Chapter 16, Solution 1. Consider the S-domain

8/14/2019 Chapter 16, Solution 1. Consider the S-domain

1/60

Chapter 16, Solution 1.

Consider the s-domain form of the circuit which is shown below.

I(s)

+

1

1/s1/s

s

222 )23()21s(

1

1ss

1

s1s1

s1)s(I

++=

++=

++=

= t

2

3sine

3

2)t(i 2t-

=)t(i A)t866.0(sine155.1 -0.5t


8/ss

s

4 2+

+

Vx

4


2/60

V)t(u)e2e24(v

3

8j3

4s

125.0

3

8j3

4s

125.0

s

25.016

)8s8s3(s

2s16V

s

32s16)8s8s3(V

0VsV)s4s2(s

)32s16()8s4(V

0

s

84

0V

2

0V

s

s

4V

t)9428.0j3333.1(t)9428.0j3333.1(x

2x

2x

x2

x2

x

xxx

+ ++=

+

+

++

+=

++

+=

+=++

=++++

+

=

+

+

+

vx = Vt3

22sine

2

6t

3

22cose)t(u4 3/t43/t4

Chapter 16, Solution 3.s

5/s 1/2

+

Vo

1/8

Current division leads to:

)625.0s(16

5

s1610

5

s8

1

2

12

1

s

5

8

1Vo

+=

+=

++

=

vo(t) = ( ) V)t(ue13125. t625.00


3/60


The s-domain form of the circuit is shown below.

6 s

10/s1/(s + 1) +

+

Vo(s)

Using voltage division,

+++=

+++=

1s

1

10s6s

10

1s

1

s106s

s10)s(V

2o

10s6s

CBs

1s

A

)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

1s

2

10s6s

10s2

1s

2)s(V

++

++

+

+=

++

+

+=

=)t(vo V)tsin(e4)tcos(e2e2-3t-3t-t


s

2

2s

1

+

2

Io

s


4/60

( )

( ) 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(

2s

1

)3229.1j5.0s)(3229.1j5.0s)(2s(

s

2

VsI

)3229.1j5.0s)(3229.1j5.0s)(2s(

s2

2ss

s2

2s

1

2

s

2

1

s

1

1

2s

1V

t2

t3229.1j2/t90t3229.1j2/t90t2o

22

2

o

2

=

++=

+

++

+

+++

++

=

++++==

++++=

+++=

+++

=


2

2s

5

+

Io

10/ss

Use current division.

t3sine3

5t3cose5)t(i

3)1s(

5

3)1s(

)1s(5

10s2s

s5

2s

5

s

102s

2sI

tto

22222o

=

++

++

+=

++=

+++

+=


5/60


The s-domain version of the circuit is shown below.

1/s1

Ix

+ 2s

1

2

+s

Z

2

2

2 s21

1s2s2

s21

s21

s2s

1

)s2(s

1

1s2//s

11Z

+

++=

++=

+

+=+=

)5.0ss(

CBs

)1s(

A

)5.0ss)(1s(

1s2

1s2s2

s21x

1s

2

Z

VI

22

2

2

2

x++

++

+=

+++

+=

++

+

+==

)1s(C)ss(B)5.0ss(A1s2 222 ++++++=+

BA2:s2 +=

2CC2CBA0:s =+=++=

-4B,6A30.5AorCA5.01:constant ===+=

222x

866.0)5.0s(

)5.0s(4

1s

6

75.0)5.0s(

2s4

1s

6I

++

+

+=

++

+

+=

[ ] A)t(ut866.0cose46)t(i t5.0x =


6/60


(a))1s(s

1s5.1s

s22

)s21(

s

1)s21//(1

s

1Z

2

+

++=

+

++=++=

(b))1s(s2

2s3s3

s

11

1

s

1

2

1

Z

1 2

+

++=

+

++=

2s3s3

)1s(s2Z

2 ++

+=


(a) The s-domain form of the circuit is shown in Fig. (a).=

++

+=+=

s1s2

)s1s(2)s1s(||2Z in 1s2s

)1s(22

2

++

1

1

2s 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

+

+=

+

+=+

2s3

6s5)s21(||21 +

+

=++

=

+

++

+

+

=

+

+=

2s3

6s5s

2s3

6s5s

2s3

6s5||sZ in 6s7s3

)6s5(s2 +

+


7/60


To find ZTh, consider the circuit below.

