chapter 18 solutions_to_exercises(engineering circuit analysis 7th)
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
1. (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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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
2. (a) ωo = 2π rad/s, f = 1 Hz, therefore T = 1 s. (b) ωo = 5.95 rad/s = 2π f rad/s, f = 0.947 Hz, therefore T = 1.056 s. (c) ) ωo = 1 rad/s = 2πf rad/s, f = 1/2π Hz, therefore T = 2π s.
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
3.
( ) 3 3cos(100 40 ) 4sin(200 10 ) 2.5cos300 Vv t t t tπ π π= − − ° + − ° + (a) V 3 0 0 0 3.000 Vav = − + + =
(b) 2 2 2 21V 3 (3 4 2.5 ) 4.962 V2eff = + + + =
(c) 2 2T 0100o
π .02 sπω π
= =
v ms = − − ° + ° + °
=
(d) (18 ) 3 3cos( 33.52 ) 4sin(2.960 ) 2.5cos(19.440 ) 2.459 V= −
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
4. (a)
t v 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 -2 (b) (c) min 3.330v =
2 2
2
max
4 sin 2 7.2 cos 4 04sin 2 7.2(cos 2 sin 2 )
4 16 414.724sin 2 7.2(1 2sin 2 ) 0.5817, 0.8595 sin 228.8
0.09881,0.83539 0.5593 for smaller max)
tt t t
t t x t
t v
v tπ π π π
π π π
3.330(
π π π
′ = − + =
∴ = −
− ± +∴ = − ∴ = = − =
∴ = ∴ =
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
5. (a) a0 = 0 (b) a0 = 0 (c) a0 = 5 (d) a0 = 5
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
6. (a) a0 = 0 (b) a0 = 0 (c) a0 = 100 (d) a0 = 100
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
7. (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 o o ocos(3 10 ) 3cos3 cos10 3sin 3 sin10t t t− = + a0 = 0, a1 = 3cos10o = 2.954, a2 = 0, b1 = 3sin10o = 0.521, b2 = 0
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
8. ao = 0
1 ( ) 2.5T
f t dtT
=∫ . a1 = a2 = 0 since function has odd symmetry
2
2
1 00 11
2 2 5b ( )sin 5sin cos 102 0 2
Tf t tdt tdt t
Tω π π
π π= = = − = −
−∫ ∫
2
2
2 00 11
2 2 5b ( )sin 2 5sin 2 cos 2 02 0 2
Tf t tdt tdt t
Tω π π
π= = = − =
−∫ ∫
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
9. ao = 2
2
0 00
1 1 2( ) 2 43 3 3
Tf t dt dt t
T= =∫ ∫ = .
a1 =2
2
00 00
2 2∫ ∫ 2 4 3( ) cos 2cos sin
3 3 3 2 3T
f t tdt t dt tT
π πωπ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2 0.551=
a2 =2
2
00 00
2 2∫ ∫ 4 4 3( ) cos 2 2cos sin
3 3 3 4 3T
f t tdt t dt tT
π πωπ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
4 0.276=
a3 =2
2
00 00
2 2∫ ∫ 6 4 3( )cos3 2cos sin
3 3 3 6 3T
f t tdt t dt tT
π πωπ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
6 0
2
2
1 00 00
2 2 2 4 3 2( )sin 2sin cos3 3 3 2 3
Tf t tdt t dt t
Tπ πω
π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠∫ ∫ 0.955= b
2
2
2 00 00
2 2 4 4 3 4( )sin 2 2sin cos3 3 3 4 3
Tf t tdt t dt t
Tπ πω
π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠∫ ∫ 0.477 b
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
10. h(t) = –3 + 8 sin πt + f(t) Use linearity and superposition. T = 2 s.
ao = 0
1 13 ( ) 3 2.52
Tf t dt
T= − + = −∫− + .
a2 = 0
b1 = 2
00 0
2 2 2( )sin 8 (1)sin 8 7.362
Tf t tdt tdt
Tω π8
π+ = + = − =∫ ∫
b2 = 2
00 1
2∫ ∫ 2( )sin 2 (1)sin 2 0
2T
f t tdt tdtT
ω π= =
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
11. (a) T 10 , F 0.1(2 4 2 2) 1.200av os a= = = × + × = (b)
2 22 2
0 0
222 2 3
000
1
(c)
F = −f (4 ) 0.2 (16 8 )5
1 80.2 16 4 0.2 32 163 3
= − +
⎡ ⎤ ⎛ ⎞= − + = − + =⎢ ⎥ ⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
∫ ∫ef t dt t t dt
t t t 1.9322
2 2 2
30 0 0
2 2
200
2
2 22 (4 )cos3 0.4 4cos 0.6 0.4 cos 0.610 10
1 11.6 sin 0.6 0.4 cos 0.6 sin 0.60.6 0.36 0.6
8 10 4sin1.2 (cos1.2 1) sin 1.2 0.045813 9 3
= × − × = −
⎛ ⎞= − +⎜ ⎟⎝ ⎠
= − − − = −
∫ ∫ ∫ta t dt t dt t t dt
tt t t
π π π
π π ππ π π
π ππ π π
π
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
12. (a) T = 8 − 2 = 6 s
(b) 1 Hz6of =
(c) 2 rad/so of 3π ω π= =
(d) 1 (10 1 5 1) 2.5= 6oa = × + ×
(e) 3 4
22 3
43
2 3
2
2 2 210sin 5sin6 3 3
1 30 2 15 2cos cos3 2 3 2 3
1 15 4 7.5 8 1 15 7.5cos 2 cos cos cos 2 (1.5) ( 1.5) 1.19373 3 3 3
⎡ ⎤= +⎢ ⎥
⎣ ⎦⎡ ⎤
= − −⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤∴ = − − − − = − − − = −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎣ ⎦⎣ ⎦
∫ ∫t tb dt dt
t t
b
π π
π ππ π
π ππ ππ π π π
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
13.
433 4
3322 3
433 4
32 32 3
2 6 6 1 10 510cos 5cos sin sin6 6 6 3
10 1 1sin 3 sin 2 sin 4 sin 3 03 2 2
1 1 1010sin 5sin cos cos3 3
10 1cos3 cos 2 cos3 2
5
⎡ ⎤⎡ ⎤= + = −⎢ ⎥⎢ ⎥
⎢ ⎥⎣ ⎦ ⎣ ⎦⎛ ⎞= − + − =⎜ ⎟⎝ ⎠
⎡ ⎤⎡ ⎤= + = − −⎢ ⎥⎢ ⎥
⎢ ⎥⎣ ⎦ ⎣ ⎦
= − − +
∫ ∫
∫ ∫
t ta dt dt t
b tdt t dt t t
π π π ππ π
π π π
t
π
π π π ππ π
π ππ
π
2 23 3
1 104 cos3 ( 1)2 3
⎛ ⎞− = − − =⎜ ⎟⎝ ⎠
+ =a b
π ππ
1.0610
1.0610
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
14.
(a) 2 23.8cos 80 1.9 1.9cos160 , T 12.5 ms, ave value 1.9160
ππ = + π = = =
πt t
(b)
(c) 23.8cos70 3.8sin80 ; , , T 2 ; ave value 0ππ − π ω = π ω = π = = =
πo ot t t t s
33.8cos 80 (3.8cos80 )(0.5 0.5cos 160 )1.9cos80 0.95cos 240 0.95cos80 2.85cos80 0.95cos 240
2T 25ms, ave value 80
π = π + π= π + π + π = π + π
π= =
π
t t tt t t t t
0=
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
15. T = 2 s (a)
( )
11
400
4 1
1 1
2 4 2 1sin cos 42 2 4
1 1 cos 44
max when 4 , 0.1252
× π= = −
ππ
∴ = − ππ
ππ = =
∫tt tb dt t
b t
t t s
(b) 41
4=
πb
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
16.
( ) 5 8cos10 5cos15 3cos 20 8sin10 4sin15 2sin 20= + − + − − +g t t t t t t t
(a) 25 T 1.25665π
ω = ∴ = =o s
(b) 5 104 3.183 Hz2
= β = = =π πo of f
(c) G 5= −av
(d) 2 2 2 2 2 2 21G ( 5) (8 5 3 8 4 2 ) 116 10.7702
= − + + + + + + = =eff
(e)
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
17.
