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13The Laplace Transform in Circuit
Analysis
Assessment Problems
AP 13.1 [a] Y = 1
R+
1
sL+ sC=
C[s2 + (1/RC)s+ (1/LC)
s
1
RC =
106
(500)(0.025) = 80,000;
1
LC = 25 108
Therefore Y =25 109(s2 + 80,000s+ 25 108)
s
[b] z1,2= 40,000
16 108 25 108 = 40,000j30,000 rad/sz1= 40,000j30,000 rad/sz2= 40,000 +j30,000 rad/s
p1 = 0 rad/s
AP 13.2 [a] Z= 2000 + 1
Y= 2000 +
4 107ss2 + 80,000s+ 25 108
=2000(s2 + 105s + 25 108)
s2 + 80,000s+ 25
108
= 2000(s+ 50,000)2
s2 + 80,000s+ 25
108
[b]z1= z2= 50,000 rad/sp1 = 40,000j30,000 rad/sp2 = 40,000 +j30,000 rad/s
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132 CHAPTER 13. The Laplace Transform in Circuit Analysis
AP 13.3 [a] Att = 0, 0.2v1= (0.8)v2; v1= 4v2; v1+v2= 100 V
Therefore v1(0) = 80V =v1(0
+); v2(0) = 20V =v2(0
+)
I= (80/s) + (20/s)
5000 + [(5 106
)/s] + (1.25 106
/s)
=20 103
s + 1250
V1 =80
s 5 10
6
s
20 103s + 1250
=
80
s + 1250
V2 =20
s 1.25 10
6
s
20 103s + 1250
=
20
s + 1250
[b] i= 20e1250tu(t)mA; v1= 80e1250tu(t) V
v2= 20e1250tu(t) V
AP 13.4 [a]
I= Vdc/s
R+sL + (1/sC)=
Vdc/L
s2 + (R/L)s+ (1/LC)
Vdc
L = 40;
R
L = 1.2;
1
LC = 1.0
I= 40
(s + 0.6 j0.8)(s+ 0.6 +j0.8)= K1
s + 0.6j0.8+ K1
s + 0.6 +j0.8
K1 = 40
j1.6= j25 = 25/ 90; K1 = 25/90
[b] i= 50e0.6t cos(0.8t 90) = [50e0.6t sin0.8t]u(t) A
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Problems 133
[c] V =sLI= 160s
(s + 0.6 j0.8)(s+ 0.6 +j0.8)
= K1
s + 0.6 j0.8+ K1
s + 0.6 +j0.8
K1 = 160(0.6 +j0.8)j1.6
= 100/36.87
[d] v(t) = [200e0.6t cos(0.8t+ 36.87)]u(t) V
AP 13.5 [a]
The two node voltage equations are
V1 V2s
+V1s=5
s and
V23
+V2 V1
s +
V2 (15/s)15
= 0
Solving for V1 and V2 yields
V1 = 5(s + 3)
s(s2 + 2.5s + 1), V2 =
2.5(s2 + 6)
s(s2 + 2.5s+ 1)
[b] The partial fraction expansions ofV1 andV2 are
V1 =15
s 50/3
s + 0.5+ 5/3
s + 2 and V2 =15
s 125/6
s + 0.5+ 25/3
s + 2
It follows that
v1(t) =
15 503
e0.5t +5
3e2t
u(t) V and
v2(t) =
15 1256
e0.5t +25
3e2t
u(t) V
[c] v1(0+) = 15 50
3 +
5
3= 0
v2(0+) = 15 1256
+253
= 2.5 V
[d] v1() = 15 V; v2() = 15 V
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134 CHAPTER 13. The Laplace Transform in Circuit Analysis
AP 13.6 [a]
With no load across terminals a b Vx= 20/s:1
2
20
s VTh
s +
1.2
20
s
VTh
= 0
therefore VTh =20(s + 2.4)
s(s + 2)
Vx = 5IT and ZTh=
VT
IT
Solving for IT gives
IT =(VT 5IT)s
2 +VT 6IT
Therefore
14IT =VTs + 5sIT+ 2VT; therefore ZTh=5(s+ 2.8)
s + 2
[b]
I= VTh
ZTh+ 2 +s=
20(s + 2.4)
s(s + 3)(s + 6)
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Problems 135
AP 13.7 [a] i2 = 1.25et 1.25e3t; therefore di2
dt = 1.25et + 3.75e3t
Therefore di2
dt = 0 when
1.25e
t
= 3.75e
3t
or e
2t
= 3, t= 0.5(ln 3) = 549.31msi2(max) = 1.25[e
0.549 e3(0.549)] = 481.13mA[b] From Eqs. 13.68 and 13.69, we have
= 12(s2 + 4s + 3) = 12(s+ 1)(s+ 3) and N1 = 60(s+ 2)
Therefore I1=N1
=
5(s+ 2)
(s + 1)(s + 3)
A partial fraction expansion leads to the expression
I1= 2.5
s + 1+
2.5
s + 3Therefore we get
i1= 2.5[et +e3t]u(t) A
[c] di1
dt = 2.5[et + 3e3t]; di1(0.54931)
dt = 2.89 A/s
[d] Wheni2 is at its peak value,
di2dt
= 0
Therefore L2di2
dt
= 0 and i2 = M
12di1
dt
[e] i2(max) =2(2.89)
12 = 481.13mA (checks)
AP 13.8 [a] The s-domain circuit with the voltage source acting alone is
V (20/s)2
+ V
1.25s+
V s
20 = 0
V = 200
(s + 2)(s + 8)=
100/3
s + 2 100/3
s + 8
v =100
3 [e2t e8t]u(t) V
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Problems 137
[b] Vo(s) = H(s) 1s
= 9600
s2 + 140s+ 62,500
= K1
s + 70j240+ K1
s+ 70 +j240
K1 = 9600j480
= j20 = 20/ 90
Therefore
vo(t) = [40e70t cos(240t 90)]u(t) V = [40e70t sin 240t]u(t) V
AP 13.12 From Assessment Problem 13.9:
H(s) = 10(s+ 2)
s2 + 2s+ 10
Therefore H(j4) = 10(2 +j4)
10 16 +j8= 4.47/
63.43
Thus,
vo= (10)(4.47) cos(4t 63.43) = 44.7cos(4t 63.43) VAP 13.13 [a]
Let R1 = 10k, R2 = 50k, C= 400 pF, R2C= 2 105
then V1=V2 = Vg R2
R2+ (1/sC)
Also
V1
VgR1 +
V1
VoR1 = 0
therefore Vo = 2V1 Vg
Now solving for Vo/Vg,we get H(s) =R2Cs 1R2Cs + 1
It follows that H(j50,000) = j 1j+ 1
=j1 = 1/90
Therefore vo= 10 cos(50,000t+ 90) V
[b] ReplacingR2 byRx gives us H(s) =RxCs 1RxCs + 1
Therefore
H(j50,000) =j20 106Rx 1j20 106Rx+ 1 =
Rx+j50,000
Rx j50,000Thus,
50,000
Rx= tan 60 = 1.7321, Rx= 28,867.51
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138 CHAPTER 13. The Laplace Transform in Circuit Analysis
Problems
P 13.1 i= 1
L t
0vd+I0; therefore I=
1
LV
s +
I0s
= V
sL+
I0s
P 13.2 VTh= Vab = C V0
1
sC
=
V0s
; ZTh= 1
sC
P 13.3 Iscab =IN=LI0
sL =I0
s ; ZN= sL
Therefore, the Norton equivalent is the same as the circuit in Fig. 13.4.