1/s Vx

+

1V2 Vo 2Vo

-

Applying KCL gives

s/12VV21 xo+=+

But xo Vs/12

2V

+= . Hence

s3

)1s2(V

s/12

V

s/12

V41 x

xx +=+

=+

+

s3

)1s2(

1

VZ xTh

+==

To find VTh, consider the circuit below.1/s Vy

+

1

2

+s2 Vo 2Vo

-

Applying KCL gives

)1s(3

4V

2

VV2

1s

2o

oo

+==+

+


8/60

But 0Vs

1V2V ooy =++

)1s(s3

)2s(4

s

2s

)1s(3

4)

s

21(VVV oyTh

+

+=

+

+=+==


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 I2Is

42

s

1

+= (1)

21 I)2s(I-22s

4-++=

+(2)

Put equations (1) and (2) into matrix form.

+

+=

+ 2

1

I

I

2s2-

2-s42

2)(s4-

s1

)4s2s(s

22 ++= ,

)2s(s

4s4s2

1 +

+= ,

s

6-2 =

4s2s

CBs

2s

A

)4s2s)(2s(

)4s4s(21I

22

21

1

++

++

+

=

+++

+=

=

)2s(C)s2s(B)4s2s(A)4s4s(21 222 ++++++=+

Equating coefficients :2s : BA21 += 1s : CB2A22- ++=


9/60

0s : C2A42 +=

Solving these equations leads to A 2= , 23-B = , -3C =

221 )3()1s(

3s23-

2s

2

I ++

++=

22221 )3()1s(

3

32

3-

)3()1s(

)1s(

2

3-

2s

2I

+++

++

++

+=

=)t(i1 [ ] A)t(u)t732.1sin(866.0)t732.1cos(e5.1e2 -t-2t 222

22

)3()1s(

3-

)4s2s(2

s

s

6-I

++=

++=

=

== )t3sin(e3

3-)t(i t-2 A)t(u)t732.1sin(e1.732-

-t


We apply nodal analysis to the s-domain form of the circuit below.

Vo

10/(s + 1) +

1/(2s) 4

s

3/s

o

oo

sV24

V

s

3

s

V1s

10

+=+

+

1s15s151015

1s10V)ss25.01( o

2

+

++=++=++

1s25.0s

CBs

1s

A

)1s25.0s)(1s(

25s15V

22o ++

++

+=

+++

+=


10/60

7

40V)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

21s

2

3

3

2

7

155

43

21s

2

1s

7

40

1s

1

7

40

43

21s

7

135s

7

40-

1s

7

40

V 222o

+

+

+

+

+

+

+=

+

+

++

+=

+

= t

2

3sine

)3)(7(

)2)(155(t

2

3cose

7

40e

7

40)t(v 2t-2t-t-o

=)t(vo V)t866.0sin(e57.25)t866.0cos(e714.5e714.52-t2-t-t +


Consider the following circuit.

Vo

1/(s + 2)

1/s 2s

Io

2 1

Applying KCL at node o,

o

ooV

1s2

1s

s12

V

1s2

V

2s

1

+

+=

++

+=

+


11/60

)2s)(1s(

1s2Vo ++

+=

2s

B

1s

A

)2s)(1s(

1

1s2

VI

o

o ++

+=

++=

+=

1A = , -1B =

2s

1

1s

1Io +

+

=

=)t(io ( ) A)t(uee -2t-t Chapter 16, Solution 14.

We first find the initial conditions from the circuit in Fig. (a).

1 4

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,

0s44

0V

s

5

s2

V

1

s15V ooo=

+

+++


12/60

oV)1s(4

s

s2

11

s

5

s

15

+++=

o

2

o

22

V)1s(s4

2s6s5V

)1s(s4

s2s2s4s4

s

10

+

++=+

++++=

2s6s5

)1s(40V

2o ++

+=

s

5

)4.0s2.1s(s

)1s(4

s

5

s2

VI

2

o

o +++

+=+=

4.0s2.1s

CBs

s

A

s

5I

2o ++

+++=

sCsB)4.0s2.1s(A)1s(4 s2 ++++=+

Equating coefficients :0s : 10AA4.04 ==1s : -84-1.2ACCA2.14 =+=+=2s : -10-ABBA0 ==+=

4.0s2.1s

8s10

s

10

s

5I

2o ++

++=

2222o 2.0)6.0s(

)2.0(10

2.0)6.0s(

)6.0s(10

s

15I

++

++

+=

=)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.

2s

5

+

Vo

+

+

5/s

s/4

+ Vx

10

3Vx


13/60

2s

5VV

2s

5VV,But

2s

s5V120V)40ss2(0

2s

s5sVVs2V120V40

0

10

2s

5V

s/5

0V

4/s

V3V

xoox

xo2

oo2

xo

ooxo

++=

+=

+++==

+++

=+

+

+

We can now solve for Vx.

