PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Find the transfer function V
o/V
iof 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.
RC
j
1
RC
j
C
j
1
R
R
)
(
i oω
+
ω
=
ω
+
=
=
ω
V
V
H
=
ω
)
(
H
0 0j
1
j
ω
ω
+
ω
ω
,
where
RC
1
0=
ω
2 0 0)
(
1
)
(
H
ω
ω
+
ω
ω
=
ω
=
H
⎟
⎠
⎞
⎜
⎝
⎛
ω
ω
−
π
=
ω
∠
=
φ
0 1-tan
2
)
(
H
This is a highpass filter. The frequency response is the same as that for P.P.14.1 except
that
ω
0=
1
RC
. Thus, the sketches of H and
φ
are shown below.
0
90
°
45
°
ω
0= 1/RC
ω
φ
0
1
0.7071
ω
0= 1/RC
ω
H
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Obtain the transfer function V
o(
s
)/V
iof the circuit in Fig. 14.69.
Figure 14.69
For Prob. 14.2.
Chapter 14, Solution 2.
6667
.
0
s
4
s
6
1
s
/
8
12
s
/
8
2
8
/
s
1
20
10
8
/
s
1
2
V
V
)
s
(
H
i o+
+
=
+
+
=
+
+
+
=
=
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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
0.2
(0.2)
F
j C
ω
s
s
⎯⎯
→
=
=
1
10
0.1
(0.1)
F
s
s
⎯⎯
→
=
The circuit becomes that shown below.
Let
10
5
10 1
(5
)
5(
)
10
5
10(
1)
//(5
)
15
5
(
3)
5
(3
)
s
s
s
s
s
s
Z
s
s
s
s s
s
s
+
+
+
=
+
=
=
=
+
+
+
12
iZ
V
V
Z
=
+
1 15
5 5 /
1
1
2
o is
s
Z
V
V
V
V
s
s
s
Z
=
=
=
•
+
+
+
+
210(
1)
10
5
(
3)
( )
10(
1)
1
2
2 (
3) 10(
1)
8
5
(
3)
o is
s
s s
s
s
V
H s
V
s
s
s s
s
s
s
s s
+
+
=
=
•
=
=
+
+
+
+ +
+
+
+
+
2
+
_
10
5
s
V
i+
_
V
oV
15
s
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Find the transfer function H(
ω
) = V
O/V
iof the circuits shown in Fig. 14.71.
Figure 14.71
For Prob. 14.4.
Chapter 14, Solution 4.
(a)
RC
j
1
R
C
j
1
||
R
ω
+
=
ω
)
RC
j
1
(
L
j
R
R
RC
j
1
R
L
j
RC
j
1
R
)
(
i oω
+
ω
+
=
ω
+
+
ω
ω
+
=
=
ω
V
V
H
=
ω
)
(
H
L
j
R
RLC
-R
2+
+
ω
ω
(b)
)
L
j
R
(
C
j
1
)
L
j
R
(
C
j
C
j
1
L
j
R
L
j
R
)
(
ω
+
ω
+
ω
+
ω
=
ω
+
ω
+
ω
+
=
ω
H
=
ω
)
(
H
RC
j
LC
1
RC
j
LC
-2 2
ω
+
ω
−
ω
+
ω
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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
sRL
R
sL
=
=
+
o s s
Z
V
V
Z
R
=
+
( )
(
)
o s s s s ssRL
V
Z
R
sL
sRL
H s
sRL
V
Z
R
RR
s R
R L
R
R
sL
+
=
=
=
=
+
+
+
+
+
(b)
Let
1
1
//
1
1
Rx
R
sC
Z
R
sC
R
sRC
sC
=
=
=
+
+
o sZ
V
V
Z
sL
=
+
R
sL
LRC
s
R
sRC
1
R
sL
sRC
1
R
sL
Z
Z
V
V
)
s
(
H
2 i o+
+
=
+
+
+
=
+
=
=
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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.
1
H
⎯⎯
→
j L
ω
=
sL
=
s
Let
//1
1
s
Z
s
s
=
=
+
We convert the current source to a voltage source as shown below.
(
1)
1
2 21
1
(
1)
3
1
1
s s o s ss
sI
sI
Z
s
V
I x
I
s
Z
s
s
s
s
s
s
s
+
=
=
=
=
+ +
+ +
+
+
+
+
+
21
3
1
o s oV
sI
I
s
s
=
=
+
+
2( )
3
1
o sI
s
H s
I
s
s
=
=
+
+
+
_
V
oI
s⋅
1
Z
1
S
+
_
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Calculate
H
(
ω
)
if
H
dBequals
(a) 0.05dB
(b) -6.2 dB
(c) 104.7 dB
Chapter 14, Solution 7.
