Bài giải mạch P16

The s-domain form of the circuit is shown below... We first find the initial conditions from the circuit in Fig... We first need to find the initial conditions... Consider the following

s

2 2

2

)23()21s(

11

ss

1s

1s1

s1)

s

(

++

=++

=++

2)

2e

24(v

3

8j3

4s

125.03

8j3

4s

125.0s

25.016)8s8s3(s

2s16V

s

32s16)8s8s3(V

0VsV)s4s2(s

)32s16()8s4(V

0s

84

0V2

0Vs

s

4V

t ) 9428 0 j 3333 1 ( t

) 9428 0 j 3333 1 ( x

2 x

2 x

x

2 x

2 x

x x

x

−

− +

−++

+

−+

−

=++

=+

++

+

−+

=+

−+

−+

−

3

22sine

2

6t3

22cose

)t(

5s

1610

5s

8

12

1 2

1s

58

1

Vo

+

=+

=

vo(t) = 0.3125(1−e−0.625t)u(t) V

Chapter 16, Solution 4

The s-domain form of the circuit is shown below

10/s 1/(s + 1) +

−

+

V o (s)

−Using voltage division,

+

=

1s

110s6s

101

s

1s106s

s10)

s

(

10s6s

CBs1

s

A)10s6s)(

1s(

10)

s

(

++

++

+

=+++

=

)1s(C)ss(B)10s6s(A

2 2

41

)3s(

)3s(21s

210s6s

10s21s

2)

s

(

V

++

−++

+

−+

=++

+

−+

s

1+

2

I o

s

( )

(e 0.7559sin1.3229t)u(t)A

or

A)t(ue

ee3779.0e

ee3779.0e

)646.2j)(

3229.1j5.1(

)3229.1j5.0(3229

.1j5.0s

)646.2j)(

3229.1j5

1

(

)3229.1j5.0(

2

s

1

)3229.1j5.0s)(

3229.1j5.0s)(

2s

(

s2

Vs

I

)3229.1j5.0s)(

3229.1j5.0s)(

2s(

s22

ss

s22s12

s2

1s

1

12

3229 1 j 2 / 90 t

2

o

2 2

2 o

2

−

=

++

=

−+

++

+

−++

+

=

−++

++

=

=

−++

++

5+

5t3cos

53

)1s(

)1s(510s2s

s52

s5s

o

2 2 2

2 2

−++

+

=++

=++

+

+

=

)s2(s

11s

=++

=+

+

=+

=

)5.0ss(

CBs)

1s(

A)5.0ss)(

1s(

1s21

s2s2

s21x1

++

+

=+++

+

=++

++

=

=

)1s(C)ss(B)5.0ss

2CC

2CBA0

:

-4B ,6A3

0.5A

or CA5.01:

2 2

2 x

866.0)5.0s(

)5.0s(41

s

675.0)5.0s

(

2s41

s

6

I

++

+

−+

=+

Chapter 16, Solution 8

(a)

)1s(s

1s5.1ss22

)s21(s

1)s21//(

1s

1

+

++

=+

++

=++

=

(b)

)1s(s2

2s3s3s

11

1s

12

1Z

+

++

=+

++

=

2s3s3

)1s(s2

++

+

=+

=

s1s2

)s1s(2)s1s(

||

2

Zin

1 s 2 s

) 1 s ( 2

2

2

+++

)2s(2s23

)s21(2)s21(

+

=+

2s3

6s5)s21(

||

21

+

+

=++

++

+

=

2s3

6s5s

2s3

6s5s2s3

6s5

||

s

Zin

6 s 7 s 3

) 6 s 5 ( s

2 + ++

+

s3

)1s2(Vs

/12

Vs

=

+

+

s3

)1s2(1

4V

2

VV

But V 0

s

1V2

Vy+ o• + o =

−

)1s(s3

)2s(4s

2s)1s(3

4)

s

21(V

−

=+

42

-22

s

4

-++

I2s2-

2-s422)(s

4

-s1

)4s2s(s

2 2

++

=

)2s(s

4s4

2 =

∆

4s2s

CBs2

s

A)4s2s)(

2s(

)4s4s(21

2 1

+++

=+++

s : 2=4A+2C

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

2 2

1

)3()1s(

3s23-2s

2I

++

−+

+

=

2 2

2 2

1

)3()1s(

33

2

3-)3()1s(

)1s(2

3-2s

2I

++

⋅++

+

+

⋅++

2

2 2

)3()1s(

3-)

