16, Chapter Solution 1. Consider the s-domain form of the circuit which is shown below. 1 I(s)
1/s
+ -
1/s
s
I(s) = 1s 1 1 = 2 = 1 + s + 1 s s + s + 1 (s + 1 2) 2 + ( 3 2) 2
i( t ) =
2
3 e - t 2 sin 2 t 3
i( t ) = 1.155 e -0.5t sin (0.866t ) A
Chapter 16, Solution 2. s + Vx - 2 4 8/s
4 s
+ -
Vx -
4 s + Vx - 0 + Vx - 0 = 0 8 s 2 4+ s (16s + 32) + (2s 2 + 4s)Vx + s 2 Vx = 0 s 16s + 32 s
Vx (4s + 8) -
Vx (3s 2 + 8s + 8) =
s+2 0.25 + - 0.125 + - 0.125 Vx = -16 = -16 s 8 4 8 4 s(3s 2 + 8s + 8) s+ - j s+ + j 3 3 3 3 v x = (-4 + 2e - (1.3333 + j0.9428) t + 2e - (1.3333 - j0.9428) t )u ( t ) V 2 2 vx = 4u ( t ) - e - 4 t / 3 cos 3 6 - 4t / 3 2 2 t - e sin 3 2 t V
Chapter 16, Solution 3. s + 5/s 1/2 Vo 1/8
- Current division leads to: 1 5 5 1 5 2 = = Vo = 8s1 1 10 + 16s 16(s + 0.625) + +s 2 8 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.
6 s + 10/s Vo(s)
1/(s + 1)
+ -
-
Using voltage division,
Vo (s) = 10 s s + 6 + 10 1 1 10 = 2 s s + 1 s + 6s + 10 s + 1
Vo (s) =
10 A Bs + C = + 2 (s + 1)(s + 6s + 10) s + 1 s + 6s + 10
2
10 = A (s 2 + 6s + 10) + B (s 2 + s) + C (s + 1) Equating coefficients : s2 : 0= A+B B = -A
s1 :
0 = 6A + B + C = 5A + C C = -5A 10 = 10A + C = 5A A = 2, B = -2, C = -10 2 (s + 3) 2 2s + 10 2 4 = - - 2 2 2 - s + 1 s + 6s + 10 s + 1 (s + 3) + 1 (s + 3) 2 + 12
s :
0
Vo (s) =
v o ( t ) = 2 e -t - 2 e -3t cos(t ) - 4 e -3t sin( t ) V
Chapter 16, Solution 5.
Io
1 s+2
s
2
2 s
1 1 2s 2s = 1 = V= (s + 2)(s + 0.5 + j1.3229)(s + 0.5 - j1.3229) 2 s + 2 1 1 s s + 2s + s + 2 + + s 2 2 Vs s2 = 2 (s + 2)(s + 0.5 + j1.3229)(s + 0.5 - j1.3229)
Io =
(-0.5 - j1.3229) 2 (-0.5 + j1.3229) 2 1 (1.5 - j1.3229)(- j2.646) (1.5 + j1.3229)(+ j2.646) + = + s+2 s + 0.5 + j1.3229 s + 0.5 - j1.3229
i o ( t ) = e - 2 t + 0.3779e - 90 e - t / 2 e - j1.3229 t + 0.3779e 90 e - t / 2 e j1.3229 t u ( t ) A or = e - 2 t - 0.7559 sin 1.3229 t u ( t ) A
(
)
(
)
Chapter 16, Solution 6. 2 Io
5 s+2
10/s
s
Use current division.
Io = 5 5s 5(s + 1) 5 s+2 = = - 2 2 2 10 s + 2 s + 2s + 10 (s + 1) + 3 (s + 1) 2 + 3 2 s+2+ s
5 i o ( t ) = 5e - t cos 3t - e - t sin 3t 3
Chapter 16, Solution 7.
The s-domain version of the circuit is shown below. 1/s 1 Ix
+ 2 s +1
2s
Z
1 (2s) 1 2s 2s 2 + 2s + 1 Z = 1 + // 2s = 1 + s = 1+ = 1 s 1 + 2s 2 1 + 2s 2 + 2s s
V 2 1 + 2s 2 2s 2 + 1 A Bs + C = = + Ix = = x 2 2 2 Z s + 1 2s + 2s + 1 (s + 1)(s + s + 0.5) (s + 1) (s + s + 0.5) 2s 2 + 1 = A(s 2 + s + 0.5) + B(s 2 + s) + C(s + 1) s2 : 2=A+B
s:
constant : Ix =
0 = A+B+C = 2+C
C = -2
A = 6, B = -4
1 = 0.5A + C or 0.5A = 3
6 4s + 2 6 4(s + 0.5) - = - 2 s + 1 (s + 0.5) + 0.75 s + 1 (s + 0.5) 2 + 0.866 2
i x ( t ) = 6 - 4e - 0.5t cos 0.866 t u ( t ) A
[
]
Chapter 16, Solution 8.
1 1 (1 + 2s) s 2 + 1.5s + 1 = (a) Z = + 1 //(1 + 2s) = + s s 2 + 2s s(s + 1) 1 1 1 1 3s 2 + 3s + 2 = + + = (b) 1 Z 2 s 2s(s + 1) 1+ s Z= 2s(s + 1) 3s 2 + 3s + 2
Chapter 16, Solution 9.
(a)
The s-domain form of the circuit is shown in Fig. (a).
2 (s + 1 s) 2 (s 2 + 1) Z in = 2 || (s + 1 s) = = 2 + s + 1 s s 2 + 2s + 1
1 s 2 1/s s 2 1 (a) (b) The s-domain equivalent circuit is shown in Fig. (b). 2/s
(b)
2 || (1 + 2 s) =
2 (1 + 2 s) 2 (s + 2) = 3+ 2 s 3s + 2 5s + 6 3s + 2
1 + 2 || (1 + 2 s) =
5s + 6 s 3s + 2 5s + 6 s (5s + 6) = Z in = s || = 2 3s + 2 5s + 6 3s + 7s + 6 s + 3s + 2
Chapter 16, Solution 10.
To find ZTh, consider the circuit below.
1/s
Vx
+ 1V 2 Vo 2Vo
Applying KCL gives 1 + 2Vo = But Vo = 1+ Vx 2 + 1/ s 2 Vx . Hence 2 + 1/ s Vx = - (2s + 1) 3s
4Vx Vx = 2 + 1/ s 2 + 1/ s Vx (2s + 1) =- 1 3s
ZTh =
To find VTh, consider the circuit below.
1/s
Vy
+ 2 s +1 2 Vo 2Vo
Applying KCL gives 2 V + 2Vo = o s +1 2 Vo = - 4 3(s + 1)
1 But - Vy + 2Vo + Vo = 0 s
2 4 s + 2 - 4(s + 2) VTh = Vy = Vo (1 + ) = - = s 3(s + 1) s 3s(s + 1)
Chapter 16, Solution 11.