[ ]0.1 0.1
0.1 0.10.1
0.1
T 0.2, ( ) V cos5 , 0.1 0.1
2 V cos5 cos10 5V cos(5 10 ) cos(10 5 )0.2
1 15V sin(10 5 ) sin(10 5 )10 5 10 5
V 2 2sin(10 5 )0.1 sin(102 1 2 1
− −
−
= = π − < <
= π π = π + π + π −
⎡ ⎤= π + π + π − π⎢ ⎥π + π π − π⎣ ⎦
= π + π +π + −
∫ ∫
m
n m m
m
m
f t t t
a t n t dt n t n t dt
n t n tn n
nn n
π
2 2
5 )0.1
V 2 2sin( 0.5 ) sin( 0.5 )2 1 2 1
V 2V2 2 1 1cos ( cos ) cos2 1 2 1 2 1 2 1
2V 4V2 1 2 1 coscos4 1 4 1
1 1V cos5 5V sin sin0.2 5 2 2
⎡ ⎤π − π⎢ ⎥⎣ ⎦⎡ ⎤= π + π + π − π⎢ ⎥π + −⎣ ⎦⎡ ⎤ ⎛= π + − π = π −⎜ ⎟⎢ ⎥π + − π + −⎣ ⎦ ⎝
− − − π= π = −
π − π −π π⎛= π = − −⎜π ⎝
m
m m
m m
o m m
n
n nn n
n n nn n n n
n n nnn n
a t dt
⎞⎠
0.1
0.1
2V
−
⎡ ⎤⎞ =⎟⎢ ⎥ π⎠⎣ ⎦
+
∫ m
2V 4V 4V 4V 4V( ) cos10 cos 20 cos30 cos 40 ...3 15 35 63
∴ = + π − π + π − ππ π π π π
m m m m mv t t t t t
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
18.
(a) 1even, wave2
−
(b) 0 for all ; 0; 0= = =n even ob n a a (c) 1 2 3 2
22
11
1 3
0, 0
8 10 6 205cos sin sin sin12 6 3 6 3 6
20 20 20sin sin 2.330, sin sin 2.1223 6 3 2 3
= = = =
π π π π⎛ ⎞= = = −⎜ ⎟π π ⎝ ⎠π π π⎛ ⎞ ⎛ ⎞∴ = − = = π − = − = −⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠
∫n
b b
π
n t n t n na dtn n
b a
a a
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
19. (a) 0
( ) 0.2sin1000 0.6sin 2000 0.4sin 3000= =
∴ = π + π +o na a
πy t t t t (b) 2 2 2Y 0.5(0.2 0.6 0.4 ) 0.5(0.56) 0.5292= + + = =eff (c) (2ms) 0.2sin 0.2 0.6sin 0.4 0.4sin 0.6 1.0686= π + π + π =y
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
20. (a) (b) (c) (d) (e)
33
5 522
3
5 52
5 5
4 2 5 32 6 5 3.2 15 10[ ] 0, 8cos sin sin sin6 6 6 10 3 3
4 2 5 32 6 15 10 3.2[ ] 0, 8sin cos cos ( 0.5)6 6 6 10 3 3
8 2 5 64 12 15[ ] 0, 8cos sin12 12 12 10
π π π π⎛ ⎞= = = = − =⎜ ⎟π π ⎝ ⎠
π − π π⎛ ⎞⎛ ⎞= = = − = − −⎜ ⎟⎜ ⎟π π⎝ ⎠⎝ ⎠
π= = =
π
∫
∫
t ta b a dt
tb a b dt
tc b a dt
0.88213
0.5093=
3
23
5 52
10sin6 6
8 10 64 12 15 10[ ] 0, 8sin cos cos12 12 12 10 6 6
π π⎛ ⎞− =⎜ ⎟⎝ ⎠
π π⎛ ⎞⎛= = = − −⎜ ⎟⎜π⎝ ⎠⎝
∫
∫td a b dt
3.801
1.0186π ⎞ =⎟⎠
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
21. T = 4 ms (a)
0.0040.004
00
1000 250 88sin125 cos1254 125
16 16cos 1 5.0932
×= π =
− π
π⎛ ⎞= − − = =⎜ ⎟π π⎝ ⎠
∫o πdt ta t (b)
0.004
10
0.004 0.004
10 0
0.004
0
1
24000 sin125 cos0.004
4000 sin125 cos500 2000 (sin 625 sin 375 )
cos625 cos375 3.2 5.3332000 (1 cos 2.5 ) (1 cos1.5 ) 0.6791625 375
4000 sin125
π= π
∴ = π π = π − π
π π⎛ ⎞= − + = − π − − π = −⎜ ⎟π π π π⎝ ⎠
= π
∫
∫ ∫
ta t dt
a t t dt t t dt
t t
b
0.004 0.004
0 0
sin 500 2000 (cos375 cos 625 )
1 1 12000 (sin1.5 ) (sin 2.5 ) 2000375 625 375 625
π = π − π
−⎡ ⎤ ⎛ ⎞= π − π = = −⎜ ⎟⎢ ⎥π π π⎣ ⎦ ⎝ ⎠
∫ ∫t t dt t t dt
1 2.716−π
(c) 4 0 : 8sin125− < < πt t (d)
[ ]
( )
0.004
1 10
0.0040.004
100
40000, 8sin125 cos 2508
cos375 cos1252000 sin 375 sin125 2000375 125
5.333 161 cos1.5 cos 1 3.3952
+
= = π π
π π⎡ ⎤∴ = π − π = − +⎢ ⎥π π⎣ ⎦
π⎛ ⎞= − π + − = −⎜ ⎟π π ⎝ ⎠
∫
∫
b a t t dt
t ta t t dt
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
22.
0.0010.001
0 0
1 3 5 7 9
1odd and wave 0, 0, 02
T 10 0.01
8 110sin 200 8000 cos 2000.01 200
40 40(cos 0.2 1) (1 cos 0.2 )
2.432, 5.556, 5.093, 2.381, 0.27
− ∴ = = =
= =
⎡ ⎤ −⎛ ⎞= π =⎢ ⎥ ⎜ ⎟π⎝ ⎠⎣ ⎦π
∴ = − π − = − ππ π
∴
∫
o n even
odd
odd
a a b
ms s
b n t dt n tn
b n nn n
= = = = =b b b b b 02
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
23.