4s2s(2

ss

6-I

++

=++

o

sV24

Vs

3s

V1

s

10

+

=+

−

+

1s

15s1510151s

10V)ss25

=++

=+

+

1s25.0s

CBs1

s

A)1s25.0s)(

1s(

25s15

++

++

+

=++

+

+

=

40V

)1s

(

A= + o s=-1 =

)1s(C)ss(B)1s25.0s(A25

4

32

1s233

271554

32

1s2

1s7

401s

17404

32

1s

7

135s7

40-1s

+

−+

=+

++

7(

)2)(

155(t2

3cose7

40e7

40)

o

V1s2

1ss12

V1s2

V2s

1

+

+

=+

++

=+

1s(

1s2

Vo

++

+

=

2s

B1s

A)2s)(

1s(

11

s2

V

+

++

=++

=+

=

1

A= , B= -1

2s

11s

1

Io

+

−+

=

=)t(

io (e -t −e -2t)u ( ) A Chapter 16, Solution 14

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

0Vs

5s2

V1

s15

=+

−+++

−

V)1s(4

ss2

11s

=

−

o

2 o

2 2

V)1s(s4

2s6s5V)

1s(s4

s2s2s4s4

s

10

+

++

=+

++++

=

2s6s5

)1s(40

++

+

=

s

5)4.0s2.1s(s

)1s(4s

5s2

V

++

+

=+

=

4.0s2.1s

CBss

As

5

++

++

+

=

sCsB)4.0s2.1s(A)1

8s10s

10s

5

++

+

−+

=

2 2 2

2

)2.0(102

.0)6.0s(

)6.0s(10s

15

I

++

−++

5+

5VV2s

5VV,But

2s

s5V120V

)40ss2(02s

s5sVVs2V120V

40

010

2s

5Vs/5

0V4

/s

V3V

x o o

x

x o

2 o

o

2 x o

o o

x o

++

=

→+

+

=

=+

−++

−

=+

−+

−+

−

We can now solve for Vx

)40s5.0s)(

2s(

)20s(5

V

2s

)20s(10V

)40s5.0s(2

02s

s5V1202

s

5V)40ss2(

2

2 x

2 x

2

x x

2

−++

=+

+

Chapter 16, Solution 16

We first need to find the initial conditions For t<0, 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,

A1-3

3-i)

0

(

iL = o = = , Vvo =-1

V5.22

1-)1-)(

2()

2

-1 V 5/(s + 2)

(b)

For mesh 1,

02

Vs

5.2Is

1Is

122

+

But, Vo =Io =I2

s

5.22s

5Is

12

1Is

5.22

V1Is

1Is

1s

5.2Is

1s2

1I

Put (1) and (2) in matrix form

−+

1s

5.2s

5.22s5

I

I

s

1s2

1s

1

-s

12

1s

1

2

2 1

s

32s

2 + +

=

)2s(s

5s

42-

2 = + + +

∆

3s2s2

CBs2

s

A)3s2s2)(

2s(

132s-I

2 2

2

++

+

=+++

0

5.1ss

714.2s7145.12s

7143.03s2s2

429.5s429.32s

7143.0

++

−

−+

=++

−

−+

=

25.1)5.0s(

)25.1)(

194.3(25.1)5.0s(

)5.0s(7145.12s

7143.0

++

+++

+

−+

1s1

s

2

2 1

For the supermesh,

0Iss

1Is1(Is

−++

1sIs

1s

1s

2s

4-Is

1sIs

24I

12

I2

+

−

=

=i (t))

t(

es

1t(u)e

22()t(u)e22[(

2s

2)e1()5.1s(s

3V

VV)5.1s(02

VsV1

V

V

) 1 t ( 5 1 t

5 1 o

s s

o

s o

o o s o

−

=

−+

=

=

→

=++

+

−

2

2 4/(s + 2)