(a)
0
.
05
=
20
log
10H
H
log
10
5
.
2
×
-3=
10=
=
2.5×10-310
H
1
.
005773
(b)
-
6.2
=
20
log
10H
H
log
0.31
-
=
10=
=
10
-0.31H
0
.
4898
(c)
104
.
7
=
20
log
10H
H
log
235
.
5
=
10=
=
10
5.235H
1
.
718
×
10
5Chapter 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)
ω
ω
j
j
+
2
10
(d)
ω
j
+
1
3
+
ω
j
+
2
6
Chapter 14, Solution 8.
(a)
H
=
0
.
05
=
=
20
log
0
.
05
H
dB 10-
26.02
,
φ
=
0
°
(b)
H
=
125
=
=
20
log
125
H
dB 1041.94
,
φ
=
0
°
(c)
=
∠
°
+
=
4
.
472
63
.
43
j
2
10
j
)
1
(
H
=
=
20
log
4
.
472
H
dB 1013
.
01
,
φ
=
63
.
43
°
(d)
=
−
=
∠
°
+
+
+
=
3
.
9
j
2
.
7
4
.
743
-
34.7
j
2
6
j
1
3
)
1
(
H
=
=
20
log
4
.
743
H
dB 1013.521,
φ
=
–34.7
˚
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A ladder network has a voltage gain of
H(
ω
) =
)
10
)(
1
(
10
ω
ω
j
j
+
+
Sketch the Bode plots for the gain.
Chapter 14, Solution 9.
)
10
j
1
)(
j
1
(
1
)
(
ω
+
ω
+
=
ω
H
10
/
j
1
log
20
j
1
log
20
-H
dB=
10+
ω
−
10+
ω
)
10
/
(
tan
)
(
tan
-
-1ω
−
-1ω
=
φ
The magnitude and phase plots are shown below.
H
dB0.1
-40
ω
100
1
10
-20
1 j /10 1 log 20 10 + ω ω +j 1 1 log 20 10φ
-135
°
-45
°
ω
100
1
10
0.1
-180
°
-90
°
10 / j 1 1 arg ω + ω +j 1 1 argPROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Sketch the Bode magnitude and phase plots of:
H(
j
ω
) =
)
5
(
50
ω
ω
j
j
+
Chapter 14, Solution 10.
⎟
⎠
⎞
⎜
⎝
⎛
+
ω
ω
=
ω
+
ω
=
ω
5
j
1
j
1
10
)
j
5
(
j
50
)
j
(
H
φ
-135
°
-45
°
ω
100
1
10
0.1
-180
°
-90
°
5 / j 1 1 arg ω + ω j 1 argH
dB-20
20
ω
100
1
10
0.1
-40
40
20 log1
⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ω j 1 log 20 ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ω + 5 j 1 1 log 20PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Sketch the Bode plots for
H(
ω
) =
)
2
(
10
ω
ω
ω
j
j
j
+
+
Chapter 14, Solution 11.
)
2
j
1
(
j
)
10
j
1
(
5
)
(
ω
+
ω
ω
+
=
ω
H
2
j
1
log
20
j
log
20
10
j
1
log
20
5
log
20
H
dB=
10+
10+
ω
−
10ω
−
10+
ω
2
tan
10
tan
90
-
°
+
-1ω
−
-1ω
=
φ
The magnitude and phase plots are shown below.
φ
-45
°
45
°
ω
100
1
10
0.1
-90
°
90
°
H
dB-20
20
ω
100
1
10
0.1
-40
40
34
14
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
A transfer function is given by
T
(
s
) =
)
10
(
1
+
+
s
s
s
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
20
1
.
0
log
20
,
)
10
/
1
(
)
1
(
1
.
0
)
(
=
−
+
+
=
ω
ω
ω
j
j
j
w
T
The plots are shown below.
|T| (db)
20
0
ω
0.1
1 10 100
-20
-40
arg T
90
o0
ω
0.1
1 10 100
-90
oPROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Construct the Bode plots for
G
(
s
) =
)
10
(
1
2+
+
s
s
s
,
s
=
j
ω
Chapter 14, Solution 13.
)
10
j
1
(
)
j
(
)
j
1
)(
10
1
(
)
j
10
(
)
j
(
j
1
)
(
2 2ω
+
ω
ω
+
=
ω
+
ω
ω
+
=
ω
G
10
j
1
log
20
j
log
40
j
1
log
20
20
-G
dB=
+
10+
ω
−
10ω
−
10+
ω
10
tan
tan
-180
°
+
-1ω
−
-1ω
=
φ
The magnitude and phase plots are shown below.
