Solucionario Capitulo 16 Sadiku
88
PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 16, Problem 1. Determine i(t) in the circuit of Fig. 16.35 by means of the Laplace transform. Figure 16.35 For Prob. 16.1. Chapter 16, Solution 1. Consider the s-domain form of the circuit which is shown below. 2 2 2 ) 2 3 ( ) 2 1 s ( 1 1 s s 1 s 1 s 1 s 1 ) s ( I + + = + + = + + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = t 2 3 sin e 3 2 ) t ( i 2 t - = ) t ( i A ) t 866 . 0 ( sin e 155 . 1 -0.5t 1/s 1 + − 1/s s I(s)
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Transcript of Solucionario Capitulo 16 Sadiku
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 1. Determine i(t) in the circuit of Fig. 16.35 by means of the Laplace transform.
Figure 16.35 For Prob. 16.1. Chapter 16, Solution 1. Consider the s-domain form of the circuit which is shown below.
222 )23()21s(1
1ss1
s1s1s1
)s(I++
=++
=++
=
⎟⎟⎠
⎞⎜⎜⎝
⎛= t
23sine
32)t(i 2t-
=)t(i A)t866.0(sine155.1 -0.5t
1/s
1
+ −
1/s
s
I(s)
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 2. Find v x in the circuit shown in Fig. 16.36 given v s .= 4u(t)V.
Figure 16.36 For Prob. 16.2.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Solution 2.
V)t(u)e2e24(v
38j
34s
125.0
38j
34s
125.0s25.016
)8s8s3(s2s16V
s32s16)8s8s3(V
0VsV)s4s2(s
)32s16()8s4(V
0
s84
0V2
0Vs
s4V
t)9428.0j3333.1(t)9428.0j3333.1(x
2x
2x
x2
x2
x
xxx
−−+− ++−=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−+
−+
++
−+−=
++
+−=
+=++
=++++
−+
=+
−+
−+
−
vx = Vt3
22sine2
6t3
22cose)t(u4 3/t43/t4⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛− −−
4 +
Vx
−
s 8/s
s4
+ − 2
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 3. Find i(t) for t > 0 for the circuit in Fig. 16.37. Assume i s = 4u(t) + 2δ (t)mA. (Hint: Can we use superposition to help solve this problem?)
Figure 16.37 For Prob. 16.3.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Solution 3. In the s-domain, the circuit becomes that shown below. 1 I 4 2s+
2 0.2s We transform the current source to a voltage source and obtain the circuit shown below. 2 1 I 8 4s+
0.2s
8 4 20 403 0.2 ( 15) 15
s A BsIs s s s s
+ += = = +
+ + +
40 8 15 20 40 52,15 3 15 3
xA B − += = = =
−
8 / 3 52 / 315
Is s
= ++
158 52( ) ( )3 3
ti t e u t−⎡ ⎤= +⎢ ⎥⎣ ⎦
+ _
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 4. The capacitor in the circuit of Fig. 16.38 is initially uncharged. Find v 0 (t) for t > 0.
Figure 16.38 For Prob. 16.4.
Chapter 16, Solution 4. The circuit in the s-domain is shown below. I 2 4I + 5 Vo 1/s 1 –
4 51/
oo
VI I I sVs
+ = ⎯⎯→ =
But 52
oVI −=
5 12.552 5 / 2
oo o
V sV Vs
−⎛ ⎞ = ⎯⎯→ =⎜ ⎟ +⎝ ⎠
2.5( ) 12.5 Vt
ov t e−=
+ _
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Chapter 16, Problem 5.
If i s (t) = e t2− u(t) A in the circuit shown in Fig. 16.39, find the value of i 0 (t).
Figure 16.39 For Prob. 16.5. Chapter 16, Solution 5.
( )
( ) A)t(ut3229.1sin7559.0e
or
A)t(ueee3779.0eee3779.0e)t(i
3229.1j5.0s)646.2j)(3229.1j5.1(
)3229.1j5.0(
3229.1j5.0s)646.2j)(3229.1j5.1(
)3229.1j5.0(
2s1
)3229.1j5.0s)(3229.1j5.0s)(2s(s
2VsI
)3229.1j5.0s)(3229.1j5.0s)(2s(s2
2sss2
2s1
2s
21
s1
12s
1V
t2
t3229.1j2/t90t3229.1j2/t90t2o
22
2o
2
−=
++=
−+++
+−
+++
−−−−
++
=
−++++==
−++++=⎟⎟
⎠
⎞⎜⎜⎝
⎛
+++=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
+++=
−
−°−−°−−
or io(t) = A)t(ut27sin
72e t2
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−−
s s2
2s1+
Io
2
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 6. Find v(t), t > 0 in the circuit of Fig. 16.40. Let v s = 20 V.
Figure 16.40 For Prob. 16.6.
Chapter 16, Solution 6. For t<0, v(0) = vs = 20 V For t>0, the circuit in the s-domain is as shown below.
10s
1 10100 0.1mF FsC s
= ⎯⎯→ =
20 2
10 110sI
ss= =
++
20101
V Is
= =+
( ) 20 ( )tv t e u t−=
+ _
+
_ 10 Ω v
20s
I
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 7. Find v 0 (t), for all t > 0, in the circuit of Fig. 16.41.
Figure 16.41 For Prob. 16.7.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Solution 7. The circuit in the s-domain is shown below. Please note, iL(0) = 0 and vo(0) = o because both sources were equal to zero for all t<0. 1 1 Vo
2/s s 2/s 1/s At node 1
11 11
2 / 2 (2 1/ )1 1
oo
V Vs V V V s Vs s
−−= + ⎯⎯→ = + − (1)
At node O, 1
11 (1 / 2) 1/
1 2 / 2o o
o oV V V s V V s V s
s s−
+ = = ⎯⎯→ = + − (2)
Substituting (2) into (1) gives 1 12 / (2 1/ )(1 / 2) (2 )o os s s V Vs s
= + + − + −
2 2
(4 1)( 1.5 1) 1.5 1o
s A Bs CVs s s s s s
+ += = +
+ + + +
2 24 1 ( 1.5 1)s A s s Bs Cs+ = + + + + We equate coefficients. s2 : 0 = A+ B or B = - A s: 4=1.5A + C constant: 1 = A, B=-1, C = 4-1.5A = 2.5
2 22
2 2
3.25 747
1 2.5 1 3/ 4 41.5 1 7 7( 3/ 4) ( 3 / 4)
4 4
o
x
s sVs s s s
s s
− + += + = − +
+ + ⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
3 / 4 3 / 47 7( ) ( ) cos 4.9135 sin4 4
t tv t u t e t e t− −= − +
+ _
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Chapter 16, Problem 8. If v 0 (0) = -1V,obtain v 0 (t) in the circuit of Fig. 16.42.
Figure 16.42 For Prob. 16.8.
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Chapter 16, Solution 8.
1 1 22
FsC s
⎯⎯→ =
We analyze the circuit in the s-domain as shown below. We apply nodal analysis.
0
3 1 414021 1 ( 4)
o o oV V V ss s s Vs s
s
− − − − −+ + = ⎯⎯→ =
+
4oA BVs s
= ++
14 187 / 2, 9 / 24 4
A B= = = = −−
7 / 2 9 / 24oV
s s= −
+
47 9( ) ( )2 2
tov t e u t−⎛ ⎞= −⎜ ⎟
⎝ ⎠
1 Vo 1
+ _
+ _
+ _
2s
1s−
4s
3s
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Chapter 16, Problem 9. Find the input impedance Z in (s) of each of the circuits in Fig. 16.43.
Figure 16.43 For Prob. 16.9. Chapter 16, Solution 9.
(a) The s-domain form of the circuit is shown in Fig. (a).
=+++
=+=s1s2)s1s(2
)s1s(||2Zin 1s2s)1s(2
2
2
+++
(b) The s-domain equivalent circuit is shown in Fig. (b).
