Tipler and Mosca Physics for Scientists and Engineers Solutions Manual chapter 13

8/13/2019 Tipler and Mosca Physics for Scientists and Engineers Solutions Manual chapter 13

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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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3/102

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






4/102


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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7/102

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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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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11/102

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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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






13/102

Problems 1313

[c] vo(t) = 76.2e5t cos(5t 156.8)u(t) V

P 13.12 [a] For t


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15/102

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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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

vo(t) = [18 30e4000t + 12e10,000t]u(t) V 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be





17/102

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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[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






19/102

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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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

s(s + 1 j7)(s + 1 +j7) 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be





21/102

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






22/102


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





23/102

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)






24/102


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)






25/102


26/102


[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






27/102

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






28/102


29/102

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)






30/102


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)






31/102

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






32/102


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


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





33/102

Problems 1333


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





34/102


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


i(0+) = 0 (checks)






35/102

Problems 1335


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






36/102


[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


37/102

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






38/102


39/102

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





40/102


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


41/102

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






42/102


43/102

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






44/102


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






45/102

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






46/102


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





47/102

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






48/102


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)






49/102

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






50/102


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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[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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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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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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[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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[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





61/102

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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[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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[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






67/102

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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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


69/102

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()






70/102


[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






71/102

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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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






73/102

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





74/102


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()






75/102

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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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


78/102


79/102

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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[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





81/102

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

Tipler and Mosca Physics for Scientists and Engineers Solutions Manual chapter 13

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Transcript of Tipler and Mosca Physics for Scientists and Engineers Solutions Manual chapter 13