3⎣ π ⎝
3

⎤
⎥
3⎥
⎦
4

8π
7.5
⎞ 7.5 ⎛
⎞ ⎤ 1 ⎡ 15
⎤
− cos 2π ⎟ ⎥ = ⎢ − (1.5) −
(−1.5) ⎥ = −1.1937
⎟−
⎜ cos
3
π
⎠ π ⎝
⎠⎦ 3 ⎣ π
⎦




10 March 2006

13.
a3 =
=

4
3
3
4
⎤
2⎡
6π t
6π t ⎤ 1 ⎡10
5
10 cos
dt + ∫ 5cos
dt ⎥ = ⎢ sin π t − sin π t ⎥
⎢∫
6 ⎣2
6
6
3⎥
2 π
3
⎦ 3⎢π
⎣
⎦

10 ⎛
1
1
⎞
⎜ sin 3π − sin 2π + sin 4π − sin 3π ⎟ = 0
3π ⎝
2
2
⎠

4
3
3
4
⎤
⎤ 1 ⎡ 10
1⎡
5
b3 = ⎢ ∫ 10sin π tdt + ∫ 5sin π t dt ⎥ = ⎢ − cos π t − cos π t ⎥
3 ⎣2
2 π
3⎥
3
⎦ 3⎢ π
⎣
⎦

=−

10 ⎛
1
1
10
⎞
⎜ cos 3π − cos 2π + cos 4π − cos 3π ⎟ = − (−1) = 1.0610
3π ⎝
2
2
3π
⎠

2
a3 + b32 = 1.0610



14.


10 March 2006

2π
= 12.5 ms, ave value = 1.9
160π

(a)

3.8cos 2 80πt = 1.9 + 1.9 cos160πt , T =

(b)

3.8cos3 80πt = (3.8cos80πt )(0.5 + 0.5cos 160πt )
= 1.9 cos80πt + 0.95cos 240πt + 0.95cos80πt = 2.85cos80πt + 0.95cos 240πt
2π
= 25 ms, ave value = 0
T=
80π

(c)

3.8cos 70πt − 3.8sin 80πt; ωot = πt , ωo = π, T =

2π
= 2 s; ave value = 0
π




10 March 2006

15. T = 2 s
1
21
4 × 2πt
1
b4 = ∫ sin
dt = − cos 4πt
20
2
4π
0

t

(a)

t

1
(1 − cos 4πt1 )
4π
π
max when 4πt1 = , t1 = 0.125 s
2

∴ b4 =

(b)

b4 =

1
4π




10 March 2006

16.
g (t ) = 5 + 8cos10t − 5cos15t + 3cos 20t − 8sin10t − 4sin15t + 2sin 20t
2π
= 1.2566 s
5

(a)

ωo = 5 ∴ T =

(b)

fo =

(c)

G av = −5

(d)

1
G eff = (−5) 2 + (82 + 52 + 32 + 82 + 42 + 22 ) = 116 = 10.770
2

5
10
β = 4 fo =
= 3.183 Hz
2π
π

(e)




10 March 2006

17.
T = 0.2, f (t ) = Vm cos 5πt , −0.1 < t < 0.1
0.1

an =

0.1

2
Vm cos 5πt cos10 nπt dt = 5Vm ∫ [ cos(5π + 10nπ)t + cos(10nπ − 5π)t ] dt
0.2 −∫
−0.1
0.1
0.1

1
1
⎡
⎤
= 5Vm ⎢
sin(10nπ + 5π) t +
sin(10nπ − 5π)t ⎥
10nπ − 5π
⎣10nπ + 5π
⎦ −0.1
V ⎡ 2
2
⎤
= m⎢
sin(10nπ + 5π) 0.1 +
sin(10nπ − 5π) 0.1⎥
π ⎣ 2n + 1
2n − 1
⎦
Vm ⎡ 2
2
⎤
⎢ 2n + 1 sin(nπ + 0.5π) + 2n − 1 sin(nπ − 0.5π) ⎥
π ⎣
⎦
V ⎡ 2
2
1 ⎞
⎤ 2V
⎛ 1
cos nπ +
(− cos nπ) ⎥ = m cos nπ ⎜
= m⎢
−
⎟
2n − 1
π ⎣ 2n + 1
π
⎦
⎝ 2n + 1 2n − 1 ⎠
2V
2n − 1 − 2n − 1
4V cos nπ
= m cos nπ
=− m 2
2
4n − 1
π
π 4n − 1
0.1
1
1 ⎡ π
⎛ π ⎞ ⎤ 2V
ao =
∫ Vm cos 5πt dt = 5Vm 5π ⎢sin 2 − sin ⎜ − 2 ⎟⎥ = πm
0.2 −0.1
⎠⎦
⎝
⎣
=

∴ v(t ) =

2Vm 4Vm
4V
4V
4V
+
cos10πt − m cos 20πt + m cos 30πt − m cos 40πt + ...
π
3π
15π
35π
63π




10 March 2006

18.
1
− wave
2

(a)

even,

(b)

bn = 0 for all n; aeven = 0; ao = 0

(c)

b1 = b2 = b3 = 0, a2 = 0
nπt
nπt
nπ
nπ ⎞
8
10 6
20 ⎛
∫ 5cos 6 dt = 3 nπ sin 6 1 = nπ ⎜ sin 3 − sin 6 ⎟
12 1
⎝
⎠
2

an =

∴ a1 =

2

20 ⎛
π
π⎞
20 ⎛
π⎞
20
= −2.122
⎜ sin − sin ⎟ = 2.330, a3 = ⎜ sin π − sin ⎟ = −
π⎝
3
6⎠
3π ⎝
2⎠
3π



19.
(a)


10 March 2006

ao = an = 0
∴ y (t ) = 0.2sin1000πt + 0.6sin 2000πt + 0.4sin 3000πt

(b)

Yeff = 0.5(0.22 + 0.62 + 0.42 ) = 0.5(0.56) = 0.5292

(c)

y (2ms) = 0.2sin 0.2π + 0.6sin 0.4π + 0.4sin 0.6π = 1.0686




10 March 2006

20.
(a)

(b)

(c)

(d)

[a]b5 = 0, a5 =

4
2π5t
32 6
5πt
3.2 ⎛ 15π
10π ⎞
∫ 8cos 6 dt = 6 10π sin 3 2 = π ⎜ sin 3 − sin 3 ⎟ = 0.8821
62
⎝
⎠

[b]a5 = 0, b5 =

4
2π5t
32 ⎛ −6 ⎞ ⎛
15π
10π ⎞
3.2
∫ 8sin 6 dt = 6 ⎜ 10π ⎟ ⎜ cos 3 − cos 3 ⎟ = − π (−0.5) = 0.5093
62
⎝
⎠⎝
⎠

3

3

(e)

3

10π ⎞
8
2π5t
64 12 ⎛ 15π
dt =
− sin
[c]b5 = 0, a5 = ∫ 8cos
⎜ sin
⎟ = 3.801
6
6 ⎠
12 2
12
12 10π ⎝
3

8
10πt
64 ⎛ 12 ⎞⎛
15π
10π ⎞
∫ 8sin 12 dt = 12 ⎜ − 10π ⎟⎜ cos 6 − cos 6 ⎟ = 1.0186
12 2
⎠
⎝
⎠⎝
3

[d ]a5 = 0, b5 =




10 March 2006

21.
T = 4 ms

(a)

1000
ao =
4
=−

0.004

∫
0

0.004

250 × 8
8sin125πt dt =
cos125πt
−125π
0

16 ⎛
π ⎞ 16
⎜ cos − 1⎟ = = 5.093
2 ⎠ π
π⎝
0.004

a1 = 4000

(b)