( )
( )
/ 4
00.001
0
2
60.0013
2 2 2 0
2 2
1 8odd and wave, T 8 ( )sin2 T
2 250 1000 1000 sin 250T
1Now, sin sin cos , 250
10( ) 10 sin 250 250 cos 250250
16 sin 04
− = ∴ = ω
πω = = π ∴ = π
= − = π
= ∴ = π − π ππ
π∴ = −
π
∫
∫
∫
T
n o
o n
x
n
n
ms b f t n t dt
b t nt dt
x ax dx a ax ax a na
f t t b n t n t n tn
nbn 1 2
3 52 2
16cos 0 sin cos 0.24604 4 4 4 4
16 3 3 3 16 5 5 5sin cos 0.4275 ; sin cos 0.134219 4 4 4 25 4 4 4
0
−
π π π π π⎛ ⎞ ⎛ ⎞− + ∴ = − =⎜ ⎟ ⎜ ⎟π⎝ ⎠ ⎝ ⎠π π π π π π⎛ ⎞ ⎛ ⎞= − = = − =⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠
=even
n n b
b b
b
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
24. (a) odd, T = 4 (b) even, T = 4:
(c) 1odd, wave: T 82
− =
(d) 1even, wave, T 8 :2
− =
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
25. (a)
1,
2 2
2 2
20 1 2 20 205 sin sin 5 , V ( 1)0.4
V 20 5Z 4 5 2 4 10 , IZ (4 10 ) 1 2.5
5 1 2.5 12.5 5I1 6.25 (1 6.25 )
12.5 1 5 1cos5 sin 51 6.25 1 6.25
∞ π= + ∴ = = −
π π π π
−= + = + = = = −
π + +− +
∴ = − = −π + π +
∴ = − +π + π +
∑s sn snodd
snn fn
n
fn
fn
ntv v ntn n n
j jj n j nn j n j
j j n jn n n n
i ntn n n
j
n
nt
21,
1 12.5 51.25 cos5 sin 51 6.25
∞ ⎡ ⎤∴ = + − +⎢ ⎥+ π π⎣ ⎦∑fodd
i ntn n
(b) 2
21,
21,
22
1,
1 12.5Ae , , (0) 0, (0) 1.251 6.25
2 1 2(0) 1.25 1.25 tanh 0.2 0.553880.16 4 0.4
1 12.5 5A 0.55388, 0.55388 1.25 cos5 sin 51 6.25
∞−
∞
∞−
⎛ ⎞= = + = = + −⎜ ⎟+ π⎝ ⎠π
∴ = − = − π =π + π ×
⎡ ⎤∴ = − = − + + − +⎢ ⎥+ π π⎣ ⎦
∑
∑
∑
tn f n f
odd
fodd
t
odd
i i i i i in
in
i e nt nt
n n
nt
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
26. (a) 2 0.40 0.2 : 2.5(1 ) (0.2 ) 2.5(1 ) 1.78848 A− − π< < π = − ∴ π = − =tt i e i e (b) 2( 0.2 )0.2 0.4 : 1.78848 (0.4 ) 0.50902 A− − ππ < < π = ∴ π =tt i e i (c) 2( 0.4 )0.4 0.6 : 2.5 (2.5 0.50902) , (0.6 ) 1.9335− − π −π < < π = − − π =tt i e i
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
27. (a)
( )
1,
2 2
2
20 15 sin 5
20 sin 5
20V
1 1 20 / 1 20 / 1 20Z 2 2 V5 2 10 2 1/ 10 10 1 20 1 20
20 1 20 20 1V , 20 cos5 sin 51 400 1 400
20 1 15 sin 51 400
∞
= +π
=π
= −π
− π − π −= + = + ∴ = × = ×
+ +− −
−
∴ = × = − ++ π π +
∴ = +π +
∑sodd
sn
sn
n cn
cn cn
cf
v ntn
v ntn
jn
j n j n j n j n j n j n j n j n j n
n j v n nt ntn n n n
vn n1,
20cos5∞ ⎛ ⎞−⎜ ⎟
⎝ ⎠∑odd
nt nt (b) / 4Ae−= t
nv (c)
2 2 21, 1,
2 21,
/ 42
1,
20 20 1 1(0) A 5 A 51 400 (1/ 20)
1 tanh 5 tanh 1.23117(1/ 20) 4(1/ 20) 20 2 40
1A 0 5 1.23117 4.60811
∞ ∞
∞
−= + + = + −
π + π +
π π π= = π =
+ ×
∴ 20 1 1( ) 4.60811 5 sin 5 20cos5
1 400
∞−
= − + × = −π
⎛ ⎞∴ = − + + −⎜ ⎟π + ⎝ ⎠∑t
codd
∑ ∑
∑
codd odd
odd
n n
n
v t e nt ntn n
v
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
28. At the frequency ω = 10nπ
( )
( )3
3
10 10 10 5 10
20 10 5 10n
j n
j n
π
π
−
−
⎡ ⎤+ ×⎣ ⎦= Ω+ ×
Z and ( )8Sn j
nπ= −I
Therefore ( )80 10 0.0520 0.05n
j njn j n
ππ π
⎡ ⎤+= − ⎢ ⎥+⎣ ⎦
V .
In the time domain, this becomes
( )2
o 1 11 2
1 ( )
1 (0.005 )40( ) cos 10 90 tan 0.005 tan 0.00251 (0.0025 )n odd
nv t n n n
n n
πππ π π
π ππ
∞− −
=
+⎛ ⎞= − +⎜ ⎟⎝ ⎠ +
∑ −
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
29. At the frequency ω = nπ
( ) ( )
( )1
223
10 32 and 120 5 10
n
Ln Sn Sn jjn nπ π
−
−= = −
+ ×I I I −
Thus, in the time domain, we can write
( )( )
( )( )
1o 12
2 21 (odd)
320 1( ) 1 cos 90 tan 0.00025n 20 1 0.00025
n
Ln
i t n t nn
π ππ π
∞ −−
=
⎡ ⎤= − − −⎢ ⎥
⎢ ⎥ +⎣ ⎦∑
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
30.
( )
30.001 0.0053
3 2 / 6 10 1003
0 0.003
0.001 0.00551000 1000
0 0.003
5 3
3 3
3
10 100 1006
10 1 16 1000 1000
100 1001 (1 1 1 1) 10.6106 6
10.610; 10.610
2
−− × π × − π
− π − π
− π − π − π
−
⎡ ⎤= −⎢ ⎥
⎣ ⎦⎡ ⎤−
= +⎢ ⎥π π⎢ ⎥⎣ ⎦
= + + − = + − + = −π π
∴ = =
=
∫ ∫j t j t
j t j t
j j j
c e e
e ej j
e e e jj j
c j c
a
( )
0.001 0.0053
0 0.003
5
2 23 3 3 3 3 3 3
10 100cos100 100cos10006
2 10 1 sin 0 sin 5 sin 3 06 1000
1 1( ) 21.22 and 21.222 2
⎡ ⎤×π − π⎢ ⎥
⎣ ⎦×
= π − − π + π =π
= − = − ∴ = + =
∫ ∫t dt t dt
c a jb j b b a b
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
31. 0.001 0.002
5 400 400
0 0.001
0.001 0.002400 400
0 0.001
0.004000.001 40002 2
0.001
1T 5 10 1000.005
20,000 1000
120,000 ( 400 1)160 400
j nt j ntm
j nt j ntn
j ntj nt
n
ms c te dt e dt
c t e dt e dt
ec j nt en j n
− π − π
− π − π
− π− π
⎡ ⎤= = +⎢ ⎥
⎣ ⎦⎡ ⎤
∴ = +⎢ ⎥⎣ ⎦
∴ = π + +π − π
∫ ∫
∫ ∫
( )
2
3 3
0.4 0.8 0.41 2 2
2
1(50 10 100 10 ) 0.15 200 300.005
1 1 120,000 (1 0.4 )160 160 400
125 (1 72 ) (1.60597 51.488 ) 12.66515 15.91548 90 (1 144 1 72 )
12.665(1 72
o o
j j j
c a
c e j e ej
− −
− π − π − π
⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦
∴ = = × + × = × =
⎡ ⎤= + π − − −⎢ ⎥π π π⎣ ⎦
= ∠ − ° ∠ ° − + ∠ ° ∠ − ° − ∠ − °π
= ∠ −
2
) (1 1.2566) 12.665 15.915(1 144 1 72 )20.339 20.513 12.665 18.709 108 24.93 88.613.16625 144 (1 2.5133) 3.16625 7.9575(1 288 1 144 )8.5645 75.697 3.16625 15.1361 144 13.309 1
j j
c j j
° + − + ∠ − ° − ∠ − °= ∠ − ° − + ∠ − ° = ∠ − °= ∠ − ° + − + ∠ − ° − ∠ − °= ∠ − ° − + ∠ ° = ∠ 77.43°
(a)
(b)
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
32. Fig. 17-8a: V 8 V, 0.2 , 6000o os f ppsτ μ= = =
(a) 1 1T , 6000, 0.2 5 MHz6000 of s fτ μ
τ= = = ∴ = =
(b) 6000 Hzof = (c)
6 6
3
3
8 0.2 10 sin(1/ 2 3 12,000 0.2 10
(d) 6 6 6
7.270 mV× ×=
× ×3333 6
2 10 8 0.2 10 sin(1/ 2 333 12,000 0.2 10333.36 10 1/ 6000 1/ 2 333 12,000 0.2 10
c ππ
− −
−
× × × × ×= ∴ =
× × ×
(e) 1/ 5 MHzβ τ= = (f) (g)
6000 3 18,000 (closest) 1/ 6000 0.0036
c
c
ππ
− −× × × × × ×× = ∴ =
9.5998 mV∴ =
2000 22002 < < 2.2 Mrad/s kHz or 318.3 350.1 kHz2 2
6 kHz 6 53 318; 324,330,336,342,348kHz o
f f
f f n
ωπ π
∴ < < < <
= ∴ = × = ∴ = 5
6 6
2278 0.2 10 sin(1/ 2 227 12,000 0.2 10 8.470 mV
1/ 6000 ( )227 6 1362 kHz
c
f
π− −× × × × × ×= =
′′
= × =
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
33.
