Obtain the transfer function Vos/Vi of the circuit in Fig... No part Find the transfer function H ω with the Bode magnitude plot shown in Fig... Calculate the impedance at resonance and
Trang 1PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
RC j C j 1 R
R )
(
i
o
ω +
ω
= ω +
=
= ω
V
V H
= ω) (
H
0
0
j 1
j
ω ω +
ω ω
RC
1
0 = ω
2 0
0
) ( 1 ) ( H
ω ω +
ω ω
= ω
ω
−
π
= ω
∠
= φ
0
1 -
tan 2 ) (
H
Trang 2Obtain the transfer function Vo(s)/Vi of 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 )
+
= + +
+
=
=
Trang 3PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
10( 1)
( 3)
o i
Trang 4Find the transfer function H( ω ) = VO/Vi of the circuits shown in Fig 14.71
R C
= ω
) RC j 1 ( L j R
R RC
j 1
R L
j
RC j 1
R )
(
i
o
ω + 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
Trang 5PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
//
Rx
R sC
R sL LRC s
R sRC
1
R sL
sRC 1 R sL
Z
Z V
V ) s ( H
2i
o
+ +
= +
+
+
= +
=
=
Trang 6For the circuit shown in Fig 14.73, find H(s) = Io(s)/Is(s)
Trang 7PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
5
10 (d)
ω
j
+ 1
3 +
ω
j
+ 2 6
2
10 j ) 1 ( H
= 3 9 j 2 7 4 743 - 34.7
j 2
6 j 1
3 ) 1 ( H
=
= 20 log 4 743
HdB 10 13.521, φ = –34.7˚
Trang 8A ladder network has a voltage gain of
H ( ω ) =
) 10 )(
j 1 (
1 )
(
ω + ω +
= ω
H
10 / j 1 log 20 j
1 log 20 -
) 10 / ( tan ) ( tan
- -1 ω − -1 ω
= φ
The magnitude and phase plots are shown below
HdB
0.1
-40
ω 100
ω+ j1
1log
1argω+
ω+ j11arg
Trang 9PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
Sketch the Bode magnitude and phase plots of:
H (j ω ) =
) 5 (
50 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/j1
1argω+
ωj
1arg
HdB
-20
20
ω 100
1
10 0.1
1 log 20
j 1
1 log 20
Trang 10Sketch 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
2 tan 10 tan 90
- ° + -1ω − -1ω
= φ
The magnitude and phase plots are shown below
φ
-45°
45°
ω 100
Trang 11PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
A transfer function is given by
T(s) =
) 10
Trang 1220 1 0 log 20 , ) 10 / 1
(
) 1
ω
j j
j w
-90o
Trang 13
PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
G(s) =
) 10 (
) j 1 )(
10 1 ( ) j 10 ( ) j (
j 1 )
ω + ω
ω +
= ω + ω
ω +
= ω
G
10 j 1 log 20 j
log 40 j
1 log 20 20 -
10 tan tan
-180 ° + -1ω − -1ω
= φ
The magnitude and phase plots are shown below
φ
-90°
90°
ω 100
0.1
-40
40
Trang 14Draw the Bode plots for
H( ω ) =
) 25 10
(
) 1 ( 50
−
+ ω ω
ω
ω
j j
ω + ω
ω +
=
5
j 25
10 j 1 j
j 1 25
50 ) (
H
ω
− ω + +
= 20 log 2 20 log 1 j 20 log j
2
10 1 j 2 5 ( j 5 ) log
−
ω
− ω +
°
= φ
5 1
25 10 tan tan
Trang 15PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
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
10 tan 2 tan tan-1ω − -1ω − -1ω
= φ
The magnitude and phase plots are shown below
φ
-45°
45°
ω 100
Trang 16Sketch Bode magnitude and phase plots for
H(s) =
) 16 (
−
–40
–4.082
Trang 17PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
Sketch the Bode plots for
G(s) =
) 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
2 tan 2 tan - -90 ° -1ω − -1ω
= φ
The magnitude and phase plots are shown below
φ
-90°
90°
ω 100
Trang 18A 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
Trang 19PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
Trang 20Sketch the asymptotic Bode plots of the magnitude and phase for
H(s) =
) 40 )(
20 )(
10
(
100
+ +
20 log j ω
ω
Trang 21PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
Sketch the magnitude Bode plot for the transfer function
H( ω ) =
) 40 (
) 5 )(
1 (
20 log 1
100
20 log 1 2
1 5
Trang 22Sketch the magnitude Bode plot for
H(s) =
) 400 ( ) 60 )(
1
(
) 20 (
+
+
s s
s
s s
Chapter 14, Solution 21