The s-domain form of the circuit is shown below.
4/s s
1/s
+ -
I1
2
I2
+ -
4/(s + 2)
Write the mesh equations. 1 4 = 2 + I1 - 2 I 2 s s -4 = -2 I1 + (s + 2) I 2 s+2 Put equations (1) and (2) into matrix form. 1 s 2 + 4 s - 2 I1 - 4 (s + 2) = - 2 s + 2 I 2
= 2 2 (s + 2s + 4) , s
1 =
(1)
(2)
s 2 - 4s + 4 , s (s + 2)
2 =
-6 s
I1 =
1 1 2 (s 2 - 4s + 4) A Bs + C = = + 2 2 (s + 2)(s + 2s + 4) s + 2 s + 2s + 4
1 2 (s 2 - 4s + 4) = A (s 2 + 2s + 4) + B (s 2 + 2s) + C (s + 2) Equating coefficients : s2 : 1 2= A+B 1 s : - 2 = 2A + 2B + C
s0 :
2 = 4 A + 2C
Solving these equations leads to
I1 =
I1 =
A = 2,
B = -3 2,
C = -3
- 3 2s - 3 2 + s + 2 (s + 1) 2 + ( 3 ) 2
2 -3 (s + 1) -3 3 + + 2 2 2 s + 2 2 (s + 1) + ( 3 ) 2 3 (s + 1) + ( 3 ) 2
i1 ( t ) = [ 2 e -2t - 1.5 e -t cos(1.732t ) - 0.866 sin(1.732t )] u(t ) A
I2 =
s -3 2 - 6 = = 2 2 s 2 (s + 2s + 4) (s + 1) + ( 3 ) 2 -3 3 e - t sin( 3t ) = - 1.732 e -t sin(1.732t ) u(t ) A
i 2 (t) =
Chapter 16, Solution 12.
We apply nodal analysis to the s-domain form of the circuit below.
s Vo
10/(s + 1)
+ -
1/(2s)
4
3/s
10 - Vo 3 V s +1 o + = + 2sVo s s 4 (1 + 0.25s + s 2 ) Vo = Vo = 10 10 + 15s + 15 + 15 = s +1 s +1
15s + 25 A Bs + C = + 2 2 (s + 1)(s + 0.25s + 1) s + 1 s + 0.25s + 1
A = (s + 1) Vo
s = -1
=
40 7
15s + 25 = A (s 2 + 0.25s + 1) + B (s 2 + s) + C (s + 1) Equating coefficients : s2 : 0= A+B B = -A 1 s : 15 = 0.25A + B + C = -0.75A + C 0 25 = A + C s :
A = 40 7 , B = - 40 7 , C = 135 7
- 40 135 40 3 1 s+ s+ 155 2 40 1 40 7 7 7 2 2 + = - + Vo = 2 2 7 s +1 7 1 s +1 1 1 2 3 7 3 3 3 s + + s + + s + + 2 2 2 4 4 4
v o (t) =
3 (155)(2) 3 40 - t 40 - t 2 e - e cos t + e - t 2 sin t 7 7 2 (7)( 3 ) 2
v o ( t ) = 5.714 e -t - 5.714 e -t 2 cos(0.866t ) + 25.57 e -t 2 sin( 0.866t ) V
Chapter 16, Solution 13.
Consider the following circuit.
1/s Vo 2s Io 2 1/(s + 2) 1
Applying KCL at node o,
Vo Vo 1 s +1 = + = V s + 2 2s + 1 2 + 1 s 2s + 1 o
Vo =
2s + 1 (s + 1)(s + 2) Vo 1 A B = = + 2s + 1 (s + 1)(s + 2) s + 1 s + 2
Io =
A = 1,
Io =
B = -1
1 1 - s +1 s + 2
i o ( t ) = ( e -t - e -2t ) u(t ) A
Chapter 16, Solution 14.
We first find the initial conditions from the circuit in Fig. (a).
1 4 + 5V
+ -
(a)
io
vc(0)
-
i o (0 - ) = 5 A , v c (0 - ) = 0 V We now incorporate these conditions in the s-domain circuit as shown in Fig.(b).
1 Vo Io 15/s 4
+ -
2s
5/s
4/s
(b)
At node o, Vo - 15 s Vo 5 Vo - 0 + + + =0 1 2s s 4 + 4 s
1 s 15 5 V - = 1 + + s s 2s 4 (s + 1) o 5s 2 + 6s + 2 10 4s 2 + 4s + 2s + 2 + s 2 Vo Vo = = 4s (s + 1) s 4s (s + 1)
Vo =
Io =
40 (s + 1) 5s 2 + 6s + 2
Vo 5 4 (s + 1) 5 + + = 2 2s s s (s + 1.2s + 0.4) s 5 A Bs + C Io = + + 2 s s s + 1.2s + 0.4
4 (s + 1) = A (s 2 + 1.2s + 0.4) + B s s + C s Equating coefficients : s0 : 4 = 0.4A A = 10
s1 :
s : Io = Io =
2
4 = 1.2A + C C = -1.2A + 4 = -8 0= A+B B = -A = -10 5 10 10s + 8 + - 2 s s s + 1.2s + 0.4 10 (s + 0.6) 10 (0.2) 15 - 2 2 - s (s + 0.6) + 0.2 (s + 0.6) 2 + 0.2 2
i o ( t ) = [ 15 - 10 e -0.6t ( cos(0.2 t ) - sin( 0.2 t )) ] u(t ) A
Chapter 16, Solution 15.
First we need to transform the circuit into the s-domain.
s/4 Vo + 3Vx
+ -
10 Vx
-
+ -
5/s
5 s+2
5 Vo - Vo - 3Vx Vo - 0 s+2 =0 + + s/4 5/s 10 5s 5s = 0 = (2s 2 + s + 40)Vo - 120Vx - 40Vo - 120Vx + 2s 2 Vo + sVo - s+2 s+2 But, Vx = Vo - 5 5 Vo = Vx + s+2 s+2
We can now solve for Vx. 5 5s (2s 2 + s + 40) Vx + =0 - 120Vx - s + 2 s+2 2(s 2 + 0.5s - 40)Vx = -10 (s 2 + 20) (s + 2)(s 2 + 0.5s - 40) (s 2 + 20) s+2
Vx = - 5
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.
2 + Vo 1
+ -
-
1F Vo/2
+ -
3V
1H
io
(a)
Hence, i L (0) = i o = -3 = -1 A , 3 v o = -1 V
- 1 v c (0) = -(2)(-1) - = 2.5 V 2 We now incorporate the initial conditions for t > 0 as shown in Fig. (b).