)40s5.0s)(2s(

)20s(5V

2s

)20s(10V)40s5.0s(2

02s

s5V120

2s

5V)40ss2(

2

2

x

2

x2

xx

2

++

+=

+

+=+

=+

++++


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

+ Vo

1 F

Vo/2+

1 H io

+

2

3 V

(a)


14/60

Hence,

A1-3

3-i)0(i oL === , V1-vo =

V5.22

1-

)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,

0

2

V

s

5.2I

s

1I

s

12

2s

5- o21 =++

++

+

But, 2oo IIV ==

s

5.2

2s

5I

s

1

2

1I

s

12 21 +

=

+

+ (1)

For mesh 2,

0s

5.2

2

V1I

s

1I

s

1s1

o

12 =+

++

1s

5.2I

s

1s

2

1I

s

1- 21 =

+++ (2)


15/60

Put (1) and (2) in matrix form.

+

=

++

+

1

s

5.2

s

5.2

2s

5

I

I

s

1s

2

1

s

1-

s

1

2

1

s

12

2

1

s

32s2 ++= ,

)2s(s

5

s

42-2 +

++=

3s2s2

CBs

2s

A

)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 =

5.1ss

714.2s7145.1

2s

7143.0

3s2s2

429.5s429.3

2s

7143.0I

22o ++

+=

++

+=

25.1)5.0s()25.1)(194.3(

25.1)5.0s()5.0s(7145.1

2s7143.0I

22o +++

+++

+=

=)t(io [ ] A)t(u)t25.1sin(e194.3)t25.1cos(e7145.1e7143.0 -0.5t-0.5t-2t +


16/60


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,

0IsIs

1I

s

1s

1s

2213 =

++

+(1)

For the supermesh,

0Iss

1I)s1(I

s

11 321 =

+++

+ (2)

But (3)4II 21 =

Substituting (3) into (1) and (2) leads to

+=

+

++

s

114I

s

1sI

s

1s2 32 (4)

1s

2

s

4-I

s

1sI

s

1s- 32 +

=

++

+ (5)

Adding (4) and (5) gives

1s

24I2 2 +=

1s

12I2 +=


17/60

== )t(i)t(i 2o A)t(u)e2(-t


vs(t) = 3u(t) 3u(t1) or Vs = )e1(s

3

s

e

s

3 ss

=

1

+

Vo

+

1/sVs 2

V)]1t(u)e22()t(u)e22[()t(v

)e1(5.1s

2

s

2)e1(

)5.1s(s

3V

VV)5.1s(02

VsV

1

VV

)1t(5.1t5.1

o

sso

soo

oso

=

+=

+=

==++

+


We incorporate the initial conditions in the s-domain circuit as shown below.

I

1/s

+

2 IV1 Vo

1/s

s

+

2

24/(s + 2)


18/60

At the supernode,

o

11sV

s

1

s

V2

2

V)2s(4++=+

+

o1 Vss

1

Vs

1

2

1

22s

2

++

+=++ (1)

But andI2VV 1o += s

1VI

1 +=

2s

2Vs

s)2s(

s2VV

s

)1V(2VV

oo

1

1

1o +

=

+

=

++= (2)

Substituting (2) into (1)

oo Vs2s

2V2s

s

s

1s2

s

122s

2

+

+

+

+

=++

oVs2s

1s2

)2s(s

)1s2(2

s

12

2s

2

+

+

+=

+

+++

+

o

22

V2s

1s4s

2s

9s2

)2s(s

s9s2

+

++=

+

+=

+

+

732.3s

B

2679.0s

A

1s4s

9s2

V 2o ++

+=

++

+=

443.2A = , 4434.0-B =

732.3s

4434.0

2679.0s

443.2Vo +

+

=

Therefore,

=)t(vo V)t(u)e4434.0e443.2(-3.732t-0.2679t


19/60


We incorporate the initial conditions and transform the current source to a voltage source

as shown.