∫

sin125πt cos

0

2πt
dt
0.004

0.004

∴ a1 = 4000

∫

0.004

sin125πt cos 500πt dt = 2000

0

0.004

=

0.004

∫

(sin 625πt − sin 375πt ) dt

0

⎛ cos 625πt cos 375πt ⎞
= 2000 ⎜ −
+
⎟
625π
375π ⎠0
⎝
b1 = 4000

∫

3.2
5.333
(1 − cos 2.5π) −
(1 − cos1.5π) = −0.6791
π
π
0.004

sin125πt sin 500πt dt = 2000

0

∫

(cos 375πt − cos 625πt ) dt

0

1
1 ⎞
⎡ 1
⎤
⎛ −1
(sin1.5π) −
(sin 2.5π) ⎥ = 2000 ⎜
= 2000 ⎢
−
⎟ = −2.716
625π
⎣ 375π
⎦
⎝ 375π 625π ⎠

(c)

−4 < t < 0 : 8sin125πt

(d)

b1 = 0, a1 =

4000
8
0.004

∴ a1 = 2000

∫
0

=

0.004

∫

8sin125πt cos 250πt dt

0

⎡ cos 375πt cos125πt ⎤
+
[sin 375πt − sin125πt ] dt = 2000 ⎢−
375π
125π ⎥ 0
⎣
⎦

0.004

π
5.333
16
(1 − cos1.5π ) + ⎛ cos − 1⎞ = −3.395+
⎜
⎟
π
π⎝
2 ⎠




10 March 2006

22.
1
− wave ∴ ao = 0, an = 0, beven = 0
2
T = 10ms = 0.01 s
odd and

8 ⎡
=
⎢
0.01 ⎣

⎤
⎛ −1 ⎞
10sin 200nπt dt ⎥ = 8000 ⎜
bodd
⎟ cos 200nπt
∫
⎝ 200nπ ⎠
0
⎦
40
40
∴ bodd = − (cos 0.2nπ − 1) =
(1 − cos 0.2nπ)
nπ
nπ
∴ b1 = 2.432, b3 = 5.556, b5 = 5.093, b7 = 2.381, b9 = 0.2702
0.001

0.001

0




10 March 2006

23.
odd and

1
8
− wave, T = 8 ms ∴ bn =
2
T

2π
ωo =
= 250π ∴ bn = 1000
T
Now,

1
∫ x sin ax dx = a ( sin a
2

T /4

∫

f (t ) sin nωot dt

0

0.001

∫

1000 t sin 250πnt dt

0

x

− ax cos ax ) , a = 250 nπ

106
0.001
sin 250nπt − 250nπt cos 250nπt )0
2 2 2 (
250 n π
nπ
nπ
16 ⎛
π π
π⎞
16 ⎛
nπ
⎞
+ 0 ⎟ ∴ b1 = 2 ⎜ sin − cos ⎟ = 0.2460
∴ bn = 2 2 ⎜ sin
− 0 − cos
4
4
π ⎝
4 4
4⎠
4
nπ ⎝
⎠
f (t ) = 103 t ∴ bn =

b3 =
beven

16 ⎛
3π 3π
3π ⎞
16 ⎛
5π 5π
5π ⎞
sin − cos ⎟ = 0.4275− ; b5 =
sin
− cos ⎟ = 0.13421
2 ⎜
2 ⎜
9π ⎝
4
4
4 ⎠
25π ⎝
4
4
4 ⎠
=0



24.
(a)

even, T = 4:

(c)

odd,

(d)

even,

10 March 2006

odd, T = 4

(b)


1
− wave: T = 8
2

1
− wave, T = 8 :
2



25.
(a)

vs = 5 +


10 March 2006

20 ∞ 1
2πnt
20
20
∑ n sin 0.4π ∴ vsn = nπ sin 5nt , Vsn = nπ (− j1)
π 1,odd

Zn = 4 + j 5n 2 = 4 + j10n, I fn =

Vsn
− j 20
j5
=
=−
Zn nπ(4 + j10n)
1 + j 2.5n

12.5 + j 5
j 5 1 − j 2.5n
=−
2
nπ 1 + 6.25n
nπ(1 + 6.25n 2 )
12.5
1
5
1
∴ i fn = −
cos 5nt +
sin 5nt
2
π 1 + 6.25n
nπ 1 + 6.25n 2
∞
1
5
⎡ 12.5
⎤
∴ i f = 1.25 + ∑
−
cos 5nt + sin 5nt ⎥
2 ⎢
π
nπ
⎦
1,odd 1 + 6.25n ⎣
∴ I fn = −

(b)

in = Ae −2t , i = i f + in , i (0) = 0, i f (0) = 1.25 +

∞

1

∑ 1 + 6.25n

1, odd

∴ i f (0) = 1.25 −

2
π

∞

∑

1,odd

2

⎛ 12.5 ⎞
⎜−
⎟
⎝ π ⎠

1
2 π
= 1.25 −
tanh 0.2π = 0.55388
n + 0.16
π 4 × 0.4
2

∴ A = −0.55388, i = −0.55388e−2t + 1.25 +

∞

1

∑ 1 + 6.25n

1,odd

2

5
⎡ 12.5
⎤
⎢ − π cos 5nt + nπ sin 5nt ⎥
⎣
⎦




10 March 2006

26.
(a)

0 < t < 0.2π : i = 2.5(1 − e −2t ) ∴ i (0.2π) = 2.5(1 − e−0.4 π ) = 1.78848 A

(b)

0.2π < t < 0.4π : i = 1.78848 e−2( t −0.2 π ) ∴ i (0.4π) = 0.50902 A

(c)

0.4π < t < 0.6π : i = 2.5 − (2.5 − 0.50902)e−2( t −0.4 π ) , i (0.6π) = 1.9335−




10 March 2006

27.
(a)

vs = 5 +

20 ∞ 1
∑ sin 5nt
π 1,odd n

20
sin 5nt
nπ
20
Vsn = − j
nπ
1
1
1
− j 20 / nπ
− j 20 / nπ 1 − j 20n
Zn = 2 +
= 2+
∴ Vcn =
×
=
×
2 + 1/ j10n j10n 1 + j 20n 1 − j 20n
j 5n 2
j10n
20
1
−20n − j1 20
∴ Vcn =
× , vcn =
( −20n cos 5nt + sin 5nt )
2
1 + 400n nπ
nπ 1 + 400n 2
20 ∞
1
⎛1
⎞
∴ vcf = 5 +
∑ 1 + 400n2 ⎜ n sin 5nt − 20 cos 5nt ⎟
π 1,odd
⎝
⎠
vsn =

(b)

vn = Ae −t / 4

(c)

vc (0) = A + 5 +
∞

∑

1,odd

20 ∞
−20
1
∑ 1 + 400n2 = A + 5 − π
π 1,odd

∞

∑
1,odd

1
n + (1/ 20) 2
2

π
π
π
1
=
= 5π tanh
= 1.23117
tanh
2
4(1/ 20)
20 × 2
40
n + (1/ 20)
2

1
∴ A = 0 − 5 + × 1.23117 = −4.60811
π
20 ∞
1
⎛1
⎞
∴ vc (t ) = −4.60811e− t / 4 + 5 +
∑ 1 + 400n2 ⎜ n sin 5nt − 20 cos 5nt ⎟
π 1,odd
⎝
⎠



28.


10 March 2006

At the frequency ω = 10nπ

(

)

10 ⎡10 + j10nπ 5 ×10−3 ⎤
⎦ Ω and I = 8 − j
Zn = ⎣
( )
Sn
πn
20 + j10nπ 5 ×10−3

(

Therefore Vn =

)

⎡ 10 + j 0.05nπ ⎤
80
(− j) ⎢
⎥.
nπ
⎣ 20 + j 0.05nπ ⎦

In the time domain, this becomes
v1 (t ) =

2
⎛ 40 ⎞ 1 + (0.005nππ )
cos 10nπ − 90o + tan −1 0.005nπ − tan −1 0.0025nπ
∑ ) ⎜ nπ ⎟
2
⎠ 1 + (0.0025nππ )
n =1 ( odd ⎝
∞

(


)


29.