P 13.4 [a] Y = 1
R+
1
sL+sC=
C[s2 + (1/RC)s+ (1/LC)]
s
Z= 1
Y
= s/C
s2
+ (1/RC)s+ (1/LC)
= 8 107s
s2
+ 40,000s+ 256 106
[b] zero at z1= 0poles atp1 = 8000 rad/s andp2= 32,000 rad/s
P 13.5 [a]
Z= (R + 1/sC)(sL)
R +sL+ (1/sC)=
(Rs)(s + 1/RC)
s2 + (R/L)s+ (1/LC)
R
L = 10,000;
1
RC= 1600;
1
LC = 16 106
Z= 1000s(s+ 1600)
s2 + 10,000s+ 16 106
[b] Z= 1000s(s+ 1600)
(s + 2000)(s+ 8000)
z1= 0; z2= 1600 rad/sp1 = 2000 rad/s; p2 = 8000 rad/s
P 13.6 [a] Z=R +sL + 1
sC =
L[s2 + (R/L)s+ (1/LC)]
s
=[s2 + 8000s+ 25 106]
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Problems 139
[b] s1,2= 4000 j3000 rad/sZeros at4000 +j3000 rad/s and4000 j3000 rad/sPole at 0.
P 13.7 Zab= 1
[s + (1/s
1)] = 1
[s + (1/(s+ 1))] =
s + (1/(s + 1))
1 +s + (1/(s+ 1))
= s2 +s + 1
s2 + 2s + 2=
(s + 0.5 +j0.866)(s+ 0.5j0.866)(s + 1 +j1)(s + 1 j1)
Zeros at0.5 +j0.866 rad/s and0.5j0.866 rad/s; poles at1 +j1 rad/sand1j1 rad/s.
P 13.8 Transform the Y-connection of the two resistors and the inductor into theequivalent delta-connection:
where
Za=(s)(1) + (1)(s) + (1)(1)
s
=2s+ 1
s
Zb= Zc=(s)(1) + (1)(s) + (1)(1)
1 = 2s+ 1
Then
Zab= Za[(1/sZc) + (1/sZb)] =Za2(1/sZb)
1/sZb =1s (2s + 1)
1s + 2s + 1
= 2s + 1
2s2 +s+ 1
Zab =
2s + 1
s
2(2s+ 1)
2s2 +s + 1
= 2(2s+ 1)2
(2s + 1)(2s2 +s + 1) + 2s(2s+ 1)=
2
s + 1
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Problems 1311
Vc = 75 V
iL(0) =
75 137.5500
= 0.125A
Fort >0:
[b] Vo=5 105
s I+
75
5
0 = 137.5s
+ 100I+ 5 105
s I+75
s 1.25 103 + 0.01sI
I
100 +
5 105s
+ 0.01s
=
62.5
s + 1.25 103
. . I= 6250 + 0.125ss2 + 104s + 5 107
Vo =5 105
s
6250 + 0.125s
s2 + 104s + 5 107
+75
s
=75s2
+ 812,500s+ 6875 106
s(s2 + 104s + 5 107)
[c] Vo =K1
s +
K2s + 5000 j5000+
K2s + 5000 +j5000
K1 =75s2 + 812,500s+ 6875 106
s2 + 104s + 5 107
s=0= 137.5
K2 =75s2 + 812,500s+ 6875 106
s(s + 5000 +j5000
s=5000+j5000= 40.02/141.34
vo(t) = [137.5 + 80.04e5000t cos(5000t+ 141.34)]u(t) V
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1312 CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.11 [a] Fort 0:
[b] (20 + 2s+ 100/s)I= 10 +70
s
. . I= 5(s+ 7)s2 + 10s+ 50
Vo =100
s I 70
s
= 70s2 200ss(s2 + 10s+ 50)
=70(s+ 20/7)
s2 + 10s + 50
= K1
s + 5 j5+ K1
s + 5 +j5
K1 =70(s+ 20/7)
s + 5 +j5
s=5+j5
= 38.1/ 156.8
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Problems 1313
[c] vo(t) = 76.2e5t cos(5t 156.8)u(t) V
P 13.12 [a] For t
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Problems 1315
= 20/L+s
s2 +sR/L+ 1/LC =
40 +s(0.005)
s2 + 8000s+ 16 106
Vo =RIo L+sLIo= 4000(40 + 0.005s)s2 + 8000s+ 16 106 0.0025 +
0.0025s(s+ 8000)
s2 + 8000s+ 16 10
=20s + 120,000(s + 4000)2
= 20(s + 4000)2
+ 40,000s + 4000
vo(t) = [20te4000t + 40,000e4000t]u(t) V
[b] Io= 0.005(s+ 8000)
s2 + 8000s+ 16 106
= K1
(s + 4000)2+
K2s + 4000
K1 = 20 K2 = 0.005
io(t) = [20te4000t + 0.005e4000t]u(t) A
P 13.14 Fort 0 :
Vo+ 400
25 + 25s+
Vo100
+Vo (400/s)
100/s = 0
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1316 CHAPTER 13. The Laplace Transform in Circuit Analysis
Vo
1
25 + 25s+
1
100+
s
100
= 4 400
25 + 25s
. . Vo = 400(s 3)s2 + 2s+ 5
Io= Vo (400/s)
100/s =20s 20
s2 + 2s+ 5
= K1
s + 1 j2+ K1
s + 1 +j2
K1 =20(s + 1)
s + 1 +j2
s=1+j2
= 10
io(t) = [20et cos2t]u(t) A
P 13.15
Vo = (18/s)(8 106/s)
2800 + 0.2s + (8 106/s)
= 720 106
s(s2
+ 14,000s+ 40 106
)
= 720 106
s(s + 4000)(s+ 10,000)
= K1
s +
K2s + 4000
+ K3
s + 10,000
K1 =720 106
4 107 = 18
K2 = 720 106
(4000)(6000) =
30
K3 = 720 106
(6000)(10,000) = 12
Vo=18
s 30
s + 4000+
12
s + 10,000
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Problems 1317
P 13.16 With a non-zero initial voltage on the capacitor, the s-domain circuit becomes:
Vo 18/s0.2s + 2800
+(Vo 30/s)s
8 106 = 0
Vo
5
s + 14,000+
s
8 106
= 30
80 106 + 90
s(s + 14,000)
. . Vo = 30s2 + 420,000s+ 720 106
s(s + 4000)(s+ 10,000)
= K1
s +
K2s + 4000
+ K3
s + 10,000
K1=720 106
40 106 = 18
K2=30s2 + 420,000s+ 720 106
s(s + 10,000)
s=4000
= 20
K3=30s2 + 420,000s+ 720 106
s(s + 4000)
s=10,000
= 8
Vo=18
s +
20
s + 4000 8
s + 10,000
vo(t) = [18 + 20e4000t 8e10,000t]u(t) V
P 13.17 [a] For t
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1318 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] V1 = 25(450/s)
(125,000/s) + 25 + 1.25 103s
= 9 106
s2 + 20, 000s+ 108 =
9 106(s + 10,000)2
v1(t) = (9 106te10,000t)u(t) V
[c] V2=90
s (25,000/s)(450/s)
(125,000/s) + 1.25 103s + 25
= 90(s+ 20,000)
s2 + 20,000s+ 108
= 900,000
(s + 10,000)2+
90
s + 10,000
v2(t) = [9 105te10,000t + 90e10,000t]u(t) V
P 13.18 [a] iL(0) =iL(0+) =243
= 8 A directed upward
VT = 25I+
20(10/s)
20 + (10/s)
IT =
25IT(10/s)
20 + (10/s)+
200
10 + 20s
IT
VTIT
=Z=250 + 200
20s+ 10 =
45
2s+ 1
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Problems 1319
Vo5
+Vo(2s+ 1)
45 +
Vo5.625s
=8
s
[9s + (2s+ 1)s + 8]Vo45s
=8
s
Vo[2s2 + 10s + 8] = 360
Vo = 360
2s2 + 10s + 8=
180
s2 + 5s+ 4
[b] Vo= 180
(s + 1)(s + 4)=
K1s + 1
+ K2s + 4
K1 =180
3 = 60; K2 =
180
3 = 60
Vo = 60
s + 1 60
s + 4vo(t) = [60e
t 60e4t]u(t) V
P 13.19 vC(0) =vC(0
+) = 0
0.01s
= Vo50,000
+ Vo5000
+ Vos50 106 6V
o
50,000
500 103s
= (1000 + 10,000 +s 6000)Vo
Vo = 500 103s(s + 5000)
=K1
s +
K2s + 5000
= 100
s 100
s + 5000
vo(t) = [100 100e5000t]u(t) V
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1320 CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.20
5 103s
= Vo
200 + 4 106/s+ 3.75 103V+
Vo0.04s
V= 4 106/s
200 + 4 106/sVo = 4 106Vo
200s+ 4 106
. .