1 2 3T 5 ; 1, 0.2 0.2, 0.5 0.25, 1 2, 0, 4= = = − = + = − − =o nms c c j c j c j c n ≥ (a) (b)
1 1 1 2 2 3 32 1, 0.4 0.4 0.4, 1 0.5, 2( )
n n n o ojb c a c a jb jb j a jb j a jbv t t t t t t t
a1 0.4cos 400 cos800 2cos1200 0.4sin 400 0.5sin800 4sin1200π π π π π π
= − = ∴ = = − = − = − − = + − = −∴ = + + − + − +
(1 ) 1 0.4cos 72 cos144 2cos 216 0.4sin 72 0.5sin144 4sin 216s = + ° + ° − ° + ° − ° + °v m ∴
(1 ) 0.332Vv ms∴ = −
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
34. (a)
( )
( )
6
6
0.6 106
60.4 10
65
10T 5 2 1cos 25 5 10
5 104 10 sin 43.2 sin 28.82
n
n
ts c n dt
c n nn
n
μ π
π
−
−
×
−×
−
= ∴ = ××
×
(b) 41 (sin172.8 sin115.2 ) 0.06203
4c
π= ° − ° = −
(c) 6 6
6
0.2 10 0.2 10 0.085 10o oc a
− −
−
× + ×= = =
×
(d) maxa little testing shows is max 0.08oc c∴ = (e)
(f) 6740 10740 148 MHz
5×
= = =ofβ
1 sin 43.2 sin 28.8nc nnπ
∴ = × ° − °
∴ = ° − °
∫
( )
( )
3 310.01 0.08 0.8 10 sin 43.2 sin 28.8 0.8 10
125 sin 43.2 sin 28.8 1
− −× = × ∴ ° − ° ≤ ×
ok for 740
° − ° ≤
n nn
n nn
π
π>n
∴
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
35. T 1/16, 32oω π= = (a)
1/961/9696 96
30 0
3
16 4016 4096
20 40( 1) 4.244 V3 3
j t j
j
c e dt ej
c j e j
π π
π
π
j (b)
π π
− −
−
×= −
−
= − = −
∫
= −∴
3 3 3 3 3
23 3
642
Near harmonics are 2 32 Hz, 3 48 HzOnly 32 and 48 Hz pass filter 2
2 8.488 0, 8.488 V8.488 1I 1.4536 31.10 A; P 1.4536 5 5.283 W
5 0.01 96 2
1 640401/16 6
o o
n n n
j t
f fa jb c
a jb c j a b
j
c e dtj
π
π
−
= =− =
− = = − ∴ = =
= = ∠ − ° = × × =+ ×
= =−
1/9664 /96
0
2 2 2
2
22
( 1) 2.7566 4.7746 V4
2 5.5132 9.5492 11.026 6011.026 60I 2.046 65.39 A5 0.01 641P 2.046 5 10.465 W P2
j
tot
e j
a b c j
j
π
π
π
− − = −
− = = − = ∠ − °∠ − °
15.748 W
∴ = = ∠ − °+ ×
= × × = ∴
∫
=∴
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
36. ( ) 5[ ( 3) ( 2) ( 2) ( 3)]f t u t u t u t u t= + + + − − − − (a)
-3 -2 -1 0 1 2 3
f(t)
t
∞(b)
2 2 3
3 2 2
2 3 2 2 3 2
3 3 2 2 2 2
F( ) ( )
F( ) 5 10 5
5 10 5F( ) ( ) ( ) ( )
5 5 10( ) ( ) ( )
5 5( 2)sin 3 (
− ω
−∞
−− ω − ω − ω
− −
ω ω − ω ω − ω − ω
ω − ω ω − ω ω − ω
ω =
∴ ω = + +
∴ ω = − + − + −− ω − ω − ω
= − + + − + − +− ω − ω − ω
= − ω +− ω − ω
∫
∫ ∫ ∫
j t
j t j t j t
j j j j j j
j j j j j j
t e dt
j e dt e dt e dt
j e e e e e ej j j
e e e e e ej j j
jj j
j f
102)sin 2 ( 2)sin 2ω + − ω− ω
10 10 20 10F( ) sin 3 sin 2 sin 2 (sin 3 sin 2 )∴ ω = ω− ω+ ω = ω+ ωω ω ω ω
j
j jj
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
37. (a) (b) (c)
0
( )
0
( ) ( ), 0 F( ) ( )
F( )
∞ ∞− − − −
−∞
∞− +
= > ∴ = =
−1 1∴ = =
∫ ∫at j t at j t
a j t
+ +
t e u t a j f t e dt e e dt
j ea j a j
ω ω
ω
ω
ωω ω
f
6 ( )
( )( )
( ) ( ), 0 F( )
1 1F( )
∞− − +
∞− + −− +
= − > ∴ =
− − ⎡ ⎤ 1∴ = = − =⎣ ⎦+ + +
o j t
∫o
o
o o o
o
at atat a j to
t
at at a j ta j t
t
e e u t t a j e e dt
j e e e e ea j a j a j
ωf t
ω ωω
ω
ωω ω ω
[ ]
( )
0( )
2 20 2
( ), 0 F( )
F( ) ( ) 1 0 [ 1]( ) ( )
∞− − +
− +∞
= > ∴ =( )
1 1( )
∴ = − + − = −+ + +
∫at a j t
a j t
t te u t a j te dt
ej a j ta j a j a j
ω
ω
ω
ω ω − =ω ω
f
ω
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
38.
0 4
4 00
14
4 4
10 0
4 0 : ( ) 2.5( 4); 0 4 : ( ) 2.5(4 )
F( ) 2.5( 4) 2.5(4 5)
ln 1 , let I 2.5(4 ) ( )
I 2.5(4 ) F( ) 2.5 (4 )( )
F( ) 5 (4 )cos 20
− ω − ω
−
ωτ
ωτ ω − ω
− < < = + < < = −
∴ ω = + + −
= τ ∴ = − τ − τ
∴ = − τ τ ∴ ω = − +
∴ ω = − ω = ×
∫ ∫
∫
∫ ∫
j t j t
st j
j j t j
t f t t t f t t
j t e dt e dt
t e d
e d j t e e dt
j t t dt
t
44 4
0 00
402
2 2 2
2sin 210× ω⎛ ⎞= ⎜ ⎟ω ω⎝ ⎠2
2
1 sin 5 cos
20 5F( ) sin 4 (cos sin )
20 5 5 5sin 4 (cos 4 1) 4 sin 4 (1 cos 4 )
2 5or, F( ) sin 2
ω − ωω
∴ ω = ω− ω + ω ωω ω
= ω − ω − − ω ω = − ωω ω ω ω
ω = ω
∫ ∫t t dt
j t t t
j
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
39.
(1 ) (1 )
(1 ) (1 ) (1 ) (1 )
( ) 5sin F( ) 5sin
5F( ) ( )2
5 [ ]2
5 1 1F( ) ( ) ( )2 (1 ) (1 )2.5 2.5( ) (
1 1
π− ω
−π
π− − ω
−π
π−ω − +ω
−π
π −ω − π −ω − π +ω π +ω
− πω πω
= − π < < π ∴ ω =
∴ ω = −
= −
⎡ ⎤ω = − − −⎢ ⎥− ω − + ω⎣ ⎦
−= − + −
− ω + ω
∫
∫
∫
j t
jt jt j t
jt jt
j j j j
j j
f t t t j t e dt
j e e e dtj
e e dtj
j e e e ej j j
e e
2 2 2
)
2.5 2.5 1 1( 2sin ) ( 2sin ) 5sin1 1 1
,
1
1
− ⎛ ⎞−⎜ ⎟+ ω⎝ ⎠
j 1 1 10sin 10sin5sin ( 1)1 1
− πω πω− +
= πω − πω = πω −− ω + ω − ω
+ ω + − ω πω πω⎛ ⎞= πω − = − =⎜ ⎟− ω − ω ω −⎝ ⎠
j je e
j j j
j j
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
40.
/ 2/ 2
/ 2 / 2
/ 2 / 2/ 2
(1 ) (1 )
/ 2
/ 2
/ 2
/ 2
( ) 8cos [ ( 0.5 ) ( 0.5 )]
F( ) 8cos 4 ( )
4
1 14(1 ) (1 )
14 ( )(1 )
π− π
π π− ω − − ω
−π −π
π−ω − +ω
−π
π− ω − − ω
−π
− πω
= + π − − π
∴ ω = = +
⎡ ⎤= +⎣ ⎦
⎧ ⎫⎪ ⎪= −⎨ ⎬− ω + ω⎪ ⎪⎩ ⎭
= − −− ω
∫ ∫
∫
j t jt jt j t
jt jt
j tjt jt j t
j
f t t u t u t
j te dt e e e dt
e e dt
e e e ej j
je j ej
/ 2 / 2 / 2
2 2
1(1 )
1 1 14 2cos 2cos 8cos1 2 1 2 2 1 1
2 cos / 28cos 162 1 1
πω − πω πω⎧ ⎫⎡ ⎤ ⎡− − −⎨ ⎬⎣ ⎦ ⎣+ ω⎩ ⎭πω πω πω⎧ ⎫ ⎛= × + × = +⎨ ⎬ ⎜ ⎟− ω + ω − ω + ω⎩ ⎭ ⎝
πω πω= =
− ω − ω
j jje jej
1
⎤⎦
⎞⎠
j
ω = ∴ j
(a) 0 F( 0) 16=
(b) 16cos 720.8, F( 0.8)0.36
°ω = =j 13.734=
(c) 16cos(3.1 90 )3.1, F( 3.1) × 0.29071 3.12
°ω = =
−j = −
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
41.