1 s

+ 1/s2/s Vo

+

1

1/(s + 1) 1/s

At the main non-reference node, KCL gives

s

1

s

V

1

V

s11

Vs2)1s(1 ooo++=

+

+

s

1sV)s1s)(1s(Vs2

1s

soo

++++=

+

oV)s12s2(2s

1s

1s

s++=

+

+

)1s2s2)(1s(

1s4s2-V

2

2

o +++

=

5.0ss

CBs

1s

A

)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

1s

1

5.0ss

1s2

1s

1V

++

+

+=

++

+

+=

=)t(vo [ ] V)t(u)2tcos(e2e 2-t-t


20/60



1 sV1 Vo

+ 2/s 2 1/s

-

At node 1,

10/s

ooo V

sVsV

s

s

VVV

s )12

()1(1021

102

11

1

++=+=

(1)

At node 2,

)12

(2

2

11 ++=+=

s

sVVsV

V

s

VVoo

oo (2)

Substituting (2) into (1) gives

oooVsssV

sVsss )5.12()1

2()12/)(1(10 2

22 ++=++++=

5.12)5.12(

1022 ++

++=

++=

ss

CBs

s

A

sssV

o

CsBsssA ++++= 22 )5.12(10

BAs +=0:2

CAs += 20:

-40/3C-20/3,B,3/205.110:constant ==== AA

++++

+

=

++

+

= 22222 7071.0)1(

7071.0

414.17071.0)1(

11

3

20

5.12

21

3

20

ss

s

sss

s

sVo

Taking the inverse Laplace tranform finally yields

[ ] V)t(ut7071.0sine414.1t7071.0cose13

20)t(v tto

=


21/60



4s

V1 V2

1s +

121 2 3/s

At node 1,

s4

V

s4

11V

1s

12

s4

VV

1

V

1s

12 2

1

211

+=

+

+=

+(1)

At node 2,

++=+=

1s2s

3

4VVV

3

s

2

V

s4

VV 2212

221 (2)

Substituting (2) into (1),

222

2 V2

3s

3

7s

3

4

s4

1

s4

111s2s

3

4V

1s

12

++=

+

++=

+

)8

9s

4

7s(

CBs

)1s(

A

)8

9s

4

7s)(1s(

9V

222

++

++

+=

+++=

)1s(C)ss(B)8

9s

4

7s(A9 22 ++++++=

Equating coefficients:

BA0:s2 +=

A4

3CCA

4

3CBA

4

70:s =+=++=

-18C-24,B,24AA8

3CA

8

99:constant ====+=


22/60

64

23)

8

7s(

3

64

23)

8

7s(

)8/7s(24

)1s(

24

)8

9s

4

7s(

18s24

)1s(

24V

2222

++

+

++

+

+=

++

+

+=

Taking the inverse of this produces:

)t(u)t5995.0sin(e004.5)t5995.0cos(e24e24)t(v t875.0t875.0t2 +=

Similarly,

)8

9s

4

7s(

FEs

)1s(

D

)8

9s

4

7s)(1s(

1s2s3

49

V22

2

1

++

++

+=

+++

++

=

)1s(F)ss(E)8

9s

4

7s(D1s2s

3

49 222 ++++++=

++

Equating coefficients:

ED12:s2 +=

D4

36FFD

4

36orFED

4

718:s =+=++=

0F4,E,8DD833orFD

899:constant ====+=

64

23)

8

7s(

2/7

64

23)

8

7s(

)8/7s(4

)1s(

8

)8

9s

4

7s(

s4

)1s(

8V

2221

++

++

++

+=

++

++

=

Thus,

)t(u)t5995.0sin(e838.5)t5995.0cos(e4e8)t(v t875.0t875.0t1 +=


The s-domain form of the circuit with the initial conditions is shown below.

V

I

1/sCsLR -2/s4/s 5C


23/60


24/60


At t = 0-, the circuit is equivalent to that shown below.

+

9A 4 5 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.

4 16 Vo

+

36V 10A 2/s 5

-

Applying KCL gives

8.12B,2.7A,5.0s

B

s

A

)5.0s(s

s206.3V

5

V

2

sV10

20

V36o

ooo ==+

+=+

+=+=+

Thus,

)t(ue8.122.7)t(v t5.0

o

=


25/60


For , the circuit in the s-domain is shown below.0t >

s6 I

Applying KVL,

+

V

+

(2s)/(s

2+ 16)

+

9/s

2/s

0s

2I

s

9s6

16s

s22

=+

+++

+

)16s)(9s6s(

32s4I

22

2

+++

+=

)16s()3s(s

288s36

s

2

s

2I

s

9V

22

2

++

++=+=

16s

EDs

)3s(

C

3s

B

s

A

s

222 +

++

++

+++=

)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 =


26/60

16s

)4)(6912.0(

16s

s2016.0

)3s(

16.8

3s

7984.1

s

4)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

V0

R

0V

o2

o

1

s

+

=

( )o

2

1s V-sCR

1RV

+=

211s

o

RRCsR

1-

V

V

+=

But 2

10

20

R

R

2

1 == , 1)1050)(1020(CR 6-31 ==

So,2s

1-

V

V

s

o

+=

)5s(3Ve3V s-5t

s +==


27/60

5)2)(s(s

3-Vo ++

=

5s

B

2s

A

5)2)(s(s

3V- o +

++

=++

=

1A = , -1B =

2s

1

5s

1Vo +

+

=

=)t(vo ( ) )t(uee -2t-5t Chapter 16, Solution 27.