2 + 1/s 2.5/s
+ -
+ -
Vo 1
-
s 5/(s + 2)
+ -
I1
I2
- +
Vo/2
-1 V
Io (b)
For mesh 1, - 5 1 1 2.5 Vo + 2 + I1 - I 2 + + =0 s+2 s s s 2 But, Vo = I o = I 2 1 1 1 5 2.5 2 + I1 + - I 2 = - 2 s s s+2 s For mesh 2, V 1 1 2.5 1 + s + I 2 - I1 + 1 - o - =0 s s 2 s 1 1 1 2.5 - I1 + + s + I 2 = -1 2 s s s (2) (1)
Put (1) and (2) in matrix form. 1 5 1 1 2.5 - I1 - 2 + s 2 s s+2 s = -1 1 1 2.5 + s + I 2 -1 s s 2 s
3 = 2s + 2 + , s Io = I2 = 2 = -2 + 4 5 + s s (s + 2)
2 - 2s 2 + 13 A Bs + C = = + 2 2 (s + 2)(2s + 2s + 3) s + 2 2s + 2s + 3
- 2s 2 + 13 = A (2s 2 + 2s + 3) + B (s 2 + 2s) + C (s + 2) Equating coefficients : s2 : - 2 = 2A + B 1 0 = 2A + 2 B + C s : 0 s : 13 = 3A + 2C Solving these equations leads to A = 0.7143 , B = -3.429 , C = 5.429 Io = 0.7143 3.429 s - 5.429 0.7143 1.7145 s - 2.714 - = - s+2 2s 2 + 2s + 3 s+2 s 2 + s + 1.5 0.7143 1.7145 (s + 0.5) (3.194)( 1.25 ) Io = - + s+2 (s + 0.5) 2 + 1.25 (s + 0.5) 2 + 1.25
i o ( t ) = 0.7143 e -2t - 1.7145 e -0.5t cos(1.25t ) + 3.194 e -0.5t sin(1.25t ) u(t ) A
[
]
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 s
1
I1 4
I2
1
For mesh 3, 1 2 1 + s + I 3 - I1 - s I 2 = 0 s +1 s s
(1)
For the supermesh, 1 1 1 + I1 + (1 + s) I 2 - + s I 3 = 0 s s
But
I1 = I 2 - 4
(2) (3)
Substituting (3) into (1) and (2) leads to 1 1 1 2 + s + I 2 - s + I 3 = 4 1 + s s s 1 1 -4 2 - s + I 2 + s + I 3 = - s s s s +1 Adding (4) and (5) gives 2 2 I2 = 4 - s +1
(4)
(5)
I2 = 2 -
1 s +1
i o ( t ) = i 2 ( t ) = ( 2 - e -t ) u(t ) A
Chapter 16, Solution 18.
3 e -s 3 = (1 - e - s ) vs(t) = 3u(t) 3u(t1) or Vs = - s s s
1 +
+ -
Vs
1/s
Vo -
2
V Vo - Vs + sVo + o = 0 (s + 1.5)Vo = Vs 2 1 Vo = 3 2 2 -s (1 - e - s ) = - (1 - e ) s(s + 1.5) s s + 1.5
v o ( t ) = [(2 - 2e -1.5t )u ( t ) - (2 - 2e -1.5( t -1) )u ( t - 1)] V
Chapter 16, Solution 19.
We incorporate the initial conditions in the s-domain circuit as shown below.
2 I 4/(s + 2)
+ -
V1
2I - +
Vo
1/s 1/s s
2
At the supernode, V1 1 4 (s + 2) - V1 +2= + + sVo s s 2
1 1 1 2 + 2 = + V1 + + s Vo 2 s s s+2
(1)
But
Vo = V1 + 2 I and
Vo = V1 + 2 (V1 + 1) s
I=
V1 + 1 s
V1 =
Vo - 2 s s Vo - 2 = (s + 2) s s+2
(2)
Substituting (2) into (1) 2 1 2s + 1 s 2 Vo - + 2- = + s Vo s+2 s s s + 2 s + 2 2 1 2 (2s + 1) 2s + 1 +s V +2- + = s+2 s s (s + 2) s + 2 o
2s 2 + 9s 2s + 9 s 2 + 4s + 1 = = Vo s (s + 2) s+2 s+2
Vo =
2s + 9 A B = + s + 4s + 1 s + 0.2679 s + 3.732
2
A = 2.443 ,
B = -0.4434
Vo =
2.443 0.4434 - s + 0.2679 s + 3.732
Therefore, v o ( t ) = ( 2.443 e -0.2679t - 0.4434 e -3.732t ) u(t ) V
Chapter 16, Solution 20.
We incorporate the initial conditions and transform the current source to a voltage source as shown.
1 2/s
+ -
1/s
Vo
1/(s + 1)
+ -
1
s
1/s
At the main non-reference node, KCL gives 1 (s + 1) - 2 s - Vo Vo Vo 1 = + + 1+1 s 1 s s s s +1 - 2 - s Vo = (s + 1)(s + 1 s) Vo + s +1 s s s +1 - - 2 = (2s + 2 + 1 s) Vo s +1 s Vo = - 2s 2 - 4s - 1 (s + 1)(2s 2 + 2s + 1) - s - 2s - 0.5 A Bs + C = + 2 2 (s + 1)(s + s + 0.5) s + 1 s + s + 0.5
s = -1
Vo =
A = (s + 1) Vo
=1
- s 2 - 2s - 0.5 = A (s 2 + s + 0.5) + B (s 2 + s) + C (s + 1) Equating coefficients : s2 : -1 = A + B B = -2
s1 : s0 :
Vo =
-2 = A+ B+C C = -1 - 0.5 = 0.5A + C = 0.5 - 1 = -0.5 2 (s + 0.5) 1 2s + 1 1 = - - 2 s + 1 s + s + 0.5 s + 1 (s + 0.5) 2 + (0.5) 2
v o ( t ) = [ e -t - 2 e -t 2 cos(t 2)] u(t ) V
Chapter 16, Solution 21. The s-domain version of the circuit is shown below. 1 V1 + 2/s s Vo 2 1/s
10/s
At node 1, 10 - V1 V - Vo s s = 1 + Vo 1 s 2
10 = ( s + 1)V1 + (
s2 - 1)Vo 2
(1)
At node 2, V1 - Vo Vo = + sVo s 2
s V1 = Vo ( + s 2 + 1) 2
(2)
Substituting (2) into (1) gives
s2 10 = ( s + 1)( s + s / 2 + 1)Vo + ( - 1)Vo = s ( s 2 + 2s + 1.5)Vo 2
2
Vo =
A Bs + C 10 = + 2 s ( s + 2s + 1.5) s s + 2s + 1.5
2
10 = A( s 2 + 2 s + 1.5) + Bs 2 + Cs s2 : 0 = A+ B s: 0 = 2A + C constant : 10 = 1.5 A
Vo =
A = 20 / 3, B = -20/3, C = -40/3
0.7071 20 1 s+2 s +1 20 1 s - s 2 + 2 s + 1.5 = 3 s - ( s + 1) 2 + 0.70712 - 1.414 ( s + 1) 2 + 0.70712 3
Taking the inverse Laplace tranform finally yields v o (t) = 20 1 - e - t cos 0.7071t - 1.414e - t sin 0.7071t u ( t ) V 3
[
]
Chapter 16, Solution 22.