5 103
s =
Vos
200s+ 4 106+
15,000Vo
200s+ 4 106+
25Vo
s
. . Vo = s + 20,000s2 + 20,000s+ 108
= K1
(s + 10,000)2+
K2s + 10,000
K1 = 10,000; K2= 1
Vo= 10,000
(s + 10,000)2+
1
s + 10,000
vo(t) = [10,000te10,000t +e10,000t]u(t) V
P 13.21 [a]
Vo 35/s2 + 0.4V+ Vo 8Is + (250/s) = 0
V =
Vo 8Is + (250/s)
s; I =
(35/s) Vo2
Solving for Vo yields:
Vo =29.4s2 + 56s+ 1750
s(s2 + 2s + 50) =
29.4s2 + 56s + 1750
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Problems 1321
Vo =K1
s +
K2s + 1j7+
K2s + 1 +j7
K1 =29.4s2 + 56s + 1750
s2 + 2s + 50
s=0= 35
K2 = 29.4s2 + 56s+ 1750
s(s + 1 +j7)
s=1+j7
= 2.8 +j0.6 = 2.86/167.91
. . vo(t) = [35 + 5.73et cos(7t+ 167.91)]u(t) V[b] Att = 0+ vo= 35 + 5.73 cos(167.91) = 29.4 V
vo 352
+ 0.4v= 0; vo 35 + 0.8v= 0
vo= v+ 8i=v+ 8(0.4v) = 4.2 V
vo+ (0.8)vo4.2
= 35; . . vo(0+) = 29.4 V(checks)
Att =
, the circuit is
v= 0, i= 0 . . vo = 35 V(checks)
P 13.22 [a]
I1+s(I1 I2) =10s
and I2+1
sI2+s(I2 I1) = 0
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1322 CHAPTER 13. The Laplace Transform in Circuit Analysis
Solving the second equation for I1:
I1=s2 +s + 1
s2 I2
Substituting into the first equation and solving for I2:
(s + 1) s
2
+s + 1s2
s I2= 10s
. . I2 = 10s2s2 + 2s+ 1
. . I1 = s2 +s + 1
s2 10s
2s2 + 2s + 1=
10(s2 +s + 1)
s(2s2 + 2s+ 1)
Io= I1 I2= 10(s2 +s + 1)
s(2s2 + 2s + 1) 10s
2s2 + 2s+ 1=
5(s+ 1)
s(s2 +s + 0.5)
=
K1
s +
K2
s + 0.5j0.5+ K2
s + 0.5 +j0.5
K1 = 10; K2 = 5/ 180. . io(t) = [10 10e0.5t cos0.5t]u(t) A
[b] Vo= sIo= 5(s+ 1)
s2 +s + 0.5=
K1s + 0.5j0.5+
K1s + 0.5 +j0.5
K1 = 3.54/ 45. . vo(t) = 7.07e0.5t cos(0.5t 45)u(t) V
[c] Att = 0
+
the circuit is
. . vo(0+) = 5V = 7.07cos(45); Io(0+) = 0Both values agree with our solutions for vo andio.Att =
the circuit is
. . vo() = 0; io() = 10 A 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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Problems 1323
Both values agree with our solutions for vo and io.
P 13.23 [a]
Vo =(1/sC)(sL)(Ig/s)
R+sL + (1/sC) =
Ig/C
s2 + (R/L)s+ (1/LC)
IgC
=0.0150.1
= 0.15
R
L = 7;
1
LC = 10
Vo = 0.15
s2 + 7s+ 10
[b] sVo = 0.15s
s2 + 7s+ 10
lims0
sVo= 0; . . vo(
) = 0
lims
sVo = 0; . . vo(0+) = 0
[c] Vo = 0.15
(s + 2)(s + 5)=
0.05
s + 2+0.05
s + 5
vo= [50e2t 50e5t]u(t) mV
P 13.24 IL=Ig
s +
Vo1/sC
=Ig
s sCVo
IL= 15s 15s
(s + 2)(s + 5)=15
s 10
s + 2+ 25
s + 5
iL(t) = [15 + 10e2t 25e5t]u(t) mA
Check:
iL(0+) = 0 (ok); iL() = 15 mA (ok)
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1324 CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.25 [a]
[b] Zeq= 50,000 +107
3s +
20 1012/s212 106/s
= 50,000 +107
3s +
20 101212 106s
=100,000s+ 107
2s
I1=20/s
Zeq=
0.4 103s + 100
V1 =107
3sI1 =
4000/3
s(s + 100)
V2 =107
6s 0.4 10
4
s + 100 =
2000/3
s(s + 100)
[c] i1(t) = 0.4e100t
u(t) mA
V1 =40/3
s 40/3
s + 100; v1(t) = (40/3)(1 1e100t)u(t) V
V2 =20/3
s 20/3
s + 100; v2(t) = (20/3)(1 1e100t)u(t) V
[d] i1(0+) = 0.4 mA
i1(0+) =
20
50 103 = 0.44 mA(checks)
v1(0+) = 0; v2(0+) = 0(checks)
v1() = 40/3 V; v2() = 20/3 V(checks)v1() +v2() = 20 V(checks)(0.3 106)v1() = 4 C(0.6 106)v2() = 4 C(checks)
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1326 CHAPTER 13. The Laplace Transform in Circuit Analysis
[c] Att = 0+ the circuit is
vo(0+) = 0; io(0
+) = 0 Checks
Att = the circuit is
vo() = 75 V; io() = 7510 + (200/30)
2030
= 3 A Checks
P 13.27 [a]
10
s
I1+10
s
(I1
I2) + 10(I1
9/s) = 0
10
s(I2 9/s) +10
s(I2 I1) + 10I2= 0
Simplifying,
(s + 2)I1 I2= 9
I1+ (s + 2)I2 = 9s
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Problems 1327
=
(s + 2) 11 (s+ 2)
=s2 + 4s+ 3 = (s + 1)(s + 3)
N1 =
9
1
9/s(s + 2) =
9s2 + 18s+ 9
s =
9
s (s + 1)
2
I1=N1
=
9
s
(s + 1)2
(s + 1)(s+ 3)
=
9(s + 1)
s(s + 3)
N2 =
(s + 2) 9
1 9/s
=18
s(s + 1)
I2=N2
=
18(s+ 1)
s(s + 1)(s+ 3)=
18
s(s + 3)
Ia= I1=9(s + 1)
s(s + 3)=
3
s+
6
s + 3
Ib=9
s I1= 9
s 9(s+ 1)
s(s + 3) =
6
s 6
s + 3
[b] ia(t) = 3(1 + 2e3t)u(t) A
ib(t) = 6(1 e3t)u(t) A
[c] Va =
10
sIb=
10
s3
s+
6
s+ 3
= 30
s2 +
60
s(s + 3)=
30
s2 +
20
s 20
s + 3
Vb = 10
s(I2 I1) =10
s
6
s 6
s + 3
3
s+
6
s + 3
= 10
s
3
s 12
s + 3
=
30
s2 40
s +
40
s + 3
Vc = 10
s
(9/s
I2) =
10
s
9
s6
s
+ 6
s + 3
= 30
s2 +
20
s 20
s + 3
[d] va(t) = [30t+ 20 20e3t]u(t) Vvb(t) = [30t 40 + 40e3t]u(t) Vvc(t) = [30t+ 20 20e3t]u(t) V
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Problems 1329
K3 = 25d
ds
8s2 + 35s+ 50
s
s=5
= 25
s(16s+ 35) (8s2 + 35s+ 50)
s2
s=5
= 5(45) 75 = 150
. . Vo=50
s 375
(s + 5)2 +
150
s + 5
[b] vo(t) = [50 375te5t + 150e5t]u(t) V[c] Att = 0+:
vo(0+) = 50 + 150 = 200 V(checks)
Att = :
vo()10 5 + vo() 50
30 = 0
. . 3vo() 150 +vo() 50 = 0; . . 4vo() = 200. . vo() = 50 V(checks)