10 March 2006

At the frequency ω = nπ

I Ln =

n −1
10
32
I Sn and I Sn = − j
( −1) 2
2
20 + jnπ 5 × 10−3
(π n )

(

)

Thus, in the time domain, we can write
iL (t ) =

n −1 ⎤
⎡ 320
1
cos nπ t − 90o − tan −1 0.00025nπ
⎢
( −1) 2 ⎥
∑ ⎢ nπ 2
2
n =1 (odd) (
⎥ 20 1 + ( 0.00025nπ )
)
⎣
⎦
∞

(

)




10 March 2006

30.
0.001
0.005
⎤
−3
103 ⎡
100e − j 3×2 πt / 6×10 − ∫ 100e− j100 πt ⎥
⎢ ∫
6 ⎣ 0
0.003
⎦
0.001
0.005
⎤
105 ⎡ −1
1
− j1000 πt
− j1000 πt
=
+
e
e
⎢
⎥
6 ⎢ j1000π
j1000π
0
0.003 ⎥
⎣
⎦
100 − jπ
100
=
(1 + 1 − 1 + 1) = − j10.610
e + 1 + e − j 5 π − e − j 3π =
j 6π
j 6π

c3 =

(

)

∴ c−3 = j10.610; c3 = 10.610
0.001
0.005
⎤
2 ×103 ⎡
a3 =
⎢ ∫ 100 cos100πt dt − ∫ 100 cos1000πt dt ⎥
6 ⎣ 0
0.003
⎦
5
2 ×10
1
=
( sin π − 0 − sin 5π + sin 3π ) = 0
6 1000π
1
1
2
c3 = (a3 − jb3 ) = − j b3 ∴ b3 = 21.22 and a3 + b32 = 21.22
2
2




10 March 2006

31.
(a)

(b)

⎤
100e− j 400 πnt dt ⎥
∫
∫
0
0.001
⎦
0.001
0.002
⎡
⎤
∴ cn = 20, 000 ⎢ ∫ 1000t e − j 400 πnt dt + ∫ e − j 400 πnt dt ⎥
0.001
⎣ 0
⎦
0.002
⎡ e − j 400 πnt
⎤
1
0.001
− j 400 πnt
e
( j 400πnt + 1)0 +
∴ cn = 20, 000 ⎢
⎥
2 2
− j 400πn
⎢160n π
0.001 ⎥
⎣
⎦
1
∴ co = ao = (50 × 10−3 + 100 × 10−3 )
= 0.15 × 200 = 30
0.005
⎡ 1
⎤
1
1
c1 = 20, 000 ⎢
e − j 0.4 π (1 + j 0.4π) −
e− j 0.8 π − e − j 0.4 π ⎥
−
2
2
j 400π
160π
⎣160π
⎦
125
= 2 (1∠ − 72°) (1.60597∠51.488°) − 12.66515 + 15.91548 ∠90°(1∠ − 144° − 1∠ − 72°)
π
= 12.665(1∠ − 72°) (1 + j1.2566) − 12.665 + j15.915(1∠ − 144° − 1∠ − 72°)
= 20.339∠ − 20.513° − 12.665 + 18.709∠ − 108° = 24.93∠ − 88.61°
c2 = 3.16625∠ − 144° (1 + j 2.5133) − 3.16625 + j 7.9575(1∠ − 288° − 1∠ − 144°)
T = 5 ms cm =

1 ⎡
⎢
0.005 ⎣

0.001

105 te − j 400 πnt dt +

0.002

(

)

= 8.5645 ∠ − 75.697° − 3.16625 + 15.1361∠144° = 13.309∠177.43°




10 March 2006

32.
Fig. 17-8a: Vo = 8 V, τ = 0.2 μ s, f o = 6000 pps
1
1
, f o = 6000, τ = 0.2 μ s ∴ f = = 5 MHz
τ
6000

(a)

T=

(b)

f o = 6000 Hz

(c)

6000 × 3 = 18, 000 (closest) ∴ c3 =

8 × 0.2 × 10−6 sin(1/ 2 × 3 ×12, 000π × 0.2 × 10−6
1/ 6000
0.0036 π

∴ c3 = 9.5998 mV

(d)

2 ×106
8 × 0.2 ×10−6 sin(1/ 2 × 333 ×12, 000π × 0.2 × 10−6
= 333.3 ∴ c333 =
= 7.270 mV
6 × 103
1/ 6000
1/ 2 × 333 ×12, 000π × 0.2 × 10−6

(e)

β = 1/ τ = 5 MHz

(f)

2 < ω < 2.2 Mrad/s ∴

(g)

2000
2200
< f <
kHz or 318.3 < f < 350.1 kHz
2π
2π
f o = 6 kHz ∴ f = 6 × 53 = 318; 324,330,336,342,348 kHz ∴ n = 5

c227

8 × 0.2 ×10−6 sin(1/ 2 × 227 × 12, 000π × 0.2 × 10−6
=
= 8.470 mV
1/ 6000
(′′)

f = 227 × 6 = 1362 kHz




10 March 2006

33.
T = 5 ms; co = 1, c1 = 0.2 − j 0.2, c2 = 0.5 + j 0.25, c3 = −1 − j 2, cn = 0, n ≥ 4

(a)

an = − jbn = 2cn ∴ ao = co = 1, a1 − jb1 = 0.4 − jb1 = 0.4 − j 0.4, a2 − jb2 = 1 + j 0.5, a3 − jb3 = −2
∴ v(t ) = 1 + 0.4 cos 400π t + cos800π t − 2 cos1200π t + 0.4sin 400π t − 0.5sin 800π t + 4sin1200π t

(b)

v (1 ms ) = 1 + 0.4 cos 72° + cos144° − 2 cos 216° + 0.4 sin 72° − 0.5sin144° + 4 sin 216° ∴
∴ v (1 ms ) = −0.332V




10 March 2006

34.
0.6×10−6

(a)

106
t
T = 5 μ s ∴ cn =
× 2 ∫ 1cos 2π n
dt
5
5 × 10−6
0.4×10−6
∴ cn = 4 × 105
∴ cn =

5 × 10−6
( sin 43.2°n − sin 28.8°n )
2π n

1
( sin 43.2°n − sin 28.8°n )
nπ

1
(sin172.8° − sin115.2°) = −0.06203
4π

(b)

c4 =

(c)

co = ao =

(d)

a little testing shows co is max ∴ cmax = 0.08

(e)

0.01× 0.08 = 0.8 × 10−3 ∴

0.2 × 10−6 + 0.2 ×10−6
= 0.08
5 × 10−6

1
( sin 43.2°n − sin 28.8°n ) ≤ 0.8 ×10−3
nπ

125
( sin 43.2°n − sin 28.8°n ) ≤ 1
nπ
ok for n > 740

∴

(f)

β = 740 f o =

740 ×106
= 148 MHz
5




10 March 2006

35.
T = 1/16, ω o = 32π
1/ 96

(a)

c3 = 16

∫

40e

0

∴ c3 = j
(b)

− j 96π t

16 × 40 − j 96π
dt −
e
− j 96π

1/ 96

0

20 − jπ
40
(e − 1) = − j
= − j 4.244 V
3π
3π

Near harmonics are 2f o = 32 Hz, 3f o = 48 Hz
Only 32 and 48 Hz pass filter an − jbn = 2cn
a3 − jb3 = 2c3 = − j8.488 ∴ a3 = 0, b3 = 8.488 V
I3 =

8.488
1
= 1.4536 ∠ − 31.10° A; P3 = × 1.45362 × 5 = 5.283 W
5 + j 0.01× 96π
2

c2 =

1
1/16

1/ 96

∫
0

40e − j 64π t dt =

640
(e − j 64π / 96 − 1) = 2.7566 − j 4.7746 V
− j 64π

a2 − b2 = 2c2 = 5.5132 − j9.5492 = 11.026 ∠ − 60°
11.026∠ − 60°
= 2.046∠ − 65.39° A
5 + j 0.01× 64π
1
∴ P2 = × 2.0462 × 5 = 10.465 W ∴ Ptot = 15.748 W
2
∴ I2 =



36.