The s-domain version of the circuit is shown below. 4s V1
V2
12 s +1
1
2
3/s
At node 1, V V - V2 12 = 1+ 1 s +1 1 4s
12 1 V = V1 1 + - 2 s +1 4s 4s 4 V1 = V2 s 2 + 2s + 1 3
(1)
At node 2, V1 - V2 V2 s = + V2 4s 2 3 Substituting (2) into (1),
(2)
4 12 1 1 4 7 3 = V2 s 2 + 2s + 11 + - = s 2 + s + V2 s +1 3 2 4s 4s 3 3 V2 = 9 7 9 (s + 1)(s 2 + s + ) 4 8 = A Bs + C + (s + 1) (s 2 + 7 s + 9 ) 4 8
9 = A(s 2 +
7 9 s + ) + B(s 2 + s) + C(s + 1) 4 8
Equating coefficients: s2 : s: 0=A+B 0= 9= 7 3 A+B+C = A+C 4 4 9 3 A + C= A 8 8 3 C=- A 4
constant :
A = 24, B = -24, C = -18
V2 =
24 - (s + 1)
3 24s + 18 24 24(s + 7 / 8) + = - 7 23 7 9 7 23 (s + 1) (s + ) 2 + (s 2 + s + ) (s + ) 2 + 8 64 4 8 8 64
Taking the inverse of this produces:
v 2 ( t ) = 24e - t - 24e -0.875t cos(0.5995t ) + 5.004e -0.875t sin(0.5995t ) u ( t ) Similarly,
4 9 s 2 + 2s + 1 Es + F 3 = D + V1 = 7 9 7 9 (s + 1) (s + 1)(s 2 + s + ) (s 2 + s + ) 4 8 4 8
[
]
7 9 4 9 s 2 + 2s + 1 = D(s 2 + s + ) + E(s 2 + s) + F(s + 1) 4 8 3 Equating coefficients: s2 : s: 12 = D + E 18 = 9= 7 3 D + E + F or 6 = D + F 4 4 3 F = 6- D 4
constant :
9 3 D + F or 3 = D 8 8
D = 8, E = 4, F = 0
V1 =
8 + (s + 1)
4s 8 4(s + 7 / 8) 7/2 = + - 7 9 7 23 7 23 (s + 1) (s 2 + s + ) (s + ) 2 + (s + ) 2 + 4 8 8 64 8 64
Thus,
v1 ( t ) = 8e - t + 4e -0.875t cos(0.5995t ) - 5.838e -0.875t sin(0.5995t ) u ( t )
[
]
Chapter 16, Solution 23.
The s-domain form of the circuit with the initial conditions is shown below.
V I 4/s R sL -2/s 1/sC 5C
At the non-reference node, 4 2 V V + + 5C = + + sCV s s R sL s 1 6 + 5 sC CV 2 s + = + RC LC s s V= 5s + 6 C s + s RC + 1 LC
2
But
1 1 = = 8, RC 10 80 V=
1 1 = = 20 LC 4 80
5s + 480 5 (s + 4) (230)(2) = 2 2 + s + 8s + 20 (s + 4) + 2 (s + 4) 2 + 22
2
v( t ) = 5 e -4t cos( 2t ) + 230 e -4t sin( 2t ) V
I= V 5s + 480 = sL 4s (s 2 + 8s + 20) 1.25s + 120 A Bs + C = + 2 2 s (s + 8s + 20) s s + 8s + 20
B = -6 , C = -46.75
I=
A = 6,
I=
6 6s + 46.75 6 6 (s + 4) (11.375)(2) - 2 = - 2 2 - s s + 8s + 20 s (s + 4) + 2 (s + 4) 2 + 22
i( t ) = 6 u(t ) - 6 e -4t cos( 2t ) - 11.375 e -4t sin( 2t ), t > 0
Chapter 16, Solution 24.
At t = 0-, the circuit is equivalent to that shown below. + 9A 4 5 vo -
v o (0) = 5x
4 (9) = 20 4+5
For t > 0, we have the Laplace transform of the circuit as shown below after transforming the current source to a voltage source. 4 + 36V 10A 2/s 5 16 Vo
Applying KCL gives 36 - Vo sV V + 10 = o + o 20 2 5 Thus,
Vo =
3.6 + 20s A B = + , s(s + 0.5) s s + 0.5
A = 7.2, B = -12.8
v o ( t )= 7.2 - 12.8e -0.5t u ( t )
[
]
Chapter 16, Solution 25.
For t > 0 , the circuit in the s-domain is shown below.
6 s + 9/s (2s)/(s2 + 16)
+ -
I
V
+ -
2/s
-
Applying KVL,
- 2s 9 2 + 6 + s + I + = 0 s s s + 16
2
I=
4s 2 + 32 (s 2 + 6s + 9)(s 2 + 16) 9 2 2 36s 2 + 288 I+ = + s s s s (s + 3) 2 (s 2 + 16)
=
V=
2 A B C Ds + E + + + + 2 2 s s s + 3 (s + 3) s + 16
36s 2 + 288 = A (s 4 + 6s 3 + 25s 2 + 96s + 144) + B (s 4 + 3s 3 + 16s 2 + 48s) + C (s 3 + 16s) + D (s 4 + 6s 3 + 9s 2 ) + E (s 3 + 6s 2 + 9s) Equating coefficients : 288 = 144A s0 : 1 s : 0 = 96A + 48B + 16C + 9E 2 36 = 25A + 16B + 9D + 6E s : 3 0 = 6A + 3B + C + 6D + E s : 4 s : 0 = A+ B+ D
Solving equations (1), (2), (3), (4) and (5) gives A = 2 , B = -1.7984 , C = -8.16 , D = -0.2016 ,
E = 2.765
(1) (2) (3) (4) (5)
V(s) =
4 1.7984 8.16 0.2016 s (0.6912)(4) - - + 2 - s s + 3 (s + 3) s 2 + 16 s 2 + 16
v( t ) = 4 u(t ) - 1.7984 e -3t - 8.16 t e -3t - 0.2016 cos(4t ) + 0.6912 sin( 4t ) V
Chapter 16, Solution 26.
Consider the op-amp circuit below.