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1330 CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.29 [a]
0 = 2.5s(I1 6/s) +5s
(I1 I2) + 10I1
75
s =
5
s (I2 I1) + 5(I2 6/s)or
(s2 + 4s + 2)I1 2I2= 6sI1+ (s + 1)I2 = 9
=
(s2 + 4s + 2) 2
1 (s + 1)
= 5(s + 2)(s + 3)
N1 =6s 29 (s + 1)
= 6(s2 +s 3)
I1=N1
=
6(s2 +s 3)s(s + 2)(s+ 3)
N2 =
(s2 + 4s + 2) 6s
1 9
= 9s2 30s 18
I2=
N2
=9s2
30s
18
s(s + 2)(s + 3)
[b] sI1= 6(s2 +s 3)(s + 2)(s+ 3)
lims
sI1= i1(0+) = 6 A; lim
s0sI1= i1() = 3 A
sI2=9s2 30s 18
(s + 2)(s + 3)
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Problems 1331
lims
sI2= i2(0+) = 9 A; lim
s0sI2= i2() = 3 A
[c] I1= 6(s2 +s 3)s(s + 2)(s+ 3)
=K1
s +
K2s + 2
+ K3s + 3
K1 = 6(3)6
= 3; K2 = 6(4 2 3)(2)(1) = 3
K3 =6(9 3 3)
(3)(1) = 6
i1(t) = [3 + 3e2t + 6e3t]u(t) A
I2=9s2 30s 18
s(s + 2)(s+ 3) =
K1s
+ K2s + 2
+ K3s+ 3
K1 =18
6 = 3; K2 =36 + 60
18
(2)(1) = 3
K3 =81 + 90 18
(3)(1) = 3
i2(t) = [3 3e2t 3e3t]u(t) A
P 13.30 [a]
AtVo :
Vo V110s
50s
+Vo V2
5 = 0
. . Vo(2s + 1) 2sV2 V1 = 500Supernode:
V1s
5 +
V1 Vo10s
+V2
1 +
V2 V15
= 0
. . Vo(2s + 1) + 12sV2+ (2s2 + 1)V1 = 0
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1332 CHAPTER 13. The Laplace Transform in Circuit Analysis
Constraint:
V1 V2 = 4I= 4V1s
5
. . V2 = (0.8s+ 1)V1Simplifying:
Vo(2s + 1) V1(1.6s2 + 2s + 1) = 500Vo(2s + 1) V1(11.6s2 + 12s+ 1) = 0
=
2s+ 1 (1.6s2 + 2s+ 1)
(2s + 1) (11.6s2 + 12s+ 1)
= 20(s2 + 1.5s+ 0.5)
No=
500(1.6s2 + 2s+ 1)0 (11.6s2 + 12s+ 1)
= 500(11.6s2
+ 12s+ 1)
Vo =No
=
25(11.6s2 + 12s+ 1)
s(s + 0.5)(s + 1)
[b] vo(0+) = lim
ssVo = 25(11.6) = 290V
vo() = lims0
sVo= 25
0.5= 50 V
[c] Att = 0+ the circuit is
4I+ 1I1= 0; I1 I= 50. . 4I+ 50 + I= 0; 5I= 50. . I= Io(0+) = 10 A
Also I1= 50 10 = 40 AVo(0
+) = 5(I1 I) + 1I1= 6I1 5I = 240 5(10) = 290 V (checks) 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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Problems 1333
Att = the circuit is
Vo() = 50(1) = 50 V(checks)
[d] Vo=25(11.6s2 + 12s+ 1)
s(s + 0.5)(s + 1) =
K1s
+ K2s + 0.5
+ K3s + 1
K1 = 25
(0.5)(1)= 50; K2=
52.5(0.5)(0.5)= 210
K3 = 15
(1)(0.5)= 30
Vo =50
s +
210
s + 0.5+
30
s + 1
vo(t) = (50 + 210e0.5t + 30et)u(t) V
vo(
) = 50 V(checks)
vo(0+) = 50 + 210 + 30 = 290 V(checks)
P 13.31 [a]
120
s = 50(I1 0.05V) +250
s (I1 I2)
250
s = 50I1 2.5
250
s
(I2 I1) +250
s I1 250
s I2
Simplfying,
(50s+ 875)I1 875I2 = 120 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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1334 CHAPTER 13. The Laplace Transform in Circuit Analysis
250(s 1)I1+ (20s2 + 450s+ 250)I2 = 0
=
(50s+ 875) 875250(s 1) (20s2 + 450s+ 250)
= 1000s(s2 + 40s+ 625)
N1 =
120 875
0 (20s2 + 450s+ 250)
= 1200(2s2 + 45s + 25)
N2 =
(50s+ 875) 120
250(s 1) 0
= 30,000(s 1)
I1=N1
=
1200(2s2 + 45s+ 25)
s(s2 + 40s+ 625)
I2=N
2 =
30,000(s
1)
s(s2 + 40s+ 625)
Io= I2 0.05V =I2 0.05
250
s (I2 I1)
I2 I1= 2400(s+ 35)s(s2 + 40s+ 625)
250
s (I2 I1) =600,000(s+ 35)
s(s2 + 40s+ 625)
. . Io = 30,000(s
1)
s(s2 + 40s+ 625)+
30,000(s+ 35)
s(s2 + 40s+ 625) =
1080
s(s2 + 40s+ 625)
[b] sIo= 1080
(s2 + 40s + 625)
io(0+) = lim
ssIo= 0
io() = lims0
sVo =1080
625 = 1728mA
[c] Att = 0+ the circuit is
i(0+) = 0 (checks)
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Problems 1335
Att = the circuit is
120 = 50(ia i1) + 700ia= 50(ia 0.05v) + 700ia = 750ia 2.5v
v= 700ia . . 120 = 750ia+ 1750ia = 2500ia
ia= 120
2500= 48mA
v= 700ia = 33.60 Vio() = 48 103 0.05(33.60) = 48 103 + 1.68 = 1728mA (checks)
[d] Io= 1080
s(s2 + 40s + 625) =
K1s
+ K2
s + 20 j15+ K2
s + 20 +j15
K1 = 1080
625 = 1.728
K2 = 1080
(20 +j15)(j30) = 1.44/126.87
io(t) = [1728 + 2880e
20t cos(15t+ 126.87
)]u(t) mA
Check: io(0+) = 0 mA; io() = 1728 mAP 13.32 [a]
100
5s=
500s
5s+ 100
= 100s
s + 20
Vo = 100s
s + 20
50
(s + 25)2
=
5000s
(s + 20)(s+ 25)2
Io = Vo100
= 50s
(s + 20)(s+ 25)2
IL =Vo5s
= 1000
(s + 20)(s + 25)2
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1336 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] Vo= K1s + 20
+ K2
(s + 25)2+
K3s + 25
K1 = 5000s
(s + 25)2
s=20= 4000
K2 = 5000s(s + 20)
s=25
= 25,000
K3 = d
ds
5000s
s + 20
s=25
=
5000
s + 20 5000s
(s + 20)2
s=25
= 4000
vo(t) = [4000e20t + 25,000te25t + 4000e25t]u(t) V
Io= K1s+ 20
+ K2
(s + 25)2+
K3s + 25
K1 = 50s
(s + 25)2
s=20= 40
K2 = 50s
(s + 20)
s=25
= 250
K3 = d
ds
50s
s + 20
s=25
=
50
s + 20 50s
(s + 20)2
s=25
= 40
io(t) = [40e20t + 250te25t + 40e25t]u(t) V
IL
= K1
s + 20+
K2
(s + 25)2+
K3
s + 25
K1 = 1000
(s + 25)2
s=20
= 40
K2 = 1000
(s + 20)
s=25
= 200
K3 = d
ds
1000
s + 20
s=25
=
1000
(s + 20)2
s=25
= 40
iL(t) = [40e20t
200te25t
40e25t]u(t) V
P 13.33 vC= 12 105te5000t V, C= 5 F; therefore