φ
-90
°
90
°
ω
100
1
10
0.1
-180
°
G
dB-20
20
ω
100
1
10
0.1
-40
40
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Draw the Bode plots for
H(
ω
) =
)
25
10
(
)
1
(
50
2+
+
−
+
ω
ω
ω
ω
j
j
j
Chapter 14, Solution 14.
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛ ω
+
ω
+
ω
ω
+
=
ω
25
j
25
10
j
1
j
j
1
25
50
)
(
H
ω
−
ω
+
+
=
20
log
2
20
log
1
j
20
log
j
H
dB 10 10 10 2 101
j
2
5
(
j
5
)
log
20
+
ω
+
ω
−
⎟
⎠
⎞
⎜
⎝
⎛
ω
−
ω
−
ω
+
°
=
φ
5
1
25
10
tan
tan
90
-
-1 -1 2The magnitude and phase plots are shown below.
φ
-90
°
90
°
ω
100
1
10
0.1
-180
°
H
dB-20
20
ω
100
1
10
0.1
-40
40
6
26
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Construct the Bode magnitude and phase plots for
H
(
s
) =
)
10
)(
2
(
)
1
(
40
+
+
+
s
s
s
,
s=j
ω
Chapter 14, Solution 15.
)
10
j
1
)(
2
j
1
(
)
j
1
(
2
)
j
10
)(
j
2
(
)
j
1
(
40
)
(
ω
+
ω
+
ω
+
=
ω
+
ω
+
ω
+
=
ω
H
10
j
1
log
20
2
j
1
log
20
j
1
log
20
2
log
20
H
dB=
10+
10+
ω
−
10+
ω
−
10+
ω
10
tan
2
tan
tan
-1ω
−
-1ω
−
-1ω
=
φ
The magnitude and phase plots are shown below.
φ
-45
°
45
°
ω
100
1
10
0.1
-90
°
90
°
H
dB-20
20
ω
100
1
10
0.1
-40
40
6
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Sketch Bode magnitude and phase plots for
H
(
s
) =
)
16
(
10
2+
+
s
s
s
,
s=j
ω
Chapter 14, Solution 16.
2 210 /16
0.625
( )
1
1
4
4
H
j
j
j
j
j
j
ω
ω
ω
ω
ω
ω
ω
=
=
⎡
⎛
⎞
⎤
⎡
⎛
⎞
⎤
+
+
+
+
⎢
⎜
⎝
⎟
⎠
⎥
⎢
⎜
⎝
⎟
⎠
⎥
⎢
⎥
⎢
⎥
⎣
⎦
⎣
⎦
220log 0.625 20log |
| 20log |1
|
4
dBj
H
=
−
j
ω
−
+
j
ω
+ ⎜
⎛
ω
⎞
⎟
⎝
⎠
(20log0,625= –4.082)
The magnitude and phase plots are shown below.
H
φ
1
0.1
4
100
ω
20 log (
j
ω
)
20 log
21
4
j
j
ω
⎛
ω
⎞
+
+ ⎜
⎟
⎝
⎠
20
10
–20
–60
40
ω
0.4
1
4
10
40
100
-90
-180
90
°
-tan
-1 2 1 1 6ω
ω
−–40
–4.082
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Sketch the Bode plots for
G
(
s
) =
)
1
(
)
2
(
+
2+
+
s
s
s
,
s=j
ω
Chapter 14, Solution 17.
2)
2
j
1
)(
j
1
(
j
)
4
1
(
)
(
ω
+
ω
+
ω
=
ω
G
2
j
1
log
40
j
1
log
20
j
log
20
4
-20log
G
dB=
10+
10ω
−
10+
ω
−
10+
ω
2
tan
2
tan
--90
°
-1ω
−
-1ω
=
φ
The magnitude and phase plots are shown below.
φ
-90
°
90
°
ω
100
1
10
0.1
-180
°
G
dB-20
20
ω
100
1
10
0.1
-40
-12
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
A linear network has this transfer function
H
(
s
) =
)
5
14
8
(
4
7
2 3 2+
+
+
+
+
s
s
s
s
s
,
s=j
ω
Use
MATLAB
or equivalent to plot the magnitude and phase (in degrees) of the transfer
function. Take 0.1 <
ω
< 10 rads/s.
Chapter 14, Solution 18.
The MATLAB code is shown below.