R2
1/sC R1
+ -
0
- +
Vs
+ Vo -
At node 0, Vs - 0 0 - Vo = + (0 - Vo ) sC R1 R2
1 + sC ( - Vo ) Vs = R 1 R2
Vo -1 = Vs sR 1C + R 1 R 2 But R 1 20 = = 2, R 2 10 Vo -1 = Vs s + 2
Vs = 3 e -5t Vs = 3 (s + 5)
R 1C = (20 103 )(50 10-6 ) = 1
So,
Vo =
-3 (s + 2)(s + 5) 3 A B = + (s + 2)(s + 5) s + 2 s + 5
- Vo =
A = 1, Vo =
B = -1
1 1 - s+5 s+2 v o ( t ) = ( e -5t - e -2t ) u(t )
Chapter 16, Solution 27.
Consider the following circuit.
2s s 2s
10/(s + 3)
+ -
I1
1
I2
1
For mesh 1, 10 = (1 + 2s) I1 - I 2 - s I 2 s+3 10 = (1 + 2s) I1 - (1 + s) I 2 s+3 For mesh 2, 0 = (2 + 2s) I 2 - I1 - s I1 0 = -(1 + s) I1 + 2 (s + 1) I 2 (1) and (2) in matrix form, (1)
(2)
10 (s + 3) 2s + 1 - (s + 1) I1 = - (s + 1) 2 (s + 1) I 0 2 = 3s 2 + 4s + 1
1 = 2 = Thus =
20 I1 (s + 1) s+3 10 (s + 1) s+3
20 (s + 1) 1 = (s + 3)( 3s 2 + 4s + 1) 10 (s + 1) 2 I = = 1 2 (s + 3)( 3s + 4s + 1) 2
I2 =
Chapter 16, Solution 28.
Consider the circuit shown below.
s 1 + 6/s
+ -
I1
2s
s
I2
Vo
-
2
For mesh 1, 6 = (1 + 2s) I1 + s I 2 s For mesh 2, 0 = s I1 + (2 + s) I 2 2 I1 = - 1 + I 2 s Substituting (2) into (1) gives 2 6 - (s 2 + 5s + 2) I2 = -(1 + 2s)1 + I 2 + s I 2 = s s s or I2 = -6 s + 5s + 2
2
(1)
(2)
Vo = 2 I 2 =
- 12 - 12 = s + 5s + 2 (s + 0.438)(s + 4.561)
2
Since the roots of s 2 + 5s + 2 = 0 are -0.438 and -4.561, Vo = A= A B + s + 0.438 s + 4.561 - 12 = -2.91 , 4.123 - 2.91 2.91 + s + 0.438 s + 4.561 v o ( t ) = 2.91 [ e -4.561t - e 0.438t ] u(t ) V B= - 12 = 2.91 - 4.123
Vo (s) =
Chapter 16, Solution 29.
Consider the following circuit.
1 Io 1:2
10/(s + 1)
+ -
4/s
8
Let
Z L = 8 ||
4 (8)(4 s) 8 = = s 8 + 4 s 2s + 1
When this is reflected to the primary side, Zin = 1 + Zin = 1 + Io = ZL , n=2 n2 2 2s + 3 = 2s + 1 2s + 1
10 1 10 2s + 1 = s + 1 Zin s + 1 2s + 3
Io =
10s + 5 A B = + (s + 1)(s + 1.5) s + 1 s + 1.5
B = 20
A = -10 ,
I o (s) =
- 10 20 + s + 1 s + 1.5 i o ( t ) = 10 2 e -1.5t - e - t u(t ) A
[
]
Chapter 16, Solution 30.
Y(s) = H(s) X(s) ,
X(s) =
4 12 = s + 1 3 3s + 1
Y(s) =
12 s 2 4 8s + 4 3 - 2 = (3s + 1) 3 (3s + 1) 2 4 8 s 4 1 - 2 - 3 9 (s + 1 3) 27 (s + 1 3) 2 -8 s 9 (s + 1 3) 2
Y(s) =
Let G (s) =
Using the time differentiation property, -8 d - 8 -1 g( t ) = ( t e -t 3 ) = t e -t 3 + e -t 3 9 dt 93 g( t ) = Hence, y( t ) = 4 8 -t 3 8 -t 3 4 -t 3 u(t) + te - e - te 3 27 9 27
4 -t 3 8 4 te u( t ) - e - t 3 + 27 9 3
8 -t 3 8 -t 3 te - e 27 9
y( t ) =
Chapter 16, Solution 31.
x(t) = u(t) X(s) =
1 s
y( t ) = 10 cos(2t ) Y(s) = Y(s) 10s 2 = X(s) s 2 + 4
10s s2 + 4
H(s) =
Chapter 16, Solution 32.
(a)
Y(s) = H(s) X(s)
=
s+3 1 s + 4s + 5 s
2
=
s+3 A Bs + C = + 2 s (s + 4s + 5) s s + 4s + 5
2
s + 3 = A (s 2 + 4s + 5) + Bs 2 + Cs Equating coefficients : 3 = 5A A = 3 5 s0 :
s1 : s2 :
Y(s) =
1 = 4A + C C = 1 - 4A = - 7 5 0= A+B B = -A = - 3 5 35 1 3s + 7 - 2 s 5 s + 4s + 5 0.6 1 3 (s + 2) + 1 - s 5 (s + 2) 2 + 1
Y(s) =
y( t ) = [ 0.6 - 0.6 e -2t cos(t ) - 0.2 e -2t sin( t )] u(t )
(b)
x ( t ) = 6 t e -2t
X(s) =
6 (s + 2) 2
Y(s) = H(s) X(s) =
s+3 6 s + 4s + 5 (s + 2) 2
2
Y(s) =
6 (s + 3) A B Cs + D = + 2 2 2 + 2 (s + 2) (s + 4s + 5) s + 2 (s + 2) s + 4s + 5
Equating coefficients : s3 : 0= A+C C = -A 2 0 = 6 A + B + 4C + D = 2 A + B + D s : 1 s : 6 = 13A + 4B + 4C + 4D = 9A + 4B + 4D 0 18 = 10A + 5B + 4D = 2A + B s : Solving (1), (2), (3), and (4) gives A=6, B = 6, C = -6 , Y(s) = 6 6 6s + 18 + 2 - s + 2 (s + 2) (s + 2) 2 + 1 6 6 6 (s + 2) 6 - + 2 - 2 s + 2 (s + 2) (s + 2) + 1 (s + 2) 2 + 1
D = -18
(1) (2) (3) (4)
Y(s) =
y( t ) = [ 6 e -2t + 6 t e -2t - 6 e -2t cos(t ) - 6 e -2t sin( t )] u(t )
Chapter 16, Solution 33.
