chapter 16, solution 1. · 2008. 11. 10. · chapter 16, solution 1. consider the s-domain form of...
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Chapter 16, Solution 1.
Consider the s-domain form of the circuit which is shown below.
I(s)
+ −
1
1/s 1/s
s
222 )23()21s(1
1ss1
s1s1s1
)s(I++
=++
=++
=
= t
23sine
32)t(i 2t-
=)t(i A)t866.0(sine155.1 -0.5t
Chapter 16, Solution 2.
8/s s
s4 2
+ −
+
Vx
−
4
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
−
− −−
Chapter 16, Solution 3. s
5/s 1/2 +
Vo
−
1/8
Current division leads to:
)625.0s(165
s16105
s81
21
21
s5
81Vo +
=+
=
++=
vo(t) = ( ) V)t(ue13125. t625.0−−0
Chapter 16, Solution 4.
The s-domain form of the circuit is shown below.
6 s
10/s1/(s + 1) + −
+
Vo(s)
−
Using voltage division,
+++
=
+++
=1s
110s6s
101s
1s106s
s10)s(V 2o
10s6sCBs
1sA
)10s6s)(1s(10
)s(V 22o +++
++
=+++
=
)1s(C)ss(B)10s6s(A10 22 ++++++=
Equating coefficients :
2s : -ABBA0 =→+=1s : A5-CCA5CBA60 =→+=++=0s : -10C-2,B,2AA5CA1010 ===→=+=
22222o 1)3s(4
1)3s()3s(2
1s2
10s6s10s2
1s2
)s(V++
−++
+−
+=
+++
−+
=
=)t(vo V)tsin(e4)tcos(e2e2 -3t-3t-t −−
Chapter 16, Solution 5.
s2
2s1+
2
Io
s
( )
( ) 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
−=
++=
−+++
+−
+++
−−−−
++
=
−++++==
−++++=
+++=
+++=
−
−°−−°−−
Chapter 16, Solution 6.
2
2s5+
Io
10/s
s
Use current division.
t3sine35t3cose5)t(i
3)1s(5
3)1s()1s(5
10s2ss5
2s5
s102s
2sI
tto
22222o
−− −=
++−
++
+=
++=
+++
+=
Chapter 16, Solution 7. The s-domain version of the circuit is shown below. 1/s 1 Ix + 2s
1
2+s
– Z
2
2
2 s211s2s2
s21s21
s2s1
)s2(s1
1s2//s11Z
+
++=
++=
++=+=
)5.0ss(CBs
)1s(A
)5.0ss)(1s(1s2
1s2s2s21x
1s2
ZVI 22
2
2
2x
++
++
+=
+++
+=
++
++
==
)1s(C)ss(B)5.0ss(A1s2 222 ++++++=+
BA2:s2 +=
2CC2CBA0:s −=→+=++=
-4B ,6A3 0.5A or CA5.01:constant ==→=+=
222x866.0)5.0s()5.0s(4
1s6
75.0)5.0s(2s4
1s6I
++
+−
+=
++
+−
+=
[ ] A)t(ut866.0cose46)t(i t5.0x
−−=
Chapter 16, Solution 8.
(a) )1s(s
1s5.1ss22)s21(
s1)s21//(1
s1Z
2
+++
=++
+=++=
(b) )1s(s2
2s3s3
s11
1s1
21
Z1 2
+++
=+
++=
2s3s3)1s(s2Z 2 ++
+=
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
+++
1
1
2 s 2/s 1/s
s
2
(a) (b)(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 +++
Chapter 16, Solution 10. To find ZTh, consider the circuit below. 1/s Vx + 1V 2 Vo 2Vo - Applying KCL gives
s/12VV21 x
o +=+
But xo Vs/12
2V+
= . Hence
s3)1s2(V
s/12V
s/12V41 x
xx +−=→
+=
++
s3)1s2(
1VZ x
Th+
−==
To find VTh, consider the circuit below. 1/s Vy +
1
2+s
2 Vo 2Vo
- Applying KCL gives
)1s(34V
2VV2
1s2
oo
o +−=→=+
+
But 0Vs1V2V ooy =+•+−
)1s(s3)2s(4
s2s
)1s(34)
s21(VVV oyTh +
+−=
+
+−=+==
Chapter 16, Solution 11. The s-domain form of the circuit is shown below.
4/s s
I2I1+ −
+ −
2 4/(s + 2) 1/s
Write the mesh equations.
21 I2Is4
2s1
−
+= (1)
21 I)2s(I-22s
4-++=
+ (2)
Put equations (1) and (2) into matrix form.