2 2
ω + ω +
ω + ω
j 6 1 ) j 1 (
) 20 / j 1 ( j 05 0 )
6 j 1 log 20 j 1 log 20 20
j 1 log 20 j log 20 ) 05 0 log(
ω +
− ω +
−
ω + +
ω +
=
The magnitude plot is as sketched below
1 0.1
H&B
ω 20
+ + ⎜ ⎝ ⎟ ⎠
20 –20
20log|jω|
Trang 23PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
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 (
104
ω + ω
+
ω +
Trang 24The 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) (
) j 10 )(
j 1 (
j 100
ω + ω +
ω
Trang 25PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
The magnitude plot in Fig 14.76 represents the transfer function of a preamplifier Find
There is a zero at ω=500 giving (1 + jω/500)
There is another pole at ω=2122 giving 1/(1 + jω/2122)
ω ω
+ +
or
8488( 500) ( )
Trang 26A 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
Trang 27PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
s / krad 5 ) 10 1 )(
10 40 (
= ω
C
4 L 4 j R ) 4 (
0
0 0
= ω
) 10 1 )(
10 5 (
4 10
40 4
10 5 j 2000 )
4
3 0
Z
) 5 4000 50
( j 2000 )
= ω
C
2 L 2 j R ) 2 (
0
0 0
= ω
) 10 1 )(
10 5 (
2 )
10 40 ( 2
) 10 5 ( j 2000 )
2
3 0
Z Z(ω0/2) = 200+j(100-2000/5)
=
ω 2 ) ( 0
= ω
C 2
1 L 2 j R ) 2 (
0 0
= ω
) 10 1 )(
10 5 )(
2 (
1 )
10 40 )(
10 5 )(
2 ( j 2000 )
= ω
C 4
1 L 4 j R ) 4 (
0 0
= ω
) 10 1 )(
10 5 )(
4 (
1 )
10 40 )(
10 5 )(
4 ( j 2000 )
Z 2 + j 0 75 k Ω
Trang 28A 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 ωo, Q, and B at resonance of the resultant series RLC circuit
Trang 29PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
Design a series RLC resonant circuit with ωo = 40 rad/s and B = 10 rad/s
10 B
R
F 2 ) 5 0 ( ) 1000 (
1 L
=
50 20
1000 B
Q = ω0 = =
Therefore, if R = 10 Ω then
=
L 0 5 H , C = 2 µ , F Q = 50
Trang 30Q B
ω
Trang 31PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
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 L0
=
=
= ω
=
F 2 0 ) 05 0 )(
100 (
1 L
=
s / rad 5 0 ) 2 0 )(
10 (
1 RC
Design a parallel resonant RLC circuit with ωo= 10rad/s and Q = 20 Calculate the
bandwidth of the circuit Let R = 10 Ω
L L
X
rad/s 10 x 796 8 10
x 40
10 x 5 10 x 10 x 2 X
Trang 32A parallel RLC circuit has the following values:
A parallel resonant circuit with quality factor 120 has a resonant frequency of 6 × 106
rad/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
=
⎯→
⎯ ω
=
80 x 10 x 6 5 x 2
10 x 40 Q
f 2
R L L
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
10 x 60 x 10 x
1 LC
1
63
1 RC
1
(c) Q = ωoRC = 1 443 x 103x x 103x 60 x 10−6 = 432 9
Trang 33PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
A parallel RLC circuit has R = 5k Ω , L = 8 mH, and C = F µ Determine:
(a) the resonant frequency
1 Y
1 R R
=
⎯→
⎯ ω
=
) 40 )(
10 200 (
80 R
Q C RC
=
⎯→
⎯
= ω
) 10 10 )(
10 4 (
1 C
1 L LC
1
6 - 10
2 0
B
3 0
s / krad 5 2
=
−
=
− ω
=
2
B0
= +
= + ω
=
2
B0
Trang 34It is expected that a parallel RLC resonant circuit has a midband admittance of 25 ×
110− 3 S, 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
mS 75 18 j 5 0 L
4 C 4
j R
1 ) 4 (
= ω
Y
=
−
= ω
01875 0 j 0005 0
1 )
4 ( 0
mS 5 7 j 5 0 L
2 C 2
j R
1 ) 2 (
= ω
Y
=
−
= ω
0075 0 j 0005 0
1 )
2 ( 0
mS 5 7 j 5 0 C 2
1 L 2 j R
1 ) 2 (
00
= ω
Y
=
ω ) 2 ( 0
Z 8 85 − j 132 74 Ω
mS 75 18 j 5 0 C 4
1 L 4 j R
1 ) 4 (
00
= ω
Y
=
ω ) 4 ( 0
Z 1 4212 − j 53 3 Ω
Trang 35PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
Rework Prob 14.25 if the elements are connected in parallel
) 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
=
1 ) C R LC ( 0
) C
1 L ( R
C
1 L C
L LR )
Z
22
2
=
− ω
⎯→
⎯
= ω
− ω +
− ω
− ω
=
Thus,
2
2C R
Trang 36Find the resonant frequency of the circuit in Fig 14.78
Figure 14.78
For Prob 14.38.