[ ]
( )22
2 2
2 2
1(a) (b) (c)
F( ) 4 ( 2) ( 2) ( ) F( )2
4 2 1 2( )22 4 5( ) 2sin 2 sin 2 (0.8) sin1.6
2
∞ω
−∞
ω ω −
− −
ω = ω+ − ω ω− ∴ = ω ωπ
1.5909
∴ = ω = = −π π π
= = ∴ =π π π
∫
∫
j t
j t j t j t j t
rad
f t e j d
f t e d e e ejt j t
f t j t t ft t
=
j u
∴
2 2
0(2 ) ( 2 )
0
2
2
4F( j e) 4 ( )2
2 2( )
2 1 1 2 1 1 2 4(1 0) (0 1)2 2 2 2 48 8( ) (0.8)
(4 ) 4.64
∞− ω − ω + ω
−∞
∞+ ω − + ω
−∞
ω = ∴ = ωπ
∴ = ω + ωπ π
⎡ ⎤ ⎛ ⎞= − + − = + =⎜ ⎟⎢ ⎥π + − + π + − π +⎣ ⎦ ⎝ ⎠
∴ = ∴ =π + π×
∫
∫ ∫
j t
jt j t
f t e d
f t e d e d
0.5488=
jt jt jt jt t
f t ft
[ ]
( )
( )
0.5 0.5
0.5 0.50.5
( ) ( 0.5 0.5 )
0.5
0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
F( j u) 4cos ( 0.5) ( 0.5)
4 2 1( ) cos2 2
1
1 1 1( ) ( )
ω πω − πω ω
− −
π+ ω − π− ω
−
π+ − π− − π+ π−
ω = πω ω+ − ω −
∴ = πω× ω = + ωπ π
⎡ ⎤= + ω⎣ ⎦π
= − + −π π + −π +
∫ ∫
∫
j t j j j t
j jt j j t
j j t j j t j j t j j
u
f t e d e e e d
e e d
e e e ej t j t ( )
( ) ( )0.5 0.5 0.5 0.5
2 2 2 2
1 1 1( ) ( )
1 1 1 2cos 0.5 1 12cos 0.5 2cos0.5
2 42cos 0.5 cos 0.5 (0.8)
− −
⎡ ⎤⎢ ⎥⎣ ⎦⎡ ⎤
= + + − −⎢ ⎥π π + −π +⎣ ⎦⎡ ⎤ ⎛ ⎞= − = −⎜ ⎟⎢ ⎥π π + −π + π π + −π +⎣ ⎦ ⎝ ⎠
−⎛ ⎞= = =⎜ ⎟− π π −⎝ ⎠
t
j t j t j t j tje je je jej t j t
tt tt t t t
t tt t
0.3992∴ f
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
42. 1.5( ) 20 ( 2) V= − −tv t e u t (a)
21.5 1.5
2(1.5 ) 3 2 3
F j e( ) 20 ( 2) 20
20 20 20F ( 0)1.5 1.5 1.5
∞ −− ω − ω
−∞ −∞
−− ω − + ω −
−∞
ω = − − =
= = ∴ =− ω − ω
∫ ∫t j t t j tv
j t jv
u t e dt e dt
e e j ej j
0.6638=
3 2
3 4
20(b) F ( ) A ( ) B ( )
1.520F ( 2) 0.39830 282.31 0.08494 0.38913
1.5 2A (2)
− ω
−
ω = ω + ω =− ω
∴ = = ∠ ° = −−
0.08494∴ =
jv v v
jv
v
j e ej
j e e jj
(c) B (2) 0.3891= −v
(d) F ( 2) 0.3983=v j (e) φv(j2) = 282.3o or -77.69o
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
43. [ ]I( ) 3cos10 ( 0.05 ) ( 0.05 )ω = ω ω + π − ω − πj u u 0.05
2 2
0.05/ 20/ 20
/ 20/ 20
1 2(a) W 4 I( ) 9cos 102
18 1 1 9 9 1cos 20 0.1 sin 202 2 20
∞ π
−∞ − π
ππ
−π−π
= × ω ω = ω ωπ π
⎛ ⎞= + ω ω = × π +⎜ ⎟π π π⎝ ⎠
∫ ∫
∫
j d d
d
0.9 Jω = (b)
9 9 1(1 cos 20 ) 0.45 2 2sin 2020
0.05 2 0.1sin 20
ω
−ω
⎡ ⎤+ ω ω = = ω + × ω⎢ ⎥π π
, 0.04159 rad/s⎣
π = ω + ω
∫x
x
ω =
x⎦
x
x x
d
x∴
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
44. 4( ) 10 ( )−= tf t te u t (a) 8
2 2 8 21
0 0 0
∞
(b)
(c) ( )22
2
16100 )(+
=ω
ωjF
mJ/Hz 90.63 )( 2
0=
=ωωjF , mJ/Hz 7.669 )( 2
4=
=ωωjF
W ( ) 100 100 (64 16 2)( 512)
100 2512
0.3906 J
∞ ∞ −−
Ω = = = × + +−
= ×
∫ ∫t
t ef t dt t e dt t t
=
(4 )4 (4 )
20 0
2 2
10F( )16
ω =ω +
j
10F( ) {10 ( )} 10 [ (4 ) 1(4 )
10(4 )
∞∞ − + ω− − + ωω = = = − + ω −
+ ω
= ∴+ ω
∫j t
t j t ej te u t t e dt j tj
j
F
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
45. 2( ) 8 V−= tv t e
(a) 2 41
0
W ( ) 2 64 32 J∞ ∞
−Ω
−∞
= = × =∫ ∫ tv t dt e dt
(b) 2
0(2 ) (2 )
00
(2 ) (2 )2
0
(c)
F ( ) ( ) 8
F ( ) 8 8
8 8 82 2 2
∞ ∞−− ω − ω
−∞ −∞
∞− ω − + ω
−∞
∞− ω − + ω
−∞
ω = =
8 32 F ( )2 4
∴ ω = +
= − = + ω− ω + ω − ω + ω
∫ ∫
∫ ∫
tj t j tv
j t j tv
j t j t
j e v t dt e e dt
j e dt e dt
e e jj j j
= =+ ω vj
1
1
2 21 1
2 2 21
11 1 1 12 21 1
11 112
1
1 32 32 10.9 32 tan2 ( 4) 2 8( 4) 16 2
16 1 2 20.9 2 tan8( 4) 16 2 4 2
20.45 tan s (by SOLVE)4 2
ω−
−ω
−
−
⎡ ⎤ω ω× = ω = +⎢ ⎥π ω + π ω +⎣ ⎦
⎡ ⎤ ⎡ ⎤ω ω ω ω
2.7174 rad/
∴ = × + = +⎢ ⎥ ⎢ ⎥π ω + π ω +⎣ ⎦ ⎣ ⎦ω ω
π = +ω +
∫ d
∴ω =∴
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
46. (a) Prove: { ( )} { ( )} ( ) Let
{ ( )} ( ) { ( )}
∞− ω − ω
−∞
∞− ω − ω− ωτ
−∞
− = = − − = τ
∴ − = τ =
∫
∫
o
o o
j t j to o
j t j tjo
f t t e f t f t t e dt t t
f t t f e e dt e f t
F F
F F
o (b)
(c) (d) Prove: { ( )} F( ) Let 1 in (c) above− = − ω =f t j kF (e)
Prove: { ( )} { ( )} Let , ,
, { ( )} ( ) ( )
We assume ( ) 0 { ( )} { ( )}
∞− ω − ω − ω
−∞
∞∞− ω − ω
−∞−∞
= ω = = = − ω
= = ∴ = + ω
±∞ = ∴ = ω
∫
∫
j t j t j t
j t j t
dff t j f t e dt u e du j edt
dv df v f f t f t e j f t e dt
f f t j f t
F F
F
F F
/
1Prove: { ( )} F ( ) Let , 0
1 1{ ( )} ( ) F
1If 0, limits are interchanged and we get: F