>> w=logspace(-1,1,200);
>> s=i*w;
>> h=(7*s.^2+s+4)./(s.^3+8*s.^2+14*s+5);
>> Phase=unwrap(angle(h))*57.23;
>> semilogx(w,Phase)
>> grid on
1 0- 1 1 00 1 01 - 6 0 - 4 0 - 2 0 0 2 0 4 0 6 0 w H( jw ) P h a s ePROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
>> H=abs(h);
>> HdB=20*log10(H);
>> semilogx(w,HdB);
>> grid on
1 0- 1 1 00 1 01 - 2 5 - 2 0 - 1 5 - 1 0 - 5 0 w Hd BPROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Sketch the asymptotic Bode plots of the magnitude and phase for
H
(
s
) =
)
40
)(
20
)(
10
(
100
+
+
+
s
s
s
s
,
s=j
ω
Chapter 14, Solution 19.
100
/ 80
( )
(
10)(
20)(
40)
(1
)(1
)(1
)
10
20
40
j
j
H
j
j
j
j
j
j
ω
ω
ω
ω
ω
ω
ω
ω
ω
=
=
+
+
+
+
+
+
20log(1/ 80) 20log |
/1| 20log |1
| 20log |1
| 20log |1
|
10
20
40
dB
j
j
j
H
=
+
j
ω
−
+
ω
−
+
ω
−
+
ω
(20log(1/80) = -38.06)
The magnitude and phase plots are shown below.
φ
-90
1
0.1
H
10
20
40
100
ω
20
40
0.1
60
90
20 log
1
80
20 log 1
10
j
ω
+
20 log
j
ω
1
2
4
10
20
40
100
200
400
ω
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Sketch the magnitude Bode plot for the transfer function
H
(
ω
) =
)
40
(
)
5
)(
1
(
10
2+
+
+
ω
ω
ω
ω
j
j
j
j
Chapter 14, Solution 20.
2 210
/100
( )
(25)(40)(1
)(1
/ 5) (1
/ 40)
(1
)(1
/ 5) (1
/ 40)
j
j
H
j
j
j
j
j
j
ω
ω
ω
ω
ω
ω
ω
ω
ω
=
=
+
+
+
+
+
+
20log(1/100) = -40
The magnitude plot is shown below.
1
0.1
5
10
50
100
ω
40
20 log
1
100
20 log
1
21
5
j
ω
⎛
+
⎞
⎜
⎟
⎝
⎠
20 log
j
ω
20
-40
-20
-60
20 log
1
1
+
j
ω
20 log
1 1 1 0 jω
+PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Sketch the magnitude Bode plot for
H
(
s
) =
)
400
(
)
60
)(
1
(
)
20
(
2+
=
+
+
s
s
s
s
s
,
s=j
ω
Chapter 14, Solution 21.
2 2(
20)
20
(1
/ 20)
( )
(
1)(
60
400)
400(
1)(1 60
/ 400
)
20
j
j
j
j
H
j
j
j
j
j
ω ω
ω
ω
ω
ω
ω
ω
ω
ω
ω
+
+
=
=
+ −
+
+
⎛
⎞
+
+
+ ⎜
⎟
⎝
⎠
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛ ω
+
ω
+
ω
+
ω
+
ω
=
ω
220
j
40
j
6
1
)
j
1
(
)
20
/
j
1
(
j
05
.
0
)
(
H
2dB
20
log(
0
.
05
)
20
log
j
20
log
1
20
j
20
log
1
j
20
log
1
j
40
6
20
j
H
⎟
⎠
⎞
⎜
⎝
⎛ ω
+
ω
+
−
ω
+
−
ω
+
+
ω
+
=
The magnitude plot is as sketched below.
1
0.1
H&B
10
100
ω
20
20 log 0.05
20 log |1+j
ω
/20|
–20 log
1
+
j
ω
40
–40
–60
–80
–20 log
26
1
40
20
j
ω
⎛
j
ω
⎞
+
+ ⎜
⎟
⎝
⎠
20
–20
20log|j
ω
|
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Find the transfer function H(
ω
) with the Bode magnitude plot shown in Fig. 14.74.
Figure 14.74
For Prob. 14.22.
Chapter 14, Solution 22.
10
k
k
log
20
20
=
10⎯
⎯→
=
A zero of slope
+
20
dB
/
dec
at
ω
=
2
⎯
⎯→
1
+
j
ω
2
A pole of slope
-
20
dB
/
dec
at
20
j
1
1
20
ω
+
⎯→
⎯
=
ω
A pole of slope
-
20
dB
/
dec
at
100
j
1
1
100
ω
+
⎯→
⎯
=
ω
Hence,
)
100
j
1
)(
20
j
1
(
)
2
j
1
(
10
)
(
ω
+
ω
+
ω
+
=
ω
H
=
ω
)
(
H
)
j
100
)(
j
20
(
)
j
2
(
10
4ω
+
ω
+
ω
+
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
The Bode magnitude plot of H(
ω
) is shown in Fig. 14.75. Find H(
ω
).
