Chapter 14-Electrical Circuit
Analysis-Problem Solutions
Electrical Circuit Analysis
Punjab Engineering College
13 pag.
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Chapter 14, Problem 1.
Find the transfer function V o /V i of the RC circuit in Fig. 14.68. Express it using ω o =
1/RC.
Figure 14.68
For Prob. 14.1.
Chapter 14, Solution 1.
Vo
R
jωRC
=
=
H (ω) =
Vi R + 1 jωC 1 + jωRC
jω ω0
1
,
where ω0 =
H (ω) =
1 + jω ω0
RC
H = H (ω) =
ω ω0
1 + (ω ω0 ) 2
φ = ∠H (ω) =
⎛ω⎞
π
− tan -1 ⎜ ⎟
2
⎝ ω0 ⎠
This is a highpass filter. The frequency response is the same as that for P.P.14.1 except
that ω0 = 1 RC . Thus, the sketches of H and φ are shown below.
H
1
0.7071
0
ω0 = 1/RC
ω
φ
90°
45°
0
ω0 = 1/RC
ω
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Chapter 14, Problem 2.
Obtain the transfer function V o (s)/V i of the circuit in Fig. 14.69.
Figure 14.69
For Prob. 14.2.
Chapter 14, Solution 2.
V
H(s) = o =
Vi
2+
1
s/8
10 + 20 +
1
s/8
=
2 + 8/s 1 s + 4
=
12 + 8 / s 6 s + 0.6667
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Chapter 14, Problem 3.
For the circuit shown in Fig. 14.70, find H(s) = V o /V i (s).
Figure 14.70
For Prob. 14.3.
Chapter 14, Solution 3.
1
1
5
=
=
jωC s (0.2) s
1
10
0.1F
⎯⎯
→
=
s (0.1) s
The circuit becomes that shown below.
0.2 F
⎯⎯
→
2
V1
5
s
+
Vi
+
_
10
s
5
Vo
_
10
5
10 1 + s
(5 + )
5(
)
10
5
10( s + 1)
s
s
s
s
=
=
Let Z = //(5 + ) =
15
5
s
s
s ( s + 3)
5+
(3 + s )
s
s
Z
V1 =
Vi
Z +2
5
s
s
Z
Vo =
V1 =
V1 =
•
Vi
5+5/ s
s +1
s +1 Z + 2
10( s + 1)
s
10s
5s
s ( s + 3)
V
H (s) = o =
•
=
= 2
Vi s + 1
10( s + 1) 2s ( s + 3) + 10( s + 1) s + 8s + 5
2+
s ( s + 3)
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Chapter 14, Problem 4.
Find the transfer function H( ω ) = V O /V i of the circuits shown in Fig. 14.71.
Figure 14.71
For Prob. 14.4.
Chapter 14, Solution 4.
(a)
R ||
1
R
=
jωC 1 + jωRC
R
Vo
R
1 + jωRC
=
=
H (ω) =
R
Vi
R + jωL (1 + jωRC)
jωL +
1 + jωRC
(b)
H (ω) =
R
- ω RLC + R + jωL
H (ω) =
jωC (R + jωL)
R + jωL
=
R + jωL + 1 jωC 1 + jωC (R + jωL)
H (ω) =
- ω 2 LC + jωRC
1 − ω 2 LC + jωRC
2
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Chapter 14, Problem 5.
For each of the circuits shown in Fig. 14.72, find H(s) = V o (s)/V s (s).
Figure 14.72
For Prob. 14.5.
Chapter 14, Solution 5.
(a) Let Z = R // sL =
Vo =
sRL
R + sL
Z
Vs
Z + Rs
sRL
V
Z
sRL
H (s) = o =
= R + sL =
sRL
Vs Z + Rs R +
RRs + s ( R + Rs ) L
s
R + sL
1
Rx
1
sC = R
(b) Let Z = R //
=
sC R + 1 1 + sRC
sC
Z
Vo =
Vs
Z + sL
V
Z
H(s) = o =
=
Vi Z + sL
R
R
1 + sRC =
2
R
s LRC + sL + R
sL +
1 + sRC
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Chapter 14, Problem 6.
For the circuit shown in Fig. 14.73, find H(s) = I o (s)/I s (s).
Figure 14.73
For Prob. 14.6.
Chapter 14, Solution 6.
1H
⎯⎯
→
Let Z = s //1 =
jω L = sL = s
s
s +1
We convert the current source to a voltage source as shown below.
1
Is ⋅ 1
S
+
+
_
Vo
Z
_
s
sI s
sI
Z
s +1 I =
Vo =
= 2 s
( I s x1) =
s
2
s
( s + 1) + s s + 3s + 1
Z + s +1
s +1+
s +1
Vo
sI s
Io = = 2
1 s + 3s + 1
I
s
H (s) = o = 2
I s s + 3s + 1
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Chapter 14, Problem 7.