At the supernode,

o 1

s

1s

V22

V)2s

(

4

++

=+

−+

o

s

1Vs

12

122

s

2

++

=+

But and Vo =V1+2I

s

1V

=

2s

2Vss)2s(

s2VVs

)1V(2V

1

1 1

−

=+

2V2s

ss

1s2s

122

+

oVs2s

1s2)2s(s

)1s2(2s

122

+

=+

++

−+

+

o

2 2

V2s

1s4s2s

9s2)2

s

(

s

s9s

2

+

++

=+

+

=+

+

732.3s

B2679

.0s

A1

s4s

9s2

Vo 2

+

++

=++

+

=

443

2

732.3s

4434.02679.0s

443.2

Vo

+

−+

Vs

11

Vs2)1

s

(

++

=+

−

−+

s

1sV)s1s)(

1s(Vs21

s

s

o o

+++

1s1

s

s

++

1s(

1s4s2-

CBs1

s

A)5.0ss)(

1s(

5.0s2s-

++

++

+

=+++

−

−

=

1V

)1s

(

A= + o s=-1=

)1s(C)ss(B)5.0ss(A5.0s2

2

)5.0s(21

s

15.0ss

1s21s

1V

++

+

−+

=++

+

−+

V V

V

2()1(102

1

10

2 1 1

1

−++

=

→

+

(2

2 1

o

Substituting (2) into (1) gives

o o

V s

s

2()12/)(

1

(

5.12)

5.12

=++

=

s s

C Bs s

A s

s

s

V o

Cs Bs s

C A

-40/3C

-20/3,B

,3/205

.110:

−+

+

−

7071.0)1(

7071.0414

.17071.0)1(

11

3

205.12

21

3

20

s s

s s

s s

s s

11V1s

12s

4

VV1

V1

s

1 2

→



−+

=

−

1s2s3

4VVV

3

s2

Vs

4

V

2 1 2

2 2

Substituting (2) into (1),

2 2

2

2

3s3

7s3

4s4

1s4

111s2s3

4V1s

)8

9s4

7s(

CBs)

1s(

A)8

9s4

7s)(

1s(

9V

2 2

2

++

++

+

=+++

=

)1s(C)ss(B)8

9s4

7s(

A

Equating coefficients:

BA0:

A4

3CC

A4

3CBA4

70:

-18C -24,B ,24AA

8

3CA8

99:

23)8

7s(

364

23)8

7s(

)8/7s(24)1s(

24)

8

9s4

7s(

18s24)

1s(

24V

2 2

2

2

++

+++

+

−+

=++

+

−+

=

Taking the inverse of this produces:

[24e 24e cos(0.5995t) 5.004e sin(0.5995t)]u(t))

9s4

7s(

FEs)

1s(

D)8

9s4

7s)(

1s(

1s2s3

49V

2 2

2 1

++

++

+

=+++

9s4

7s(D1s2s

D4

36FF

D4

36

or FED4

718:

0F 4,E ,8DD

8

33

or FD8

99:

64

23)8

7s(

2/764

23)8

7s(

)8/7s(4)1s(

8)8

9s4

7s(

s4)

1s(

8V

2 2

2

1

++

−++

++

+

=++

++

1/sC

sL

At the non-reference node,

sCVsL

VR

VC5s

2s

4

++

=++

=

+

LC

1RC

sss

CVs

sC5

LC1RCss

C6s5

++

+

=

8010

1RC

1

=

804

1LC

1

=

=

2 2 2

2

)2)(

230(2)4s(

)4s(520s8s

480s5V

++

+++

+

=++

+

=

=)t(

v 5 e -4t cos( 2 )+230 e -4t sin( 2 ) V

)20s8s(s4

480s5sL

V

++

+

=

=

20s8s

CBss

A)20s8s(s

120s25.1

++

++

=++

2

)2)(375.11(2)4s(

)4s(6s

620s8s

75.46s6s

6I

++

−++

+

−

=++

+

−

=

=)t(

i 6 u ( )−6 e -4t cos( 2 )−11 375 e -4t sin( 2 t ), t >0

-

20)9(54

4x5)0(

+

=

For t > 0, we have the Laplace transform of the circuit as shown below after

transforming the current source to a voltage source

Bs

A)5.0s(s

s206.3V5

V2

sV10

20

V

36

o o

o

++

=+

+

=

→

+

=+

−

Thus,

[7.2 12.8e ]u(t))