2s3)2s(2
s23)s21(2
)s21(||2++
=++
=+
2s36s5
)s21(||21++
=++
=⎟⎠⎞
⎜⎝⎛
++
+
⎟⎠⎞
⎜⎝⎛
++
⋅=⎟
⎠⎞
⎜⎝⎛
++
=
2s36s5
s
2s36s5
s
2s36s5
||sZin 6s7s3)6s5(s
2 +++
2
s
1/s
(a) (b)
2 s 2/s
1
1
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Chapter 16, Problem 10. Use Thevenin’s theorem to determine v 0 (t), t > 0 in the circuit of Fig. 16.44.
Figure 16.44 For Prob. 16.10.
Chapter 16, Solution 10. 1 1H s⎯⎯→ and iL(0) = 0 (the sources is zero for all t<0).
1 1 44
FsC s
⎯⎯→ = and vC(0) = 0 (again, there are no source
contributions for all t<0). To find ZTh , consider the circuit below. 1 s ZTh 2
21//( 2)
3ThsZ ss+
= + =+
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To find VTh, consider the circuit below. 1 s
+
102s +
VTh 2
-
2 10 103 2 3Th
sVs s s+
= =+ + +
The Thevenin equivalent circuit is shown below ZTh + 4/s Vo VTh –
2 3 2
404 4 310 40 3
4 4 2 3 6 12 ( 3) ( 3)3
o Th
Th
s sV V s s s s sZs s s
= = = =+ + + + + ++ ++
.
3( ) 23.094 sin 3t
ov t e t−=
+ _
+ _
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Chapter 16, Problem 11. Solve for the mesh currents in the circuit of Fig. 16.45. You may leave your results in the s-domain.
Figure 16.45 For Prob. 16.11.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Solution 11. In the s-domain, the circuit is as shown below.
1 210 1(1 )
4 4s I sI
s= + − (1)
1 21 5(4 ) 04 4
sI I s− + + = (2)
In matrix form,
1
2
110 14 4
1 50 44 4
s s Is
Is s
⎡ ⎤+ −⎡ ⎤ ⎢ ⎥ ⎡ ⎤⎢ ⎥ = ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎢ ⎥− +⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦
4s49s
41 2 ++=∆
1
10 140 504
5 40 44
ss
ss
−∆ = = +
+
2
10154
1 204
ss
s
+∆ = =
−
)16s9s(s160s50
4s25.2s25.0225
s40
I 221
1++
+=
++
+=
∆∆
=
16s9s10
4s25.2s25.05.2I 22
22
++=
++=
∆∆
=
1 Ω 4
+ _ I1 I2
1 H4 1 s
10s
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Chapter 16, Problem 12. Find v o (t) in the circuit of Fig. 16.46.
Figure 16.46 For Prob. 16.12.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Solution 12. We apply nodal analysis to the s-domain form of the circuit below.
oo
osV2
4V
s3
s
V1s
10
+=+−
+
1s15s1510
151s
10V)ss25.01( o
2
+++
=++
=++
1s25.0sCBs
1sA
)1s25.0s)(1s(25s15
V 22o +++
++
=+++
+=
740
V)1s(A 1-so =+= =
)1s(C)ss(B)1s25.0s(A25s15 22 ++++++=+
Equating coefficients :
2s : -ABBA0 =⎯→⎯+= 1s : C-0.75ACBA25.015 +=++= 0s : CA25 +=
740A = , 740-B = , 7135C =
43
21
s
23
32
7155
43
21
s
21
s
740
1s1
740
43
21
s
7135
s740-
1s740
V 222o
+⎟⎠⎞
⎜⎝⎛+
⎟⎠⎞
⎜⎝⎛
⋅++⎟
⎠⎞
⎜⎝⎛+
+−
+=
+⎟⎠⎞
⎜⎝⎛+
++
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟⎠
⎞⎜⎜⎝
⎛−= t
23
sine)3)(7()2)(155(
t23
cose740
e740
)t(v 2t-2t-t-o
=)t(vo V)t866.0sin(e57.25)t866.0cos(e714.5e714.5 2-t2-t-t +−
s
3/s 4 1/(2s) + −
10/(s + 1)
Vo
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Chapter 16, Problem 13. Determine i 0 (t) in the circuit of Fig. 16.47.
Figure 16.47 For Prob. 16.13. Chapter 16, Solution 13. Consider the following circuit.
Applying KCL at node o,
ooo V
1s21s
s12V
1s2V
2s1
++
=+
++
=+
)2s)(1s(1s2
Vo +++
=
2sB
1sA
)2s)(1s(1
1s2V
I oo +
++
=++
=+
=
1A = , -1B =
2s1
1s1
Io +−
+=
=)t(io ( ) A)t(uee -2t-t −
2s
2 1
1/s
1/(s + 2)
Vo
Io
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Chapter 16, Problem 14. * Determine i 0 (t) in the network shown in Fig. 16.48.
Figure 16.48 For Prob. 16.14. * An asterisk indicates a challenging problem. Chapter 16, Solution 14. We first find the initial conditions from the circuit in Fig. (a).
A5)0(io =− , V0)0(vc =−
We now incorporate these conditions in the s-domain circuit as shown in Fig.(b).
4 Ω1 Ω
+ −
5 V
(a)
+
vc(0)
−
io
4/s
4 1
+ −
15/s
Vo
Io
(b)
5/s 2s
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
At node o,
0s440V
s5
s2V
1s15V ooo =
+−
+++−
oV)1s(4
ss2
11
s5
s15 ⎟
⎠
⎞⎜⎝
⎛
+++=−
o
2
o
22
V)1s(s42s6s5
V)1s(s4
s2s2s4s4s
10+++
=+
++++=
2s6s5)1s(40
V 2o +++
=
s5
)4.0s2.1s(s)1s(4
s5
s2V
I 2o
o +++
+=+=
4.0s2.1sCBs
sA
s5
I 2o +++
++=
sCsB)4.0s2.1s(A)1s(4 s2 ++++=+
Equating coefficients :
0s : 10AA4.04 =⎯→⎯= 1s : -84-1.2ACCA2.14 =+=⎯→⎯+= 2s : -10-ABBA0 ==⎯→⎯+=
4.0s2.1s8s10
s10
s5
I 2o +++
−+=
2222o 2.0)6.0s()2.0(10
2.0)6.0s()6.0s(10
s15
I++
−++
+−=
=)t(io ( )[ ] A)t(u)t2.0sin()t2.0cos(e1015 0.6t- −−
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 15. Find V x (s) in the circuit shown in Fig. 16.49.
Figure 16.49 For Prob. 16.15. Chapter 16, Solution 15. First we need to transform the circuit into the s-domain.
2s5VV
2s5VV,But
2ss5V120V)40ss2(0
2ss5sVVs2V120V40
010
2s5V
s/50V
4/sV3V
xoox
xo2
oo2
xo
ooxo
++=→
+−=
+−−++==
+−++−
=+−
+−
+−
We can now solve for Vx.
)40s5.0s)(2s()20s(5V
2s)20s(10V)40s5.0s(2
02s
s5V1202s
5V)40ss2(
2
2x
2x
2
xx2
−++
+−=
++
−=−+
=+
−−⎟⎠⎞
⎜⎝⎛
++++
10
3Vx
+ Vx −
s/4
5/s 2s5+
+ − +
Vo
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 16. * Find i 0 (t) for t > 0 in the circuit of Fig. 16.50.
Figure 16.50 For Prob. 16.16. * An asterisk indicates a challenging problem.
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Chapter 16, Solution 16. We first need to find the initial conditions. For 0t < , the circuit is shown in Fig. (a).
To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit. Hence,
A1-33-
i)0(i oL === , V1-vo =
V5.221-
)1-)(2(-)0(vc =⎟⎠⎞
⎜⎝⎛
−=
We now incorporate the initial conditions for 0t > as shown in Fig. (b).