1{ ( )} F
∞− ω
−∞
∞− ωτ
−∞
ω⎛ ⎞= = τ = >⎜ ⎟⎝ ⎠
ω⎛ ⎞∴ = τ τ = ⎜ ⎟⎝ ⎠
ω⎛ ⎞< − ⎜ ⎟⎝ ⎠
ω⎛ ⎞∴ = ⎜ ⎟⎝ ⎠
∫
∫
j t
j k
jf kt f kt e dt kt kk k
jf kt f e dk k k
jkk k
jf ktk k
F
F
F
rove: { ( )} F( ) Now, F( ) ( )
dF( ) ( )( ) { ( )} { ( )} ( )}
∞− ω
−∞
∞− ω
−∞
= ω ω =ω
ω= − = − ∴ = ω
ω
∫
∫
j t
j t
dtf t j j j f t e dtd
j
P
f t jt e dt j tf t tf f j f td
F
F F F∴
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
47. (a) (b) ( ) 4[sgn( 1) ( )] {4sgn( 1) ( )} { 4 ( )} 4= − δ ∴ − δ = − δ =f t t t t tF F −
f ( ) 4[sgn( ) ( 1)] {4[sgn( ) ( 1)] {4sgn(1) ( 1)} {4 ( 1)} 4 − ω= δ − ∴ δ − = δ − = δ − = j
(c) (10 30 ) (10 30 )
30 10 30 10 / 6 / 6
/ 6 / 6
4( ) 4sin(10 30 ) {4sin(10 30 )2
{ 2 2 } 2 2 ( 10) 2 2 ( 10)4 [ ( 10) ( 10)]
− ° − − °
− ° ° − − π π
− π π
⎧ ⎫⎡ ⎤= − ° ∴ − ° = − =⎨ ⎬⎣ ⎦⎩ ⎭− + = − πδ ω − + πδ ω+
= − π δ ω− − δ ω+
j t j t
j j t j j t j j
j j
f t t t e ej
j e e j e e j e j e
F F
F
j e e
t t t t t t t eF F F
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
48. (a) ( ) A cos( ) F( ) {A cos cos A sin sin }
A cos { [ ( ) ( )]} Asin [ ( ) ( )]
A{cos [ ( ) ( )] sin [ ( ) ( )]}
F( ) A[ ( ) ( )]φ − φ
= ω + φ ∴ ω = φ ω − φ ω =
⎧ ⎫πφ π δ ω + ω + δ ω − ω − φ δ ω − ω − δ ω + ω =⎨ ⎬
⎩ ⎭π φ δ ω + ω + δ ω − ω + φ δ ω − ω − δ ω + ω
∴ ω = π δ ω − ω + δ ω + ω
o o
o o o o
o o o oj j
o o
ot j t t
jj
Ff t
j e e (b) 2
2
2 1( ) 3sgn( 2) 2 ( ) ( 1) F( ) 3 2 ( )
6 1F( ) 2 ( )
− ω − ω
− ω − ω
⎡ ⎤= − − δ − − ∴ ω = × × − − πδ ω +⎢ ⎥ω ω⎣ ⎦
⎡ ⎤∴ ω = − − − πδ ω −⎢ ⎥ω ω⎣ ⎦
j j
j j
f t t t u t j e ej
(c)
j
j j e e j
2 2 2 2
1( ) sinh ( ) F( ) [ ] ( )2
1 1 1 1F( )2 2 2( )
−⎧ ⎫= ∴ ω = −⎨ ⎬⎩ ⎭
+ ω+ − ω −∴ ω = − = =
− + ω + ω − − ω ω +
kt ktkt u t j e e u tFf t
k j k j kjk j k j k k
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
49.
113
3 3
1(a) F( ) 3 ( 3) 3 ( 1) ( ) [3 ( 3) 3 ( 1)]2
3 3 1 3( ) ( )2 2 2
3(5) (1 5 1 1510
∞ω
−∞
ω ω + −
− −
ω = ω+ − ω− ∴ = ω + − ω − ωπ
(b) so f(5) = 0.1039∠73.52o
(c)
) 0.10390 106.48
∴ = = = −π π π
= − ∠ − ∠ −π
∫
∫
j t
j t j t jt j t
rad rad = ∠ − °
j u u f t u u e d
f t e dt e e ejt j t
f j∴
3
F( ) 3 ( 3 ) 3 ( 1)F( ) 3 F ( )
3( ) 3 ( ) ( ) (5) 0 0.10390 106.482
−
ω = − − ω + ω− →∴ ω = − ω
= δ − − ∴ = − ∠ − °π
a
jt j t
uj j
f t t e e fj t
j u
3
2 1F( ) 2 ( ) 3 ( 3 ) 3 ( 1) Now, {2 ( )}2
1 3 1( ) ( ) (5) 0.10390 106.482
−
ω = δ ω + − − ω + ω − δ ω = =π π
⎡ ⎤0.3618 15.985+= + − − ∴ = − ∠ − °⎢ ⎥π π π⎣ ⎦
jt j t
j u u
f t e e fj t
F
= ∠ °∴
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
50. 3 3(a) jF( ) 3 3 ( 1)
1ω = + + + δ ω −
+ ω ω 1.5( ) 3 ( ) 1.5sgn( ) 3 ( )−∴ = + + δ +
πt jt
j j
f t e u t t t e 1 sin 8 / 2(b)
(c) 32 2 2
6(3 ) 6(3 )F( ) ( ) 3 cos 2 ( )(3 ) 4 (3 ) 2
−+ ω + ωω = = ∴ =
+ ω + + ω +tj jj f t t u t
j j
F( ) 5sin 4 8 2.58 / 2ω
ω = ω = ×ω ω
j
( ) 2.5[ ( 4) ( 4)]∴ = + − −f t u t u t
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
51.
1/ 2
11
/ 22 2
1
/ 2 / 22 2 2 2
/ 2 / 22 2
T 4, periodic; find exp l form
1 104
12.5/ 2 / 4
1 1 1 12.52 / 4 / 2 / 4
1 42.5 ( ) (/ 2
− π
−
− π
−
− π π
− π π
′=
∴ =
⎡ ⎤⎛ ⎞∴ = −⎢ ⎥⎜ ⎟− π − π⎝ ⎠⎣ ⎦
⎡ ⎤⎛ ⎞ ⎛ ⎞∴ = + − +⎢ ⎥⎜ ⎟ ⎜ ⎟− π π π π⎝ ⎠ ⎝ ⎠⎣ ⎦
= − − +π π
∫ jn tn
jn tn
jn jnn
jn jn
c te dt
tc ejn n
c e ejn n jn n
e ejn n
/ 2 / 2
2 2
/ 22 2
2 2
)
5 102cos 2sin2 2
10 20( ) cos sin2 2
10 20F( ) cos sin 22 2 2
− π π
∞π
−∞
∞
−∞
⎡ ⎤−⎢ ⎥
⎣ ⎦π π⎛ ⎞= × + −⎜ ⎟π π ⎝ ⎠
π π⎡ ⎤∴ = −⎢ ⎥π π⎣ ⎦π π π⎡ ⎤ ⎛ ⎞
⎝ ⎠
∑
jn jn
jn t
e e
j n njn n
j n nf t j en n
∴ ω = − πδ ω−⎜ ⎟⎢ ⎥π π⎣ ⎦∑ j n n nj j
n n
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
53. 1F( ) 20 ( 20 )! 1
1 1 1 1 120 ( ) ( 20) ( 20) ( 40) ( 40)1 1 1 1 1 1 2 1 3
1 1( 60) ( 60) ...7 7
20 2010 ( ) [ ( 20) ( 20)] [ ( 40) ( 40)]2 3
20 20[ ( 60) ( 60) [ (7 25
∞
−∞
ω = δ ω−+
⎡= δ ω + δ ω + + δ ω− + δ ω + + δ ω−⎢ + + + +⎣⎤+ δ ω+ + δ ω− + ⎥⎦
= δ ω + πδ ω+ + πδ ω− + πδ ω+ + πδ ω − +π π
πδ ω + + πδ ω− + πδπ π
∑j nn
80) ( 80)] ...
10 20 20 20 20( ) cos 20 cos 40 cos 60 cos80 ...2 2 3 7 2520 1 1 1 10.25 cos 20 cos 40 cos 60 cos80 ...