t(

vo = − −0.5t

9s616

−

)16s)(

9s6s

(

32s4

2++

+

+

=

)16s()3s(s

288s

36s

2s

2Is

9

2++

++

=+

=

16s

EDs)3s(

C3

s

Bs

As

2

2

+++

++++

=

)s48s16s3s(B)144s96s25s6s(A288s

)s9s6s(E)s9s6s(D)s16s(

C 3+ + 4+ 3+ 2 + 3+ 2++

)4)(

6912.0(16s

s2016.0)3s(

16.83s

7984.1s

4)

−+

−+

V s

At node 0,

sC)V0(R

V0R

0V

o 2

o 1

s− = − + −

( )o

2 1

R

1R

=

2 1 1 s

o

RRCsR

1-V

R2

1 = = , R C (20 103)(50 10- 6) 1

So,

2s

1-V

Vs

o+

=

)5s(3Ve

3

3-

Vo

++

=

5s

B2s

A5)2)(s(s

3V

- o

+

++

=++

=

1

A= , B= -1

2s

15s

1

Vo

+

−+

=

=)t(

vo (e -5t −e -2t)u ( ) Chapter 16, Solution 27

Consider the following circuit

1 I sII

s21(3s

10

−

−+

=+

2

1 (1 s II

s21(3s

10

+

−+

=

For mesh 2,

1 1

2 I sII

s22(

2

1 2(s 1 II

s1(

++

I)1s(2)1s(

)1s(1s20

)3s(10

1s4s

3 2+ +

=

∆

)1s(20

)1s(10

+

=

∆Thus

=

∆

∆

= 1 1I

) 1 s 4 s 3 )(

3 s (

) 1 s ( 20

) 1 s 4 s 3 )(

3 s (

) 1 s ( 10

2 + ++

6

++

For mesh 2,

2

1 (2 s II

Substituting (2) into (1) gives

2

2 2

s

2)5s(s-IsIs

212s)-(1s

=+

=

or

2s5s

6-

I2 2

++

=

(s

12-2

s5s

12-I

2

++

=++

=

=

Since the roots of s2+5s+2=0 are -0.438 and -4.561,

561.4s

B438

.0s

A

Vo

+

++

=

-2.914.123

12-

4.123-

12-

561.4s

91.20.438s

2.91-)s(

Vo

+

++

=

=)t(

vo 2 91[e -4.561t −e 0 438 t]u ( ) V Chapter 16, Solution 29

Consider the following circuit

8s48

)s4)(

8(s

=

=

When this is reflected to the primary side,

2n,n

Z1

1s2

3s21s2

21

Zin

+

+

=++

=

3s2

1s21s

10Z

11s

10I

in

+

⋅+

=

⋅+

=

B1s

A)5.1s)(

1s(

5s10

Io

+

++

=++

201s

10-)s(

Io

+

++

=

=)t(

io 10[2 e -1.5t −e−t]u ( ) A Chapter 16, Solution 30

)s(X)s(H)

s

(

1s3

1231s

4)s(X

+

=+

=

2 2

2

)1s3(

34s83

4)1s3(

s12)

=

2

2 (s 13)

127

4)31s(

s9

83

4)

s9

8-)s(

+

⋅

=G

Using the time differentiation property,

8-)et(dt

d9

8-)t(g

3 t - 3 t

9

8et27

8)t(

Hence,

3 t - 3

t - 3 t

27

4e

9

8et27

8)t(u3

4)t(

=)t(

27

4 e

9

8 ) ( u 3

4

+

−

Chapter 16, Solution 31

s

1)s(X)

t(u)

t

(

4s

s10)s(Y)

t2cos(

10)