For mesh 1,
02
Vs5.2
Is1
Is1
22s
5- o21 =++−⎟
⎠⎞
⎜⎝⎛
+++
(a)
3 V
2 Ω
+ −
io1 H + Vo/2
1 F
+ −Vo
1 Ω
(b)
-1 V
2
− +
Io
+ Vo/2
1/s
+ −Vo
1
+ −
2.5/s
s
+ −
5/(s + 2) I1 I2
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But, 2oo IIV ==
s5.2
2s5
Is1
21
Is1
2 21 −+
=⎟⎠⎞
⎜⎝⎛
−+⎟⎠⎞
⎜⎝⎛
+ (1)
For mesh 2,
0s5.2
2V
1Is1
Is1
s1 o12 =−−+−⎟
⎠⎞
⎜⎝⎛
++
1s5.2
Is1
s21
Is1
- 21 −=⎟⎠⎞
⎜⎝⎛
+++ (2)
Put (1) and (2) in matrix form.
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−
−+
=⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
++
−+
1s5.2
s5.2
2s5
I
I
s1
s21
s1-
s1
21
s1
2
2
1
s3
2s2 ++=∆ , )2s(s
5s4
2-2 +++=∆
3s2s2CBs
2sA
)3s2s2)(2s(132s-
II 22
22
2o +++
++
=+++
+=
∆∆
==
)2s(C)s2s(B)3s2s2(A132s- 222 ++++++=+
Equating coefficients :
2s : BA22- += 1s : CB2A20 ++= 0s : C2A313 +=
Solving these equations leads to
7143.0A = , -3.429B = , 429.5C =
5.1ss714.2s7145.1
2s7143.0
3s2s2429.5s429.3
2s7143.0
I 22o ++−
−+
=++
−−
+=
25.1)5.0s()25.1)(194.3(
25.1)5.0s()5.0s(7145.1
2s7143.0
I 22o +++
+++
−+
=
=)t(io [ ] A)t(u)t25.1sin(e194.3)t25.1cos(e7145.1e7143.0 -0.5t-0.5t-2t +−
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Chapter 16, Problem 17. Calculate i 0 (t) for t > 0 in the network of Fig. 16.51.
Figure 16.51 For Prob. 16.17. Chapter 16, Solution 17. We apply mesh analysis to the s-domain form of the circuit as shown below. For mesh 3,
0IsIs1
Is1
s1s
2213 =−−⎟
⎠⎞
⎜⎝⎛++
+ (1)
For the supermesh,
0Iss1
I)s1(Is1
1 321 =⎟⎠⎞
⎜⎝⎛
+−++⎟⎠⎞
⎜⎝⎛+ (2)
s
1 1
1/s
2/(s+1)
+ −
I3
I2I1
4/s
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Adding (1) and (2) we get, I1 + I2 = –2/(s+1) (3) But –I1 + I2 = 4/s (4) Adding (3) and (4) we get, I2 = (2/s) – 1/(s+1) (5) Substituting (5) into (4) yields, I1 = –(2/s) – (1/(s+1)) (6) Substituting (5) and (6) into (1) we get,
1s
2Is
1s1s
s2)1s(s
1s2
32
2 +−=⎟
⎟⎠
⎞⎜⎜⎝
⎛ ++
++−
++
js
j5.05.1js
j5.05.1s2I3 −
++
+−
+−=
Substituting (3) into (1) and (2) leads to
)1s(s)2s2s(2I
s1sI
s1s- 2
232
+
++−=⎟
⎠⎞
⎜⎝⎛ ++⎟
⎠⎞
⎜⎝⎛ + (4)
232s
)1s(4Is1sI
s1s2 +
−=⎟⎠⎞
⎜⎝⎛ +−⎟
⎠⎞
⎜⎝⎛ ++ (5)
We can now solve for Io. Io = I2 – I3 = (4/s) – (1/(s+1)) + ((–1.5+0.5j)/(s+j)) + ((–1.5–0.5)/(s–j)) or io(t) = [4 – e–t + 1.5811e–jt+161.57˚ + 1.5811ejt–161.57˚]u(t)A This is a challenging problem. I did check it with using a Thevenin equivalent circuit and got the same exact answer.
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Chapter 16, Problem 18. (a) Find the Laplace transform of the voltage shown in Fig. 16.52(a). (b) Using that value of v s (t) in the circuit shown in Fig. 16.52(b), find the value of v 0 (t).
Figure 16.52 For Prob. 16.18. Chapter 16, Solution 18.
vs(t) = 3u(t) – 3u(t–1) or Vs = )e1(s3
se
s3 s
s−
−−=−
V)]1t(u)e22()t(u)e22[()t(v
)e1(5.1s
2s2)e1(
)5.1s(s3V
VV)5.1s(02
VsV
1VV
)1t(5.1t5.1o
sso
soo
oso
−−−−=
−⎟⎠⎞
⎜⎝⎛
+−=−
+=
=+→=++−
−−−
−−
1 Ω
1/s Vs
+ − 2 Ω
+
Vo
−
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Chapter 16, Problem 19.
In the circuit of Fig. 16.53, let i(0) = 1 A, v 0 (0) and v s = 4e t2− u(t) V. Find v 0 (t) for t > 0.
Figure 16.53 For Prob. 16.19.
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Chapter 16, Solution 19. We incorporate the initial conditions in the s-domain circuit as shown below. At the supernode,
o11 sV
s1
sV
22
V))2s(4(++=+
−+
o1 Vss1
Vs1
21
22s
2++⎟
⎠⎞
⎜⎝⎛
+=++
(1)
But I2VV 1o += and s
1VI 1 +=
2s2Vs
s)2s(s2V
Vs
)1V(2VV oo
11
1o +−
=+−
=⎯→⎯+
+= (2)
Substituting (2) into (1)
oo Vs2s
2V2s
ss22s
s12
2s2
+⎥⎦
⎤⎢⎣
⎡+
−⎟⎠⎞
⎜⎝⎛
+⎟⎠⎞
⎜⎝⎛ +
=−++
oVs21
s1
s12
2s2
⎥⎦
⎤⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛=+−+
+
oV)2/1s(2s6s2
)2s(24s2
+=++
=+++
2sB
2/1sA
)2/1s)(2s(6s2Vo +
++
=+++
=
333.3)25.0/()61(A =+−+−= , 3333.1)2/12/()64(B −=+−+−=
2s3333.1
2/1s333.3Vo +
−+
=
Therefore, =)t(vo (3.333e-t/2 – 1.3333e-2t)u(t) V
2
+ −
s
1/s
VoV1
2 I
− +
2 4/(s + 2)
1/s
I
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Chapter 16, Problem 20. Find v 0 (t) in the circuit of Fig. 16.54 if v x (0) = 2 V and i(0) = 1A.
Figure 16.54 For Prob. 16.20.
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Chapter 16, Solution 20. We incorporate the initial conditions and transform the current source to a voltage source as shown.
At the main non-reference node, KCL gives
s1
sV
1V
s11Vs2)1s(1 ooo ++=
+−−+
s1sV)s11)(1s(Vs2
1ss
oo+
+++=−−+
oV)s12s2(2s
1s1s
s++=−
+−
+
)1s2s2)(1s(1s4s2-
V 2
2
o +++−−
=
5.0ssCBs
1sA
)5.0ss)(1s(5.0s2s-
V 22o +++
++
=+++
−−=
1V)1s(A 1-so =+= =
)1s(C)ss(B)5.0ss(A5.0s2s- 222 ++++++=−−
Equating coefficients :
2s : -2BBA1- =⎯→⎯+= 1s : -1CCBA2- =⎯→⎯++= 0s : -0.515.0CA5.00.5- =−=+=
222o )5.0()5.0s()5.0s(2
1s1
5.0ss1s2
1s1
V++
+−
+=
+++
−+
=
=)t(vo [ ] V)t(u)2tcos(e2e 2-t-t −
1
+ −
Vo
2/s
1/s
1/s + −
1 s 1/(s + 1)
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Chapter 16, Problem 21. Find the voltage v 0 (t) in the circuit of Fig. 16.55 by means of the Laplace transform.
Figure 16.55 For Prob. 16.21.