2 3 7 2520 1 1 1 1(0.05) 0.25 cos1 cos 2 cos3 cos 4 ... 1.3858
2 3 7 25
ω + + πδ ω − +
∴ = + + + + +π π π π π
⎡= + + + + +⎢π ⎣⎡ ⎤∴ = + + + + + =⎢ ⎥π ⎣ ⎦
rad
f t t t t t
t t t t
f
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
54. Input ( ) 5[ ( ) ( 1)] ( ) ( ) ( )t
x t u t u t y t x z h t z d−∞
= = − − = −∫ z
u t
(a) (b) h t( ) 2 ( )h t u t= ( ) 2 ( 1)− u t= − (c) h t ( ) 2 ( 2)= x(t – z)
zt-1 t h( z)
z
2
y(t)
t
10
32
t < 0: y(t) = 0 1: ( ) 0t y t< = 2 : ( ) 0t y t< = 0 t< <1: 2 : 1 t< < 2 3t :< <
0
( ) 10 10= =∫t
y t dz t ∫ ==t
tdzty1
1) - 10( 10 )( ∫ ==t
tdzty2
2) - 10( 10 )(
t > 1: t > 2: t > 3:
∫ ==t
t
dzty1-
10 10 )( ∫ ==t
t
dzty1-
10 10 )( ∫ ==t
t
dzty1-
10 10 )(
5
x(t – z)
z t-1 t
x(t – z)
zt-1 t h( z)
z
2
h( z)
z
2
1 2
y(t)
t
10
1
y(t)
t
10
1 2
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
55. ( ) 5[ ( ) ( 2)]; ( ) 2[ ( 1) ( 2)]x t u t u t h t u t u t= − − = − − −
1
0
2
2
( ) ( ) ( )
1: ( ) 0
1 2 : ( ) 10 10( 1)
2 3: ( ) 10
3 4 : ( ) 10 10(2 2) 10(4 )
4 : ( ) 0( 0.4) 0; (0.4) 0; (1.4) 4
(2.4) 10; (3.4) 6; (4.4) 0
−∞
−
−
= −
< =
< < = = −
< < =
< < = = − + = −
> =∴ − = = =
= = =
∫
∫
∫
t
t
t
y t x z h t z dz
t y t
t y t dz t
t y t
t y t dz t
t y ty y y
y y y
t
t
or…. same answers as above
0
1
2
2
( ) ( ) ( )
1: ( ) 0
1 2 : ( ) 10 10( 1)
2 3: ( ) 10
3 4 : ( ) 10 10(2 2) 10(4 )
4 : ( ) 0
∞
−
= −
< =
< < = = −
< < =
< < = = − + = −
> =
∫
∫
∫
t
t
y t x t z h z dz
t y t
t y t dz t
t y t
t y t dz t
t y t
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
56.
2
( ) 2( )
0
2 20
02 2
2 2
( ) 3[ ], ( ) ( )
( ) ( ) ( )
3[ ]
13 [ ] 32
3 ( 1) 1.5 ( 1)( ) 3(1 ) 1.5(1 ) 1.5 3 1.5 , 0
− −
−∞
− − − −
− −
− −
− − − −
= − =
= −
= −
⎡ ⎤= − ⎢ ⎥⎣ ⎦= − − −
∴ = − − − = − + >
∫
∫
t t
t
tt z t z
tt z t t Z
t t t t
t t t t
h t e e x t u t
y t x z h t z dz
e e dz
e e e e
e e e ey t e e e e t
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
57.
0
( ) ( 2) ( )
2( ) (5 ), 2 53
∞
= −
= − < <
∫y t x t h z dz
h t t t
(a)
5 5
2 2
2 20( ) 10 (5 ) (5 )3 3
Note: ( ) is in window for 4 6
= × − = −
< <
∫ ∫ (b)
y t z dz
h z t
z dz
52
2
20 1( ) (5 )3 210 (0 9) 30 at 53
z⎛ ⎞= − −⎜ ⎟⎝ ⎠
= − − t= =
y t
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
58. ( 2)
0
( ) 5 ( 2), ( ) (4 16) [ ( 4) ( 7)], ( ) ( ) ( )∞
− −= − = − − − − = −∫tx t e u t h t t u t u t y t x t z h z dz
=
(a) 6 : ( ) 0 (5) 0t y t y< = ∴ (b)
6(8 2)
46 6
6 6
4 46
6 6 6 4
46 6 4 2 2
2
8 : (8) 5 (4 16)
(8) 20 80
20 ( 1) 80 ( )1
20 (5 3 ) 80 80 20 80 6020(1 ) 22.71
− − −
− −
− −
− −
−
= = −
2− −
∴ = −
⎡ ⎤= − − −⎢ ⎥
⎣ ⎦= − − + = + −
= + =
∫
∫ ∫
z
z z
z
t y (c)
e z dz
y e z e dz e e dz
ee z e e e
e e e e e ee
7 4
7(10 2)
47
8
47 7
8 8 8 7 84
4 48 7 4 1 4 1 4
10 : (10) 5 (4 16)
(10) 20 ( 4)
(10) 20 80 20 [ ( 1)] 80 ( )
20 (6 3 ) 80( ) 40 20 15.081
− − −
−
− − − −
− − − − −
= = −
∴ = −
∴ = − = − −
= − − − = + =
∫
∫
∫ ∫
z
z
z z z −
e z dz
y e e z dz
t y
y e ze dz e e dz e e z e e e
e e e e e e e
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
59.
0
0 0
0
( ) sin , 0 ; 0 elsewhere, Let ( ) ( )
( ) ( ) ( )
0 : ( ) 0
0 : ( ) sin sin
1( ) (sin cos )2
1 [ (sin cos ) 1]21 (sin cos )2
−
∞
− + −
−
−
−
= < < π =
= −
< =
< < π = × =
⎡ ⎤∴ = −⎢ ⎥⎣ ⎦
= − +
= − +
∫
∫ ∫
t
t tt z t z
tt z
t t
t
h t t t x t e u t
y t x t z h z dz
t y t
t y t z e dz e e z d
y t e e z z
e e t t
t t e
z
y
y
(a) (1) 0.3345+= (b) (2.5) 0.7409= (c)
0
0
: ( ) sin
1 1: ( ) (sin cos ) ( 1) 12.0702 2
(4) 0.2211
π−
π− − π
> π =
⎡ ⎤> π = − = + =⎢ ⎥⎣ ⎦−
∴ =
∫t z
t z t t
t e e z dz
y y t e e z z e e e
y
y y
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
60.
0
1
21
2
2
2 3
( ) 0.8( 1)[ ( 1) ( 3)],( ) 0.2 ( 2)[ ( 2) ( 3)]
( ) ( ) ( ) ,
3 : ( ) 0
3 4 : ( ) 0.8( 1)0.2( 2)
( ) 0.16 ( 2 2 2)
1 10.16 [ ( 1) 2 2 ] 0.16 ( 1)3 2
∞
−
−
= − − − −= − − − −
= −
< =
< < = − − −
∴ = − − + − +
= − + + + − = − + +
∫
∫
∫
t
t
x t t u t u th t t u t u t
y t x t z h z dz
t y t
t y t t z z dz
y t tz t z z z dz
z t z t dz z t11
2
22
3 2
3 2 2 2
3 2
(2 2 )
1 8 1 10.16 ( 1) ( 1) ( 1) ( 1) 4 (2 2 ) ( 1 2)3 3 2 21 1 8 1( ) 0.16 ( 1) ( 1) 2 2 2 6 2 63 3 3 2
1 1 1 10.16 1 2 1 6 3 86 2 2 2
−− ⎡ ⎤+ −⎢ ⎥⎣ ⎦
⎡ ⎤= − − + + + − − + + − − −⎢ ⎥⎣ ⎦⎡ ⎤∴ = − + − + + + − − − − + − − +⎢ ⎥⎣ ⎦⎡ ⎤⎛ ⎞ ⎛ ⎞= + − − + − − + + + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
∫tt
z t z
t t t t t t
y t t t t t t t t t t
t t t 3 21 3 9 90.166 2 2 2
⎛ ⎞= − +⎜ ⎟⎝ ⎠
−t
t t
(b)
(a)
3(4.8) 90.67 10y −∴ = ×
3
333 2
22
(3.8) 13.653 10
1 14 5 : ( ) 0.16( 1) ( 2) 0.16 ( 1) (2 2 )3 2
1 1( ) 0.16 (27 8) ( 1)5 (2 2 )13 219 110.16 2.5 2.5 2 2 0.16 0.53 6
−∴ = ×
⎡ ⎤< < = − − − = − + + + −⎢ ⎥⎣ ⎦
⎡ ⎤= − − + + + −⎢ ⎥⎣ ⎦⎡ ⎤ ⎛ ⎞= − + + + − = −⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠
∫t y t t z z dz z t z t z
y t t t
t t t
y
∴
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
61. 2 2
0
2( ) 2
0
2 2
02
( ) 10 ( ), ( ) 10 ( )
( ) ( ) ( )
( ) 10 10
100 100
( ) 100 ( )
− −
∞
− − −
− −
−
= =
= −
∴ =
= = ×
∫
∫
∫
t t
tt z z
tt t
x
∴ = t
t e u t h t e u t
y t x t z h z dz
y t e e dz
e dz e t
y t t e u t
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
62. 4( ) 5 ( )th t e u t−= (a)
0.88 0.8 6.4
10.1
25W 25 ( ) 1.3990 J8
25% 1.3990 / 100%8
− − −Ω = = − =
⎛ ⎞
44.77%= ×⎜ ⎟
⎝ ⎠
∫ te dt e e
= (b)
∴
221
1 200
11
5 1 25 25 1H( ) W tan4 16 4 4
25 1 0.9224W tan 0.9224 J % 100%4 2 25 / 8
−Ω
−Ω
ωω = ∴ = ω =
ω + π ω + π
29.52%= = ∴ = ×π
∫j
=
j d
∴
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
63.