)s(Y)

s

(

H

4 s

s 10

3s

++

+

=

5s4s

CBss

A)5s4s(s

3s

2

++

=++

+

=

CsBs)5s4s(A3

s+ = 2+ + + 2+Equating coefficients : 0

s : 3=5A → A=3 51

s : 1=4A+C → C=1−4A=- 7 52

s : 0=A+B → B=-A=-3 5

5s4s

7s35

1s

53)s(

++

1)2s(35

1s

6.0)s(

++

++

⋅

−

=

=)t(

y [0 6−0 6 e -2t cos( )−0 2 e -2t sin( )]u ( )

(b) - 2t 2

)2s(

6)

s(Xe

t6)t(x

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

+

⋅++

+

=

=

5s4s

DCs)

2s(

B2

s

A)5s4s()2s(

)3s(6)

s(

++

+++

++

=+++

+

=

Equating coefficients : 3

2

s : 0=6A+B+4C+D=2A+B+D (2) 1

s : 6=13A+4B+4C+4D=9A+4B+4D (3) 0

18s6)2s(

62s

6)s(

++

+

−+

++

=

1)2s(

61

)2s(

)2s(6)2s(

62s

6)s(

++

−++

+

−+

++

=

=)t(

y [6 e -2t +6 t e -2t −6 e -2t cos( )−6 e -2t sin( )]u ( ) Chapter 16, Solution 33

)s(X

)s(Y)

16)2s(

)4)(

3(16)2s(

s2)

3s(2

1s

4)

s

(

++

−++

−++

s 12 20

s 4 s

s 2 )

3 s ( 2

s

2

++

−++

−++

V2s

V

+

=+

−

o

2 o

10

1)2s(4

11V10

s4

12s

1)2s(

++

=

s 2s 9s 30 V20

1

=s

oV

V

30 s 9 s 2

VII

VV

I s− 1

=

Vs

2

VV

2

s3V2

s33s

V

−

=+

s

2

s3V2

s33s

s 2

2s9s3

)3s(s3V

++

+

=

s 2

1

2s9s3

s9V

3s

3V

++

=+

=

=

=s

oV

V)s(H

2 s 9 s 3

s 9

2+ +

Chapter 16, Solution 36

From the previous problem,

s 2

2s9s3

s33

s

VI3

++

=+

=

s

2s9s3

sI

++

2s9s3

=

2 2

o o

V)s(

2Is

23

For loop 2,

0Is

2Is

2s2V

Vx 1 2

s

2Is

2s2)IIs

8

1 2 2

−

2

1 2s Is

6Is

6-

I

Is2s6s6-

s2-s230V

=

∆

s

2s

6s2s

6s

23

4s6s

1

s64s18

)s2s6(I

ss3V

I

s

1

9 s 2 s 3

3 s

2

2

−+

2)IIs

2V

=

=

s

s x

2

4V-

Vs6V

I

2s

3 -

V

1 1

s

1Vs

1s1

At node o,

o o

o o

1 sV V (s 1)Vs

VV

+

=+

2

s (s 1 1s)(s s 1)V 1sV

o 2

2 s 3 s 2 s

3 1 s

s V V (s 2s 3s 2)V (s s 1)V

o 2

1 s 2 s s

I)s(

s

o s

I)s

s

o s

o 4

H

2 s 3 s 2 s

V s

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

=

sC

1RI

Vo o

sC

IVV

sC1RV

V)s(H

LRsLR

RV

V)s(H

s

o

+

=+

=

=

=)t(

Bs

A)LRs(s

LRV

LRs

LR

++

=+

=+

=

1

A= , B= -1

LRs

1s

t(

o ( 1−e -Rt L ) u ( )

Chapter 16, Solution 41

)s(X)s(H)

s

(

1s

2)s(H)

t(ue2)

t(u5)

t

(

1s

Bs

A)1s(s

10)

s

(

Y

++

=+

=

10

1s

10s

10)

s

2

)21s(2

11

s2

1)s(X

)s(Y)

C 'i 0; i vv

i)t(

−

Thus,

)t(uivi

iv

C '