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Chapter 16, Solution 21. The s-domain version of the circuit is shown below. 1 s
V1 Vo + 2/s 2 1/s
- At node 1,
ooo VsVsVs
sVVV
s )12
()1(1021
102
11
1−++=⎯→⎯+
−=
− (1)
At node 2,
)12
(2
21
1 ++=⎯→⎯+=− ssVVsVVsVV
oooo (2)
Substituting (2) into (1) gives
ooo VsssVsVsss )5.12()12
()12/)(1(10 22
2 ++=−++++=
5.12)5.12(10
22 +++
+=++
=ss
CBssA
sssVo
CsBsssA ++++= 22 )5.12(10
BAs +=0:2 CAs += 20:
-40/3C -20/3,B ,3/205.110:constant ===⎯→⎯= AA
⎥⎦
⎤⎢⎣
⎡++
−+++
−=⎥⎦⎤
⎢⎣⎡
+++
−= 22222 7071.0)1(7071.0414.1
7071.0)1(11
320
5.1221
320
sss
ssss
sVo
Taking the inverse Laplace tranform finally yields
[ ] V)t(ut7071.0sine414.1t7071.0cose1320)t(v tt
o−− −−=
10/s
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Chapter 16, Problem 22. Find the node voltages v 1 and v 2 in the circuit of Fig. 16.56 using the Laplace transform technique. Assume that i s = 12e t− u(t) A and that all initial conditions are zero.
Figure 16.56 For Prob. 16.22. Chapter 16, Solution 22. The s-domain version of the circuit is shown below. 4s V1 V2
1s
12+
1 2 3/s
At node 1,
s4V
s411V
1s12
s4VV
1V
1s12 2
1211 −⎟
⎠⎞
⎜⎝⎛ +=
+⎯→⎯
−+=
+ (1)
At node 2,
⎟⎠⎞
⎜⎝⎛ ++=⎯→⎯+=
−1s2s
34VVV
3s
2V
s4VV 2
212221 (2)
Substituting (2) into (1),
222
2 V23s
37s
34
s41
s4111s2s
34V
1s12
⎟⎠⎞
⎜⎝⎛ ++=⎥
⎦
⎤⎢⎣
⎡−⎟
⎠⎞
⎜⎝⎛ +⎟⎠⎞
⎜⎝⎛ ++=
+
)89s
47s(
CBs)1s(
A
)89s
47s)(1s(
9V22
2++
++
+=
+++=
)1s(C)ss(B)89s
47s(A9 22 ++++++=
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Equating coefficients:
BA0:s2 +=
A43CCA
43CBA
470:s −=⎯→⎯+=++=
-18C -24,B ,24AA83CA
899:constant ===⎯→⎯=+=
6423)
87s(
3
6423)
87s(
)8/7s(24)1s(
24
)89s
47s(
18s24)1s(
24V222
2++
+++
+−
+=
++
+−
+=
Taking the inverse of this produces:
[ ] )t(u)t5995.0sin(e004.5)t5995.0cos(e24e24)t(v t875.0t875.0t2
−−− +−= Similarly,
)89s
47s(
FEs)1s(
D
)89s
47s)(1s(
1s2s349
V22
2
1++
++
+=
+++
⎟⎠⎞
⎜⎝⎛ ++
=
)1s(F)ss(E)89s
47s(D1s2s
349 222 ++++++=⎟
⎠⎞
⎜⎝⎛ ++
Equating coefficients:
ED12:s2 +=
D436FFD
436or FED
4718:s −=⎯→⎯+=++=
0F 4,E ,8DD833or FD
899:constant ===⎯→⎯=+=
6423)
87s(
2/7
6423)
87s(
)8/7s(4)1s(
8
)89s
47s(
s4)1s(
8V222
1++
−++
++
+=
+++
+=
Thus,
[ ] )t(u)t5995.0sin(e838.5)t5995.0cos(e4e8)t(v t875.0t875.0t1
−−− −+=
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Chapter 16, Problem 23. Consider the parallel RLC circuit of Fig. 16.57. Find v(t) and i(t) given that v(0) = 5 and i(0) = -2 A.
Figure 16.57 For Prob. 16.23.
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Chapter 16, Solution 23. The s-domain form of the circuit with the initial conditions is shown below. At the non-reference node,
sCVsLV
RV
C5s2
s4
++=++
⎟⎠⎞
⎜⎝⎛
++=+
LC1
RCs
ss
CVs
sC56 2
LC1RCssC6s5
V 2 +++
=
But 88010
1RC1
== , 208041
LC1
==
22222 2)4s()2)(230(
2)4s()4s(5
20s8s480s5
V++
+++
+=
+++
=
=)t(v V)t(u))t2sin(e230)t2cos(e5( -4t-4t +
)20s8s(s4480s5
sLV
I 2 +++
==
20s8sCBs
sA
)20s8s(s120s25.1
I 22 +++
+=++
+=
6A = , -6B = , -46.75C =
22222 2)4s()2)(375.11(
2)4s()4s(6
s6
20s8s75.46s6
s6
I++
−++
+−=
+++
−=
=)t(i 0t),t(u))t2sin(e375.11)t2cos(e66( -4t-4t >−−
-2/s R sL 1/sC 4/s
VI
5C
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Chapter 16, Problem 24. The switch in Fig. 16.58 moves from position 1 to position 2 at t =0. Find v(t), for all t > 0.
Figure 16.58 For Prob. 16.24. Chapter 16, Solution 24. When the switch is position 1, v(0)=12, and iL(0) = 0. When the switch is in position 2, we have the circuit as shown below. s/4 + 100/s v 12/s
–
1 10010 0.01mF FsC s
= ⎯⎯→ =
2
12 / 48/ 4 100 / 400
sIs s s
= =+ +
, 2
124 400s sV sLI I
s= = =
+
( ) 12cos 20 , 0v t t t= >
+ _
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Chapter 16, Problem 25. For the RLC circuit shown in Fig. 16.59, find the complete response if v(0) = 2 V when the switch is closed.
Figure 16.59 For Prob. 16.25. Chapter 16, Solution 25. For 0t > , the circuit in the s-domain is shown below.
s
2/s
9/s
6
+ −
(2s)/(s2 + 16) + −
I
+
V
−
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Applying KVL,
0s2
Is9
s616ss2
2 =+⎟⎠⎞
⎜⎝⎛
++++−
)16s)(9s6s(32I
22 +++
−=
)16s()3s(s288
s2
s2I
s9V 22 ++
−+=+=
16sEDs
)3s(C
3sB
sA
s2
22 ++
++
++
++=
)s48s16s3s(B)144s96s25s6s(A288- 234234 ++++++++=
)s9s6s(E)s9s6s(D)s16s(C 232343 ++++++++
Equating coefficients : 0s : A144288 =− (1) 1s : E9C16B48A960 +++= (2) 2s : E6D9B16A250 +++= (3) 3s : ED6CB3A60 ++++= (4) 4s : DBA0 ++= (5)
Solving equations (1), (2), (3), (4) and (5) gives
2A −= , 202.2B = , 84.3C = , 202.0-D = , 766.2E =
16s)4)(6915.0(
16ss202.0
)3s(84.3
3s202.2)s(V 222 +
++
−+
++
=
=)t(v 2.202e-3t + 3.84te-3t – 0.202cos(4t) + 0.6915sin(4t)u(t) V
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Chapter 16, Problem 26.
For the op amp circuit in Fig. 16.60, find v 0 (t) for t > 0. Take v s = 3e t5− u(t) V.
Figure 16.60 For Prob. 16.26.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Solution 26. Consider the op-amp circuit below.
At node 0,
sC)V0(R
V0R
0Vo
2
o
1
s −+−
=−
( )o2
1s V-sCR1
RV ⎟⎠
⎞⎜⎝
⎛+=
211s
o
RRCsR1-
VV
+=
But 21020
RR
2
1 == , 1)1050)(1020(CR 6-31 =××=
So, 2s
1-VV
s
o
+=
)5s(3Ve3V s-5t
s +=⎯→⎯=
5)2)(s(s3-
Vo ++=
5sB
2sA
5)2)(s(s3
V- o ++
+=
++=
1A = , -1B =
2s1
5s1
Vo +−
+=
=)t(vo ( ) )t(uee -2t-5t −
−+
1/sC
R1
R2
+ −
Vs
0
+Vo−
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Chapter 16, Problem 27. Find I 1 (s) and I 2 (s) in the circuit of Fig. 16.61.