Figure 14.75
For Prob. 14.23.
Chapter 14, Solution 23.
A zero of slope
+
20
dB
/
dec
at the origin
⎯
⎯→
j
ω
A pole of slope
-
20
dB
/
dec
at
1
j
1
1
1
ω
+
⎯→
⎯
=
ω
A pole of slope
-
40
dB
/
dec
at
2)
10
j
1
(
1
10
ω
+
⎯→
⎯
=
ω
Hence,
2)
10
j
1
)(
j
1
(
j
)
(
ω
+
ω
+
ω
=
ω
H
=
ω
)
(
H
2)
j
10
)(
j
1
(
j
100
ω
+
ω
+
ω
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
The magnitude plot in Fig. 14.76 represents the transfer function of a preamplifier. Find
H
(
s
).
Figure 14.76
For Prob. 14.24.
Chapter 14, Solution 24.
1040 20log
=
K
⎯⎯
→
K
=
100
There is a pole at
ω
=50 giving 1/(1+j
ω
/50)
There is a zero at
ω
=500 giving (1 + j
ω
/500).
There is another pole at
ω
=2122 giving 1/(1 + j
ω
/2122).
Thus,
1
40
(
500)
40(1
/ 500)
500
( )
1
1
(1
/ 50)(1
/ 2122)
(
50)(
2122)
50 2122
x
s
j
H
j
j
x
s
s
ω
ω
ω
ω
+
+
=
=
+
+
+
+
or
8488(
500)
( )
(
50)(
2122)
s
H s
s
s
+
=
+
+
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A series
RLC
network has
R
= 2 k
Ω
,
L
= 40 mH, and
C
= 1
µ
F. Calculate the
impedance at resonance and at one-fourth, one-half, twice, and four times the resonant
frequency.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
s
/
krad
5
)
10
1
)(
10
40
(
1
LC
1
6 -3 -0=
×
×
=
=
ω
=
=
ω
)
R
(
0Z
2
k
Ω
⎟
⎠
⎞
⎜
⎝
⎛
ω
−
ω
+
=
ω
C
4
L
4
j
R
)
4
(
0 0 0Z
⎟
⎠
⎞
⎜
⎝
⎛
×
×
−
×
⋅
×
+
=
ω
)
10
1
)(
10
5
(
4
10
40
4
10
5
j
2000
)
4
(
-3 3 -6 3 0Z
)
5
4000
50
(
j
2000
)
4
(
ω
0=
+
−
Z
=
ω
4
)
(
0Z
2
−
j
0
.
75
k
Ω
⎟
⎠
⎞
⎜
⎝
⎛
ω
−
ω
+
=
ω
C
2
L
2
j
R
)
2
(
0 0 0Z
⎟
⎠
⎞
⎜
⎝
⎛
×
×
−
×
×
+
=
ω
)
10
1
)(
10
5
(
2
)
10
40
(
2
)
10
5
(
j
2000
)
2
(
-3 3 -6 3 0Z
Z(
ω
0/2) = 200+j(100-2000/5)
=
ω
2
)
(
0Z
2
−
j
0
.
3
k
Ω
⎟
⎠
⎞
⎜
⎝
⎛
ω
−
ω
+
=
ω
C
2
1
L
2
j
R
)
2
(
0 0 0Z
⎟
⎠
⎞
⎜
⎝
⎛
×
×
−
×
×
+
=
ω
)
10
1
)(
10
5
)(
2
(
1
)
10
40
)(
10
5
)(
2
(
j
2000
)
2
(
3 -3 3 -6 0Z
=
ω
)
2
(
0Z
2
+
j
0
.
3
k
Ω
⎟
⎠
⎞
⎜
⎝
⎛
ω
−
ω
+
=
ω
C
4
1
L
4
j
R
)
4
(
0 0 0Z
⎟
⎠
⎞
⎜
⎝
⎛
×
×
−
×
×
+
=
ω
)
10
1
)(
10
5
)(
4
(
1
)
10
40
)(
10
5
)(
4
(
j
2000
)
4
(
3 -3 3 -6 0Z
=
ω
)
4
(
0Z
2
+
j
0
.
75
k
Ω
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
A coil with resistance 3
Ω
and inductance 100 mH is connected in series with a capacitor
of 50 pF, a resistor of 6
Ω
and a signal generator that gives 110 V rms at all frequencies.
Calculate
ω
o,
Q
, and
B
at resonance of the resultant series
RLC
circuit.
Chapter 14, Solution 26.
Consider the circuit as shown below. This is a series RLC resonant circuit.