+
+=
+ 2
1
II
2s2-2-s42
2)(s4-s1
)4s2s(s2 2 ++=∆ ,
)2s(s4s4s2
1 ++−
=∆ , s6-
2 =∆
4s2sCBs
2sA
)4s2s)(2s()4s4s(21
I 22
21
1 +++
++
=++++−⋅
=∆∆
=
)2s(C)s2s(B)4s2s(A)4s4s(21 222 ++++++=+−⋅
Equating coefficients :
2s : BA21 += 1s : CB2A22- ++=
0s : C2A42 += Solving these equations leads to A 2= , 23-B = , -3C =
221 )3()1s(3s23-
2s2
I++−
++
=
22221 )3()1s(3
323-
)3()1s()1s(
23-
2s2
I++
⋅++++
⋅++
=
=)t(i1 [ ] A)t(u)t732.1sin(866.0)t732.1cos(e5.1e2 -t-2t −−
2222
2 )3()1s(3-
)4s2s(2s
s6-I
++=
++⋅=
∆∆
=
== )t3sin(e33-
)t(i t-2 A)t(u)t732.1sin(e1.732- -t
Chapter 16, Solution 12. We apply nodal analysis to the s-domain form of the circuit below.
Vo
10/(s + 1) + −
1/(2s) 4
s
3/s
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 +−
Chapter 16, Solution 13.
Consider the following circuit.
Vo
1/(s + 2)
1/s 2s
Io
2 1
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 −
Chapter 16, Solution 14. We first find the initial conditions from the circuit in Fig. (a).
1 Ω 4 Ω
io
+
vc(0)
−
+ −
5 V
(a)A5)0(io =− , V0)0(vc =−
We now incorporate these conditions in the s-domain circuit as shown in Fig.(b).
2s 5/s
Io
Vo
+ −
1 4
15/s 4/s
(b)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- −−
Chapter 16, Solution 15.
First we need to transform the circuit into the s-domain.
2s5+
Vo
+ −
+ − 5/s
s/4
+ Vx −
10 3Vx
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
−++
+−=
++
−=−+
=+
−−
++++
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.
1 Ω
+ −Vo
1 F
Vo/2 + − 1 H io
+ −
2 Ω
3 V
(a)
Hence,
A1-33-
i)0(i oL === , V1-vo =
V5.221-
)1-)(2(-)0(vc =
−=
We now incorporate the initial conditions for as shown in Fig. (b). 0t >
I2I1 + −
s
2.5/s + −
1
+ −Vo
1/s
Vo/2 + −
Io
− +
2
-1 V
5/(s + 2)
(b)For mesh 1,
02
Vs5.2
Is1
Is1
22s
5- o21 =++−
+++
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 : B A22- +=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 +−
Chapter 16, Solution 17. We apply mesh analysis to the s-domain form of the circuit as shown below.
2/(s+1)
I3
+ −
1/s
I2I1
4
s
1 1
For mesh 3,
0IsIs1
Is1
s1s
2213 =−−
++
+ (1)
For the supermesh,
0Iss1
I)s1(Is1
1 321 =
+−++
+ (2)
But (3) 4II 21 −= Substituting (3) into (1) and (2) leads to
+=
+−
++s1
14Is1
sIs1
s2 32 (4)
1s2
s4-
Is1
sIs1
s- 32 +−=
++
+ (5)
Adding (4) and (5) gives
1s2
4I2 2 +−=
1s1
2I2 +−=
== )t(i)t(i 2o A)t(u)e2( -t− Chapter 16, Solution 18.
vs(t) = 3u(t) – 3u(t–1) or Vs = )e1(s3
se
s3 s
s−
−−=−
1 Ω
+
Vo
−
+ − 1/s
Vs 2 Ω
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
−−−−=
−
+−=−
+=
=→=++−
−−−
−−
+
Chapter 16, Solution 19. We incorporate the initial conditions in the s-domain circuit as shown below.
I
1/s
− +
2 I V1 Vo
1/s
s
+ −
2
2 4/(s + 2)
At the supernode,
o11 sV
s1
sV
22
V)2s(4++=+
−+
o1 Vss1
Vs1
21
22s
2++
+=++
(1)
But and I2VV 1o +=s
1VI 1 +=
2s2Vs
s)2s(s2V
Vs
)1V(2VV oo
11
1o +−
=+−
=→+
+= (2)
Substituting (2) into (1)
oo Vs2s
2V
2ss
s1s2
s1
22s
2+
+−
+
+
=−++
oVs2s1s2
)2s(s)1s2(2
s1
22s
2
+
++
=++
+−++
o
22
V2s
1s4s2s9s2
)2s(ss9s2
+++
=++
=++
732.3sB
2679.0sA
1s4s9s2
V 2o ++
+=
+++
=
443.2A = , 4434.0-B =
732.3s4434.0
2679.0s443.2
Vo +−
+=
Therefore,
=)t(vo V)t(u)e4434.0e443.2( -3.732t-0.2679t −
Chapter 16, Solution 20. We incorporate the initial conditions and transform the current source to a voltage source as shown.