22 2 2F( ) ( ) (2 2 ) ( )(1 )(2 ) 1 2
− −ω = = − ∴ = −+ ω + ω + ω + ω
t tj f t e e u tj j j j
(a) 2 3 41
0
4 8 4 1W (4 8 4 )2 3 4
J∞
− − −Ω = − + = −∫ t t te e e dt
e e e e tf e e
− − −
− − ×
= − + = − + = = =
3+ =
(b) f t 2
0.69315 2 0.69315max
( ) 2 4 0, 2 4 0, 2, 0.693152( ) 0.5
t t t t
=
∴ = −
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
64.
2 3
1 1/ 6 1/ 2(a)
1/ 3)(2 )(3 ) 2 3
ω = = − +ω + ω + ω ω + ω + ω
jF(j
1 1 1( ) sgn( ) ( ) ( )12 2 3
− −∴ = − +t t
j j j j j
f t t e u t e u t
2 3
1 1/ 6 1/ 2(b) 2 / 3)(2 )(3 ) 2 3
+ ωω = = + −
ω + ω + ω ω + ω + ωjjF(
j 1 1 2( ) sgn( ) ( ) ( )
12 2 3− −∴ = + −t t
j j j j j
f t t e u t e u t 2
2 3
(1 ) 1/ 6 1/ 2 4 / 3(c) )(2 )(3 ) 2 3
+ ωω = = − +
ω + ω + ω ω + ω + ω (d)
F(
1 1 4( ) sgn( ) ( ) ( )12 2 3
− −∴ = − +t t
jj j j j j j
f t t e u t e u t
j
3
2 3
(1 ) 1/ 6 1/ 2 8 / 3) 1(2 )(3 ) 2 3
+ ωω = = + + −
ω + ω + ω ω + ω + ωF(
1 1 8( ) ( ) sgn( ) ( ) ( )12 2 3
− −∴ = δ + + −t t
jj j j j j j
f t t t e u t e u t
j
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
65. ( ) 2 ( )th t e u t−=
(a) 1 2H( ) 21 1
ω = × =+ ω + ω
jj j
(b) V1 1 1 1/V 1 1/
H( )2 1 2
ωω = =
+ ωo
i
=+ ω
j jj j
(c) Gain = 2
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
66.
2
2
2
2 2
1 1( ) 22V ( ) 1 1 ( ) 2( ) 21
2( ) 2( ) 2 2( ) 2( )V ( ) 1
( ) 2( ) 2 ( ) 2( ) 2
2 2 4 8Let V ( ) 1 ; 1 12 2 2
A B A BV ( ) 1 Let 0 01 1 1 1 1 1 1 1
Let
ω+ω +ωω = =
ω + ω ++ ω+ω
ω + ω + − ω − ω∴ ω = = +
ω + ω + ω + ω +
− ± −ω = ∴ = − = = − ±
+ +
∴ = + + = = ∴ + =+ + + − + −
o
o
o
o
jjjj
j jjj
j j j jjj j j j
xj x x x jx x
x xx j x j j j
( 1 1) ( 1 1)
A B B 2 B 1 2 A B 2, A B 21 1 1 1 1 1
B B 2 2 B B 0 B 1 1 A 1 11 1 1 1 1 1 1 1V ( ) 1 , V ( ) 11 1 1 1 ( ) 1 1 ( ) 1 1
( ) ( ) (1 1) ( ) (1 1) (− − − +
+= − ∴ + = ∴ − = = + ∴ + =
− +∴ − + + + + = ∴ = − − ∴ = − +
− + − − − +∴ = + + ω = − −
+ + + − ω + + ω + −
∴ = δ − − − +
o o
j t j to
jx j jj j j j
j j j j jj j j jx j
0−
x j x j j j j jv t t j e u t j e u
45 45
)
( ) 2 ( ) 2 ( )− °− − °+ −= δ − −j jt t j jt t
t
t e u t e u t
t( ) 2 2 cos( 45 ) ( )−= δ − + °tt e t u
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
67.
22
/ 6
5 / 10 /V ( ) 105 / 35 30( ) 1/ 7 6( )
10 10 / 6V ( ) 7 16( ) 7( ) 1 ( ) ( )6 6
49 24 1 10 / 6 2 27 / 6 / 2 , 1 V ( )36 36 6 ( 1/ 6)( 1) 1/ 6 1
( ) 2( ) ( )− −
ω ωω = =
ω + + ω ω + + ω
∴ ω = =ω + ω + ω + ω +
⎛ ⎞∴ ω = − ± − = − − ∴ ω = = −⎜ ⎟⎜ ⎟ ω + ω + ω+ ω⎝ ⎠ +
∴
c
c
c
j jjj j j j
jj j j j
j jj j j j
= −t tcv t e e u t
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
68. 2 3( ) 5 ( ), ( ) 4 ( )t tf t e u t g t e u t− −= = (a) f g
0
2 2 3 2
0 02
2 3
( ) ( )
5 4 20
20 ( 1) V( ) ( )
∞
− − − −
−
− −
∗ = −
= =
= − −
∴ ∗ = −
∫
∫ ∫t t
t z z t z
t t
t t
f t z g z dz
e e e dz e e dz
e e
f g e e u t
2 3
5 4 20(b)
F( ) , G( ) F( )G( )2 3 ( 2)( 3)
20 20F( )G( )2 3
ω = ω = ∴ ω ω =ω + ω + ω + ω +
20( 2 ) ( )− −ω ω = −ω + ω +
j jj j j j
j j f g e u tj j
∴ ∗ = −t t
j j
∴
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Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
69. 2( )4 2
jjjωω
ω=
+H
24( ) from Table 18.2i jj
ωω
=V
Therefore o2 24 24( )
4 2 2jj
j j jωω
ω ω ω⎡ ⎤ ⎛ ⎞
= =⎜ ⎟⎢ ⎥+ +⎣ ⎦ ⎝ ⎠V
In the time domain, then, we find 2( ) 24 ( ) Vt
ov t e u t−=
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Engineering Circuit Analysis, 7th Edition Chapter Eightenn Solutions 10 March 2006
70. so fromTable 18.2, ( ) 2 cos 4th t e t−=
( )( )2
2 1( )
1 1
jj
j
ωω
ω
+=
+ +H
6. Define output function f(t).
(a) ( ) 4 ( )jω πδ ω=I
Therefore F(ω) = 2
8 (1 ) 8( ) ( )(1 ) 16 17
jj
π ω πδ ω δω
⎡ ⎤+=⎢ ⎥+ +⎣ ⎦
ω .
The time domain output is then given by f(t) = 4/17. (b) ( ) 2 jj e ωω −=I
Therefore F(ω) = 2
4(1 )(1 ) 16
jj ej
ωωω
−⎡ ⎤+⎢ ⎥+ +⎣ ⎦
.
( ) ( )14 cos 4 1 ( 1)te t u t− − ⎡ ⎤− −⎣ ⎦ The time domain output is then given by f(t) =
(c) We find the response due to a unit step u t and treat as two unit steps, each shifted appropriately.
( ) ( )i t
2
2(1 ) 1( ) ( )(1 ) 16
jjj j
ωω πδ ωω ω
⎡ ⎤+= +⎢ ⎥+ + ⎣ ⎦
R
[ ]1 1( ) sgn( ) 2 cos 4 4sin 4 ( )17 17 17
ter t t t t u t−
= + − −
Therefore the system response is
( ) { }
( ) { }
0.25
0.25
2 1 cos 4( 0.25) 4sin 4( 0.25) ( 0.25)17
2 1 cos 4( 0.25) 4sin 4( 0.25) ( 0.25)17
t
t
e t t u t
e t t u t
− +
− −
⎡ ⎤− + − + +⎣ ⎦
⎡ ⎤− − − − − −⎣ ⎦
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