For mesh 1,

I2

2s

I1+

10/(s + 3) 1

2ss

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

I

I

)1s(2)1s(-

)1s(-1s2

0

)3s(10

1s4s3 2 ++=


28/60

3s

)1s(201 +

+=

3s

)1s(10

2+

+=

Thus

=

= 11I )1s4s3)(3s(

)1s(202 ++

=

= 22I )1s4s3)(3s(

)1s(102 ++

2

I1=


Consider the circuit shown below.

+

Vo

I1

+

1

2s s I2

s

6/s 2

For mesh 1,

21 IsI)s21(s

6++= (1)

For mesh 2,

21 I)s2(Is0 ++=

21 Is

21-I

+= (2)

Substituting (2) into (1) gives

2

2

22 Is

2)5s(s-IsI

s

212s)-(1

s

6 ++=+

++=

or2s5s

6-I

22 ++=


29/60

4.561)0.438)(s(s

12-

2s5s

12-I2V

22o ++=

++==

Since the roots of s are -0.438 and -4.561,02s52 =++

561.4s

B

438.0s

AVo +

++

=

-2.914.123

12-A == , 91.2

4.123-

12-B ==

561.4s

91.2

0.438s

2.91-)s(Vo +

++

=

=)t(vo [ ] V)t(uee91.2 t438.0-4.561t Chapter 16, Solution 29.


1 : 2Io

+

10/(s + 1)