H(s) =
Y(s) , X(s)
X(s) =
1 s
Y(s) =
4 1 2s (3)(4) + - - 2 s 2 (s + 3) (s + 2) + 16 (s + 2) 2 + 16
H(s) = s Y(s) = 4 +
s 2s2 12 s - 2 - 2 2 (s + 3) s + 4s + 20 s + 4s + 20
Chapter 16, Solution 34.
Consider the following circuit.
2
s
Vo +
Vs
+ -
4
10/s
Vo(s) -
Using nodal analysis, Vs - Vo Vo Vo = + s+2 4 10 s
1 1 1 s 1 Vs = (s + 2) + + Vo = 1 + (s + 2) + (s 2 + 2s) Vo 4 s + 2 4 10 10 Vs = 1 ( 2s 2 + 9s + 30) Vo 20
20 Vo = 2 Vs 2s + 9s + 30
Chapter 16, Solution 35. Consider the following circuit. I 2/s V1 s + 2I Vo - At node 1, 2I + I = V1 , s+3 where I = 3
Vs
+ -
Vs - V1 2s
3
Vs - V1 V = 1 2s s+3
3s V1 3s = Vs - V1 2 s+3 2 1 3s 3s + V1 = Vs s + 3 2 2 V1 = Vo = 3s (s + 3) V 3s 2 + 9s + 2 s 9s 3 V V1 = 2 3s + 9s + 2 s s+3
9s Vo = 2 Vs 3s + 9s + 2
H(s) =
Chapter 16, Solution 36. From the previous problem, 3I = I= V1 3s V = 2 s + 3 3s + 9s + 2 s
s V 3s + 9s + 2 s
2
But
Vs =
3s 2 + 9s + 2 Vo 9s
V s 3s 2 + 9 s + 2 I= 2 Vo = o 3s + 9 s + 2 9s 9
H(s) =
Vo =9 I
Chapter 16, Solution 37. (a) Consider the circuit shown below. 3 + I1 Vx
-
2s
Vs
+ -
2/s
I2
+ -
4Vx
For loop 1, 2 2 Vs = 3 + I1 - I 2 s s For loop 2, 2 2 4Vx + 2s + I 2 - I1 = 0 s s But, 2 Vx = (I1 - I 2 ) s 2 2 8 (I1 - I 2 ) + 2s + I 2 - I1 = 0 s s s 0= 6 -6 I1 + - 2s I 2 s s (2) (1)
So,
In matrix form, (1) and (2) become Vs 3 + 2 s - 2 s I1 0 = - 6 s 6 s - 2s I 2
6 2 2 6 = 3 + - 2s - s s s s = 18 - 6s - 4 s 2 = 6 V s s
6 1 = - 2s Vs , s
I1 =
1 (6 s - 2s) = V 18 s - 4 - 6s s
I1 3 s-s s2 - 3 = = 2 Vs 9 s - 2 - 3 3s + 2s - 9
(b)
I2 =
2 2 2 1 - 2 ( I1 - I 2 ) = s s 2 s Vs (6 s - 2s - 6 s) - 4Vs =
Vx =
Vx =
6 s Vs - 3 I2 = = Vx - 4Vs 2s
Chapter 16, Solution 38.
(a) Consider the following circuit.
Is
1
V1
s
Vo
Io +
Vs
+ -
1/s
1/s
1
Vo
-
At node 1, Vs - V1 V1 - Vo = s V1 + 1 s 1 1 Vs = 1 + s + V1 - Vo s s (1)
At node o, V1 - Vo = s Vo + Vo = (s + 1) Vo s V1 = (s 2 + s + 1) Vo Substituting (2) into (1) Vs = (s + 1 + 1 s)(s 2 + s + 1)Vo - 1 s Vo Vs = (s 3 + 2s 2 + 3s + 2)Vo
H 1 (s) = Vo 1 = 3 2 Vs s + 2s + 3s + 2
(2)
(b)
I s = Vs - V1 = (s 3 + 2s 2 + 3s + 2)Vo - (s 2 + s + 1)Vo I s = (s 3 + s 2 + 2s + 1)Vo
H 2 (s) = Vo 1 I o Vo 1 = = H 2 (s) = 3 2 Is Is s + s + 2s + 1 I o Vo 1 = = H 1 (s) = 3 2 Vs Vs s + 2s + 3s + 2 Vo 1 = 3 2 Is s + s + 2s + 1
(c)
Io =
H 3 (s) =
(d).
H 4 (s) =
Chapter 16, Solution 39.
Consider the circuit below.
Va Vb Vs
+ -
- +
R Io
+ Vo
-
1/sC
Since no current enters the op amp, I o flows through both R and C. 1 Vo = -I o R + sC
Va = Vb = Vs = H(s) = - Io sC
Vo R + 1 sC = = sRC + 1 Vs 1 sC
Chapter 16, Solution 40.
Vo R R L = = Vs R + sL s + R L
R - Rt L e u( t ) L
(a)
H(s) =
h(t) = (b)
v s (t) = u(t) Vs (s) = 1 s
Vo =
R L R L A B Vs = = + s+R L s (s + R L) s s + R L
B = -1
A = 1,
1 1 Vo = - s s+R L
v o ( t ) = u ( t ) - e -Rt L u ( t ) = (1 - e -Rt L ) u(t )
Chapter 16, Solution 41.
Y(s) = H(s) X(s)
h ( t ) = 2 e -t u ( t ) H(s) = 2 s +1
v i (t) = 5 u(t) Vi (s) = X(s) = 5 s Y(s) = 10 A B = + s (s + 1) s s + 1
B = -10
A = 10 ,
Y(s) =
10 10 - s s +1
y( t ) = 10 (1 - e -t ) u(t )
Chapter 16, Solution 42.