10 March 2006

f (t ) = 5[u (t + 3) + u (t + 2) − u (t − 2) − u (t − 3)]
f(t)

(a)

t
-3

-2
∞

(b)

F( jω) =

∫

-1

0

1

2

3

f (t ) e − jωt dt

−∞
−2

∴ F( jω) =

∫ 5e

−3

− j ωt

2

dt + ∫ 10e
−2

− jωt

3

dt + ∫ 5e − jωt dt
2

5
10 − j 2 ω
5
(e j 2 ω − e j 3ω ) +
(e
(e − j 3ω − e − j 2 ω )
− e j 2ω ) +
− jω
− jω
− jω
5
5
10
( −e j 3ω + e − j 3ω ) +
(e j 2 ω − e − j 2 ω ) +
(−e j 2 ω + e − j 2 ω )
=
− jω
− jω
− jω
5
5
10
(− j 2) sin 3ω +
( j 2) sin 2ω +
=
(− j 2) sin 2ω
− jω
− jω
− jω
10
10
20
10
(sin 3ω + sin 2ω)
∴ F( jω) = sin 3ω − sin 2ω + sin 2ω =
ω
ω
ω
ω
∴ F( jω) =




10 March 2006

37.
(a)

f (t ) = e − at u (t ), a > 0 ∴ F( jω ) =

∞

∫

−∞

∴ F( jω ) =

−1 − ( a + jω )t
e
a + jω

∞

=
0

∞

f (t )e − jωt dt = ∫ e− at e− jω t dt
0

1
a + jω
∞

(b)

f (t ) = e at6 e − at u (t − to ), a > 0 ∴ F( jω ) = e ato ∫ e − ( a + jω )t dt
to

∴ F( jω ) = e ato

−1 − ( a + jω ) t
e
a + jω

∞

= e ato
to

1
−1
⎡ − ( a + jω )to ⎦ =
⎤
e − jω to
⎣ −e
a + jω
a + jω

∞

(c)

f (t ) = te − at u (t ), a > 0 ∴ F( jω ) = ∫ te− ( a + jω ) t dt
0

∴ F( jω ) =

− ( a + jω ) t

1
1
e
∞
−(a + jω )t − 1]0 = 0 −
[−1] =
2 [
2
(a + jω ) 2
(a + jω )
(a + jω )




10 March 2006

38.
−4 < t < 0 : f (t ) = 2.5(t + 4); 0 < t < 4 : f (t ) = 2.5(4 − t )
0

∴ F( jω) =

∫

−4

4

2.5(t + 4) e− jωt dt + ∫ 2.5(4 − 5)e− jωt dt
0

0

ln 1st , let t = τ ∴ I1 = ∫ 2.5(4 − τ)e jωτ (− d τ)
4

4

4

0

0

∴ I1 = ∫ 2.5(4 − τ)e jωτ d τ ∴ F( jω) = 2.5∫ (4 − t )(e jωt + e− jωt ) dt
4

4

4

1
∴ F( jω) = 5 ∫ (4 − t ) cos ωt dt = 20 × sin ωt − 5∫ cos ωt dt
ω
0
0
0
20
5
4
sin 4ω − 2 (cos ωt + ωt sin ωt )0
ω
ω
20
5
5
5
= sin 4ω − 2 (cos 4ω − 1) − 2 4ω sin 4ω = 2 (1 − cos 4ω)
ω
ω
ω
ω

∴ F( jω) =

2×5
⎛ sin 2ω ⎞
or, F( jω) = 2 sin 2 2ω = 10 ⎜
⎟
ω
⎝ ω ⎠

2




10 March 2006

39.
π

f (t ) = 5sin t, − π < t < π ∴ F( jω) =

∫ 5sin t e

− jωt

dt

−π
π

∴ F( jω) =

5
− jt
− j ωt
jt
∫ (e − e ) e dt
j 2 −π

π

=

5
− jt (1+ω )
jt (1−ω )
] dt
∫ [e − e
j 2 −π

F( jω) =

⎤
5 ⎡ 1
1
j π (1−ω )
− e− jπ (1−ω) ) −
(e− jπ (1+ω) − e jπ (1+ω) ) ⎥
⎢ j (1 − ω) (e
j2 ⎣
− j (1 + ω)
⎦

2.5
−2.5
( −e − jπω + e jπω ) −
( −e − jπω + e jπω )
1− ω
1+ ω
2.5
1
1 ⎞
−2.5
⎛
( j 2sin πω) −
( j 2sin πω) = j 5sin πω ⎜ −
−
=
⎟
1− ω
1+ ω
⎝ 1− ω 1+ ω ⎠
=

j10sin πω j10sin πω
⎛ 1+ ω +1− ω ⎞
= j 5sin πω(−1) ⎜
=
⎟=−
2
1 − ω2
ω2 − 1
⎝ 1− ω
⎠




10 March 2006

40.
f (t ) = 8cos t [u (t + 0.5π) − u (t − 0.5π)]
π/ 2

∴ F( jω) =

∫

8cos te − jωt dt = 4

−π / 2
π/ 2

=4

∫

−π / 2

π/ 2

∫

(e jt + e − jt ) e− jωt dt

−π / 2

⎡e jt (1−ω) + e − jt (1+ω) ⎤ dt
⎣
⎦

⎧ 1
π/2
1
⎪
− j ωt
= 4⎨
e jt e − π / 2 −
e− jt e− jωt
j (1 + ω)
⎪ j (1 − ω)
⎩

⎫
⎪
⎬
−π / 2 ⎪
⎭
π/ 2

⎧ 1
⎫
1
− j πω / 2
⎡ je − jπω / 2 − (− j ) e jπω / 2 ⎤ −
− je jπω / 2 ⎤ ⎬
= 4⎨
⎦
⎣
⎦ j (1 + ω) ⎡ − je
⎣
⎩ j (1 − ω)
⎭
1 ⎞
πω
πω ⎫
πω ⎛ 1
1
⎧ 1
= 4⎨
× 2 cos
+
× 2 cos
+
⎬ = 8cos
⎜
⎟
2 1+ ω
2 ⎭
2 ⎝ 1− ω 1+ ω ⎠
⎩1 − ω
πω 2
cos πω / 2
= 8cos
= 16
2
2 1− ω
1 − ω2
(a)

ω = 0 ∴ F( j 0) = 16

(b)

ω = 0.8, F( j 0.8) =

16 cos 72°
= 13.734
0.36

(c)

ω = 3.1, F( j 3.1) =

16 cos(3.1× 90°)
= −0.2907
1 − 3.12




10 March 2006

41.
(a)

F( jω) = 4 [u (ω + 2) − ω (ω − 2) ] ∴ f (t ) =
2

4
2 1 j ωt
jωt
∴ f (t ) =
∫2 e d ω = π jt e
2π −
∴ f (t ) =
(b)

(

2
e j 2t − e− j 2t
jπt

=
−2

)

2
4
5
j 2sin 2t = sin 2t ∴ f (0.8) = sin1.6 rad = 1.5909
πt
π
2πt

F( jω) = 4e

∞

−2 ω

∴ f (t ) =

4
−2 ω + j ω t
dω
∫e
2π −∞
∞

0

∴ f (t ) =

2

∞

1
j ωt
∫ e F( jω)d ω
2π −∞

2
2
(2 + jt ) ω
d ω + ∫ e( −2+ jω) t d ω
∫e
π −∞
π 0

⎤ 2⎛ 1
2⎡ 1
1
1 ⎞ 2 4
⎢ 2 + jt (1 − 0) + −2 + jt (0 − 1) ⎥ = π ⎜ 2 + jt + 2 − jt ⎟ = π 4 + t 2
π⎣
⎦
⎝
⎠
8
8
∴ f (t ) =
∴ f (0.8) =
= 0.5488
2
π(4 + t )
π× 4.64
=

(c)

F( jω) = 4 cos πω [u (ω + 0.5) − u (ω − 0.5) ]
∴ f (t ) =

4
2π

1
=
π
=

0.5

∫

cos πω× e jωt d ω =

−0.5
0.5

∫

−0.5

2
π

0.5

∫ 2 (e
1

j πω

)

+ e − jπω e jωt d ω

−0.5

⎡ e( jπ+ jt ) ω + e( − j 0.5 π− j 0.5t ) ω ⎤ d ω
⎣
⎦

⎤
1⎡ 1
1
j 0.5 π+ j 0.5 t
e− j 0.5π+ j 0.5t − e j 0.5π− j 0.5t ⎥
− e − j 0.5 π− j 0.5t +
⎢ j (π + t ) e
j (−π + t )
π⎣
⎦

(

)

(

)

⎤
1⎡ 1
1
j 0.5t
+ je − j 0.5t +
− je j 0.5t − je− j 0.5t ⎥
⎢ j (π + t ) je
π⎣
j (−π + t )
⎦
1⎡ 1
1
1 ⎞
⎤ 2 cos 0.5t ⎛ 1
= ⎢
−
2 cos 0.5t −
2 cos 0.5t ⎥ =
⎜
⎟
π ⎣π+t
−π + t
π
⎦
⎝ π + t −π + t ⎠

=

(

)

(

)

4
⎛ −2 ⎞
= 2 cos 0.5t ⎜ 2
= 2 2 cos 0.5t ∴ f (0.8) = 0.3992
2 ⎟
⎝ t −π ⎠ π −t



42.

