Chapter 16, Solution 1.
Consider the s-domain form of the circuit which is shown below.
1 I(s)
1/s + 1/s
−
s
1s 1 1
I(s) = = 2 =
1 + s + 1 s s + s + 1 (s + 1 2) 2 + ( 3 2) 2
2 3
i( t ) = e - t 2 sin
2 t
3
i( t ) = 1.155 e -0.5t sin (0.866t ) A
Chapter 16, Solution 2.
s 8/s
+
+
4 − Vx 2 4
s −
4
Vx −
s + Vx − 0 + Vx − 0 = 0
s 2 8
4+
s
(16s + 32)
Vx (4s + 8) − + (2s 2 + 4s)Vx + s 2 Vx = 0
s
16s + 32
Vx (3s 2 + 8s + 8) =
s
s+2 0.25 + − 0.125 + − 0.125
Vx = −16 = −16
s(3s 2 + 8s + 8) s 4 8 4 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 6 − 4t / 3 2 2
vx = 4u ( t ) − e − 4 t / 3 cos
3 t −
e sin
3 t V
2
Chapter 16, Solution 3.
s
+
5/s 1/2 Vo 1/8
−
Current division leads to:
1
1 5 5 5
Vo = 2 = =
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
+
1/(s + 1) + 10/s Vo(s)
−
−
Using voltage division,
10 s 1 10 1
Vo (s) = = 2
s + 6 + 10 s s + 1 s + 6s + 10 s + 1
10 A Bs + C
Vo (s) = = + 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
0
s : 10 = 10A + C = 5A
→ A = 2, B = -2, C = -10
2 2s + 10 2 2 (s + 3) 4
Vo (s) = − 2 = − 2 −
s + 1 s + 6s + 10 s + 1 (s + 3) + 1
2
(s + 3) 2 + 12
v o ( t ) = 2 e -t − 2 e -3t cos(t ) − 4 e -3t sin( t ) V
Chapter 16, Solution 5.
Io
2 2
1 s
s
s+2
1 1
= 1
2s 2s
V= 2 =
(s + 2)(s + 0.5 + j1.3229)(s + 0.5 − j1.3229)
s + 2 1 1 s s + 2s + s + 2
+ +
s 2 2
Vs s2
Io = =
2 (s + 2)(s + 0.5 + j1.3229)(s + 0.5 − j1.3229)
(−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
10/s
5 s
s+2
Use current division.
s+2 5 5s 5(s + 1) 5
Io = = = −
10 s + 2 s + 2s + 10 (s + 1) + 3
2 2 2
(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
+ 2s
2
s +1
–
Z
1
(2s)
1 2s 2s 2 + 2s + 1
Z = 1 + // 2s = 1 + s = 1+ =
s 1 1 + 2s 2 1 + 2s 2
+ 2s
s
V 2 1 + 2s 2 2s 2 + 1 A Bs + C
Ix = = x = = +
Z s + 1 2s + 2s + 1 (s + 1)(s + s + 0.5) (s + 1) (s + s + 0.5)
2 2 2
2s 2 + 1 = A(s 2 + s + 0.5) + B(s 2 + s) + C(s + 1)
s2 : 2=A+B
s: 0 = A+B+C = 2+C
→ C = −2
constant : 1 = 0.5A + C or 0.5A = 3
→ A = 6, B = -4
6 4s + 2 6 4(s + 0.5)
Ix = − = −
s + 1 (s + 0.5) + 0.75 s + 1 (s + 0.5) 2 + 0.866 2
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) = + + =
Z 2 s 1 2s(s + 1)
1+
s
2s(s + 1)
Z=
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 2
s 2/s
1/s
1
(a) (b)
(b) The s-domain equivalent circuit is shown in Fig. (b).
2 (1 + 2 s) 2 (s + 2)
2 || (1 + 2 s) = =
3+ 2 s 3s + 2
5s + 6
1 + 2 || (1 + 2 s) =
3s + 2
5s + 6
s ⋅
5s + 6 3s + 2 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
Vx
1 + 2Vo =
2 + 1/ s
2
But Vo = Vx . Hence
2 + 1/ s
4Vx Vx (2s + 1)
1+ =
→ Vx = −
2 + 1/ s 2 + 1/ s 3s
Vx (2s + 1)
ZTh = =−
1 3s
To find VTh, consider the circuit below.
1/s Vy
+
2
2 Vo 2Vo
s +1
-
Applying KCL gives
2 V 4
+ 2Vo = o
→ Vo = −
s +1 2 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 I2 + 4/(s + 2)
2
− −
Write the mesh equations.