1

4/s 8

Let1s2

8

s48

)s4)(8(

s

4||8ZL +

=+

==

When this is reflected to the primary side,

2n,n

Z

1Z 2L

in=+=

1s2

3s2

1s2

21Zin +

+=

++=

3s2

1s2

1s

10

Z

1

1s

10I

in

o +

+

+=

+=


30/60

5.1s

B

1s

A

)5.1s)(1s(

5s10Io +

++

=++

+=

-10A = , 20B =

5.1s

20

1s

10-)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

3

4

)1s3(

s12)s(Y

+

+=

+=

22 )31s(

1

27

4

)31s(

s

9

8

3

4)s(Y

+

+=

Let 2)31s(

s

9

8-

)s( +=G

Using the time differentiation property,

+== 3t-3t-3t- eet

3

1-

9

8-)et(

dt

d

9

8-)t(g

3t-3t- e9

8et

27

8)t(g =

Hence,

3t-3t-3t- et27

4e

9

8et

27

8)t(u

3

4)t(y +=

=)t(y 3t-3t- et27

4e

9

8)t(u

3

4+


31/60


s

1)s(X)t(u)t(x ==

4s

s10)s(Y)t2cos(10)t(y

2 +==

==)s(X

)s(Y)s(H

4s

s102

2

+


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

s

1

5s4s

3s2

++

+=

5s4s

CBs

s

A

)5s4s(s

3s22 ++

++=

++

+=

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

Equating coefficients :0s : 53AA53 == 1s : 57-A41CCA41 ==+= 2s : 53--ABBA0 ==+=

5s4s

7s3

5

1

s

53)s(Y

2 ++

+=

1)2s(

1)2s(3

5

1

s

6.0

)s(Y 2 ++

++

=

=)t(y [ ] )t(u)tsin(e2.0)tcos(e6.06.0 -2t-2t


32/60

(b)2

2t-

)2s(

6)s(Xet6)t(x

+==

22 )2s(

6

5s4s

3s)s(X)s(H)s(Y

+

++

+==

5s4s

DCs

)2s(

B

2s

A

)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

2s

6)s(Y

22 ++

+

++

+=

1)2s(

6

1)2s(

)2s(6

)2s(

6

2s

6)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 = ,

s

1)s(X =

16)2s(

)4)(3(

16)2s(

s2

)3s(2

1

s

4)s(Y

22 ++

++

++=

== )s(Ys)s(H20s4s

s12

20s4s

s2

)3s(2

s4

22

2

++++


33/60


34/60

3s

V

s2

VV3

11s

+=

1s

1V

2

s3V

2

s3

3s

V=

+

s1 V2

s3V

2

s3

3s

1=

+

+

s21V

2s9s3

)3s(s3V

++

+=

s21oV

2s9s3

s9V

3s

3V

++=

+=

==s

o

V

V)s(H

2s9s3

s92 +


From the previous problem,

s2

1V

2s9s3

s3

3s

VI3

++

=

+

=

s2V

2s9s3

sI

++=

But o

2

s Vs9

2s9s3V

++=

2

2

3 9 2

3 9 2 9 9

o

o

V s s sI V

s s s

+ +=

+ +=

==I

V)s(H

o9


35/60


(a) Consider the circuit shown below.3 2s

+

Vx

2/s +I1 I2+

Vs 4Vx

For loop 1,

21s Is

2I

s

23V

+= (1)

For loop 2,

0Is

2I

s

2s2V4 12x =

++

But,

=

s

2)II(V 21x

So, 0Is

2I

s

2s2)II(

s

81221 =

++

21 Is2s

6I

s

6-0

+= (2)

In matrix form, (1) and (2) become

+=

2

1s

I

I

s2s6s6-

s2-s23

0

V

+=

s

2

s

6s2

s

6

s

23

4s6s

18=

s1 Vs2s

6

= , s2 Vs

6=


36/60

s

1

1 Vs64s18

)s2s6(I

=

=

=

=

32s9

ss3

V

I

s

1

9s2s3

3s2

2

(b)

= 22I

== 2121x s

2)II(

s

2V

=

=

ss

x

V4-)s6s2s6(Vs2V

==s

s

x

2

4V-

Vs6

V

I

2s

3-


(a) Consider the following circuit.

1

+

Vo

sIs

1/s

VoV1

1/s+

Vs

1 Io

At node 1,

s

VVVs

1

VV o11

1s +=

o1s Vs

1V

s

1s1V

++= (1)


37/60

At node o,

ooo

o1V)1s(VVs

s

VV+=+=

o2

1 V)1ss(V ++= (2)

Substituting (2) into (1)

oo2

s Vs1V)1ss)(s11s(V ++++=

o23

s V)2s3s2s(V +++=

==s

o

1 V

V)s(H

2s3s2s

123 +

(b) o2o231ss V)1ss(V)2s3s2s(VVI +++++==o

23s V)1s2ss(I +++=

==s

o

2 I

V)s(H

1s2ss

123 +

(c)1

VI

o

o =

==== )s(HI

V

I

I)s(H 2

s

o

s

o

3 1s2ss

123 +

(d). ==== )s(HV

V

V

I)s( 1

s

o

s

o

4H 2s3s2s

123 +


Consider the circuit below.

+

Vo

Io

+

1/sC

RVb

Va+

Vs


38/60

Since no current enters the op amp, flows through both R and C.oI

+=

sC

1RI-V oo

sC

I-VVV

o

sba ===

=+

==sC1

sC1R

V

V)s(H

s

o1sRC+


(a)LRs

LR

sLR

R

V

V)s(H

s

o

+=

+==

=)t(h )t(ueL

RLRt-

(b) s1)s(V)t(u)t(v ss ==

LRs

B

s

A

)LRs(s

LRV

LRs

LRV so

+

+=

+

=

+

=

1A = , -1B =

LRs

1

s

1Vo +

=

== )t(ue)t(u)t(v L-Rto )t(u)e1(L-Rt


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

1s

2)s(H)t(ue2)t(h t-

+==


39/60

s5)s(X)s(V)t(u5)t(v ii ===

1s

B

s

A

)1s(s

10)s(Y

++=

+=

10A = , -10B =

1s

10

s

10)s(Y

+=

=)t(y )t(u)e1(10 -t


)s(X)s(Y)s(Ys2 =+

)s(X)s(Y)1s2( =+

)21s(2

1

1s2

1

)s(X

)s(Y)s(H

+=

+==

=)t(h )t(ue5.0 2-t


i(t)

+

1

1Fu(t)

1H

First select the inductor current iL and the capacitor voltage vC to be the statevariables.

Applying KVL we get:'CC vi;0'ivi)t(u ==+++


40/60

Thus,

)t(uivi

iv

C'

'C

+=

=

Finally we get,

[ ] [ ] )t(u0i

v10)t(i;)t(u

1

0

i

v

11

10

i

v CCC +

=

+

=


1/8 F1H

)t(u4 2+

+

vx

4

First select the inductor current iL and the capacitor voltage vC to be the statevariables.