Fv ( jω) =

10 March 2006

v(t ) = 20e1.5t u (−t − 2) V

(a)


∞

∫

20e1.5t u (−t − 2)e − jωt dt =

−∞

=
(b)

20
e(1.5− jω)t
1.5 − jω

−2

∫ 20e

1.5 t − jωt

dt

−∞
−2

=
−∞

20
20 −3
e −3+ j 2 ω ∴ Fv ( j 0) =
e = 0.6638
1.5 − jω
1.5

Fv ( jω) = A v (ω) + Bv (ω) =

20
e −3e j 2 ω
1.5 − jω

20
e −3 e j 4 = 0.39830 ∠282.31° = 0.08494 − j 0.38913
1.5 − j 2
∴ A v (2) = 0.08494
∴ Fv ( j 2) =

(c)

Bv (2) = −0.3891

(d)

Fv ( j 2) = 0.3983

(e)

φv(j2) = 282.3o or -77.69o




43.

I( jω) = 3cos10ω [u (ω + 0.05π) − u (ω − 0.05π) ]

(a)

W = 4×

10 March 2006

=

(b)

9
π

18
π

ωx

1
2π

∞

∫

−∞

I( jω) d ω =
2

2
π

0.05 π

∫

9 cos 2 10ω d ω

−0.05 π
π / 20

π / 20

9
9 1
⎛1 1
⎞
∫/ 20 ⎜ 2 + 2 cos 20 ω ⎟ d ω = π × 0.1π + π 20 sin 20ω −π / 20 = 0.9 J
⎝
⎠
−π
9⎡

∫ (1 + cos 20ω) d ω = 0.45 = π ⎢ 2ω
⎣

−ωx

x

+

1
⎤
× 2sin 20 ωx ⎥
20
⎦

∴ 0.05π = 2ωx + 0.1sin 20ωx , ωx = 0.04159 rad/s



44.
(a)

∞

∞

∞

W1Ω = ∫ f (t ) dt = ∫ 100t e

2 −8 t

2

=

0

e−8t
(64t 2 + 16t + 2)
dt = 100 ×
(−512)
0

100
× 2 = 0.3906 J
512
∞

F( jω) = F {10te u (t )} = 10 ∫ t e
−4 t

∞

− (4 + jω ) t

0

=

(c)

10 March 2006

f (t ) = 10te −4t u (t )

0

(b)


10e − (4+ jω)t
dt =
[−(4 + jω)t − 1
(4 + jω) 2
0

10
10
∴ F( jω) = 2
2
(4 + jω)
ω + 16

F ( jω )

2

=

(ω

100
2

+ 16)

2

F ( jω ) ω = 0 = 390.6 mJ/Hz , F ( jω ) ω = 4 = 97.66 mJ/Hz
2

2



45.

v(t ) = 8e

(a)

W1Ω =

−2 t

∞

v 2 (t ) dt = 2 × 64

−∞

(b)

− j ωt
∫ e v(t ) dt = 8

−∞

0

∴ Fv ( jω) = 8

∫

e

−∞

8
=
e(2− jω)t
2 − jω
ω

(c)

∫e

−4 t

dt = 32 J

∫e

e− jωt dt

0

∞

Fv ( jω) =

10 March 2006

V

∞

∫


(2 − jω) t

∞

−2 t

−∞

∞

dt + 8∫ e− (2+ jω) t dt
0

0

8
−
e − (2+ jω)t
2 + jω
−∞

1 1
322
322
0.9 × 32 =
dω =
∫
2π −ω1 (ω2 + 4) 2
2π

∞

=
0

8
32
8
=
= Fv ( jω)
+
2 − jω 2 + jω 4 + ω2

⎡
1
ω
ω ⎤
+ tan −1 1 ⎥
⎢
2
2⎦
⎣ 8(ω1 + 4) 16

16 ⎡ ω1
1 ω1 ⎤ 2 ⎡ 2ω1
ω ⎤
×2⎢
+
+ tan −1 1 ⎥
⎥= ⎢ 2
2
2⎦
π
⎣ 8(ω1 + 4) 16 2 ⎦ π ⎣ ω1 + 4
2ω
ω
∴ 0.45π = 2 1 + tan −1 1 ∴ω1 = 2.7174 rad/s (by SOLVE)
2
ω1 + 4
∴ 0.9 =




10 March 2006

46.
(a)

∞

Prove: F { f (t − to )} = e − jωto F { f (t )} =

∫

f (t − to )e − jωt dt Let t − to = τ

−∞

∴F { f (t − to )} =

∞

∫

f (τ)e − jωτ e − jωto dt = e − jωto F { f (t )}

−∞

(b)

Prove:

F { f (t )} =

jωF { f (t )} =

∞

∫e

− jωt

−∞

dv = df , v = f ∴F { f (t )} = f (t )e− jωt

∞
−∞

df
dt Let u = e − jωt , du = − jωe− jωt ,
dt
∞

+

∫

jωf (t )e− jωt dt

−∞

We assume f (±∞) = 0 ∴F { f (t )} = jωF { f (t )}

(c)

Prove:

F { f (kt )} =

∴F { f (kt )} =

∞

∫

∞

1 ⎛ jω ⎞
− j ωt
F⎜
dt Let τ = kt , k > 0
⎟ = ∫ f (kt )e
k ⎝ k ⎠ −∞

f (τ)e − jωτ / k

−∞

1
1 ⎛ jω ⎞
dτ = F⎜
⎟
k
k ⎝ k ⎠

If k < 0, limits are interchanged and we get: −
∴F { f (kt )} =

1 ⎛ jω ⎞
F⎜
⎟
k ⎝ k ⎠

1 ⎛ jω ⎞
F⎜
⎟
k ⎝ k ⎠

(d)

Prove: F { f (−t )} = F(− jω) Let k = 1 in (c) above

(e)

d
Prove: F {tf (t )} = j
F( jω) Now, F( jω) = ∫ f (t )e − jωt dt
dω
−∞

∞

∞

∴

dF( jω)
= ∫ f (t )(− jt )e − jωt dt = − j F {tf (t )} ∴F {tf ( f )} = jωF f (t )}
dω
−∞




10 March 2006

47.
(a)

f (t ) = 4[sgn(t )δ(t − 1)] ∴F {4[sgn(t )δ(t − 1)] = F {4sgn(1) δ(t − 1)} = F {4δ(t − 1)} = 4e − jω

(b)

f (t ) = 4[sgn(t − 1) δ(t )] ∴F {4sgn(−1)δ(t )} = F {−4δ(t )} = −4

(c)

⎧4
⎫
f (t ) = 4sin(10t − 30°) ∴F {4sin(10t − 30°) = F ⎨ ⎡e j (10t −30°) − e − j (10t −30°) ⎤ ⎬ =
⎣
⎦
⎩ j2
⎭
− j 30° j10 t
j 30° − j10 t
− jπ / 6
jπ / 6
F {− j 2e e + j 2e e } = − j 2e 2πδ(ω − 10) + j 2e 2πδ(ω + 10)
= − j 4π [e − jπ / 6 δ(ω − 10) − e jπ / 6δ(ω + 10)]