R = 6 + 3 = 9
Ω
3 121
1
447.21 krad/s
100 10
50 10
oLC
x
x
x
ω
− −=
=
=
3 3447.21 10 100 10
4969
9
oL
x
x
x
Q
R
ω
=
=
=
3447.21 10
90 rad/s
4969
ox
B
Q
ω
=
=
=
+
_
100 mH
6
Ω
50
µ
F
3
Ω
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Design a series
RLC
resonant circuit with
ω
o= 40 rad/s and
B
= 10 rad/s.
Chapter 14, Solution 27.
21
1
40
40
oLC
LC
ω
=
=
⎯⎯
→
=
B
R
10
R
10
L
L
=
=
⎯⎯
→
=
If we select R =1
Ω
, then L = R/10 = 0.1 H and
2 2
1
1
6.25 mF
40
40 0.1
C
L
x
=
=
=
Chapter 14, Problem 28.
Design a series
RLC
circuit with
B
= 20 rad/s and
ω
o= 1,000 rad/s. Find the circuit’s
Q
.
Let
R
= 10
Ω
.
Chapter 14, Solution 28.
Let
R
=
10
Ω
.
H
5
.
0
20
10
B
R
L
=
=
=
F
2
)
5
.
0
(
)
1000
(
1
L
1
C
2 2 0µ
=
=
ω
=
50
20
1000
B
Q
=
ω
0=
=
Therefore, if
R
=
10
Ω
then
=
L
0
.
5
H
,
C
=
2
µ
F
,
Q
=
50
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Let
v
s= 20 cos(
at
) V in the circuit of Fig. 14.77. Find
ω
o,
Q
, and
B
, as seen by the
capacitor.
Figure 14.77
For Prob. 14.29.
Chapter 14, Solution 29.
We convert the voltage source to a current source as shown below.
20
cos
12
si
=
ω
t
, R = 12//45= 12x45/57 = 9.4737 k
Ω
3 61
1
4.082 krad/s
60 10
1 10
oLC
x
x x
ω
− −=
=
=
3 61
1
105.55 rad/s
9.4737 10 10
B
RC
x
x
−=
=
=
4082
38.674
105.55
oQ
B
ω
=
=
=
= 38.67
12 k
1
µ
F
60 mH
i
s45 k
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
A circuit consisting of a coil with inductance 10 mH and resistance 20
Ω
is connected in
series with a capacitor and a generator with an rms voltage of 120 V. Find:
(a) the value of the capacitance that will cause the circuit to be in resonance at 15 kHz
(b) the current through the coil at resonance
(c) the
Q
of the circuit
Chapter 14, Solution 30.
Select
R
=
10
Ω
.
mH
50
H
05
.
0
)
20
)(
10
(
10
Q
R
L
0=
=
=
ω
=
F
2
.
0
)
05
.
0
)(
100
(
1
L
1
C
2 0=
=
ω
=
s
/
rad
5
.
0
)
2
.
0
)(
10
(
1
RC
1
B
=
=
=
Therefore, if
R
=
10
Ω
then
=
L
50 mH,
C
=
0
.
2
F
,
B
=
0
.
5
rad
/
s
Chapter 14, Problem 31.
Design a parallel resonant
RLC
circuit with
ω
o= 10rad/s and Q = 20. Calculate the
bandwidth of the circuit. Let
R
= 10
Ω
.
Chapter 14, Solution 31.
ω
=
⎯→
⎯
ω
=
L LL
L
X
X
rad/s
10
x
796
.
8
10
x
40
10
x
6
.
5
x
10
x
10
x
2
X
R
L
R
B
6 3 3 6 L=
π
=
ω
=
=
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
A parallel
RLC
circuit has the following values:
R
= 60
Ω
,
L
= 1 mH, and
C
= 50
µ
F
.
Find the quality factor, the resonant frequency, and the bandwidth of the
RLC
circuit.
Chapter 14, Solution 32.
3 61
1
4.472 krad/s
10
50 10
oLC
x
x
ω
− −=
=
=
61
1
333.33 rad/s
60 50 10
B
RC
x
x
−=
=
=
4472
13.42
333.33
oQ
B
ω
=
=
=
Chapter 14, Problem 33.
A parallel resonant circuit with quality factor 120 has a resonant frequency of 6
×
10
6rad/s. Calculate the bandwidth and half-power frequencies.
Chapter 14, Solution 33.
pF
84
.
56
10
x
40
x
10
x
6
.
5
x
2
80
R
f
2
Q
C
RC
Q
6 3 o o=
π
=
π
=
⎯→
⎯
ω
=
80
x
10
x
6
.