Figure 16.61 For Prob. 16.27.
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Chapter 16, Solution 27. Consider the following circuit.
For mesh 1,
221 IsII)s21(3s
10−−+=
+
21 I)s1(I)s21(3s
10+−+=
+ (1)
For mesh 2,
112 IsII)s22(0 −−+=
21 I)1s(2I)s1(-0 +++= (2) (1) and (2) in matrix form,
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡++++
=⎥⎦
⎤⎢⎣
⎡ +
2
1
II
)1s(2)1s(-)1s(-1s2
0)3s(10
1s4s3 2 ++=∆
3s)1s(20
1 ++
=∆
3s)1s(10
2 ++
=∆
Thus
=∆∆
= 11I )1s4s3)(3s(
)1s(202 ++++
=∆∆
= 22I
)1s4s3)(3s()1s(10
2 ++++
2I1=
2s
1 10/(s + 3) + −
I1
2s
1 I2
s
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Chapter 16, Problem 28. For the circuit in Fig. 16.62, find v 0 (t) for t > 0.
Figure 16.62 For Prob. 16.28.
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Chapter 16, Solution 28. Consider the circuit shown below. For mesh 1,
21 IsI)s21(s6
++= (1)
For mesh 2,
21 I)s2(Is0 ++=
21 Is2
1-I ⎟⎠⎞
⎜⎝⎛+= (2)
Substituting (2) into (1) gives
2
2
22 Is
2)5s(s-IsI
s2
12s)-(1s6 ++
=+⎟⎠⎞
⎜⎝⎛++=
or 2s5s
6-I 22 ++=
4.561)0.438)(s(s12-
2s5s12-
I2V 22o ++=
++==
Since the roots of 02s5s2 =++ are -0.438 and -4.561,
561.4sB
438.0sA
Vo ++
+=
-2.914.123
12-A == , 91.2
4.123-12-
B ==
561.4s91.2
0.438s2.91-
)s(Vo ++
+=
=)t(vo [ ] V)t(uee91.2 t438.0-4.561t −
s
2 I2s 2s
1
+ −
6/s I1 +
Vo
−
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Chapter 16, Problem 29. For the ideal transformer circuit in Fig. 16.63, determine i 0 (t).
Figure 16.63 For Prob. 16.29. Chapter 16, Solution 29. Consider the following circuit.
Let 1s2
8s48)s4)(8(
s4
||8ZL +=
+==
When this is reflected to the primary side,
2n,nZ
1Z 2L
in =+=
1s23s2
1s22
1Zin ++
=+
+=
3s21s2
1s10
Z1
1s10
Iin
o ++
⋅+
=⋅+
=
5.1sB
1sA
)5.1s)(1s(5s10
Io ++
+=
+++
=
-10A = , 20B =
5.1s20
1s10-
)s(Io ++
+=
=)t(io [ ] A)t(uee210 t-1.5t −−
8 4/s
1
10/(s + 1) + −
Io 1 : 2
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Chapter 16, Problem 30. The transfer function of a system is
H(s) = 13
2
+ss
Find the output when the system has an input of 4e 3/t− u(t). Chapter 16, Solution 30.
)s(X)s(H)s(Y = , 1s3
1231s
4)s(X
+=
+=
22
2
)1s3(34s8
34
)1s3(s12
)s(Y++
−=+
=
22 )31s(1
274
)31s(s
98
34
)s(Y+
⋅−+
⋅−=
Let 2)31s(s
98-
)s(G+
⋅=
Using the time differentiation property,
⎟⎠⎞
⎜⎝⎛
+=⋅= 3t-3t-3t- eet31-
98-
)et(dtd
98-
)t(g
3t-3t- e98
et278
)t(g −=
Hence,
3t-3t-3t- et274
e98
et278
)t(u34
)t(y −−+=
=)t(y 3t-3t- et274
e98
)t(u34
+−
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Chapter 16, Problem 31. When the input to a system is a unit step function, the response is 10 cos 2tu(t). Obtain the transfer function of the system. Chapter 16, Solution 31.
s1
)s(X)t(u)t(x =⎯→⎯=
4ss10
)s(Y)t2cos(10)t(y 2 +=⎯→⎯=
==)s(X)s(Y
)s(H4s
s102
2
+
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Chapter 16, Problem 32. A circuit is known to have its transfer function as
H(s) = 54
32 ++
+ss
s
Find its output when: (a) the input is a unit step function (b) the input is 6te t2− u(t). Chapter 16, Solution 32.
(a) )s(X)s(H)s(Y =
s1
5s4s3s
2 ⋅++
+=
5s4s
CBssA
)5s4s(s3s
22 +++
+=++
+=
CsBs)5s4s(A3s 22 ++++=+
Equating coefficients :
0s : 53AA53 =⎯→⎯= 1s : 57- A41CCA41 =−=⎯→⎯+= 2s : 53--ABBA0 ==⎯→⎯+=
5s4s7s3
51
s53
)s(Y 2 +++
⋅−=
1)2s(1)2s(3
51
s6.0
)s(Y 2 ++++
⋅−=
=)t(y [ ] )t(u)tsin(e2.0)tcos(e6.06.0 -2t-2t −−
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(b) 22t-
)2s(6
)s(Xet6)t(x+
=⎯→⎯=
22 )2s(6
5s4s3s
)s(X)s(H)s(Y+
⋅++
+==
5s4sDCs
)2s(B
2sA
)5s4s()2s()3s(6
)s(Y 2222 +++
++
++
=+++
+=
Equating coefficients :
3s : -ACCA0 =⎯→⎯+= (1) 2s : DBA2DC4BA60 ++=+++= (2) 1s : D4B4A9D4C4B4A136 ++=+++= (3) 0s : BA2D4B5A1018 +=++= (4)
Solving (1), (2), (3), and (4) gives
6A = , 6B = , 6-C = , -18D =
1)2s(18s6
)2s(6
2s6
)s(Y 22 +++
−+
++
=
1)2s(6
1)2s()2s(6
)2s(6
2s6
)s(Y 222 ++−
+++
−+
++
=
=)t(y [ ] )t(u)tsin(e6)tcos(e6et6e6 -2t-2t-2t-2t −−+
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Chapter 16, Problem 33. When a unit step is applied to a system at t = 0 its response is
y(t) = ⎥⎦⎤
⎢⎣⎡ +−+ −− )4sin34cos2(
214 23 ttee tt u(t)
What is the transfer function of the system? Chapter 16, Solution 33.
)s(X)s(Y
)s(H = , s1
)s(X =
16)2s()4)(3(
16)2s(s2
)3s(21
s4
)s(Y 22 ++−
++−
++=
== )s(Ys)s(H20s4s
s1220s4s)2s(s2
)3s(2s4 22 ++
−++
+−
++
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Chapter 16, Problem 34. For the circuit in Fig. 16.64, find H(s) = V 0 (s)/V s (s). Assume zero initial conditions.
Figure 16.64 For Prob. 16.34. Chapter 16, Solution 34. Consider the following circuit.
Using nodal analysis,
s10V
4V
2sVV ooos +=
+−
o2
os V)s2s(101
)2s(41
1V10s
41
2s1
)2s(V ⎟⎠⎞
⎜⎝⎛
++++=⎟⎠⎞
⎜⎝⎛
+++
+=
( ) o2
s V30s9s2201
V ++=
=s
o
VV
30s9s220
2 ++
s
4
2
+ −
Vs 10/s
Vo
+
Vo(s)
−
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Chapter 16, Problem 35. Obtain the transfer function H(s) = V 0 /V s for the circuit of Fig. 16.65.
Figure 16.65 For Prob. 16.35. Chapter 16, Solution 35. Consider the following circuit.