10 March 2006

48.
(a)

f (t ) = A cos(ωo t + φ) ∴ F( jω) = F {A cosφ cos ωo t − A sin φ sin ωo t} =
⎧π
⎫
A cos φ{π[δ(ω + ωo ) + δ(ω − ωo )]} − A sin φ ⎨ [δ(ω − ωo ) − δ(ω + ωo )]⎬ =
⎩j
⎭
πA{cos φ [δ(ω + ωo ) + δ(ω − ωo )] + j sin φ [δ(ω − ωo ) − δ(ω + ωo )]}
∴ F( jω) = πA[e jφ δ(ω − ωo ) + e − jφ δ(ω + ωo )]

(b)

f (t ) = 3sgn(t − 2) − 2δ(t ) − u (t − 1) ∴ F( jω) = e − j 2 ω × 3 ×
∴ F( jω) = − j

(c)

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

6 − j 2ω
1⎤
⎡
− 2 − e − jω ⎢ πδ(ω) − j ⎥
e
ω
ω⎦
⎣

⎧1
⎫
f (t ) = sinh kt u (t ) ∴ F( jω) = F ⎨ [e kt − e− kt ] u (t ) ⎬
⎩2
⎭
k + jω + k − jω
−k
1
1
1
1
∴ F( jω) =
−
=
= 2
2
2
ω + k2
2 − k + jω 2 k + jω
2(− k − ω )




10 March 2006

49.
∞

(a)

F( jω) = 3u (ω + 3) − 3u (ω − 1) ∴ f (t ) =
1

3
3 1 j ωt
j ωt
∴ f (t ) =
∫3 e dt = 2π jt e
2π −
∴ f (5) = − j
(b)

1

=
−3

1
jωt
∫ [3u(ω + 3) − 3u(ω − 1)] e d ω
2π −∞
3
(e+ jt − e− j 3t )
j 2πt

3
(1∠5rad − 1∠ − 15rad ) = 0.10390 ∠ − 106.48°
10π

F( jω) = 3u (−3 − ω) + 3u (ω − 1) →
∴ F( jω) = 3 − Fa ( jω)
3
(e jt − e − j 3t ) ∴ f (5) = 0 − 0.10390 ∠ − 106.48°
j 2πt
so f(5) = 0.1039∠73.52o
f (t ) = 3δ(t ) −

(c)

F( jω) = 2δ(ω) + 3u (−3 − ω) + 3u (ω − 1) Now, F {2δ(ω)} =
∴ f (t ) =

2 1
=
2π π

⎤
1 ⎡
3
1
(e jt − e − j 3t ) ⎥ ∴ f (5) = − 0.10390 ∠ − 106.48° = 0.3618 ∠15.985+°
+ ⎢−
π ⎣ j 2πt
π
⎦




10 March 2006

50.
(a)

F( jω) =

3
3
+
+ 3 + 3δ(ω − 1)
1 + jω jω

∴ f (t ) = 3e − t u (t ) + 1.5sgn(t ) + 3δ(t ) +

(b)

(c)

1.5 jt
e
π

1
sin ω 8 / 2
5sin 4ω = 8
× 2.5
ω
ω8 / 2
∴ f (t ) = 2.5[u (t + 4) − u (t − 4)]
F( jω) =

F( jω) =

6(3 + jω)
6(3 + jω)
=
∴ f (t ) = 3−3t cos 2t u (t )
2
2
2
(3 + jω) + 4 (3 + jω) + 2




10 March 2006

51.
T = 4, periodic; find exp′l form
1

1
∴ cn = ∫ 10te − jnπt / 2 dt
4 −1
1

⎡
⎛
⎞⎤
t
1
∴ cn = 2.5 ⎢ e − jnπt / 2 ⎜
− 2 2 ⎟⎥
⎝ − jnπ / 2 −n π / 4 ⎠ ⎦ −1
⎣
⎡
⎛ 1
⎛ 1
1 ⎞
1 ⎞⎤
∴ cn = 2.5 ⎢ e − jnπ / 2 ⎜
+ 2 2 ⎟ − e jnπ / 2 ⎜
+ 2 2 ⎟⎥
⎝ − jnπ2 n π / 4 ⎠
⎝ jnπ / 2 n π / 4 ⎠ ⎦
⎣
⎡ 1
⎤
4
(−e − jnπ / 2 − e jnπ / 2 ) + 2 2 (e − jnπ / 2 − e jnπ / 2 ) ⎥
= 2.5 ⎢
nπ
⎣ jnπ / 2
⎦
=

j5
nπ 10 ⎛
nπ ⎞
× 2 cos
+ 2 2 ⎜ − j 2sin ⎟
nπ
2 nπ ⎝
2 ⎠

∞
nπ
20
nπ ⎤
⎡ j10
cos
∴ f (t ) = ∑ ⎢
− j 2 2 sin ⎥ e jnπt / 2
2
nπ
2⎦
−∞ ⎣ nπ
∞
nπ
nπ ⎤
nπ ⎞
20
⎡ j10
⎛
∴ F( jω) = ∑ ⎢
− j 2 2 sin ⎥ 2πδ ⎜ ω − ⎟
cos
nπ
2
2⎦
2 ⎠
⎝
−∞ ⎣ nπ


Chapter 18 solutions_to_exercises(engineering circuit analysis 7th)



10 March 2006

53.
∞

F( jω) = 20∑
−∞

1
δ(ω − 20n)
n !+ 1

1
1
1
1
⎡ 1
= 20 ⎢
δ(ω) +
δ(ω + 20) +
δ(ω − 20) +
δ(ω + 40) + δ(ω − 40)
1+1
1+1
2 +1
3
⎣1 + 1
1
1
⎤
+ δ(ω + 60) + δ(ω − 60) + ...⎥
7
7
⎦
20
20
= 10δ(ω) + [πδ(ω + 20) + πδ(ω − 20)] + [πδ(ω + 40) + πδ(ω − 40)] +
2π
3π
20
20
[πδ(ω + 60) + πδ(ω − 60) +
[πδ(ω + 80) + πδ(ω − 80)] + ...
7π
25π
10 20
20
20
20
cos 20t + cos 40t +
cos 60t +
cos80t + ...
∴ f (t ) =
+
2π 2π
3π
7π
25π
20 ⎡
1
1
1
1
=
⎢0.25 + 2 cos 20t + 3 cos 40t + 7 cos 60t + 25 cos80t + ...
π ⎣
∴ f (0.05) =

20 ⎡
1
1
1
1
⎤
rad
⎢0.25 + 2 cos1 + 3 cos 2 + 7 cos 3 + 25 cos 4 + ...⎥ = 1.3858
π ⎣
⎦




10 March 2006

t

Input = x(t ) = 5[u (t ) − u (t − 1)]

54.

∫ x( z ) h(t − z ) dz

−∞

h(t ) = 2u (t )

(a)

y (t ) =

(b) h(t ) = 2u (t − 1)

x(t – z)

(c) h(t ) = 2u (t − 2)
x(t – z)

x(t – z)

5
t-1

z

t
h( z)

t-1

z

t
h( z)

2

t-1

2

2
z

y(t)

z

1

z

2
y(t)

y(t)

10

z

t
h( z)

10

10
t

t

t

1

1

2

2

t < 0:
y(t) = 0

t < 1: y (t ) = 0

t < 2 : y (t ) = 0

0 < t < 1:

1< t < 2:

3

2 < t < 3:

t

t

t

y (t ) = ∫ 10dz = 10t

y (t ) = ∫ 10dz = 10(t - 1)

y (t ) = ∫ 10dz = 10(t - 2)

t > 1:

t > 2:

t > 3:

0

1

t

y (t ) =

∫ 10dz = 10

t -1

2

t

y (t ) =

∫ 10dz = 10

t -1

t

y (t ) =

∫ 10dz

= 10

t -1



55.