2s Y(s) + Y(s) = X(s) (2s + 1) Y(s) = X(s)
H(s) = Y(s) 1 1 = = X(s) 2s + 1 2 (s + 1 2)
h ( t ) = 0.5 e -t 2 u(t )
Chapter 16, Solution 43. 1 i(t)
u(t)
+ -
1F
1H
First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KVL we get:
- u ( t ) + i + v C + i' = 0; i = v 'C
Thus,
v 'C = i i ' = -v C - i + u(t)
Finally we get, v 0 1 v C 0 v C C = i + 1 u ( t ) ; i( t ) = [0 1] i + [0]u ( t ) i - 1 - 1
Chapter 16, Solution 44. 1H + vx
-
1/8 F
4u ( t )
+ -
2
4
First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KCL we get:
v - iL + x + 2 v 'C = 0; or v 'C = 8i L - 4v x 8
i 'L = 4u ( t ) - v x v 'C v 'C v x = vC + 4 = vC + = v C + 4i L - 2v x ; or v x = 0.3333v C + 1.3333i L 8 2
v 'C = 8i L - 1.3333v C - 5.333i L = -1.3333v C + 2.666i L i 'L = 4u ( t ) - 0.3333v C - 1.3333i L
Now we can write the state equations. v 'C - 1.3333 2.666 v C 0 0.3333 v C ' = i + 4 u ( t ); v x = 1.3333 i L i L - 0.3333 - 1.3333 L
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:
v 'C v o - + + i L = 0 or v 'C = 4i L + 2 v o 4 2 i 'L = v o - v 2 v o = - v C + v1
v 'C = 4i L - 2 v C + 2 v1 i 'L = - v C + v1 - v 2
i 0 - 1 i L 1 - 1 v1 ( t ) iL v1 ( t ) L = v + 2 0 v ( t ) ; v o ( t ) = [0 - 1] v + [1 0] v ( t ) 2 2 C v C 4 - 2 C 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: v - i L + v 'C + C - i s = 0 or v 'C = -0.25v C + i L + i s 4 i 'L = - v C + v s Thus, v ' - 0.25 1 v ' 0 1 v s 1 v 0 0 v s ; v o (t) = C + 'C = 'C + 0 i L 1 0 i s iL -1 0 i L 0 0 i s
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 = Loop 1:
v' - v1 + v C + 2 C - i L = 0 or v 'C = 4i L - 2 v C + 2 v1 4
v 'C and i2 = iL and applying KVL we get: 4
Loop 2:
' i - v C + i ' + v = 0 or 2 L L 2 4 4i - 2v C + 2v1 - v 2 = - v C + v1 - v 2 i 'L = -2i L + L 2
i1 =
4i L - 2 v C + 2 v1 = i L - 0.5v C + 0.5v1 4
i L = v C
0 - 1 i L 1 - 1 v1 ( t ) 4 - 2 v + 2 0 v ( t ) ; C 2
i1 ( t ) 1 - 0.5 i L 0.5 0 v1 ( t ) + i ( t ) = 1 0 v C 0 0 v 2 ( t ) 2
Chapter 16, Solution 48.
' Let x1 = y(t). Thus, x1 = y ' = x 2 and x '2 = y = -3x1 - 4 x 2 + z( t )
This gives our state equations.
' x1 0 1 x 1 0 x1 ' = x + 1 z( t ); y( t ) = [1 0] x + [0]z( t ) x 2 - 3 - 4 2 2
Chapter 16, Solution 49.
' Let x1 = y( t ) and x 2 = x1 - z = y ' - z or y ' = x 2 + z
Thus, x '2 = y - z ' = -6x1 - 5( x 2 + z) + z ' + 2z - z ' = -6x1 - 5x 2 - 3z This now leads to our state equations,
' x1 0 1 x1 1 x1 ' = x + - 3 z( t ); y( t ) = [1 0] x + [0]z( t ) x 2 - 6 - 5 2 2
Chapter 16, Solution 50.
' Let x1 = y(t), x2 = x1 , and x 3 = x '2 . Thus,
x " = -6x1 - 11x 2 - 6x 3 + z( t ) 3 We can now write our state equations.
' x1 0 1 0 x 1 0 x1 ' x + 0 z( t ); y( t ) = [1 0 0] x + [0]z( t ) 0 1 2 x 2 = 0 2 x ' - 6 - 11 - 6 x 1 x 3 3 3
Chapter 16, Solution 51. We transform the state equations into the s-domain and solve using Laplace transforms.
1 sX(s) - x (0) = AX(s) + B s Assume the initial conditions are zero.
1 (sI - A)X(s) = B s s + 4 - 4 X(s) = s 2
Y(s) = X1 (s) =
-1
4 0 0 1 s 1 2 s = 2 2 s + 4 ( 2 / s ) s + 4s + 8
8
1 -s-4 = + s(s 2 + 4s + 8) s s 2 + 4s + 8 1 1 -2 - (s + 2) -s-4 + = + = + 2 2 2 2 s (s + 2) + 2 s (s + 2) + 2 (s + 2) 2 + 2 2
y(t) = 1 - e - 2 t (cos 2t + sin 2t ) u ( t )
(
)
Chapter 16, Solution 52.
Assume that the initial conditions are zero. Using Laplace transforms we get,
1 s + 2 X(s) = - 2 s + 4 X1 = 3s + 8 s((s + 3) + 1 )
2 2 -1
s + 4 - 1 3 / s 1 1 1 / s 1 4 0 2 / s = 2 s + 2 4 / s s + 6s + 10 2 0.8 - 0.8s - 1.8 + s (s + 3) 2 + 12
=
=
0.8 s+3 1 - 0.8 + .6 s (s + 3) 2 + 12 (s + 3) 2 + 12
x1 ( t ) = (0.8 - 0.8e -3t cos t + 0.6e -3t sin t )u ( t )
X2 =
4s + 14 s((s + 3) 2 + 12
=
1.4 - 1.4s - 4.4 + s (s + 3) 2 + 12
=
1.4 s+3 1 - 1.4 - 0.2 2 2 s (s + 3) + 1 (s + 3) 2 + 12
x 2 ( t ) = (1.4 - 1.4e -3t cos t - 0.2e -3t sin t )u ( t ) y1 ( t ) = -2x1 ( t ) - 2x 2 ( t ) + 2u ( t ) = (-2.4 + 4.4e - 3t cos t - 0.8e - 3t sin t )u ( t )
y 2 ( t ) = x1 ( t ) - 2u ( t ) = (-1.2 - 0.8e -3t cos t + 0.6e -3t sin t )u ( t )
Chapter 16, Solution 53.
If Vo is the voltage across R, applying KCL at the non-reference node gives Is = Vo V 1 1 + sC Vo + o = + sC + Vo R sL R sL
Is = sRL Is sL + R + s 2 RLC
Vo =
1 1 + sC + R sL
Io =
Vo sL Is = 2 R s RLC + sL + R Io sL s RC = 2 = 2 Is s RLC + sL + R s + s RC + 1 LC
H(s) = The roots
s1, 2 =
-1 1 1 2 - 2RC (2RC) LC
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)
H1 (s) =
3 , s +1
H 2 (s) =
1 s+4
H(s) = H1 (s) H 2 (s) =
3 (s + 1)(s + 4)
A B + h ( t ) = L-1 [ H(s)] = L-1 s +1 s + 4
A = 1, B = -1 -t -4t h ( t ) = ( e - e ) u( 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
Vo1 be the voltage at the output of the first op amp. Vo1 - 1 sC - 1 = = , Vs R sRC H(s) = Vo 1 = 2 2 2 Vs s R C t R C2
2
Vo -1 = Vo1 sRC
h(t) =
lim h ( t ) = , i.e. the output is unbounded.
t
Hence, the circuit is unstable.
Chapter 16, Solution 56.