1 s
+ −1/s 2/s
Vo
+ −
1
1/(s + 1) 1/s
At the main non-reference node, KCL gives
s1
sV
1V
s11Vs2)1s(1 ooo ++=
+−−+
s1s
V)s1s)(1s(Vs21s
soo
++++=−−
+
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 −
Chapter 16, Solution 21. The s-domain version of the circuit is shown below. 1 s
V1 Vo + 2/s 2 1/s
At n
s
s1
10−
At n
1 −s
V
Subs
10 =
=Vo
=10
cons
=Vo
Taki
(vo
10/
-
ode 1,
ooo VsVsVs
sVVV
)12
()1(102
2
11
1−++=→+
−= (1)
ode 2,
)12
(2
21 ++=→+= ssVVsVVV
oooo (2)
tituting (2) into (1) gives
ooo VsssVsVsss )5.12()12
()12/)(1( 22
2 ++=−++++
5.12)5.12(10
22 +++
+=++ ss
CBssA
sss
CsBsssA ++++ 22 )5.12( BAs +=0:2 CAs += 20:
-40/3C -20/3,B ,3/205.110:tant ===→= AA
++
−+++
−=
+++
− 22222 7071.0)1(7071.0414.1
7071.0)1(11
320
5.1221
320
sss
ssss
s
ng the inverse Laplace tranform finally yields
[ ] V)t(ut7071.0sine414.1t7071.0cose1320)t tt −− −−=
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 ++++++=
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 t875.0t875.0t
1−−− −+=
Chapter 16, Solution 23.
The s-domain form of the circuit with the initial conditions is shown below. V
I
1/sC sL R -2/s
4/s 5C
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)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),t2sin(e375.11)t2cos(e6)t(u6 -4t-4t >−−
Chapter 16, Solution 24. At t = 0-, the circuit is equivalent to that shown below. + 9A 4Ω 5Ω vo -
20)9(54
4x5)0(vo =+
=
For t > 0, we have the Laplace transform of the circuit as shown below after transforming the current source to a voltage source. 4Ω 16Ω Vo + 36V 10A 2/s 5
Ω
- Applying KCL gives
8.12B,2.7A,5.0s
BsA
)5.0s(ss206.3V
5V
2sV
1020
V36o
ooo −==+
+=++
=→+=+−
Thus,
[ ] )t(ue8.122.7)t(v t5.0o
−−=
Chapter 16, Solution 25. For , the circuit in the s-domain is shown below. 0t >
s 6 I
Applying KVL,
+
V
−
+ −
(2s)/(s2 + 16)
+ −
9/s
2/s
0s2
Is9
s616ss2
2 =+
++++−
)16s)(9s6s(32s4
I 22
2
++++
=
)16s()3s(s288s36
s2
s2I
s9V 22
2
+++
+=+=
16sEDs
)3s(C
3sB
sA
s2
22 ++
++
++
++=
)s48s16s3s(B)144s96s25s6s(A288s36 2342342 ++++++++=+
)s9s6s(E)s9s6s(D)s16s(C 232343 ++++++++
Equating coefficients : 0s : A144288 = (1) 1s : E9C16B48A960 +++= (2) 2s : E6D9B16A2536 +++= (3) 3s : ED6CB3A60 ++++= (4) 4s : (5) DBA0 ++=
Solving equations (1), (2), (3), (4) and (5) gives
2A = , B , C7984.1-= 16.8-= , D 2016.0-= , 765.2E =
16s)4)(6912.0(
16ss2016.0
)3s(16.8
3s7984.1
s4
)s(V 222 ++
+−
+−
+−=
=)t(v V)t4sin(6912.0)t4cos(2016.0et16.8e7984.1)t(u4 -3t-3t +−−−
Chapter 16, Solution 26.
Consider the op-amp circuit below.
R2
+Vo −
0
+ −
R1
1/sC
−+
Vs
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
-5ts +=→=
5)2)(s(s3-
Vo ++=
5sB
2sA
5)2)(s(s3
V- o ++
+=
++=
1A = , -1B =
2s1
5s1
Vo +−
+=
=)t(vo ( ) )t(uee -2t-5t −
Chapter 16, Solution 27.