10 March 2006

x(t ) = 5[u (t ) − u (t − 2)]; h(t ) = 2[u (t − 1) − u (t − 2)]
t

y (t ) =

∫ x( z ) h(t − z ) dz

−∞

t < 1: y (t ) = 0
t −1

1 < t < 2 : y (t ) =

∫ 10 dz = 10(t − 1)
0

2 < t < 3 : y (t ) = 10
2

3 < t < 4 : y (t ) =

∫ 10 dz = 10(2 − t + 2) = 10(4 − t )

t −2

t > 4 : y (t ) = 0
∴ y (−0.4) = 0; y (0.4) = 0; y (1.4) = 4
y (2.4) = 10; y (3.4) = 6; y (4.4) = 0
∞

or….

y (t ) = ∫ x(t − z ) h( z ) dz
0

t < 1: y (t ) = 0
t

1 < t < 2 : y (t ) = ∫ 10 dz = 10(t − 1)
1

2 < t < 3 : y (t ) = 10
2

3 < t < 4 : y (t ) =

∫ 10 dz = 10(2 − t + 2) = 10(4 − t )

t −2

t > 4 : y (t ) = 0

same answers as above




10 March 2006

56.
h(t ) = 3[e − t − e −2t ], x(t ) = u (t )
t

y (t ) =

∫ x( z )h(t − z) dz

−∞
t

= ∫ 3[e − ( t − z ) − e −2( t − z ) ] dz
0

t

−t

= 3e [e ] − 3e
z t
0

−2 t

⎡ 1 2Z ⎤
⎢2 e ⎥
⎣
⎦0

= 3e −t (et − 1) − 1.5e−2t (e 2t − 1)
∴ y (t ) = 3(1 − e − t ) − 1.5(1 − e −2t ) = 1.5 − 3e− t + 1.5e−2t , t > 0




10 March 2006

57.
∞

y (t ) = ∫ x(t − 2)h( z )dz
0

(a)

2
h(t ) = (5 − t ), 2 < t < 5
3
5
5
2
20
y (t ) = ∫ 10 × (5 − z ) dz =
(5 − z ) dz
3
3 ∫
2
2
Note: h( z ) is in window for 4 < t < 6
5

(b)

20 ⎛ 1 ⎞
y (t ) = ⎜ − ⎟ (5 − z ) 2
3 ⎝ 2⎠
2
=−

10
(0 − 9) = 30 at t = 5
3



58.

x(t ) = 5e

− ( t − 2)


10 March 2006

∞

u (t − 2), h(t ) = (4t − 16) [u (t − 4) − u (t − 7)], y (t ) = ∫ x(t − z ) h( z ) dz
0

(a)

t < 6 : y (t ) = 0 ∴ y (5) = 0

(b)

t = 8 : y (8) = ∫ 5e − (8− z − 2) (4 z − 16) dz

6

4

∴ y (8) = 20e

−6

6

∫ze

z

dz − 80e

4

−6

6

∫e

z

dz

4

6

⎡ ez
⎤
= 20e ⎢ ( z − 1) ⎥ − 80e −6 (e6 − e4 )
⎣1
⎦4
−6

= 20e −6 (5e6 − 3e 4 ) − 80 + 80e −2 = 20 + 80e−2 − 60e−2
= 20 (1 + e −2 ) = 22.71
7

(c)

t = 10 : y (10) = ∫ 5e− (10− z − 2) (4 z − 16) dz
4

7

∴ y (10) = ∫ 20e−8e z ( z − 4)dz
4

7

7

∴ y (10) = 20e −8 ∫ ze z dz − 80e−8 ∫ e z dz = 20e−8 [e z ( z − 1)]7 − 80e−8 (e7 − e 4 )
4
4

−8

4

= 20e (6e − 3e ) − 80(e−1 − e −4 ) = 40e−1 + 20e −4 = 15.081
7

4




10 March 2006

59.
h(t ) = sin t , 0 < t < π; 0 elsewhere, Let x(t ) = e −t u (t )
∞

y (t ) = ∫ x(t − z ) h( z ) dz
0

t < 0 : y (t ) = 0
t

t

0 < t < π : y (t ) = ∫ sin z × e − t + z dz = e − t ∫ e z sin z dz
0

0

t

⎡1
⎤
∴ y (t ) = e −t ⎢ e z (sin z − cos z ) ⎥
⎣2
⎦0
1
= e − t [et (sin t − cos t ) + 1]
2
1
= (sin t − cos t + e − t )
2

(a)

y (1) = 0.3345+

(b)

y (2.5) = 0.7409

(c)

y > π : y (t ) = e − t ∫ e z sin z dz

π

0

π

1
⎡1
⎤
y > π : y (t ) = e ⎢ e z (sin z − cos z ) ⎥ = e − t (e π + 1) = 12.070e − t
⎣2
⎦0 2
∴ y (4) = 0.2211
−t



60.


10 March 2006

x(t ) = 0.8(t − 1)[u (t − 1) − u (t − 3)],
h(t ) = 0.2 (t − 2)[u (t − 2) − u (t − 3)]
∞

y (t ) = ∫ x(t − z ) h( z ) dz,
0

t < 3 : y (t ) = 0
t −1

∫

(a) 3 < t < 4 : y (t ) =

0.8(t − z − 1) 0.2( z − 2) dz

2

t −1

∴ y (t ) = 0.16 ∫ (tz − 2t − z 2 + 2 z − z + 2) dz
2

t −1

t −1

1
⎡ 1
⎤
= 0.16 ∫ [− z 2 + (t + 1) z + 2 − 2t ] dz = 0.16 ⎢ − z 3 + (t + 1) z 2 + (2 − 2t ) z ⎥
2
⎣ 3
⎦2
2
8 1
1
⎡ 1
⎤
= 0.16 ⎢ − (t − 1)3 + + (t + 1) (t − 1) 2 − (t + 1) 4 + (2 − 2t ) (t − 1 − 2) ⎥
3 2
2
⎣ 3
⎦
1 8 1
⎡ 1
⎤
∴ y (t ) = 0.16 ⎢ − t 3 + t 2 − t + + + (t 2 − 1) (t − 1) − 2t − 2 + 2t − 6 − 2t 2 + 6t ⎥
3 3 2
⎣ 3
⎦

⎡1
1
1 ⎤
3
9 9⎞
⎛ 1
⎞ ⎛
⎞
⎛1
= 0.16 ⎢ t 3 + t 2 ⎜1 − − 2 ⎟ + t ⎜ −1 − + 6 ⎟ + 3 + − 8⎥ = 0.16 ⎜ t 3 − t 2 + t − ⎟
2
2 ⎦
2
2 2⎠
⎝ 2
⎠ ⎝
⎠
⎝6
⎣6

∴ y (3.8) = 13.653 × 10−3
3

1
⎡ 1
⎤
4 < t < 5 : y (t ) = ∫ 0.16 (t − z − 1) ( z − 2) dz = 0.16 ⎢ − z 3 + (t + 1) z 2 + (2 − 2t ) z ⎥
2
⎣ 3
⎦2
2
3

(b)

1
⎡ 1
⎤
∴ y (t ) = 0.16 ⎢ − (27 − 8) + (t + 1) 5 + (2 − 2t )1⎥
2
⎣ 3
⎦
11 ⎞
⎡ 19
⎤
⎛
= 0.16 ⎢ − + 2.5t + 2.5 + 2 − 2t ⎥ = 0.16 ⎜ 0.5t − ⎟
6⎠
⎣ 3
⎦
⎝
∴ y (4.8) = 90.67 × 10−3




10 March 2006

61.
x(t ) = 10e −2t u (t ), h(t ) = 10e −2t u (t )
∞

y (t ) = ∫ x(t − z ) h( z ) dz
0

t

∴ y (t ) = ∫ 10e −2( t − z ) 10e −2 z dz
0

t

= 100e −2t ∫ dz = 100 e−2t × t
0

−2 t

∴ y (t ) = 100t e u (t )



62.