1 4
= 2 + I1 − 2 I 2 (1)
s s
-4
= -2 I1 + (s + 2) I 2 (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 2 − 4s + 4 -6
∆= (s + 2s + 4) , ∆1 = , ∆2 =
s s (s + 2) s
∆1 1 2 ⋅ (s 2 − 4s + 4) A Bs + C
I1 = = = + 2
∆ (s + 2)(s + 2s + 4) s + 2 s + 2s + 4
2
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 A = 2, B = -3 2, C = -3
2 - 3 2s − 3
I1 = +
s + 2 (s + 1) 2 + ( 3 ) 2
2 -3 (s + 1) -3 3
I1 = + ⋅ + ⋅
s + 2 2 (s + 1) + ( 3 )
2 2
2 3 (s + 1) + ( 3 ) 2
2
i1 ( t ) = [ 2 e -2t − 1.5 e -t cos(1.732t ) − 0.866 sin(1.732t )] u(t ) A
∆2 - 6 s -3
I2 = = ⋅ =
∆ s 2 (s + 2s + 4) (s + 1) + ( 3 ) 2
2 2
-3
i 2 (t) = e - t sin( 3t ) = - 1.732 e -t sin(1.732t ) u(t ) A
3
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
10 10 + 15s + 15
(1 + 0.25s + s 2 ) Vo = + 15 =
s +1 s +1
15s + 25 A Bs + C
Vo = = + 2
(s + 1)(s + 0.25s + 1) s + 1 s + 0.25s + 1
2
40
A = (s + 1) Vo s = -1 =
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
s : 25 = A + C
A = 40 7 , B = - 40 7 , C = 135 7
40 - 40 135 1 3
s+ 40 1 40 s+ 155 2
7 7 7 2 2
Vo = + = − + ⋅
s +1 1 2
3 7 s +1 7 1 2
3 7 3 1 2 3
s + + s + + s + +
2 4 2 4 2 4
40 - t 40 - t 2 3 (155)(2) 3
v o (t) = 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 2s
Vo
Io
2 1/(s + 2) 1
Applying KCL at node o,
1 Vo Vo s +1
= + = V
s + 2 2s + 1 2 + 1 s 2s + 1 o
2s + 1
Vo =
(s + 1)(s + 2)
Vo 1 A B
Io = = = +
2s + 1 (s + 1)(s + 2) s + 1 s + 2
A = 1, B = -1
1 1
Io = −
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).
1Ω 4Ω
+
5V +
io vc(0)
−
−
(a)
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 4
Vo
Io
15/s + 2s 5/s 4/s
−
(b)
At node o,
Vo − 15 s Vo 5 Vo − 0
+ + + =0
1 2s s 4 + 4 s
15 5 1 s
− = 1 + + V
s s 2s 4 (s + 1) o
10 4s 2 + 4s + 2s + 2 + s 2 5s 2 + 6s + 2
= Vo = Vo
s 4s (s + 1) 4s (s + 1)
40 (s + 1)
Vo =
5s 2 + 6s + 2
Vo 5 4 (s + 1) 5
Io = + = +
2s s s (s + 1.2s + 0.4) s
2
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 : 4 = 1.2A + C
→ C = -1.2A + 4 = -8
2
s : 0= A+B
→ B = -A = -10
5 10 10s + 8
Io = + − 2
s s s + 1.2s + 0.4
15 10 (s + 0.6) 10 (0.2)
Io = − 2 −
s (s + 0.6) + 0.2
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 10
Vo
+ Vx −
+ + 5
3Vx − 5/s − s+2
5
Vo −
Vo − 3Vx Vo − 0 s+2 =0
+ +
s/4 5/s 10
5s 5s
40Vo − 120Vx + 2s 2 Vo + sVo − = 0 = (2s 2 + s + 40)Vo − 120Vx −
s+2 s+2
5 5
But, Vx = Vo − → Vo = Vx +
s+2 s+2
We can now solve for Vx.
5 5s
(2s 2 + s + 40) Vx + − 120Vx − =0
s + 2 s+2
(s 2 + 20)
2(s 2 + 0.5s − 40)Vx = −10
s+2
(s 2 + 20)
Vx = − 5
(s + 2)(s 2 + 0.5s − 40)
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.