Consider the following circuit.
For mesh 1,
I2
2s
I1+ −
10/(s + 3) 1
2s s
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=
Chapter 16, Solution 28.
Consider the circuit shown below.
VI1 +
−
1
2s s I2
s 6/s
For mesh 1,
21 IsI)s21(s6
++=
For mesh 2,
21 I)s2(Is0 ++=
21 Is2
1-I
+=
Substituting (2) into (1) gives
2
2
22 Is
2)5s(s-IsI
s2
12s)-(1s6 ++
=+
++=
or 2s5s
6-I 22 ++=
+
o
−2
(1)
(2)
4.561)0.438)(s(s12-
2s5s12-
I2V 22o ++=
++==
Since the roots of s are -0.438 and -4.561, 02s52 =++
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 −
Chapter 16, Solution 29.
Consider the following circuit.
1 : 2Io
+ −
10/(s + 1)
1
4/s 8
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 −−
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- e
98
et278
)t(g −=
Hence,
3t-3t-3t- et274
e98
et278
)t(u34
)t(y −−+=
=)t(y 3t-3t- et274
e98
)t(u34
+−
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
+
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 −−
(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 : (1) -ACCA0 =→+=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 −−+
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
s2)3s(2
s4 22
2
++−
++−
++
Chapter 16, Solution 34.
Consider the following circuit.
+
Vo(s)
−
Vo
10/s
2
4+ −
Vs
s
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 ++
Chapter 16, Solution 35.
Consider the following circuit.
+
Vo
−
I V1
+ −
2/s
2I
s
Vs 3
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 ++
Chapter 16, Solution 36.
From the previous problem,
s21 V
2s9s3s3
3sV
I3++
=+
=
s2 V2s9s3
sI
++=
But o
2
s Vs9
2s9s3V
++=
2
2
3 9 23 9 2 9 9
ooVs s sI V
s s s+ +
= ⋅+ +
=
==I
V)s(H o 9
Chapter 16, Solution 37.
(a) Consider the circuit shown below.
3 2s
+
Vx
−
2/s + −
I1 I2+ −
Vs 4Vx
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)
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-
Chapter 16, Solution 38.
(a) Consider the following circuit.
1+
Vo
−
s Is
1/s
VoV1
1/s+ −
Vs
1 Io
At node 1,
sVV
Vs1
VV o11
1s −+=
−
o1s Vs1
Vs1
s1V −
++= (1)
At node o,
oooo1 V)1s(VVs
sVV
+=+=−
o
21 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( 1s
o
s
o4H
2s3s2s1
23 +++
Chapter 16, Solution 39.
Consider the circuit below.
+
Vo
−
Io
+ −
1/sC
R Vb
Va−+
Vs
Since no current enters the op amp, flows through both R and C. oI
+=sC1
RI-V oo
sCI-
VVV osba ===
=+
==sC1
sC1RVV
)s(Hs
o 1sRC +
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− Chapter 16, Solution 41.
)s(X)s(H)s(Y =
1s2
)s(H)t(ue2)t(h t-
+=→=
s5)s(X)s(V)t(u5)t(v ii ==→=
1sB
sA
)1s(s10
)s(Y+
+=+
=
10A = , -10B =
1s10
s10
)s(Y+
−=
=)t(y )t(u)e1(10 -t−
Chapter 16, Solution 42.
)s(X)s(Y)s(Ys2 =+
)s(X)s(Y)1s2( =+
)21s(21
1s21
)s(X)s(Y
)s(H+
=+
==
=)t(h )t(ue5.0 2-t
Chapter 16, Solution 43.
i(t)
+ −
1Ω
1F u(t)
1H
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 +
=
+
−−
=
′
′
Chapter 16, Solution 44. 1/8 F 1H
)t(u4 2Ω
+ −
+
vx
−
4Ω
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
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
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
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
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, 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 +
=
−
+
−−
=
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
+
=
+
−−−=
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 +− −
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 t3t3
12−− +−−=−=
Chapter 16, Solution 53.
If is the voltage across R, applying KCL at the non-reference node gives oV
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.
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.
Chapter 16, Solution 55.
Let be the voltage at the output of the first op amp. 1oV
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.
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
Chapter 16, Solution 57. The circuit in the s-domain is shown below.