' C

;)t(u1

0i

v11

10i

)t(

x x

L C

' C C

' C C

' C

' C x L

i3333.1v3333.0vor

;v2i4v2

vv8

v4v

v

v)t(ui

vi8vor

;08

v2

vi

+

=

−+

=+

=+

−

L C

' L

L C

L C

L

' C

i3333.1v3333.0)t(ui

i666.2v3333.1i

333.5v3333.1i8v

L

C '

L

' C

i

v3333.1

3333.0v

;)t(u4

0i

v3333.13333.0

666.23333.1i

v

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:

2 o

'

L

o L

' C L

o

'

C

vv

i

vi4vor0i2

v4

−

1 C

2 1 C

'

L

1 C L

'

C

vvvi

v2vi4

v

−+

)t(v01v

i10)t(v

;)t(v

)t(v02

11v

i24

10v

i

2

1 C

L o

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:

s C

' L

s L C

' C s

C ' C L

vvi

iiv25.0v

or0i4

vvi

+

−

=

++

−

=

=

−++

C o

s

s '

L

' C '

L

' C

i

v00

00i

v0

1)t(v

;i

v01

10i

v01

125.0i

' C L

' C C

4

v2v

+

−

Loop 2:

2 1 C 2

1 C L L

'

L

2

' L

' C L

vvvv2

v2v2i4i2i

or0vi4

vi

2

−+

−

=

−+

−+

−

=

=++

1 C

L 1 C L

4

vv2i

)t(v00

05.0v

i01

5.01)t(i

)t(i

;)t(v

)t(v02

11v

i24

10

v

i

2

1 C

L 2

1 2

1 C

L C

L

Chapter 16, Solution 48

Let x1 = y(t) Thus, x1' =y' =x2 and x'2 =y′′=− x1−4x2 +z(t)

This gives our state equations

[ ] [ ]0z(t)x

x01)t(y

;)t(z1

0x

x43

10x

x

2

1 2

1 '

Thus,

z3xxz

z2z)zx(5xz

y

x'2 = ′′− ' =− 1− 2 + + '+ − ' =− 1− 2 −This now leads to our state equations,

[ ] [ ]0 z(t)x

x01)t(y

;)t(z3

1x

x56

10x

x

2

1 2

1 '

x"3=− 1− 2− 3+

We can now write our state equations

[ ] [ ]0 z(t)

xx

x001)t(y

;)t(z100x

xx6116

100

010x

xx

3 2 1 3

2 1 '

3

' 2

' 1

−

s

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

Assume the initial conditions are zero

04s2

4s8s4s

1s

12

0s

2

44s)s(X

s

1B)s(X)AsI(

2 1

2 2 2

2 2

2

2 2

1

2)2s(

22

)2s(

)2s(s

12)2s(

4ss

1

8s4s

4ss

1)8s4s(s

8)

s(X)s(Y

++

−+

++

+

−+

=++

−

−+

=

++

−

−+

=++

s/32s2

14s10s6s

1s

/2

s/104

114s2

12s)s(X

2 1

2 2 2

2 1

1)3s(

8.1s8.0s

8.0)1)3s((

s

8s3X

++

−

−+

=++

+

=

1)3s(

16

.1)3s(

3s8.0s

8.0

++

+++

+

−

=

)t(u)tsine6.0tcose

8.08.0()t(

2 2 2

2 2

1)3s(

4.4s4.1s

4.11)3s((

s

14s4X

++

−

−+

=++

+

=

2 2 2

12

.01)3s(

3s4.1s

4.1

++

−++

+

−

=

)t(u)tsine2.0tcose

4.14.1()t(

)t(u)tsine8.0tcose

4.44.2(

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

t 3 t

3 2 1

8.02.1()t(u)t(x)t(

o

sL

1sCR

1sL

VVsCR

=++

=

RLCsRsL

IsRLsL

1sCR1

I

RsLRLCs

IsLR

V

LC1RCss

RCsR

sLRLCs

sLI

I)s(

s

o

++

=++

=

=The roots

LC

1)RC2(

1RC

2

1-

s1,2 = ± 2 −

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

+

4s

1)s(

H2

+

=

)4s)(

1s(

3)