Calculate H (ω ) if H dB equals
(a) 0.05dB
(b) -6.2 dB
(c) 104.7 dB
Chapter 14, Solution 7.
(a)
0.05 = 20 log10 H
2.5 × 10 -3 = log10 H
H = 10 2.5×10 = 1.005773
-3
(b)
- 6.2 = 20 log10 H
- 0.31 = log10 H
H = 10 -0.31 = 0.4898
(c)
104.7 = 20 log10 H
5.235 = log10 H
H = 10 5.235 = 1.718 × 10 5
Chapter 14, Problem 8.
Determine the magnitude (in dB) and the phase (in degrees) of H( ω ) = at ω = 1 if
H (ω ) equals
(a) 0.05 dB
(b) 125
(c)
Chapter 14, Solution 8.
(a)
H = 0.05
H dB = 20 log10 0.05 = - 26.02 ,
(b)
(c)
H = 125
H dB = 20 log10 125 = 41.94 ,
(d)
3
6
+
2 + jω
1 + jω
φ = 0°
φ = 0°
H(1) =
H dB
(d)
j10
= 4.472∠63.43°
2+ j
= 20 log10 4.472 = 13.01 ,
10 jω
2 + jω
H(1) =
φ = 63.43°
3
6
+
= 3.9 − j2.7 = 4.743∠ - 34.7°
1+ j 2 + j
=
,
φ = –34.7˚
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Chapter 14, Problem 9.
A ladder network has a voltage gain of
H( ω ) =
10
(1 + jω )(10 + jω )
Sketch the Bode plots for the gain.
Chapter 14, Solution 9.
1
(1 + jω)(1 + jω 10)
H (ω) =
H dB = -20 log10 1 + jω − 20 log10 1 + jω / 10
φ = - tan -1 (ω) − tan -1 (ω / 10)
The magnitude and phase plots are shown below.
HdB
0.1
1
10
ω
100
20 log 10
-20
1
1 + jω / 10
20 log10
-40
1
1 + jω
φ
0.1
-45°
1
10
ω
100
arg
1
1 + jω / 10
-90°
arg
-135°
1
1 + jω
-180°
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Chapter 14, Problem 10.
Sketch the Bode magnitude and phase plots of:
H(j ω ) =
50
jω (5 + jω )
Chapter 14, Solution 10.
H( jω) =
50
=
jω(5 + jω)
10
jω ⎞
⎛
1 jω⎜1 + ⎟
5 ⎠
⎝
HdB
40
20 log1
20
10
0.1
-20
1
100
⎛
⎜
1
20 log⎜
⎜
jω
⎜ 1+
5
⎝
⎛ 1 ⎞
⎟
20 log⎜⎜
⎟
⎝ jω ⎠
-40
φ
0.1
-45°
⎞
⎟
⎟
⎟
⎟
⎠
ω
1
10
ω
100
arg
1
1 + jω / 5
-90°
arg
-135°
1
jω
-180°
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ω)=
10 + jω
jω ( 2 + jω )
Chapter 14, Solution 11.
5 (1 + jω 10)
H (ω) =
jω (1 + jω 2)
H dB = 20 log10 5 + 20 log10 1 + jω 10 − 20 log10 jω − 20 log10 1 + jω 2
φ = -90° + tan -1 ω 10 − tan -1 ω 2
The magnitude and phase plots are shown below.
HdB
40
34
20
14
0.1
-20
1
10
100
ω
1
10
100
ω
-40
φ
90°
45°
0.1
-45°
-90°
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s +1
s ( s + 10)
Sketch the magnitude and phase Bode plots.
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Chapter 14, Solution 12.
T ( w) =
0.1(1 + jω )
,
jω (1 + jω / 10)
20 log 0.1 = −20
The plots are shown below.
|T|
(db)
20
ω
0
0.1
1
10
100
1
10
100
-20
-40
arg T
90o
ω
0
0.1
-90o
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Chapter 14, Problem 13.
Construct the Bode plots for
G(s) =
s +1
,
s ( s + 10)
s=j ω
2
Chapter 14, Solution 13.
G (ω) =
(1 10)(1 + jω)
1 + jω
=
2
( jω) (10 + jω) ( jω) 2 (1 + jω 10)
G dB = -20 + 20 log10 1 + jω − 40 log10 jω − 20 log10 1 + jω 10
φ = -180° + tan -1ω − tan -1 ω 10
The magnitude and phase plots are shown below.
GdB
40
20
0.1
-20
1
10
100
ω
1
10
100
ω
-40
φ
90°
0.1
-90°
-180°
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