iC=C
dvCdt
= 6e5000t(1 5000t) A
iC>0 when 1> 5000t or iC>0 when 0< t
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Problems 1337
and iC 200 s
iC= 0 when 1 5000t= 0, or t= 200 s
dvCdt = 12 10
5
e5000t
[1 5000t]
. . iC= 0 when dvCdt
= 0
P 13.34
VTh= 10s
10s+ 1000 40
s =
400
10s+ 1000=
40
s + 100
ZTh= 1000 + 100010s= 1000 + 10,000s10s+ 1000
= 2000(s+ 50)
s + 100
I= 40/(s+ 100)
(5 105)/s + 2000(s+ 50)/(s+ 100) = 40s
2000s2 + 600,000s+ 5 107
= 0.02s
s2 + 300s+ 25,000=
K1s + 150
j50
+ K1
s + 150 +j50
K1= 0.02s
s + 150 +j50
s=150+j50
= 31.62 103/71.57
i(t) = 63.25e150t cos(50t+ 71.57)u(t) mA
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Problems 1339
Laplace transform the equations to get
120I1+ 8(sI1 15) 6(sI2+ 10) = 0270I2+ 18(sI2+ 10) 6(sI1 15) = 0In standard form,
(8s+ 120)I1 6sI2= 1806sI1+ (18s + 270)I2 = 270
=
8s + 120 6s6s 18s+ 270
= 108(s+ 10)(s+ 30)
N1 =
180 6s270 18s+ 270
= 1620(s+ 30)
N2 =
8s + 120 180
6s 270 = 1080(s+ 30)
I1=N1
=
1620(s+ 30)
108(s+ 10)(s + 30) =
15
s + 10
I2=N2
=
1080(s+ 30)108(s+ 10)(s + 30)
= 10s + 10
[c] i1(t) = 15e10tu(t) A; i2(t) = 10e10tu(t) A
[d] W120=
0 (225e
20t
)(120) dt= 27,000
e20t
20
0 = 1350J
W270=
0(100e20t)(270) dt= 27,000
e20t
20
0= 1350J
W120+W270= 2700J
[e] W =1
2L1i
21+
1
2L2i
22+Mi1i2= 900 + 900 900 = 900 J
With the dot reversed the s-domain equations are
(8s+ 120)I1+ 6sI2 = 60
6sI1+ (18s+ 270)I2 =
90
As before, = 108(s+ 10)(s+ 30). Now,
N1 =
60 6s90 18s+ 270
= 1620(s+ 10)
N2 =
8s + 120 60
6s 90
= 1080(s+ 10) 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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1340 CHAPTER 13. The Laplace Transform in Circuit Analysis
I1=N1
=
15
s + 30; I2=
N2
= 10s + 30
i1(t) = 15e30tu(t) A; i2(t) = 10e30tu(t) A
W270=
0
(100e60t)(270) dt= 450 J
W120=
0(225e60t)(120) dt= 450 J
W120+W270= 900J
P 13.37 The s-domain equivalent circuit is
V1 48/s4 + (100/s)
+V1+ 9.6
0.8s +
V10.8s+ 20
= 0
V1 = 1200
s2 + 10s+ 125
Vo= 20
0.8s + 20
V1 = 30,000
(s + 25)(s+ 5 j10)(s+ 5 +j10)
= K1s + 25
+ K2
s + 5 j10+ K2
s + 5 +j10
K1 = 30,000
s2 + 10s + 125
s=25
= 60
K2 = 30,000
(s + 25)(s+ 5 +j10)
s=5+j10
= 67.08/63.43
vo(t) = [60e25t + 134.16e5t cos(10t+ 63.43)]u(t) V
P 13.38 Fort
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Problems 1341
Fort >0+:
Note that because of the dot locations on the coils, the sign of the mutualinductance is negative! (See Example C.1 in Appendix C.)
L1 M= 3 + 1 = 4 H; L2 M= 2 + 1 = 3 H
18 4 = 72; 18 3 = 54
V 724s + 20
+ V
s + 10+V + 54
3s = 0
V
1
4s + 20+
1
s + 10+ 1
3s
=
72
4s+ 20 54
3s
V
3s(s + 10) + 3s(4s + 20) + (4s+ 20)(s + 10)
3s(
s + 10)(4s+ 20)
=
72(3s) 54(4s+ 20)3s(4s+ 20)
V =[72(3s) 54(4s+ 20)](s+ 10)
5s2 + 110s+ 200
Io= V
s + 10 = 108
(s + 2)(s+ 20)=1.2s + 2
+ 1.2
s+ 20
io(t) = 1.2[e20t e2t]u(t) A
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Problems 1343
[b] Reversing the dot on the 1.25 H coil will reverse the sign ofM, thus thecircuit becomes
The two simulanteous equations are
150
s = (25 + 0.9375s)I1 0.625sI2
0 = 0.625sI1+ (1.25s+ 50)I2When these equations are compared to those derived in Problem 13.39we see the only difference is the algebraic sign of the 0.625s term. Thusreversing the dot will have no effect on I1 and will reverse the sign ofI2.
Hence,i2(t) = (2e
20t 2e80t)u(t) AP 13.41 [a] s-domain equivalent circuit is
Note: i2(0+) = 2010
= 2 A
[b] 24
s = (120 + 3s)I1+ 3sI2+ 6
0 = 6 + 3sI1+ (360 + 15s)I2+ 36In standard form,
(s + 40)I1+sI2= (8/s) 2
sI1+ (5s + 120)I2 = 10
=
s + 40 s
s 5s + 120
= 4(s+ 20)(s + 60)
N1 =
(8/s) 2 s10 5s + 120
=200(s 4.8)
s
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1344 CHAPTER 13. The Laplace Transform in Circuit Analysis
I1=N1
=
50(s 4.8)s(s + 20)(s+ 60)
[c] sI1= 50(s 4.8)(s + 20)(s+ 60)
lims
sI1= i1(0+) = 0 A
lims0
sI1= i1() =(50)(4.8)(20)(60)
= 0.2 A
[d] I1=K1
s +
K2s + 20
+ K3s + 60
K1 = 240
1200= 0.2; K2 =
50(20) + 240(20)(40) = 1.55
K3 =50(60) + 240(60)(40) = 1.35
i1(t) = [0.2 1.55e20t + 1.35e60t]u(t) A
P 13.42 [a] Voltage source acting alone:
Vo1 60/s10
+V01s
80 +
V0120 + 10s
= 0
. . V01 = 480(s+ 2)s(s + 4)(s + 6)
Vo210
+V02s
80 +
V02 30/s10(s + 2)
= 0
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Problems 1345
. . V02= 240s(s + 4)(s + 6)
Vo =Vo1+Vo2 =480(s+ 2) + 240
s(s + 4)(s + 6) =
480(s+ 2.5)
s(s + 4)(s + 6)
[b] Vo= K1
s + K
2
s + 4+ K
3
s+ 6
K1 =(480)(2.5)
(4)(6) = 50; K2 =
480(1.5)(4)(2) = 90; K3 =
480(3.5)(6)(2) = 140
vo(t) = [50 + 90e4t 140e6t]u(t) V
P 13.43 =
Y11 Y12
Y12 Y22
=Y11Y22 Y212
N2 =
Y11 [(Vg/R1) +C (/s)]Y12 (Ig C)
V2 =N2
Substitution and simplification lead directly to Eq. 13.90.