W1Ω = 25 ∫ e−8t dt =

10 March 2006

h(t ) = 5e −4t u (t )

(a)


0.8

0.1

25 −0.8 −6.4
(e − e ) = 1.3990 J
8

⎛ 25 ⎞
∴ % = 1.3990 / ⎜ ⎟ ×100% = 44.77%
⎝ 8 ⎠
(b)

5
1
25
25 1
ω
∴ W1Ω = ∫ 2
dω =
H( jω) =
tan −1
π 0 ω + 16
π 4
40
jω + 4
2

∴ W1Ω =

2

25
1
0.9224
tan −1 = 0.9224 J ∴ % =
× 100% = 29.52%
4π
2
25 / 8




10 March 2006

63.
F( jω) =

2
2
2
=
−
∴ f (t ) = (2e− t − 2e−2t ) u (t )
(1 + jω)(2 + jω) 1 + jω 2 + jω
∞

(a)

W1Ω = ∫ (4e−2t − 8e−3t + 4e −4t ) dt =
0

(b)

4 8 4 1
− + = J
2 3 4 3

f (t ) = −2e −t + 4e−2t = 0, − 2 + 4e− t = 0, et = 2, t = 0.69315
∴ f max = 2(e −0.69315 − e−2×0.69315 ) = 0.5




10 March 2006

64.
(a)

(b)

(c)

(d)

1
1/ 6 1/ 2
1/ 3
=
−
+
jω(2 + jω)(3 + jω) jω 2 + jω 3 + jω
1
1
1
∴ f (t ) = sgn(t ) − e−2t u (t ) + e−3t u (t )
12
2
3
F( jω) =

1 + jω
1/ 6
1/ 2
2/3
=
+
−
jω(2 + jω)(3 + jω) jω 2 + jω 3 + jω
1
1
2
∴ f (t ) = sgn(t ) + e−2t u (t ) − e −3t u (t )
12
2
3
F( jω) =

(1 + jω) 2
1/ 6 1/ 2
4/3
=
−
+
F( jω) =
jω(2 + jω)(3 + jω) jω 2 + jω 3 + jω
1
1
4
∴ f (t ) = sgn(t ) − e−2t u (t ) + e −3t u (t )
12
2
3
(1 + jω)3
1/ 6 1/ 2
8/3
= 1+
+
−
jω(2 + jω)(3 + jω)
jω 2 + jω 3 + j ω
1
1
8
∴ f (t ) = δ(t ) + sgn(t ) + e−2t u (t ) − e −3t u (t )
12
2
3
F( jω) =



65.

H( jω) = 2 ×

(b)

1
1
1 Vo
1/ jω
=
=
H( jω) =
2
1 + jω 2 Vi 1 + 1/ jω

(c)

10 March 2006

h(t ) = 2e −t u (t )

(a)


Gain = 2

1
2
=
1 + jω 1 + jω




10 March 2006

66.
1
1
jω +
( jω) 2 + 2
2
jω
=
Vo ( jω) =
1
1
( jω) 2 + 2( jω) + 2
1 + jω +
2
jω
−2( jω)
( jω) 2 + 2( jω) + 2 − 2( jω)
∴ Vo ( jω) =
= 1+
2
2
( jω) + 2( jω) + 2
( jω) + 2( jω) + 2
−2 ± 4 − 8
2x
= −1 ± j1
; x=
2
x + 2x + 2
A
B
A
B
∴ Vo ( x) = 1 +
+
= Let x = 0 ∴
+
=0
1 + j1 1 − j1
x + 1 + j1 x + 1 − j1
A
B
B + j2
B
Let x = −1 ∴ +
= 2 ∴ A − B = j 2, A = B + j 2 ∴
+
=0
j1 − j1
1 + j1 1 − j1
∴ B − jB + j 2 + 2 + B + jB = 0 ∴ B = −1 − j1 ∴ A = −1 + j1
−1 + j1
−1 − j1
1 − j1
1 + j1
∴ Vo ( x) = 1 +
+
, Vo ( jω) = 1 −
−
( jω) + 1 + j1 ( jω) + 1 − j1
x + 1 + j1 x + 1 − j1
Let jω = x ∴ Vo ( x) = 1 −

∴ vo (t ) = δ(t ) − (1 − j1) e( −1− j1) t u (t ) − (1 + j1)e( −1+ j1)t u (t )
= δ(t ) − 2 e − j 45°− jt −t u (t ) − 2 e j 45°+ jt −t u (t )
= δ(t ) − 2 2 e −t cos(t + 45°) u (t )




10 March 2006

67.
5 / jω
10 / jω
=
5 / jω + 35 + 30( jω) 1/ jω + 7 + 6( jω)
10
10 / 6
∴ Vc ( jω) =
=
2
6( jω) + 7( jω) + 1 ( jω) 2 + 7 ( jω) + 1
6
6
⎛
49 24 ⎞
1
10 / 6
2
2
∴ jω = ⎜ −7 / 6 ±
− ⎟ / 2 = − , − 1 ∴ Vc ( jω) =
=
−
⎜
⎟
36 36 ⎠
6
( jω + 1/ 6)( jω + 1) jω + 1/ 6 jω + 1
⎝
∴ vc (t ) = 2(e −t / 6 − e− t ) u (t )
Vc ( jω) = 10



68.

10 March 2006

f (t ) = 5e −2t u (t ), g (t ) = 4e−3t u (t )

(a)


f ∗ g = ∫ f (t − z ) g ( z ) dz

∞

0
t

t

= ∫ 5e −2t e 2 z 4e −3 z dz = 20e−2t ∫ e− z dz
0

0

−2 t

= −20 e (e − 1) V
t

∴ f ∗ g = (e −2t − e−3t ) u (t )
(b)

5
4
20
, G( jω) =
∴ F( jω)G( jω) =
jω + 2
jω + 3
( jω + 2)( jω + 3)
20
20
∴ F( jω)G( jω) =
−
∴ f ∗ g = 20(e −2t − 2−3t ) u (t )
jω + 2 jω + 3
F( jω) =



H ( jω ) =

24
jω

10 March 2006

j 2ω
4 + j 2ω

Vi ( jω ) =

69.


from Table 18.2

⎡ j 2ω ⎤ ⎛ 24 ⎞
24
Therefore Vo ( jω ) = ⎢
⎥ ⎜ jω ⎟ = 2 + jω
⎣ 4 + j 2ω ⎦ ⎝
⎠
In the time domain, then, we find
vo (t ) = 24e −2t u (t ) V



70.

h(t ) = 2e − t cos 4t

H ( jω ) =

10 March 2006

so fromTable 18.2,

2 (1 + jω )

(1 + jω )


2

+ 16

. Define output function f(t).

(a) I ( jω ) = 4πδ (ω )
⎡ 8π (1 + jω ) ⎤
8π
Therefore F(ω) = ⎢
⎥ δ (ω ) = 17 δ (ω ) .
2
⎣ (1 + jω ) + 16 ⎦
The time domain output is then given by f(t) = 4/17.
(b) I ( jω ) = 2e − jω
⎡ 4(1 + jω ) ⎤ − jω
Therefore F(ω) = ⎢
⎥e .
2
⎣ (1 + jω ) + 16 ⎦
The time domain output is then given by f(t) = 4e −(t −1) cos ⎡ 4 ( t − 1) ⎤ u (t − 1)
⎣
⎦

(c) We find the response due to a unit step u (t ) and treat i (t ) as two unit steps, each
shifted appropriately.

R ( jω ) =

r (t ) =

2(1 + jω ) ⎡
1 ⎤
⎢πδ (ω ) + jω ⎥
2
(1 + jω ) + 16 ⎣
⎦

1 1
e−t
+ sgn(t ) − 2 [ cos 4t − 4sin 4t ] u (t )
17 17
17

Therefore the system response is
2 ⎡
−( t + 0.25 )
{cos 4(t + 0.25) − 4sin 4(t + 0.25)}⎤ u (t + 0.25)
⎣1 − e
⎦
17
2
− t − 0.25 )
− ⎡1 − e (
{cos 4(t − 0.25) − 4sin 4(t − 0.25)}⎤ u (t − 0.25)
⎣
⎦
17


Chapter 18 solutions_to_exercises(engineering circuit analysis 7th)

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