5
x
2
10
x
40
Q
f
2
R
L
L
R
Q
36 o oπ
=
π
=
⎯→
⎯
ω
=
= 14.21 µH
Chapter 14, Problem 34.
A parallel
RLC
circuit is resonant at 5.6 MHz, has a
Q
of 80, and has a resistive branch of
40 k
Ω
. Determine the values of
L
and
C
in the other two branches.
Chapter 14, Solution 34.
(a)
1
.
443
krad/s
10
x
60
x
10
x
8
1
LC
1
6 3 o=
=
=
ω
− −(b)
3
.
33
rad/s
10
x
60
x
10
x
5
1
RC
1
B
=
=
3 −6=
(c)
Q
=
ω
oRC
=
1
.
443
x
10
3x
5
x
10
3x
60
x
10
−6=
432
.
9
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
A parallel
RLC
circuit has
R
= 5k
Ω
,
L
= 8 mH, and
C
= F
µ
. Determine:
(a) the resonant frequency
(b) the bandwidth
(c) the quality factor
Chapter 14, Solution 35.
At
resonance,
=
×
=
=
⎯→
⎯
=
-310
25
1
Y
1
R
R
1
Y
40
Ω
=
×
=
ω
=
⎯→
⎯
ω
=
)
40
)(
10
200
(
80
R
Q
C
RC
Q
3 0 010
µ
F
=
×
×
=
ω
=
⎯→
⎯
=
ω
)
10
10
)(
10
4
(
1
C
1
L
LC
1
6 -10 2 0 02
.
5
µ
H
=
×
=
ω
=
80
10
200
Q
B
3 0s
/
krad
5
.
2
=
−
=
−
ω
=
ω
200
1
.
25
2
B
0 1198.75 krad/s
=
+
=
+
ω
=
ω
200
1
.
25
2
B
0 2201.25 krad/s
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
It is expected that a parallel
RLC
resonant circuit has a midband admittance of 25
×
110
−3S, quality factor of 80, and a resonant frequency of 200 krad/s. Calculate the
values of
R
,
L
, and
C
. Find the bandwidth and the half-power frequencies.
Chapter 14, Solution 36.
s
/
rad
5000
LC
1
0=
=
ω
=
=
ω
⎯→
⎯
=
ω
(
)
R
R
1
)
(
0Z
0Y
2
k
Ω
mS
75
.
18
j
5
.
0
L
4
C
4
j
R
1
)
4
(
0 0 0⎟⎟
=
−
⎠
⎞
⎜⎜
⎝
⎛
ω
−
ω
+
=
ω
Y
=
−
=
ω
01875
.
0
j
0005
.
0
1
)
4
(
0Z
1
.
4212
+
j
53
.
3
Ω
mS
5
.
7
j
5
.
0
L
2
C
2
j
R
1
)
2
(
0 0 0⎟⎟
=
−
⎠
⎞
⎜⎜
⎝
⎛
ω
−
ω
+
=
ω
Y
=
−
=
ω
0075
.
0
j
0005
.
0
1
)
2
(
0Z
8
.
85
+
j
132
.
74
Ω
mS
5
.
7
j
5
.
0
C
2
1
L
2
j
R
1
)
2
(
0 0 0⎟⎟
=
+
⎠
⎞
⎜⎜
⎝
⎛
ω
−
ω
+
=
ω
Y
=
ω
)
2
(
0Z
8
.
85
−
j
132
.
74
Ω
mS
75
.
18
j
5
.
0
C
4
1
L
4
j
R
1
)
4
(
0 0 0⎟⎟
=
+
⎠
⎞
⎜⎜
⎝
⎛
ω
−
ω
+
=
ω
Y
=
ω
)
4
(
0Z
1
.
4212
−
j
53
.
3
Ω
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Rework Prob. 14.25 if the elements are connected in parallel.
Chapter 14, Solution 37.
2 2)
C
1
L
(
R
)
C
1
L
(
j
R
LR
j
C
L
L
j
C
j
1
R
)
C
j
1
R
(
L
j
)
C
j
1
R
//(
L
j
Z
ω
−
ω
+
⎟
⎠
⎞
⎜
⎝
⎛
ω
−
ω
−
⎟
⎠
⎞
⎜
⎝
⎛
+
ω
=
ω
+
ω
+
ω
+
ω
=
ω
+
ω
=
1
)
C
R
LC
(
0
)
C
1
L
(
R
C
1
L
C
L
LR
)
Z
Im(
2 2 2 2 2 2=
−
ω
⎯→
⎯
=
ω
−
ω
+
⎟
⎠
⎞
⎜
⎝
⎛
ω
−
ω
−
ω
=
Thus,
2 2C
R
LC
1
−
=
ω
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Find the resonant frequency of the circuit in Fig. 14.78.