At node 1,
3sV
II2 1
+=+ , where
s2VV
I 1s −=
3sV
s2VV
3 11s
+=
−⋅
1s1 V
2s3
V2s3
3sV
−=+
s1 V2s3
V2s3
3s1
=⎟⎠⎞
⎜⎝⎛
++
s21 V2s9s3
)3s(s3V
+++
=
s21o V2s9s3
s9V
3s3
V++
=+
=
==s
o
VV
)s(H2s9s3
s92 ++
s
3 2I
2/s
+ −
Vs
V1I
+
Vo
−
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Chapter 16, Problem 36. The transfer function of a certain circuit is
H(s) = 1
5+s
- 2
3+s
+4
6+s
Find the impulse response of the circuit. Chapter 16, Solution 36.
Taking the inverse Laplace transform of each term gives
( )2 4( ) 5 3 6 ( )t t th t e e e u t− − −= − +
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Chapter 16, Problem 37. For the circuit in Fig. 16.66, find:
(a) I 1 /V s (b) I 2 /V x
Figure 16.66 For Prob. 16.37. Chapter 16, Solution 37.
(a) Consider the circuit shown below.
For loop 1,
21s Is2
Is2
3V −⎟⎠⎞
⎜⎝⎛+= (1)
For loop 2,
0Is2
Is2
s2V4 12x =−⎟⎠⎞
⎜⎝⎛
++
But, ⎟⎠⎞
⎜⎝⎛
−=s2
)II(V 21x
So, 0Is2
Is2
s2)II(s8
1221 =−⎟⎠⎞
⎜⎝⎛
++−
21 Is2s6
Is6-
0 ⎟⎠⎞
⎜⎝⎛
−+= (2)
+ −
Vs I2I1
2s 3
+ 4Vx 2/s
+
Vx
−
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In matrix form, (1) and (2) become
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
+=⎥⎦
⎤⎢⎣
⎡
2
1s
II
s2s6s6-s2-s23
0V
⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛
−⎟⎠⎞
⎜⎝⎛
−⎟⎠⎞
⎜⎝⎛+=∆
s2
s6
s2s6
s2
3
4s6s
18−−=∆
s1 Vs2s6
⎟⎠⎞
⎜⎝⎛
−=∆ , s2 Vs6
=∆
s1
1 Vs64s18
)s2s6(I
−−−
=∆∆
=
=−−
−=
32s9ss3
VI
s
1
9s2s33s
2
2
−+−
(b) ∆∆
= 22I
⎟⎠⎞
⎜⎝⎛
∆∆−∆
=−= 2121x s
2)II(
s2
V
∆=
∆−−
= ssx
V4-)s6s2s6(Vs2V
==s
s
x
2
4V-Vs6
VI
2s3-
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Chapter 16, Problem 38. Refer to the network in Fig. 16.67. Find the following transfer functions:
(a) H 1 (s) = V 0 (s)/V s (s) (b) H 2 (s) = V 0 (s)/I s (s) (c) H 3 (s) = I 0 (s)/I s (s) (d) H 4 (s) = I 0 (s)/V s (s)
Figure 16.67 For Prob. 16.38.
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Chapter 16, Solution 38. (a) Consider the following circuit.
At node 1,
sVV
Vs1
VV o11
1s −+=
−
o1s Vs1
Vs1
s1V −⎟⎠⎞
⎜⎝⎛
++= (1)
At node o,
oooo1 V)1s(VVs
sVV
+=+=−
o2
1 V)1ss(V ++= (2) Substituting (2) into (1)
oo2
s Vs1V)1ss)(s11s(V −++++=
o23
s V)2s3s2s(V +++=
==s
o1 V
V)s(H
2s3s2s1
23 +++
(b) o
2o
231ss V)1ss(V)2s3s2s(VVI ++−+++=−=
o23
s V)1s2ss(I +++=
==s
o2 I
V)s(H
1s2ss1
23 +++
(c) 1
VI o
o =
==== )s(HIV
II
)s(H 2s
o
s
o3 1s2ss
123 +++
(d) ==== )s(HVV
VI
)s(H 1s
o
s
o4 2s3s2s
123 +++
s
1
1
+ −
Vs 1/s
V1 Vo Io
+
Vo
−
1/s
Is
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Chapter 16, Problem 39. Calculate the gain H(s) = V 0 /V s in the op amp circuit of Fig. 16.68.
Figure 16.68 For Prob. 16.39. Chapter 16, Solution 39. Consider the circuit below.
Since no current enters the op amp, oI flows through both R and C.
⎟⎠⎞
⎜⎝⎛
+=sC1
RI-V oo
sCI-
VVV osba ===
=+
==sC1
sC1RVV
)s(Hs
o 1sRC +
−+
Va
Vb R
1/sC
+ −
Vs Io
+
Vo
−
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Chapter 16, Problem 40. Refer to the RL circuit in Fig. 16.69. Find: (a) the impulse response h(t) of the circuit. (b) the unit step response of the circuit.
Figure 16.69 For Prob. 16.40. Chapter 16, Solution 40.
(a) LRs
LRsLR
RVV
)s(Hs
o
+=
+==
=)t(h )t(ueLR LRt-
(b) s1)s(V)t(u)t(v ss =⎯→⎯=
LRsB
sA
)LRs(sLR
VLRs
LRV so +
+=+
=+
=
1A = , -1B =
LRs1
s1
Vo +−=
=−= )t(ue)t(u)t(v L-Rt
o )t(u)e1( L-Rt−
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Chapter 16, Problem 41.
A parallel RL circuit has R = 4Ω and L = 1 H. The input to the circuit is i s (t) = 2e t− u(t)A. Find the inductor current i L (t) for all t > 0 and assume that i L (0) = -2 A.
Chapter 16, Solution 41. Consider the circuit as shown below.
24o o
sV VI
s s= + −
But 21SI
s=
+
2 1 1 2 4 2 2 4 2( )1 4 4 1 ( 1)o o
s sV Vs s s s s s s s
+ +⎛ ⎞= + − ⎯⎯→ = + =⎜ ⎟+ + +⎝ ⎠
8(2 1)( 1)( 4)o
sVs s
+=
+ +
8(2 1)( 1)( 4) 1 4
oL
V s A B CIs s s s s s s
+= = = + +
+ + + +
8(1) 8( 2 1) 8( 8 1)2, 8 / 3, 14 / 3
(1)(4) ( 1)(2) ( 4)( 3)A B C− + − += = = = = = −
− − −
2 8 / 3 14 / 31 4
oL
VIs s s s
−= = + +
+ +
48 14( ) 2 ( )3 3
t tLi t e e u t− −⎛ ⎞= + −⎜ ⎟
⎝ ⎠ = A)t(ue
314e
382 t4t ⎟
⎠⎞
⎜⎝⎛ −+ −−
4 Is s
I2
Vo
2s−
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 42. A circuit has a transfer function
H(s) = 2)2)(1(4++
+ss
s
Find the impulse response.
Chapter 16, Solution 42.
2 2
4( )( 1)( 2) 1 2 ( 2)
s A B CH ss s s s s
+= = + +
+ + + + +
2 2 24 ( 2) ( 1)( 2) ( 1) ( 2 4) ( 3 2) ( 1)s A s B s s C s A s s B s s C s+ = + + + + + + = + + + + + + + We equate coefficients. s2 : 0=A+B or B=-A s: 1=4A+3B+C=B+C constant: 4=4A+2B+C =2A+C Solving these gives A=3, B=-3, C=-2
2
3 3 2( )1 2 ( 2)
H ss s s
= − −+ + +
2 2( ) (3 3 2 ) ( )t t th t e e te u t− − −= − −
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 43. Develop the state equations for Prob. 16.1. Chapter 16, Solution 43.
First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KVL we get: '
CC vi;0'ivi)t(u ==+++− Thus,
)t(uivi
iv
C'
'C
+−−=
=
Finally we get,
[ ] [ ] )t(u0i
v10)t(i;)t(u
10
iv
1110
iv CCC +⎥
⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡+⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−−
=⎥⎥⎦
⎤
⎢⎢⎣
⎡
′
′
1F
1Ω
+ −
u(t)
1H
i(t)
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 44. Develop the state equations for Prob. 16.2. Chapter 16, Solution 44.