P 13.44 [a] Vo =
Zf
ZiVg
Zf= 104(80 106/s)104 + 80 106/s =
80 106s + 8000
Zi = 4000 + 109
62.5s=
4000(s+ 4000)
s
Vg =16,000
s2
. . Vo= 320 106
s(s + 4000)(s+ 8000)
[b] Vo=K1
s +
K2s + 4000
+ K3
s + 8000
K1 =20,000(16,000)
(4000)(8000) = 10
K2 = 320 106(4000)(4000) = 20
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1346 CHAPTER 13. The Laplace Transform in Circuit Analysis
K3 = 320 106(8000)(4000) = 10
. . vo(t) = (10 + 20e4000t 10e8000t)u(t) V[c]
10 + 20e4000ts
10e8000ts =
5
. . 20e4000ts 10e8000ts = 5Letx= e4000ts. Then
20x 10x2 = 5; or x2 2x+ 0.5 = 0Solving,
x= 1
0.5 so x= 0.2929
. . e4000ts = 0.2929; . . ts= 306.99 s
[d] vg= m tu(t); Vg =ms2
Vo = 20,000m
s(s + 4000)(s+ 8000)
K1 = 20,000m(4000)(8000)
=20,000m
32 106
. . 5 =20,000m32 106
. . m = 8000 V/s
Thus, m must be less than or equal to 8000 V/s to avoid saturation.P 13.45 [a] Letva be the voltage across the 0.5 F capacitor, positive at the upper
terminal.Letvb be the voltage across the 100 k resistor, positive at the upperterminal.Also note
106
0.5s=
2 106s
and 106
0.25s=
4 106s
; Vg =0.5
s
sVa
s 106
+Va (0.5/s)
200,000
+ Va
200,000
= 0
sVa+ 10Va 5s
+ 10Va = 0
Va = 5
s(s + 20)
0 Va200,000
+(0 Vb)s
4 106 = 0 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems 1347
. . Vb= 20s
Va = 100s2(s + 20)
Vb100,000
+(Vb 0)s
4 106 +(Vb Vo)s
4 106 = 0
40Vb+sVb+sVb=sVo
. . Vo =2(s + 20)Vbs
; Vo = 2100
s3
=200
s3
[b] vo(t) = 100t2u(t) V[c]100t2 = 4; t= 0.2 s = 200 ms
P 13.46
Va 0.016/s2000
+ Vas
50 106 +(Va Vo)s
50 106 = 0
(0 Va)s50 106 +
(0 Vo)10,000
= 0
Va =5000Vo
s
. . 5000Vos
(2s+ 25,000) sVo = 25,000
0.016
s
Vo= 4000
(s + 5000 j10,000)(s+ 5000 +j10,000)
K1 = 400j10,000
=j0.02 = 0.02/90
vo(t) = 40e5000t cos(10,000t+ 90) = 40e5000t sin(10,000t)u(t) mV
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1348 CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.47 [a]
Vp = 50/s
5 + 50/sVg2=
50
5s+ 50Vg2
Vp 40/s20
+ Vp Vo
5 +V
p Vo100/s
= 0
Vp
1
20+
1
5+
s
100
Vo
1
5+
s
100
=
2
s
s + 25
100
50
5s + 50
16
s 2
s=Vo
1
5+
s
100
= Vo
s + 20
100
Vo = 100
s + 20
16(s+ 25)
10(s + 10)(s) 2
s
=
40s+ 2000s(s + 10)(s+ 20)
= K1s
+ K2s + 10
+ K3s + 20
K1 = 10; K2 = 24; K3 = 14. . vo(t) = [10 24e10t + 14e20t]u(t) V
[b] 10 24x+ 14x2 = 514x2 24x+ 5 = 0x= 0 or 0.242691
e10t
= 0.242691 . . t= 141.60msP 13.48 Let vo1 equal the output voltage of the first op amp. Then
Vo1 =Zf1
ZA1Vg where Zf1= 25 103
ZA1 = 25,000 + 25,000(20 104/s)25,000 + (20 104/s)
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Problems 1349
=25,000(s+ 16)
(s + 8)
. . Vo1 =(s + 8)(s + 16)
Vg
vg(t) = 16u(t)mV; . . Vg =16 103
s
Vo1 =16 103(s + 8)
s(s + 16) =
0.008s
+0.008
s + 16
. . vo1= 0.008(1 + e16t) V
The op amp will saturate when vo1=
6 V. Hence, saturation will occur when
0.008(1 + e16t) = 6 so e16t = 749
Thus t=ln749
16 = 0.414s
Thus, the first op amp never saturates. We must investigate the output of thesecond op amp:
Vo=Zf2
ZA2Vo1 where Zf2=
2 108s
andZA2= 25,000
. . Vo =8000s
Vo1 =8000
s
(s + 8)(s + 16)
Vg
= 8000(s+ 8)
s(s + 16) Vg
vg(t) = 16u(t)mV; . . Vg =16 103
s
Vo= 128(s+ 8)
s2(s + 16) =
K1s2
+K2
s +
K3s + 16
K1= 128(8)
16 = 64
K2= 128d
ds
s + 8
s + 16
s=0
= 4
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1350 CHAPTER 13. The Laplace Transform in Circuit Analysis
K3 = 128(8)
256 = 4
vo(t) = [64t+ 4 4e16t]u(t) V
The op amp will saturate when vo= 6 V. Hence, saturation will occur when
64t+ 4 4e16t = 6 or 16t 0.5 =e16t
This equation can be solved by trial and error. First note that t >0.5/16 ort > 31.25 ms.Try 40 ms:
0.64 0.5 = 0.14; e0.64 = 0.53
Try 50 ms:
0.80 0.5 = 0.30; e0.80 = 0.45
Try 60 ms:
0.96 0.5 = 0.46; e0.96 = 0.38
Further trial and error gives
tsat=56.5 ms
P 13.49 [a]
420
Vi= Vo+ 100,000Vi100,000 + (4 108/s)
0.2Vi sVis + 4000
=Vo
. . VoVi
=H(s) =0.8(s 1000)
(s + 4000)
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Problems 1351
[b]z1 = 1000 rad/sp1 = 4000rad/s
P 13.50 [a] Vo
Vi=
1/sC
R + 1/sC
= 1
RCs+ 1
H(s) = (1/RC)
s + (1/RC)=
250
s + 250; p1 = 250rad/s
[b] Vo
Vi=
R
R + 1/sC =
RCs
RCs+ 1=
s
s + (1/RC)
= s
s + 250; z1= 0, p1 = 250rad/s
[c] Vo
Vi=
sL
R+sL=
s
s +R/L=
s
s + 8000
z1= 0; p1= 8000rad/s
[d] Vo
Vi=
R
R +sL=
R/L
s + (R/L) =
8000
s + 8000
p1 = 8000rad/s[e]
Vos
4 106 + Vo10,000
+Vo Vi40,000
= 0
sVo+ 400Vo+ 100Vo = 100Vi
H(s) =VoVi
= 100
s + 500
p1 =
500rad/s
P 13.51 [a] 1/sC
R + 1/sC =
1
RsC+ 1=
1/RC
s + 1/RC
There are no zeros, and a single pole at1/RCrad/sec.
[b] R
R +sL=
R/L
s +R/L
There are no zeros, and a single pole atR/Lrad/sec.
