Chapter 16, Solution 1. - WordPress.comChapter 16, Solution 1. Consider the s-domain form of the circuit which is shown below. I(s) + − 1 1/s 1/s s 2 s 1 2( 32) 2 1 s s 1 1 1 s 1

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


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


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 −−


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

−=

++=

−+++

+−

+++

−−−−

++

=

−++++==

−++++=

+++=

+++=

−

−°−−°−−


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

−−=


(a) )1s(s

1s5.1ss22)s21(

s1)s21//(1

s1Z

2

+++

=++

+=++=

(b) )1s(s2

2s3s3

s11

1s1

21

Z1 2

+++

=+

++=

2s3s3)1s(s2Z 2 ++

+=


(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 ++++++=+−⋅


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 ++++++=+


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 +−


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 ++++=+


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- −−


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 ++++++=+


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−−− −+=


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 +−−−


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 −



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=


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 −



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 −−


)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

+−


s1

)s(X)t(u)t(x =→=

4ss10

)s(Y)t2cos(10)t(y 2 +=→=

==)s(X)s(Y

)s(H4s

s102

2

+


(a) )s(X)s(H)s(Y =

s1

5s4s3s

2 ⋅++

+=

5s4s

CBssA

)5s4s(s3s

22 +++

+=++

+=

CsBs)5s4s(A3s 22 ++++=+


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 +++

++

++

=+++

+=


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 −−+


)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

++−

++−

++



+

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 ++



+

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 ++


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


(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-


(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 +++


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 +


(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−


)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


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


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


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


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


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 +

=

+

−−

=


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 +

=

−

+

−−

=


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

+

=

+

−−−=


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 +− −


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−− +−−=−=


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.


(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.


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.


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

++

++

=


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 =→=


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


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

+=

+=


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)


)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)


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-−


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 ==

Chapter 16, Solution 1. - WordPress.comChapter 16, Solution 1. Consider the s-domain form of the circuit which is shown below. I(s) + − 1 1/s 1/s s 2 s 1 2( 32) 2 1 s s 1 1 1 s 1

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