First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KCL we get:
LCxxLC'C
C'C
Cx
x'L
xL'C
'Cx
L
i3333.1v3333.0vor;v2i4v2
vv
8v
4vv
v)t(u4i
v4i8vor;08
v2
vi
+=−+=+=+=
−=
−==++−
LC
'L
LCLCL'C
i3333.1v3333.0)t(u4i
i666.2v3333.1i333.5v3333.1i8v
−−=
+−=−−=
Now we can write the state equations.
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−−
−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
L
Cx
L
C'L
'C
iv
3333.13333.0
v;)t(u40
iv
3333.13333.0666.23333.1
iv
4Ω +
vx
−
1H 1/8 F
)t(u4
+ − 2Ω
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 45. Develop the state equations for the circuit shown in Fig. 16.70.
Figure 16.70 For Prob. 16.45. Chapter 16, Solution 45.
First select the inductor current iL (current flowing left to right) and the capacitor voltage vC (voltage positive on the left and negative on the right) to be the state variables. Applying KCL we get:
2o'L
oL'CL
o'C
vvi
v2i4vor0i2
v4
v
−=
+==++−
1Co vvv +−=
21C
'L
1CL'C
vvvi
v2v2i4v
−+−=
+−=
[ ] [ ] ⎥⎦
⎤⎢⎣
⎡+⎥⎦
⎤⎢⎣
⎡−=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡ −+⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−−
=⎥⎥⎦
⎤
⎢⎢⎣
⎡′
′
)t(v)t(v
01vi
10)t(v;)t(v)t(v
0211
vi
2410
vi
2
1
C
Lo
2
1
C
L
C
L
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 46. Develop the state equations for the circuit shown in Fig. 16.71.
Figure 16.71 For Prob. 16.46. Chapter 16, Solution 46.
First select the inductor current iL (left to right) and the capacitor voltage vC to be the state variables. Letting vo = vC and applying KCL we get:
sC'L
sLC'Cs
C'CL
vvi
iiv25.0vor0i4
vvi
+−=
++−==−++−
Thus,
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡+
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
s
s
L
Co
s
s'L
'C
'L
'C
iv
0000
iv
01
)t(v;iv
0110
iv
01125.0
iv
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 47. Develop the state equations for the circuit shown in Fig. 16.72.
Figure 16.72 For Prob. 16.47.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Solution 47.
First select the inductor current iL (left to right) and the capacitor voltage vC (+ on the left) to be the state variables.
Letting i1 = 4
v'C and i2 = iL and applying KVL we get:
Loop 1:
1CL'CL
'C
C1 v2v2i4vor0i4
v2vv +−==
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−++−
Loop 2:
21C21CL
L'L
2'L
'C
L
vvvv2
v2v2i4i2i
or0vi4
vi2
−+−=−+−
+−=
=++⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−
1CL1CL
1 v5.0v5.0i4
v2v2i4i +−=
+−=
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡+⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡ −=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡ −+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−−
=⎥⎥⎦
⎤
⎢⎢⎣
⎡′
′
)t(v)t(v
0005.0
vi
015.01
)t(i)t(i
;)t(v)t(v
0211
vi
2410
vi
2
1
C
L
2
1
2
1
C
L
C
L
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 48. Develop the state equations for the following differential equation.
2
2 )(dt
tyd + dt
tdy )(4 + 3y(t) = z(t)
Chapter 16, Solution 48. Let x1 = y(t). Thus, )t(zx4x3yxandxyx 21
'22
''1 +−−=′′===
This gives our state equations.
[ ] [ ] )t(z0xx
01)t(y;)t(z10
xx
4310
xx
2
1
2
1'2
'1 +⎥
⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−−
=⎥⎥⎦
⎤
⎢⎢⎣
⎡
Chapter 16, Problem 49. * Develop the state equations for the following differential equation.
2
2 )(dt
tyd + dt
tdy )(5 + 6y(t) = dt
tdz )( z(t)
* An asterisk indicates a challenging problem. Chapter 16, Solution 49.
zxyorzyzxxand)t(yxLet 2'''
121 +=−=−== Thus, z3x5x6zz2z)zx(5x6zyx 21
''21
''2 −−−=−+++−−=−′′=
This now leads to our state equations,
[ ] [ ] )t(z0xx
01)t(y;)t(z3
1xx
5610
xx
2
1
2
1'2
'1 +⎥
⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡−
+⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−−
=⎥⎥⎦
⎤
⎢⎢⎣
⎡
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 50. * Develop the state equations for the following differential equation.
3
3 )(dt
tyd + 2
2 )(6dt
tyd + dt
tdy )(11 + 6y(t) = z(t)
* An asterisk indicates a challenging problem. Chapter 16, Solution 50. Let x1 = y(t), x2 = .xxand,x '
23'1 =
Thus, )t(zx6x11x6x 321
"3 +−−−=
We can now write our state equations.
[ ] [ ] )t(z0xxx
001)t(y;)t(z100
xxx
6116100010
xxx
3
2
1
3
2
1
'3
'2
'1
+⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡+
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 51. * Given the following state equation, solve for y(t):
.x = ⎥
⎦
⎤⎢⎣
⎡−−
0244
x+ ⎥⎦
⎤⎢⎣
⎡20
u(t)
y(t) = [ ]01 x
* An asterisk indicates a challenging problem. Chapter 16, Solution 51. We transform the state equations into the s-domain and solve using Laplace transforms.
⎟⎠⎞
⎜⎝⎛+=−
s1B)s(AX)0(x)s(sX
Assume the initial conditions are zero.
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡+++
=⎟⎠⎞
⎜⎝⎛⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡ −+=
⎟⎠⎞
⎜⎝⎛=−
−
)s/2(0
4s24s
8s4s1
s1
20
s244s
)s(X
s1B)s(X)AsI(
2
1
222222
221
2)2s(2
2)2s()2s(
s1
2)2s(4s
s1
8s4s4s
s1
)8s4s(s8)s(X)s(Y
++
−+
++
+−+=
++
−−+=
++
−−+=
++==
y(t) = ( )( ) )t(ut2sint2cose1 t2 +− −
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 52. * Given the following state equation, solve for y 1 (t).
.x = ⎥
⎦
⎤⎢⎣
⎡−−−
4212
x + ⎥⎦
⎤⎢⎣
⎡0411
⎥⎦
⎤⎢⎣
⎡)(2
)(tu
tu
y = ⎥⎦
⎤⎢⎣
⎡−−−
0122
x + ⎥⎦
⎤⎢⎣
⎡−1002
⎥⎦
⎤⎢⎣
⎡)(2
)(tu
tu
* An asterisk indicates a challenging problem. Chapter 16, Solution 52. Assume that the initial conditions are zero. Using Laplace transforms we get,
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡+−+
++=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡+−
+=
−
s/4s/3
2s214s
10s6s1
s/2s/1
0411
4s212s
)s(X 2
1
222211)3s(8.1s8.0
s8.0
)1)3s((s8s3X
++
−−+=
++
+=
2222 1)3s(16.
1)3s(3s8.0
s8.0
+++
++
+−=
)t(u)tsine6.0tcose8.08.0()t(x t3t3
1−− +−=
222221)3s(4.4s4.1
s4.1
1)3s((s14s4X
++
−−+=
++
+=
2222 1)3s(12.0
1)3s(3s4.1
s4.1
++−
++
+−=
)t(u)tsine2.0tcose4.14.1()t(x t3t3
2−− −−=
)t(u)tsine8.0tcose4.44.2(
)t(u2)t(x2)t(x2)t(yt3t3
211−− −+−=
+−−=
)t(u)tsine6.0tcose8.02.1()t(u2)t(x)t(y t3t312
−− +−−=−=
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 53. Show that the parallel RLC circuit shown in Fig. 16.73 is stable.