+
Vx
−
V1
+ −
R1
C R2
L
Vi
Z
CsRLCs1sLR
sC1sLR)sLR()sC1(
)sLR(||sC1
Z2
22
2
22 ++
+=
+++⋅
=+=
i1
1 VZR
ZV
+=
i12
21
2
2o V
ZRZ
sLRR
VsLR
RV
+⋅
+=
+=
CsRLCs1sLR
R
CsRLCs1sLR
sLRR
ZRZ
sLRR
VV
22
21
22
2
2
2
12
2
i
o
+++
+
+++
⋅+
=+
⋅+
=
sLRRCRsRLCRsR
VV
212112
2
i
o
++++=
LCRRR
CR1
LR
ss
LCRR
VV
1
21
1
22
1
2
i
o
++
++
=
Comparing this with the given transfer function,
LCRR
51
2= CR
1L
R6
1
2 += LCR
RR25
1
21 +=
Since and , Ω= 4R1 Ω= 1R 2
201
LCLC41
5 =→= (1)
C41
L1
6 += (2)
201
LCLC45
25 =→=
Substituting (1) into (2),
01C24C80C41
C206 2 =+−→+=
Thus, 201
,41
C =
When 41
C = , 51
C201
L == .
When 201
C = , 1C20
1L == .
Therefore, there are two possible solutions. =C F25.0 =L H2.0 or =C F05.0 =L H1
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
Chapter 16, Solution 59. Consider the circuit shown below. We notice that o3 VV = and o32 VVV == .
Y4
Y3
Y1
V1
V2Y2
+ −
−+ Vo
Vin
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 42
2243323142
2221o21in −−+++++=
43323121
21
in
o
YYYYYYYYYY
VV
+++=
1Y and must be resistive, while and must be capacitive. 2Y 3Y 4Y
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 R , then Ω= k11
6
212110
CCRR1
= and 100CRRR
221
21 =R
+
We have three equations and four unknowns. Thus, there is a family of solutions. One such solution is
=2R Ωk1 , C =1 nF50 , =2C F20 µ
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);
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);
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);
Chapter 16, Solution 63. We use the following commands to obtain the unit step as shown below. num=[1 2]; den= [1 4 3]; step(num,den);
Chapter 16, Solution 64. With the following commands, we obtain the response as shown below. t=0:0.01:5; x=10*exp(-t); num=4; den= [1 5 6]; y=lsim(num,den,x,t); plot(t,y)
Chapter 16, Solution 65. We obtain the response below using the following commands. t=0:0.01:5; x=1 + 3*exp(-2*t); num=[1 0]; den= [1 6 11 6]; y=lsim(num,den,x,t); plot(t,y)
Chapter 16, Solution 66. We obtain the response below using the following MATLAB commands. t=0:0.01:5; x=5*exp(-3*t); num=1; den= [1 1 4]; y=lsim(num,den,x,t); plot(t,y)
Chapter 16, Solution 67.
Using the result of Practice Problem 16.14,
)YYY(YYYYY-
VV
321432
21
i
o
+++=
When , 11 sCY = F5.0C1 µ=
12 R
1Y = , Ω= k10R1
23 YY = , 24 sCY = , F1C2 µ=
)RsC2(RsC1RsC-
)R2sC(sCR1RsC-
VV
1112
11
11221
11
i
o
++=
++=
1RC2sRCCsRsC-
VV
122121
211
i
o
+⋅+=
1)10)(1010(1)2(s)10)(1010)(110(0.5s)10)(1010(0.5s-
VV
36-236-6-2
3-6
i
o
+××+×××××
=
42i
o
102s400ss100-
VV
×++=
Therefore,
=a 100- , =b 400 , =c 4102× Chapter 16, Solution 68.
(a) Let 3s
)1s(K)s(Y
++
=
Ks31
)s11(Klim
3s)1s(K
lim)(Yss
=++
=++
=∞∞→∞→
i.e. . K25.0 =
Hence, Y =)s()3s(4
1s++
(b) Consider the circuit shown below.
I
Vs = 8 V + −
t = 0
YS
s8V)t(u8V ss =→=
)3s(s)1s(2
3s1s
s48
)s(V)s(YZV
I ss
++
=++
⋅===
3sB
sA
I+
+=
32A = , 34-B =
=)t(i [ ] A)t(ue4231 3t-−
Chapter 16, Solution 69.
The gyrator is equivalent to two cascaded inverting amplifiers. Let be the voltage at the output of the first op amp.
1V
ii1 -VVRR-
V ==
i1o VsCR
1V
RsC1-
V ==
CsRV
RV
I 2oo
o ==
CsRIV 2
o
o =
CRLwhen,sLIV 2
o
o ==