Applying KCL we get:

LCxxLC

'C

C

'C

Cx

x'L

xL'C

'CxL

i3333.1v3333.0vor;v2i4v2

vv

8

v4vv

v)t(u4i

v4i8vor;08

v

2

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

i

v

3333.1

3333.0v;)t(u

4

0

i

v

3333.13333.0

666.23333.1

i

v


41/60


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(v01

v

i10)t(v;

)t(v

)t(v

02

11

v

i

24

10

v

i

2

1

C

Lo

2

1

C

L

C

L



42/60

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'CsC'CL

vvi

iiv25.0vor0i4

vvi

+=

++==++

Thus,

+

=

+

=

s

s

L

Co

s

s

'L

'C

'L

'C

i

v

00

00

i

v

0

1)t(v;

i

v

01

10

i

v

01

125.0

i

v


First select the inductor current iL (left to right) and the capacitor voltage vC (+ on theleft) 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

+=+

+=

=++


43/60

1CL1CL

1 v5.0v5.0i4

v2v2i4i +=

+=

+

=

+

=

)t(v

)t(v

00

05.0

v

i

01

5.01

)t(i

)t(i;

)t(v

)t(v

02

11

v

i

24

10

v

i

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(z0x

x01)t(y;)t(z

1

0

x

x

43

10

x

x

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(z0x

x01)t(y;)t(z

3

1

x

x

56

10

x

x

2

1

2

1

'2

'1 +

=

+

=


Let x1 = y(t), x2 = .xxand,x'23

'1 =

Thus,

)t(zx6x11x6x321

"

3+=

We can now write our state equations.

[ ] [ ] )t(z0

x

x

x

001)t(y;)t(z

1

0

0

x

x

x

6116

100

010

x

x

x

3

2

1

3

2

1

'3

'2

'1

+

=

+

=


44/60


We transform the state equations into the s-domain and solve using Laplace

transforms.

+=

s

1B)s(AX)0(x)s(sX

Assume the initial conditions are zero.

+++=

+=

=

)s/2(

0

4s2

4s

8s4s

1

s

1

2

0

s2

44s)s(X

s

1B)s(X)AsI(

2

1

222222

221

2)2s(

2

2)2s(

)2s(

s

1

2)2s(

4s

s

1

8s4s

4s

s

1

)8s4s(s

8)s(X)s(Y

++

+

++

++=

++

+=

++

+=

++==

y(t) = ( )( ) )t(ut2sint2cose1 t2 +


Assume that the initial conditions are zero. Using Laplace transforms we get,

+

+

++=

+

+=

s/4

s/3

2s2

14s

10s6s

1

s/2

s/1

04

11

4s2

12s)s(X

2

1

22221

1)3s(

8.1s8.0

s

8.0

)1)3s((s

8s3X

++

+=

++

+=

2222 1)3s(

16.

1)3s(

3s8.0

s

8.0

+++

++

+=

)t(u)tsine6.0tcose8.08.0()t(x t3t31 +=


45/60

22222 1)3s(

4.4s4.1

s

4.1

1)3s((s

14s4X

++

+=

++

+=

2222

1)3s(

12.0

1)3s(

3s4.1

s

4.1

++

++

+=

)t(u)tsine2.0tcose4.14.1()t(x t3t32 =

)t(u)tsine8.0tcose4.44.2(

)t(u2)t(x2)t(x2)t(y

t3t3

211

+=

+=

)t(u)tsine6.0tcose8.02.1()t(u2)t(x)t(y t3t312 +==


If is the voltage across R, applying KCL at the non-reference node givesoV

o

o

o

o

s VsL

1sC

R

1

sL

VVsC

R

VI

++=++=

RLCsRsL

IsRL

sL1sC

R1

IV

2

ss

o

++

=

++

=

RsLRLCs

IsL

R

VI

2

so

o ++==

LC1RCss

RCs

RsLRLCs

sL

I

I)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.


46/60


47/60


LCs1

sL

sC1sL

sC

1sL

sC

1||sL

2

+

=

+

=

RsLRLCs

sL

LCs1

sLR

LCs1

sL

V

V2

2

2

1

2

++=

++

+=

LC

1

RC

1ss

RC

1s

V

V

21

2

++

=

Comparing this with the given transfer function,

RC

12 = ,

LC

16 =

If ,= k1R ==R2

1C F500

== C6

1

L H3.333


The circuit in the s-domain is shown below.