Figure 14.78
For Prob. 14.38.
Chapter 14, Solution 38.
2 2 2L
R
L
j
R
C
j
C
j
L
j
R
1
Y
ω
+
ω
−
+
ω
=
ω
+
ω
+
=
At resonance,
Im(
Y
)
=
0
, i.e.
0
L
R
L
C
2 2 0 2 0 0+
ω
=
ω
−
ω
C
L
L
R
2 2 0 2+
ω
=
2 -3 6 --3 2 2 010
40
50
)
10
1
)(
10
40
(
1
L
R
LC
1
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
−
×
×
=
−
=
ω
=
ω
04841
rad
/
s
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
For the “tank” circuit in Fig. 14.79, find the resonant frequency.
Figure 14.79
For Probs. 14.39 and 14.91.
Chapter 14, Solution 39.
(a)
B
=
ω
2−
ω
1=
2
π
(
f
2−
f
1)
=
2
π
(
90
−
86
)
x
10
3=
8
π
krad
/
s
3 3 2 1 o=
2
1
(
ω
+
ω
)
=
2
π
(
88
)
x
10
=
176
π
X
10
ω
nF
89
.
19
10
x
2
x
10
x
8
1
BR
1
C
RC
1
B
3 3=
π
=
=
⎯→
⎯
=
(b)
9 2 3 o 2 o10
x
89
.
19
x
)
10
X
176
(
1
C
1
L
LC
1
−π
=
ω
=
⎯→
⎯
=
ω
= 164.45 µH
(c )
ω
o=
176
π
=
552
.
9
krad
/
s
(d)
B
=
8
π
=
25
.
13
krad
/
s
(e)
22
8
176
B
Q
o=
π
π
=
ω
=
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
A parallel resonance circuit has a resistance of 2 k
Ω
and half-power frequencies of 86
kHz and 90 kHz. Determine:
(a) the capacitance
(b) the inductance
(c) the resonant frequency
(d) the bandwidth
(e) the quality factor
Chapter 14, Solution 40.
(a)
L = 5 + 10 = 15 mH
=
=
=
ω
− −3 6 010
x
20
x
10
x
15
1
LC
1
1.8257 k rad/sec
=
=
ω
=
0RC
1
.
8257
x
10
3x
25
x
10
3x
20
x
10
−6Q
912.8
=
=
=
− 6 320
x
10
10
x
25
1
RC
1
B
2 rad/s
(b) To increase B by 100% means that B’ = 4.
=
=
′
=
′
4
x
10
x
25
1
B
R
1
C
310 µF
Since
10
F
C
C
C
C
C
2 1 2 1=
µ
+
=
′
and C
1= 20 µF, we then obtain C
2= 20 µF.
Therefore, to increase the bandwidth, we merely add another 20 µF in series
with the first one.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
For the circuit shown in Fig. 14.80, next page:
(a) Calculate the resonant frequency
ω
o, the quality factor
Q
, and the bandwidth
B
.
(b) What value of capacitance must be connected in series with the 20- F
µ
capacitor in
order to double the bandwidth?
Figure 14.80
For Prob. 14.41.
Chapter 14, Solution 41.
(a)
This is a series RLC circuit.
Ω
=
+
=
2
6
8
R
,
L
=
1
H
,
C
=
0
.
4
F
=
=
=
ω
4
.
0
1
LC
1
01
.
5811
rad
/
s
=
=
ω
=
8
5811
.
1
R
L
Q
00
.
1976
=
=
L
R
B
8
rad
/
s
(b)
This is a parallel RLC circuit.
F
2
6
3
)
6
)(
3
(
F
6
and
F
3
=
µ
+
⎯→
⎯
µ
µ
F
2
C
=
µ
,
R
=
2
k
Ω
,
L
=
20
mH
=
×
×
=
=
ω
)
10
20
)(
10
2
(
1
LC
1
3 -6 -05
krad
/
s
=
×
×
×
=
ω
=
)
10
20
)(
10
5
(
10
2
L
R
Q
3 -3 3 020
=
×
×
=
=
)
10
2
)(
10
2
(
1
RC
1
B
3 -6250 rad/s
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
For the circuits in Fig. 14.81, find the resonant frequency
ω
o, the quality factor
Q
, and
the bandwidth
B
.
Figure 14.81
For Prob. 14.42.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 14, Solution 42.
(a)
Z
in=
(
1
j
ω
C
)
||
(
R
+
j
ω
L
)
RC
j
LC
1
L
j
R
C
j
1
L
j
R
C
j
L
j
R
2