Figure 16.73 For Prob. 16.53. Chapter 16, Solution 53. If oV is the voltage across R, applying KCL at the non-reference node gives
oo
oo
s VsL1
sCR1
sLV
VsCRV
I ⎟⎠⎞
⎜⎝⎛
++=++=
RLCsRsLIsRL
sL1
sCR1
IV 2
sso ++
=++
=
RsLRLCsIsL
RV
I 2so
o ++==
LC1RCssRCs
RsLRLCssL
II
)s(H 22s
o
++=
++==
The roots
LC1
)RC2(1
RC21-
s 22,1 −±=
both lie in the left half plane since R, L, and C are positive quantities. Thus, the circuit is stable.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 54. A system is formed by cascading two systems as shown in Fig. 16.74. Given that the impulse response of the systems are
h 1 (t)= 3e t− u(t), h 2 (t) = e t4− u(t)
(a) Obtain the impulse response of the overall system. (b) Check if the overall system is stable.
Figure 16.74 For Prob. 16.54. Chapter 16, Solution 54.
(a) 1s
3)s(H1 += ,
4s1
)s(H2 +=
)4s)(1s(3
)s(H)s(H)s(H 21 ++==
[ ] ⎥⎦⎤
⎢⎣⎡
++
+==
4sB
1sA
)s(H)t(h 1-1- LL
1A = , 1-B =
=)t(h )t(u)ee( -4t-t −
(b) Since the poles of H(s) all lie in the left half s-plane, the system is stable.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 55. Determine whether the op amp circuit in Fig. 16.75 is stable.
Figure 16.75 For Prob. 16.55. Chapter 16, Solution 55. Let 1oV be the voltage at the output of the first op amp.
sRC1
RsC1
VV
s
1o −=
−= ,
sRC1
VV
1o
o −=
222s
o
CRs1
VV
)s(H ==
22CRt
)t(h =
∞=
∞→)t(hlim
t, i.e. the output is unbounded.
Hence, the circuit is unstable.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 56. It is desired to realize the transfer function
)()(
1
2
sVsV =
622
2 ++ sss
using the circuit in Fig. 16.76. Choose R = 1 kΩ and find L and C.
Figure 16.76 For Prob. 16.56. Chapter 16, Solution 56.
LCs1sL
sC1
sL
sC1
sL
sC1
||sL 2+=
+
⋅=
RsLRLCssL
LCs1sL
R
LCs1sL
VV
2
2
2
1
2
++=
++
+=
LC1
RC1
ss
RC1
s
VV
21
2
+⋅+
⋅=
Comparing this with the given transfer function,
RC1
2 = , LC1
6 =
If Ω= k1R , ==R21
C F500 µ
==C61
L H3.333
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 57.
Design an op amp circuit, using Fig. 16.77, that will realize the following transfer function:
)()(0
sVsV
i
= -)000,4(2
000,1++
ss
Choose C 1 =10 ;Fµ determine R 1 , R 2 , and C 2
Figure 16.77 For Prob. 16.57.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Solution 57. The circuit is transformed in the s-domain as shown below.
1/sC2 1/sC1 R2 Vi – Vo R1 +
Let 1
111 1
1 1 11
1
11// 1 1
RRsCZ R
sC sR CRsC
= = =++
222
2 22 2 2
22
11// 1 1
RRsCZ R
sC sR CRsC
= = =++
This is an inverting amplifier.
2
2 2 1 1 12 2 1 1 1 1
11 1 2 2 2
1 1 2 2 2 2
1 11( ) 1 11
o
i
R s sZ R R C CsR C R C R CVH s V RZ R R C Cs s
sR C R C R C
⎡ ⎤ ⎡ ⎤− + +⎢ ⎥ ⎢ ⎥−+⎢ ⎥ ⎢ ⎥= = − = = − =⎢ ⎥ ⎢ ⎥+ +⎢ ⎥ ⎢ ⎥+ ⎣ ⎦ ⎣ ⎦
Comparing this with ( 1000)( )
2( 4000)sH ss+
= −+
we obtain: 1
2 12
1/ 2 2 20C C C FC
µ= ⎯⎯→ = =
1 3 61 1 1
1 1 11000 1001000 10 10 10
RR C C x x −= ⎯⎯→ = = = Ω
2 3 62 2 2
1 1 14000 12.84000 4 10 20 10
RR C C x x x −= ⎯⎯→ = = = Ω 12.5Ω
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 58. Realize the transfer function
)()(0
sVsV
s
= 10+s
s
using the circuit in Fig. 16.78. Let Y 1 = sC 1 , Y 2 = 1/R 1 , Y 3 = sC 2 . Choose R 1 = 1kΩ and determine C 1 and C 2 .
Figure 16.78 For Prob. 16.58. Chapter 16, Solution 58. We apply KCL at the noninverting terminal at the op amp.
)YY)(V0(Y)0V( 21o3s −−=−
o21s3 V)YY(-VY +=
21
3
s
o
YYY-
VV
+=
Let 11 sCY = , 12 R1Y = , 23 sCY =
11
12
11
2
s
o
CR1sCsC-
R1sCsC-
VV
+=
+=
Comparing this with the given transfer function,
1CC
1
2 = , 10CR
1
11=
If Ω= k1R1 ,
=== 421 101
CC F100 µ
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 59. Synthesize the transfer function
)()(0
sVsV
in
= 62
6
1010010
++ ss
using the topology of Fig. 16.79. Let Y 1 = 1/ 1R , Y 2 = 1/R 2 , Y 3 = sC 1 , Y 4 sC 2 . Choose R 1 = 1kΩ and determine C 1 , C 2 , and R 2 .
Figure 16.79 For Prob. 16.59. Chapter 16, Solution 59. Consider the circuit shown below. We notice that o3 VV = and o32 VVV == .
−+ Vo
+ −
Vin
V2
V1
Y1 Y2
Y4
Y3
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
At node 1, 4o12o111in Y)VV(Y)VV(Y)VV( −+−=−
)YY(V)YYY(VYV 42o42111in +−++= (1) At node 2,
3o2o1 Y)0V(Y)VV( −=−
o3221 V)YY(YV +=
o2
321 V
YYY
V+
= (2)
Substituting (2) into (1),
)YY(VV)YYY(Y
YYYV 42oo421
2
321in +−++⋅
+=
)YYYYYYYYYYYYYY(VYYV 422243323142
2221o21in −−+++++=
43323121
21
in
o
YYYYYYYYYY
VV
+++=
1Y and 2Y must be resistive, while 3Y and 4Y must be capacitive.
Let 1
1 R1
Y = , 2
2 R1
Y = , 13 sCY = , 24 sCY =
212
2
1
1
1
21
21
in
o
CCsRsC
RsC
RR1
RR1
VV
+++=
2121221
212
2121
in
o
CCRR1
CRRRR
ss
CCRR1
VV
+⎟⎠
⎞⎜⎝
⎛ +⋅+
=
Choose Ω= k1R1 , then
6
212110
CCRR1
= and 100CRRRR
221
21 =+
We have three equations and four unknowns. Thus, there is a family of solutions. One such solution is
=2R Ωk1 , =1C nF50 , =2C F20 µ
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 60. Obtain the transfer function of the op amp circuit in Fig. 16.80 in the form of
)()(0
sVsV
i
= cbss
as++2
where a, b, and c are constants. Determine the constants.
Figure 16.80 For Prob. 16.67.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Solution 60. With the following MATLAB codes, the Bode plots are generated as shown below. num=[1 1]; den= [1 5 6]; bode(num,den);
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 61. A certain network has an input admittance Y(s). The admittance has a pole at s = -3, a zero at s = -1, and Y(∞ ) = 0.25 S. (a) Find Y(s). (b) An 8-V battery is connected to the network via a switch. If the switch is closed at t = 0, find the current i(t) through Y(s) using the Laplace transform. Chapter 16, Solution 61. We use the following codes to obtain the Bode plots below. num=[1 4]; den= [1 6 11 6]; bode(num,den);
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 16, Problem 62.
A gyrator is a device for simulating an inductor in a network. A basic gyrator circuit is shown in Fig. 16.81. By finding V i (s)/I 0 (s), show that the inductance produced by the gyrator is L = CR 2 .
Figure 16.81 For Prob. 16.69. Chapter 16, Solution 62. The following codes are used to obtain the Bode plots below. num=[1 1]; den= [1 0.5 1]; bode(num,den);