+Vx

V1

+

R1

C R2

L

Vi

Z

CsRLCs1

sLR

sC1sLR

)sLR()sC1()sLR(||

sC

1Z

22

2

2

2

2 ++

+=

++

+=+=


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i

1

1 VZR

ZV

+=

i

12

2

1

2

2

o VZR

Z

sLR

R

VsLR

R

V ++=+=

CsRLCs1

sLRR

CsRLCs1

sLR

sLR

R

ZR

Z

sLR

R

V

V

22

2

1

22

2

2

2

12

2

i

o

++

++

++

+

+

=+

+

=

sLRRCRsRLCRs

R

V

V

212112

2

i

o

++++=

LCR

RR

CR

1

L

Rss

LCR

R

V

V

1

21

1

22

1

2

i

o

++

++

=


LCR

R5

1

2= CR

1

L

R6

1

2 += LCR

RR25

1

21 +=

Since and ,= 4R1 = 1R2

20

1LC

LC4

15 == (1)

C4

1

L

16 += (2)

20

1LC

LC4

525 ==


01C24C80C4

1C206 2 =++=

Thus,20

1,

4

1C =


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When4

1C = ,

5

1

C20

1L == .

When 20

1

C = , 1C20

1

L == .

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

YY

Y-

V

V

+=

Let ,11 sCY = 12 R1Y = , 23 sCY =

11

12

11

2

s

o

CR1s

CsC-

R1sC

sC-

V

V

+=

+=


1C

C

1

2 = , 10CR

1

11

=

If ,= k1R1

===421 10

1CC F100


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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 +=

o

2

32

1 VY

YYV

+= (2)


)YY(VV)YYY(Y

YYYV 42oo421

2

32

1in ++++

=

)YYYYYYYYYYYYYY(VYYV 422243323142

2221o21in +++++=

43323121

21

in

o

YYYYYYYY

YY

V

V

+++=

1Y and must be resistive, while and must be capacitive.2Y 3Y 4Y

Let1

1 R

1Y = ,

2

2 R

1Y = , 13 sCY = , 24 sCY =


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212

2

1

1

1

21

21

in

o

CCsR

sC

R

sC

RR

1

RR

1

V

V

+++

=

2121221

212

2121

in

o

CCRR

1

CRR

RRss

CCRR

1

V

V

+

++

=

Choose R , then= k11

6

2121

10CCRR

1= and 100

CR

RR

221

21 =R

+

We have three equations and four unknowns. Thus, there is a family of solutions. Onesuch solution is

=2R k1 , C =1 nF50 , =2C F20


52/60


With the following MATLAB codes, the Bode plots are generated as shown below.

num=[1 1];den= [1 5 6];

bode(num,den);


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We use the following codes to obtain the Bode plots below.

num=[1 4];den= [1 6 11 6];

bode(num,den);


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The following codes are used to obtain the Bode plots below.

num=[1 1];den= [1 0.5 1];

bode(num,den);


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We use the following commands to obtain the unit step as shown below.

num=[1 2];den= [1 4 3];

step(num,den);


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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)


57/60


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)


58/60


59/60

When ,11 sCY = F5.0C1 =

1

2 R

1Y = , = k10R1

23 YY = , 24 sCY = , F1C2 =

)RsC2(RsC1

RsC-

)R2sC(sCR1

RsC-

V

V

1112

11

11221

11

i

o

++=

++=

1RC2sRCCs

RsC-

V

V

122121

2

11

i

o

++=

1)10)(1010(1)2(s)10)(1010)(110(0.5s

)10)(1010(0.5s-

V

V

36-236-6-2

3-6

i

o

++

=

42i

o

102s400s

s100-

V

V

++=

Therefore,

=a 100- , =b 400 , =c 4102


(a) Let3s

)1s(K)s(Y

+

+=

Ks31

)s11(Klim

3s

)1s(Klim)(Y

ss=

+

+=

+

+=

i.e. .K25.0 =

Hence, Y =)s()3s(4

1s++

(b) Consider the circuit shown below.I

Vs = 8 V+

t = 0

YS


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s8V)t(u8V ss ==

)3s(s

)1s(2

3s

1s

s4

8)s(V)s(Y

Z

VI s

s

+

+=

+

+===

3s

B

s

AI

++=

32A = , 34-B =

=)t(i [ ] A)t(ue423

13t-


The gyrator is equivalent to two cascaded inverting amplifiers. Let be the

voltage at the output of the first op amp.1V

ii1 -VVR

R-V ==

i1o VsCR

1

VR

sC1-

V ==

CsR

V

R

VI

2

oo

o ==

CsRI

V2

o

o=

CRLwhen,sLI

V2

o

o =

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Chapter 16, Solution 1. Consider the S-domain