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1352 CHAPTER 13. The Laplace Transform in Circuit Analysis
[c] There are several possible solutions. One is
R= 10 ; L= 10 mH; C= 100 F
P 13.52 [a] R
R + 1/sC =
RsC
RsC+ 1=
s
s + 1/RC
There is a single zero at 0 rad/sec, and a single pole at1/RCrad/sec.[b]
sL
R +sL=
s
s +R/L
There is a single zero at 0 rad/sec, and a single pole atR/Lrad/sec.[c] There are several possible solutions. One is
R= 100; L= 10 mH; C= 1 F
P 13.53 [a] R
1/sC+sL +R =
(R/L)s
s2 + (R/L)s+ 1/LC
There is a single zero at 0 rad/sec, and two poles:p1 = (R/2L) +
(R/2L)2 (1/LC); p2 = (R/2L)
(R/2L)2 (1/LC)
[b] There are several possible solutions. One is
R= 250; L= 10 mH; C= 1 F
These component values yield the following poles:
p1 = 5000 rad/sec and p2 = 20,000 rad/sec[c] There are several possible solutions. One is
R= 200; L= 10 mH; C= 1 FThese component values yield the following poles:
p1 = 10,000 rad/sec and p2 = 10,000 rad/sec[d] There are several possible solutions. One is
R= 120; L= 10 mH; C= 1 F
These component values yield the following poles:
p1 = 6000 +j8000 rad/sec and p2 = 6000 j8000 rad/sec
P 13.54 [a] Zi= 1000 +5 106s
=1000(s+ 5000)s
Zf =40 106
s 40,000 =40 10
6
s + 1000
H(s) = ZfZi
=40 106/(s+ 1000)
1000(s+ 5000)/s =
40,000s(s + 1000)(s+ 5000)
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Problems 1353
[b] Zero at z1 = 0; Poles atp1= 1000 rad/s andp2 = 5000 rad/sP 13.55 [a] LetR1 = 250k; R2= 125k; C2 = 1.6nF; and Cf= 0.4nF. Then
Zf=(R2+ 1/sC2)1/sCf
R2+ 1sC2
+ 1
sCf
= (s + 1/R2C2)
Cfs s + C2+CfC2CfR2
1
Cf= 2.5 109
1
R2C2=
62.5 107125 103 = 5000 rad/s
C2+CfC2CfR2
= 2 109
(0.64 1018)(125 103)= 25,000rad/s
. . Zf =
2.5 109(s + 5000)s(s + 25,000)
Zi =R1 = 250 103
H(s) =VoVg
=Zf
Zi=104(s + 5000)
s(s + 25,000)
[b]z1 = 5000rad/sp1 = 0; p2 = 25,000rad/s
P 13.56 [a]
Va Vg1000
+ sVa5 106 +
(Va Vo)s5 106 = 0
5000Va 5000Vg+ 2sVa sVo = 0(5000 + 2s)Va sVo = 5000Vg(0 Va)s
5 106 +0 Vo
5000 = 0
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1354 CHAPTER 13. The Laplace Transform in Circuit Analysis
sVa 1000Vo = 0; . . Va 1000s
Vo
(2s+ 5000)1000
s
Vo sVo= 5000Vg
1000Vo(2s+ 5000) + s2
Vo= 5000sVgVo(s
2 + 2000s+ 5 106) = 5000sVgVoVg
= 5000s
s2 + 2000s+ 5 106
s1,2= 1000
106 5 106 = 1000 j2000VoVg
= 5000s
(s + 1000 j2000)(s+ 1000 +j2000)
[b] z1= 0; p1 = 1000 +j2000; p2 = 1000j2000P 13.57 [a]
Vo5000
+ Vo0.2s
+ Vo(107)s= Ig
. . Vo= 10 106
ss2 + 2000s+ 50 106 Ig
Ig = 0.1s
s2 + 108; Io = 10
7sVo
. . H(s) = s2
s2 + 2000s+ 50 106
[b] Io= (s2)(0.1s)
(s + 1000j7000)(s+ 1000 +j7000)(s2 + 108)
Io= 0.1s3
(s + 1000 j7000)(s+ 1000 +j7000)(s+j104)(sj104)[c] Damped sinusoid of the form
Me1000t cos(7000t+1)
[d] Steady-state sinusoid of the form
Ncos(104t +2)
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1356 CHAPTER 13. The Laplace Transform in Circuit Analysis
I2=N2
=
35sVg25(s+ 2000)(s+ 8000)
H(s) = I2Vg
= 1.4s
(s + 2000)(s+ 8000)
. . z1= 0; p1 = 2000 rad/s; p2= 8000 rad/s
P 13.59 [a]
2000(Io Ig) + 8000Io+(Ig Io)(2000) + 2sIo= 0
. . Io= 1000(1 )s + 1000(5 ) Ig
. . H(s) = 1000(1 )s + 1000(5 )
[b]
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Problems 1357
= 5:
Io =20,000
s2 ; io= 20,000t u(t) A
= 6:
Io = 25s 25
s 1000 ; io= 25[1 e1000t]u(t) A
P 13.60 [a]
y(t) = 0 t
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1358 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b]
y(t) = 0 t < 0
0 t 10 : y(t) = t
0312.5 d= 312.5t
10 t 20 : y(t) = 100
312.5 d= 3125
20 t 30 : y(t) = 10
t20312.5 d= 312.5(30 t)
30 t < : y(t) = 0
[c]
y(t) = 0 t < 0
0 t 1 : y(t) = t
0625 d= 625t
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1360 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] 0 t 10:
y(t) = t
040 d= 40
t0= 40t
10
t
40:
y(t) = t
t1040 d= 40
tt10
= 400
40 t 50:
y(t) = 40
t1040 d= 40
40t10
= 40(50 t)
t 50 : y(t) = 0 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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Problems 1361
[c] The expressions are
0 t 1 : y(t) = t
0400 d= 400
t0= 400t
1 t 40 : y(t) = t
t1 400 d= 400t
t1= 400
40 t 41 : y(t) = 40
t1400 d= 400
40t1
= 400(41 t)
41 t < : y(t) = 0[d]
[e] Yes, note that h(t) is approaching 40(t), thereforey(t) must approach40x(t), i.e.
y(t) = t
0h(t )x() d
t0
40(t )x() d
40x(t)
This can be seen in the plot, e.g., in part (c), y(t) =40x(t).P 13.62 H(s) =
VoVi
= 1
s + 1; h(t) = et
For 0 t 1:
vo= t
0e d= (1 et) V
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1364 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b]
0 t 0.5:
vo = t
0
100(1
2) d= 100(
2)
t
0
= 100t(1
t)
0.5 t :
vo = 0.5
0100(1 2) d= 100( 2)
0.50
= 25
[c]
P 13.67 [a]1 t 4:
vo = t+1
010 d= 52
t+1
0= 5t2 + 10t+ 5 V
4 t 9:vo =
t+1t4
10 d= 52t+1t4
= 50t 75 V
9 t 14:
vo = 10 10
t4 d+ 10
t+110
10 d
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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1366 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b]
P 13.68 [a] h() =2
5 0 5
h() = 42
5 5 10
0 t 5:
vo = 10 t
0
2
5 d= 2t2
5 t 10:
vo = 10 5
0
2
5 d+ 10
t5
4 2
5
d
=42
2
50
+40t
542
2
t5
= 100 + 40t 2t2
10 t :
vo = 10 5
0
2
5 d+ 10
105
4 2
5
d
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Problems 1367
=42
2
50
+4010
54
2
2
105
= 50 + 200 150 = 100vo= 2t
2 V 0
t
5
vo= 40t 100 2t2 V 5 t 10vo= 100 V 10 t
[b]
[c] Area =1
2(10)(2) = 10 . . 1
2(4)h= 10 so h= 5
h() =5
2 0 2
h() =
10 52
2 4
0 t 2:
vo= 10 t
0
5
2 d= 12.5t2
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1368 CHAPTER 13. The Laplace Transform in Circuit Analysis
2 t 4:
vo = 10 2
0
5
2 d+ 10
t2
10 5
2
d
=252
2
2
0
+100 t
2 252
2
t
2
= 100 + 100t 12.5t2
4 t :
vo = 10 2
0
5
2 d+ 10
42
10 5
2
d
=252
2
20
+100
4225
2
2
42
= 50 + 200
150 = 100
vo = 12.5t2 V 0 t 2
vo = 100t 100 12.5t2 V 2 t 4vo = 100 V 4 t
[d] The waveform in part (c) is closer to replicating the input waveformbecause in part (c) h() is closer to being an ideal impulse response.That is, the area was preserved as the base was shortened.