s(H)s(H)s(

++

=

=

4s

B1s

A)

s(H)

t(

h L- 1 L- 1

1

A= , B= -1

=)t(

sC1V

Vs

V1

=

2 2 2 s

o

CRs

1V

V)s(

2

2CR

t)t(

t , i.e the output is unbounded

Hence, the circuit is unstable

Chapter 16, Solution 56

LCs1sLsC

1sLsC

1sLsC

⋅

=

RsLRLCssLLC

s1

sLR

LCs1sLV

V

2 2

2 1

2

++

=+

1ss

RC

1sV

V2 1

2

+

⋅+

1

=

=C6

sLRsC

1sLR

)sLR()sC1()sLR(

2 2

2

+

=+

+

+

⋅

=+

=

i 1

ZR

ZV

+

=

i 1 2

2 1

2

2

ZR

ZsLR

RV

sLR

RV

+

⋅+

=+

=

CsRLCs1

sLRR

CsRLCs1

sLRsL

R

RZ

R

ZsLR

RV

V

2 2

2 1

2 2

2

2

2 1

2

2 i

o

++

++

++

+

⋅+

=+

⋅+

=

sLRRCRsRLCRs

RV

V

2 1 2 1 1

2

2 i

o

++++

=

LCR

RRCR

1L

Rss

LCRRV

V

1

2 1 1

2 2

1 2

i

o

++

=

Comparing this with the given transfer function,

LCR

1L

R6

1

2 +

=

LCR

RR25

4

1

C4

1L

1

20

1LCLC

4

1C20

Thus,

20

1,4

1

C=

V0(Y)0V( s − 3 = − o 1− 2

o 2 1 s

3V (Y Y )V

2 1

3 s

o

YY

Y-V

1 2 1

1

2 s

o

CR1s

CsC-R1sC

sC-V

V

+

=+

=

Comparing this with the given transfer function,

1C

C

1

CR

11 1

in V )Y (V V )Y (V V )Y

V

)YY(V)YYY(VY

At node 2,

3 o

2 o

1 V )Y (V 0)Y

V

o 3 2 2

1Y (Y Y )V

o 2

3 2

Y

YY

Substituting (2) into (1),

)YY(VV)YYY(Y

YYY

2

3 2 1

)YYYYYYYYYYYYYY(VY

Y

2 4 3 3 2 3 1 4 2

2 2 2 1 o 2 1

4 3 3 2 3 1 2 1

2 1 in

o

YYYYYYYY

YYV

V

++

Y = , Y3 =sC1, Y4 =sC2

2 1 2 2

1 1

1 2 1

2 1 in

o

CCsR

sCR

sCRR1

RR1V

V

+++

=

2 1 2 1 2 2 1

2 1 2

2 1 2 1 in

o

CCRR

1C

RR

RRss

CCRR1V

=

Choose R1 = k1 Ω, then

6 2 1 2

1

10CC

RR

2 2 1

2

R+

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

=

2

R 1 kΩ, C1 = 50 nF, C2 =20 µ F

V

3 2 1 4 3 2

2 1 i

o

+++

=

R2sC(sCR1

RsC-V

V

1 1 1

2

1 1 1

1 2

2 1

1 1 i

o

++

=+

+

=

1RC2sRCCs

RsC-V

V

1 2

2 1 2 1 2

1 1 i

o

+

⋅+

=

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

)10)(1010(0.5s-V

V

3 6

2

-3 6

6

2

-3 -6

i

o

102s400s

s100-V

V

×++

)1s(K)s(Y

+

+

=

Ks31

)s11(Klim3

s

)1s(Klim)(Y

Hence, Y(s)=

) 3 s ( 4

1 s

++

(b) Consider the circuit shown below

t(u8

)3s(s

)1s(23s

1ss4

8)s(V)s(YZ

Bs

AI

++

=

32

=)t(

i [2 4 e ]u ( ) A 3

R

R-

i 1

sCR

1VR

sC1-

CsR

VR

V

o = =

CsRI

o

o =

C R L when ,

sL I

o