1 sL 1 sC = sL sL || = 1 1 + s 2 LC sC sL + sC sL 2 V2 sL = 1 + s LC = 2 sL V1 s RLC + sL + R R+ 2 1 + s LC V2 = V1 1 RC 1 1 s2 + s + RC LC s
Comparing this with the given transfer function, 1 1 2= , 6= RC LC If R = 1 k , C= L= 1 = 500 F 2R 1 = 333.3 H 6C
Chapter 16, Solution 57.
The circuit in the s-domain is shown below.
R1 V1 C L + R2 Vx
Vi
+ -
Z
-
Z=
(1 sC) (R 2 + sL) R 2 + sL 1 || (R 2 + sL) = = sC R 2 + sL + 1 sC 1 + s 2 LC + sR 2 C
V1 =
Z V R1 + Z i
R2 R2 Z V1 = V R 2 + sL R 2 + sL R 1 + Z i
Vo =
R 2 + sL Vo R2 R2 1 + s 2 LC + sR 2 C Z = = R 2 + sL Vi R 2 + sL R 1 + Z R 2 + sL R1 + 1 + s 2 LC + sR 2 C Vo R2 = 2 Vi s R 1 LC + sR 1 R 2 C + R 1 + R 2 + sL
R2 Vo R 1 LC = R2 Vi 1 R1 + R 2 + s 2 + s + L R 1C R 1 LC
Comparing this with the given transfer function, R2 R2 R1 + R 2 1 5= 6= 25 = + R 1 LC L R 1C R 1 LC Since R 1 = 4 and R 2 = 1 , 1 1 5= LC = 4 LC 20 6= 1 1 + L 4C 5 4 LC LC = 1 20
(1)
(2)
25 =
Substituting (1) into (2), 1 6 = 20 C + 80 C 2 - 24 C + 1 = 0 4C Thus, C = 1 , 4 1 20
When C =
1 , 4 1 , 20
L=
1 1 = . 20 C 5 1 = 1. 20 C
When C =
L=
Therefore, there are two possible solutions. C = 0.25 F L = 0.2 H or C = 0.05 F
L = 1H
Chapter 16, Solution 58.
We apply KCL at the noninverting terminal at the op amp. (Vs - 0) Y3 = (0 - Vo )(Y1 - Y2 ) Y3 Vs = - (Y1 + Y2 )Vo
Vo - Y3 = Vs Y1 + Y2
Let
Y1 = sC1 ,
Y2 = 1 R 1 ,
Y3 = sC 2
Vo - sC 2 - sC 2 C1 = = Vs sC1 + 1 R 1 s + 1 R 1C1
Comparing this with the given transfer function, C2 1 = 1, = 10 R 1 C1 C1 If R 1 = 1 k , C1 = C 2 = 1 = 100 F 10 4
Chapter 16, Solution 59.
Consider the circuit shown below. We notice that V3 = Vo and V2 = V3 = Vo .
Y4 Y1 Y2 V1 Y3
V2
Vin
+ -
- +
Vo
At node 1, (Vin - V1 ) Y1 = (V1 - Vo ) Y2 + (V1 - Vo ) Y4 Vin Y1 = V1 (Y1 + Y2 + Y4 ) - Vo (Y2 + Y4 ) At node 2, (V1 - Vo ) Y2 = (Vo - 0) Y3 V1 Y2 = (Y2 + Y3 ) Vo
V1 = Y2 + Y3 Vo Y2
(1)
(2)
Substituting (2) into (1), Y2 + Y3 Vin Y1 = (Y1 + Y2 + Y4 ) Vo - Vo (Y2 + Y4 ) Y2 Vin Y1 Y2 = Vo (Y1 Y2 + Y22 + Y2 Y4 + Y1 Y3 + Y2 Y3 + Y3 Y4 - Y22 - Y2 Y4 )
Vo Y1 Y2 = Vin Y1 Y2 + Y1 Y3 + Y2 Y3 + Y3 Y4
Y1 and Y2 must be resistive, while Y3 and Y4 must be capacitive.
Let
Y1 =
1 , R1
Y2 =
1 , R2
Y3 = sC1 ,
Y4 = sC 2
1 Vo R 1R 2 = sC1 sC1 1 Vin + + + s 2 C1 C 2 R 1R 2 R 1 R 2 1 Vo R 1 R 2 C1C 2 = R1 + R 2 Vin 1 + s2 + s R 1 R 2 C 2 R 1 R 2 C1 C 2
Choose R 1 = 1 k , then 1 = 10 6 R 1 R 2 C1 C 2 and
R1 + R 2 = 100 R 1R 2 C 2
We have three equations and four unknowns. Thus, there is a family of solutions. One such solution is R 2 = 1 k , C1 = 50 nF , C 2 = 20 F
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, Vo - Y1 Y2 = Vi Y2 Y3 + Y4 (Y1 + Y2 + Y3 )
When Y1 = sC1 ,
C1 = 0.5 F R 1 = 10 k Y4 = sC 2 , C 2 = 1 F
Y2 =
1 , R1
Y3 = Y2 ,
Vo - sC1 R 1 - sC1 R 1 = = 2 Vi 1 R 1 + sC 2 (sC1 + 2 R 1 ) 1 + sC 2 R 1 (2 + sC1 R 1 )
Vo - sC1 R 1 = 2 2 Vi s C1C 2 R 1 + s 2C 2 R 1 + 1 Vo - s (0.5 10 -6 )(10 10 3 ) = Vi s 2 (0.5 10 -6 )(1 10 -6 )(10 10 3 ) 2 + s (2)(1 10 -6 )(10 10 3 ) + 1 Vo - 100 s = 2 Vi s + 400 s + 2 10 4
Therefore, a = - 100 ,
b = 400 ,
c = 2 10 4
Chapter 16, Solution 68.
K (s + 1) s+3
(a)
Let
Y(s) =
Y() = lim i.e.
K (s + 1) K (1 + 1 s) = lim =K s s 1 + 3 s s+3
0.25 = K .
Hence, Y(s) =
s+1 4 (s + 3)
(b)
Consider the circuit shown below.
t=0 I
Vs = 8 V
+ -
YS
Vs = 8 u ( t ) Vs = 8 s I= Vs 8 s + 1 2 (s + 1) = Y(s) Vs (s) = = Z 4s s + 3 s (s + 3)
I=
A B + s s+3
B= -4 3
A = 2 3,
i( t ) =
1 [ 2 - 4 e -3t ] u(t ) A 3
Chapter 16, Solution 69.
The gyrator is equivalent to two cascaded inverting amplifiers. Let V1 be the voltage at the output of the first op amp. V1 = Vo =
Io =
-R V = -Vi R i - 1 sC 1 V1 = V R sCR i
Vo Vo = R sR 2 C
Vo = sR 2 C Io
Vo = sL, when L = R 2 C Io