P 13.69 [a] Vo=16
20Vg
. . H(s) = VoVg
=4
5
h() = 0.8()
[b]
0< t
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Problems 1369
0.5 s t 1.0 s:
vo= t0.5
075[0.8()] d= 60 V
1 s< t < : vo= 0[c]
Yes, because the circuit has no memory.
P 13.70 [a]
Vo Vg5
+Vos
4 +
Vo20
= 0
(5s+ 5)Vo= 4Vg
H(s) =VoVg
= 0.8
s + 1; h() = 0.8eu()
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1370 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b]
0 t 0.5 s;vo =
t0
75(0.8e) d= 60e
1t
0
vo = 60 60et V, 0 t 0.5 s0.5 s t 1 s:
vo = t0.5
0(75)(0.8e) d+
tt0.5
75(0.8e) d
= 60e
1t0.5
0+60
e
1tt0.5
= 120e(t0.5) 60et 60 V, 0.5 s t 1 s
1 s t ;vo =
t0.5t1
(75)(0.8e) d+ t
t0.575(0.8e) d
= 60e
1t0.5
t1+60
e
1tt0.5
= 120e(t0.5) 60e(t1) 60et V, 1 s t
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Problems 1371
[c]
[d] No, the circuit has memory because of the capacitive storage element.
P 13.71 vi= 25 sin10 [u() u( /10)]H(s) =
32
s + 32
h() = 32e32
h(t ) = 32e32(t) = 32e32te32
vo = 800e32t t
0
e32 sin 10 d
= 800e32t
e32
322 + 102(32 sin 10 10cos 10
t0
= 800e32t
1124 [e32t(32 sin 10t 10 cos 10t) + 10]
= 800
1124[32sin10t 10cos10t+ 10e32t]
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1372 CHAPTER 13. The Laplace Transform in Circuit Analysis
vo(0.075) = 10.96 V
P 13.72 H(s) = 16s
40 + 4s + 16s =
0.8s
s + 2= 0.8
1 2
s + 2
= 0.8 1.6
s + 2
h() = 0.8() 1.6e2u()
vo= t
075[0.8() 1.6e2] d=
t0
60() d 120 t
0e2 d
= 60 120e2
2t
0= 60 + 60(e2t 1)
= 60e2tu(t) V
P 13.73
Vo= 5 103Ig
25 103 + 2.5 106/s(20 103)
VoIg
=H(s) = 4000ss + 100
H(s) = 4000
1 100s + 100
= 4000 4 10
5
s + 100
h(t) = 4000(t) 4 105e100t
vo = 103
0(20 103)[4000() 4 105e100] d
+
5103
103(10 103)[4 105e100] d
= 80 + 8000 103
0e100 d 4000
5103103
e100 d
= 80 80(e0.1 1) + 40(e0.5 e0.1)= 40e0.5 120e0.1 = 84.32 V
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Problems 1373
Alternate:
Ig = 4103
0(10 103)est dt+
61034103
(20 103)est dt
=10
s 30
se4103s
+
20
se6103s 103
Vo =IgH(s) = 40
s + 100[1 3e4103s + 2e6103s]
= 40
s + 100 120e
4103s
s+ 100 +
80e6103s
s + 100 ]
vo(t) = 40e100t 120e100(t4103)u(t 4 103)
+80e100(t6103)u(t 6 103)
vo(5 103) = 40e0.5 120e0.1 + 80(0) = 84.32V (checks)
P 13.74 [a] H(s) =VoVi
= 1/LC
s2 + (R/L)s+ (1/LC)
= 100
s2 + 20s+ 100=
100
(s + 10)2
h() = 100e10u()
0 t 0.5:vo = 500
t0
e10 d
= 500
e10
100 (10 1)
t0
= 5[1 e10t(10t+ 1)] 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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1374 CHAPTER 13. The Laplace Transform in Circuit Analysis
0.5 t :
vo = 500 t
t0.5e10 d
= 500e10
100 (
10
1) t
t0.5
= 5e10t[e5(10t 4) 10t 1][b]
P 13.75 [a] Io= Vo105
+ Vos
5 106 =Vo(s + 50)
5 106VoIg
=H(s) =5 106
s + 50
h() = 5 106e50u()
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Problems 1375
0 t 0.1 s:
vo = t
0(50 106)(5 106)e50 d= 250e
50
50t0
= 5(1
e50t) V
0.1 s t 0.2 s:
vo = t0.1
0(50 106)(5 106e50 d)
+ t
t0.1(50 106)(5 106e50 d)
= 250e5050t0.1
0+250e
50
50t
t0.1
= 5
e50(t0.1) 1 5
e50t e50(t0.1)
vo= [10e50(t0.1) 5e50t 5] V
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1376 CHAPTER 13. The Laplace Transform in Circuit Analysis
0.2 s t :
vo =
t0.1
t0.2 250e50 d +
t
t0.1250e50 d
=
5e50
t0.1t0.2
5e50t
t0.1
vo = [10e50(t0.1) 5e50(t0.2) 5e50t] V
[b] Io= Vos
5 106 = s
5 1065 106Ig
s + 50
Io
Ig =H(s) =
s
s + 50 = 1 50
s + 50
h() =() 50e50
0< t
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Problems 1377
0.1 s< t
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Problems 1379
i100k= 49.33 A
io = 50 + 49.33 = 0.67 AFrom the solution for io,
vo(0.2
) = 50e
10 100e
5 = 0.67 A (checks)At t= 0.2+:
vo(0.2+) =vo(0.2
) = 4.93V; i100k = 49.33 Aio = 0 + 49.33 = 49.33 A
From the solution for io,
io(0.2+) = 50e10 100e5 + 50 = 49.33 A(checks)
P 13.76 [a] Y(s) =
0
y(t)est dt
Y(s) =
0est
0h()x(t ) d
dt
=
0
0esth()x(t ) ddt
=
0h()
0estx(t ) dtd
But x(t ) = 0 when t < .
Therefore Y(s) =
0h()
estx(t ) dtd
Let u= t ; du= dt; u= 0 when t= ; u= when t= .
Y(s) =
0h()
0es(u+)x(u) dud
=
0h()es
0esux(u) dud
=
0h()esX(s) d= H(s) X(s)
Note on x(t ) = 0, t < We are using one-sided Laplace transforms; therefore h(t) and x(t) areassumed zero for t
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1380 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] F(s) = a
s(s +a)2 =
1
s a
(s +a)2 =H(s)X(s)
. . h(t) = u(t), x(t) = at eatu(t)
. . f(t) = t
0(1)aea d= a
ea
a2 (a 1)
t
0
= 1
a[eat(at 1) 1(1)] =1
a[1 eat ateat]
=
1
a1
aeat teat
u(t)
Check:
F(s) = a
s(s +a)2 =
K0s
+ K1
(s +a)2+
K2s +a
K0 = 1
a; K1 =
1; K2=
d
dsa
s
s=a=
1
a
f(t) =
1
a teat 1
aeat
u(t)
P 13.77 H(j3) = 4(3 +j3)
9 +j24 + 41 = 0.42/8.13
. . vo(t) = 16.97cos(3t+ 8.13) V 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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Problems 1381
P 13.78 Vo= 50
s + 8000 20
s + 5000 =
30(s + 3000)
(s + 5000)(s+ 8000)
Vo= H(s)Vg =H(s)
30
s
. . H(s) = s(s + 3000)(s + 5000)(s+ 8000)
H(j6000) = (j6