Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 14, Problem 6(6).
Obtain the transfer function H(ω) = Io/Is of the circuits shown in Fig. 14.70.
Figure 14.70
Chapter 14, Solution 6(6).
Using current division,
Io
R
H (ω) =
=
I i R + jωL + 1 jωC
H (ω) =
jω (20)(0.25)
jωRC
=
2
1 + jωRC − ω LC 1 + jω(20)(0.25) − ω2 (10)(0.25)
H (ω) =
jω5
1 + jω5 − 2.5 ω 2
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
We apply nodal analysis to the circuit below.
Io
Vx
0.5 Vx
+ -
R
Is
Is =
1/jwC
jwL
Vx Vx − 0.5Vx
+
R jωL + 1 jωC
Io =
But
0.5 Vx
jωL + 1 jωC

→ Vx = 2 I o ( jωL + 1 jωC)
Is
1
0 .5
= +
Vx R jωL + 1 jωC
Is
1
1
= +
2 I o ( jωL + 1 jωC) R 2 ( jωL + 1 jωC)
I s 2 ( jωL + 1 jωC)
=
+1
Io
R
Io
1
jωRC
=
=
I s 1 + 2 ( jωL + 1 jωC) R jωRC + 2 (1 − ω 2 LC)
jω
H (ω) =
jω + 2 (1 − ω2 0.25)
H (ω) =
H (ω) =
jω
2 + jω − 0.5 ω 2
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 14, Problem 29.
For the circuit in Fig. 14.74, find the frequency ω for which v(t) and i(t) are in
phase.
Figure 14.74
Chapter 14, Solution 29.
jω
1/jω
Z
1
Z = jω +
jω
1
jω
+
j ω 1 + jω

1  ω 2 + jω
Z = j ω −  +

ω  1 + ω2
Since v( t ) and i( t ) are in phase,
1
ω
Im(Z) = 0 = ω − +
ω 1 + ω2
ω4 + ω2 − 1 = 0
ω2 =
-1 ± 1+ 4
= 0.618
2
ω = 0.7861 rad / s
Copyright ©2004 The McGraw-Hill Companies Inc.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 14, Problem 40.
For the circuit shown in Fig. 14.77: (a) Calculate the resonant frequency ωo, the
quality factor Q, and the bandwidth B. (b) What value of capacitance must be
connected in series with the 20-µF capacitor in order to double the bandwidth?
Figure 14.77
(b)
(c)
Chapter 14, Solution 40.
(a)
L = 5 + 10 = 15 mH
ω0 =
1
LC
1
=
15x10
−3
x 20x10
−6
= 1.8257 k rad/sec
Q = ω0 RC = 1.8257 x10 3 x 25x10 3 x 20x10 −6 = 912.8
1
1
=
= 2 rad
RC 25x10 3 20x10 −6
B=
(b)
To increase B by 100% means that B’ = 4.
C′ =
Since C′ =
1
1
=
= 10 µF
RB′ 25x10 3 x 4
C1C 2
= 10µF and C1 = 20 µF, we then obtain C2 = 20 µF.
C1 + C 2
Therefore, to increase the bandwidth, we merely add another 20 µF in
series with the first one.
Copyright ©2004 The McGraw-Hill Companies Inc.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 14, Problem 42.
Calculate the resonant frequency of each of the circuits in Fig. 14.79.
Figure 14.79
Chapter 14, Solution 42.
Z in = (1 jωC) || (R + jωL)
Z in =
Z in =
R + jωL
jωC
R + jωL +
1
jωC
=
R + jωL
1 − ω2 LC + jωRC
(R + jωL)(1 − ω2 LC − jωRC)
(1 − ω2 LC) 2 + ω2 R 2 C 2
At resonance, Im(Z in ) = 0 , i.e.
0 = ωL(1 − ω2 LC) − ωR 2 C
ω2 LC = L − R 2 C
ω0 =
L − R 2C
=
LC
1 R2
−
C L
Copyright ©2004 The McGraw-Hill Companies Inc.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 14, Problem 45.
For the network illustrated in Fig. 14.82, find
(a) the transfer function H(ω) = Vo(ω)/I(ω),
(b) the magnitude of H at ω0 = 1 rad/s.
Figure 14.82
(b)
(c)
Chapter 14, Solution 45.
jω
,
1 + jω
1 || jω =
1 ||
1 jω
1
1
=
=
jω 1 + 1 jω 1 + jω
Transform the current source gives the circuit below.
jω
I
1 + jω
+
-
jω
1 + jω
1
1
1 + jω
+
Vo
-
1
jω
1 + jω
Vo =
I
⋅
1
jω 1 + j ω
1+
+
1 + jω 1 + jω
H (ω) =
Vo
jω
=
I
2 (1 + jω) 2
H (1) =
1
2 (1 + j) 2
H (1) =
1
2 ( 2)2
= 0.25
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