2Ω Vo
+ −
1Ω
1F
+
3V
−
+
Vo/2 −
1H io
(a)
Hence,
-3
i L (0) = i o = = -1 A , v o = -1 V
3
- 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 Vo
+ −
1
1/s
s
+
+ 2.5/s
5/(s + 2) I1 − I2
−
− -1 V
+
Vo/2 − +
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 = − (1)
s 2 s s+2 s
For mesh 2,
1 1 V 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 2 s s
Put (1) and (2) in matrix form.
1 1 1 5 2.5
2 + s − I1 −
2 s s+2 s
=
-1 1 1 2.5
s + s + I 2 −1
2 s s
3 4 5
∆ = 2s + 2 + , ∆ 2 = -2 + +
s s s (s + 2)
∆2 - 2s 2 + 13 A Bs + C
Io = I2 = = = + 2
∆ (s + 2)(2s + 2s + 3) s + 2 2s + 2s + 3
2
- 2s 2 + 13 = A (2s 2 + 2s + 3) + B (s 2 + 2s) + C (s + 2)
Equating coefficients :
s2 : - 2 = 2A + B
1
s : 0 = 2A + 2 B + C
0
s : 13 = 3A + 2C
Solving these equations leads to
A = 0.7143 , B = -3.429 , C = 5.429
0.7143 3.429 s − 5.429 0.7143 1.7145 s − 2.714
Io = − = −
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 I2 1
4
For mesh 3,
2 1 1
+ s + I 3 − I1 − s I 2 = 0 (1)
s +1 s s
For the supermesh,
1 1
1 + I1 + (1 + s) I 2 − + s I 3 = 0 (2)
s s
But I1 = I 2 − 4 (3)
Substituting (3) into (1) and (2) leads to
1 1 1
2 + s + I 2 − s + I 3 = 4 1 + (4)
s s s
1 1 -4 2
- s + I 2 + s + I 3 = − (5)
s s s s +1
Adding (4) and (5) gives
2
2 I2 = 4 −
s +1
1
I2 = 2 −
s +1
i o ( t ) = i 2 ( t ) = ( 2 − e -t ) u(t ) A
Chapter 16, Solution 18.
3 e −s 3
vs(t) = 3u(t) – 3u(t–1) or Vs = − = (1 − e − s )
s s s
1Ω
+
Vs +
1/s 2Ω
− Vo
−
Vo − Vs V
+ sVo + o = 0 → (s + 1.5)Vo = Vs
1 2
3 2 2
Vo = (1 − e − s ) = − −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.
2I
2
V1 Vo
− +
I
+
4/(s + 2) 1/s 2
−
1/s s
At the supernode,
4 (s + 2) − V1 V1 1
+2= + + sVo
2 s s
2 1 1 1
+ 2 = + V1 + + s Vo (1)
s+2 2 s s
V1 + 1
But Vo = V1 + 2 I and I=
s
2 (V1 + 1) Vo − 2 s s Vo − 2
Vo = V1 +
→ V1 = = (2)
s (s + 2) s s+2
Substituting (2) into (1)
2 1 2s + 1 s 2
+ 2− = Vo − + s Vo
s+2 s s s + 2 s + 2
2 1 2 (2s + 1) 2s + 1
+2− + = +s V
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
2s + 9 A B
Vo = = +
s + 4s + 1 s + 0.2679 s + 3.732
2
A = 2.443 , B = -0.4434
2.443 0.4434
Vo = −
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.
2/s
1 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
- 2s 2 − 4s − 1
Vo =
(s + 1)(2s 2 + 2s + 1)
- s − 2s − 0.5 A Bs + C
Vo = = + 2
(s + 1)(s + s + 0.5) s + 1 s + s + 0.5
2
A = (s + 1) Vo s = -1 =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 : -2 = A+ B+C → C = -1
s0 : - 0.5 = 0.5A + C = 0.5 − 1 = -0.5
1 2s + 1 1 2 (s + 0.5)
Vo = − 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 s
V1 Vo
+ 2/s 2 1/s
10/s
-
At node 1,
10
− V1
s V − Vo s s2
= 1 + Vo
→ 10 = ( s + 1)V1 + ( − 1)Vo (1)
1 s 2 2
At node 2,
V1 − Vo Vo s
= + sVo
→ V1 = Vo ( + s 2 + 1) (2)
s 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
10 A Bs + C
Vo = = + 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 → A = 20 / 3, B = -20/3, C = -40/3
20 1 s+2 20 1 s +1 0.7071
Vo = 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
3
[ ]
1 − e − t cos 0.7071t − 1.414e − t sin 0.7071t u ( t ) V
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