Chapter 14, Solution 38
22
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
0 2
0
ω +
ω
− ω
C
L L
0
2-36
-3
-2
20
10 40
50 )
10 1 )(
10 40 (
1 L
R LC
=
ω0 4841 rad / s
Trang 37PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
For the “tank” circuit in Fig 14.79, find the resonant frequency
21
o ( ) 2 ( 88 ) x 10 176 X 10 2
= ω
nF 89 19 10 x 2 x 10 x 8
1 BR
1 C RC
3o
2o
10 x 89 19 x ) 10 X 176 (
1 C
1 L LC
1
−
π
= ω
π
π
= ω
=
Trang 38A parallel resonance circuit has a resistance of 2 k Ω and half-power frequencies of 86 kHz and 90 kHz Determine:
(a) the capacitance
−
0
10 x 20 x 10 x 15
1 LC
1
1.8257 k rad/sec
=
= ω
x 25
1 RC
1 B
R
1 C
3 10 µF
C C
C C C
21
2
+
=
′ and C1 = 20 µF, we then obtain C2 = 20 µF
Therefore, to increase the bandwidth, we merely add another 20 µF in series
with the first one
Trang 39PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
(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- F µ capacitor in order to double the bandwidth?
4 0
1 LC
L
Q 0 0 1976
=
= L
R
B 8 rad / s
(b) This is a parallel RLC circuit
F 2 6 3
) 6 )(
3 ( F
6 and F
+
⎯→
⎯ µ µ
F 2
) 10 20 )(
10 2 (
1 LC
1
3 - 6
=
) 10 20 )(
10 5 (
10 2 L
10 2 (
1 RC
1
Trang 40For the circuits in Fig 14.81, find the resonant frequency ωo, the quality factor Q, and the bandwidth B
Figure 14.81
For Prob 14.42.
Trang 41PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
(a) Zin = ( 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
ω +
= ω + ω + ω
ω +
=
Z
2 2 2 2 2
2
) RC j LC 1
)(
L j R (
ω + ω
−
ω
− ω
− ω +
=
Z
At resonance, Im( Zin) = 0 , i.e
C R )
LC 1
( L
0 = ω0 − ω02 − ω0 2
C R L C
C L
C R L
(
) LC 1
( R C
j 1 L j R
) C j 1 L j ( R
ω + ω
=
Z
2 2 2 2 2
2 2
] RC j ) LC 1
)[(
LC 1
( R
ω + ω
−
ω
− ω
− ω
( R
0 LC
1 − ω2 =
=
ω0
LC 1
Trang 42Calculate the resonant frequency of each of the circuits in Fig 14.82
ω
=
C j
1 R
||
L j R
L j R
2 1
1 in
Z
L j R
L jR C j
1 R
C j
1 R L j R
L R j
1
1 2
2 1
1
in
ω +
ω +
ω +
⋅ ω +
1
2 1
) C R j 1 ( L R j
ω
− ω
+ ω +
ω + ω
=
Z
) C R R L ( j LCR LCR
R
L R j LC R R -
2 1 2
2 1
2 1
1 2
1 2
ω + ω
=
Z
2 2 1 2
2 2
2 1
2 1
2 1 2
2 1
2 1 1 2
1 2
)] C R R L ( j LCR LCR
R )[
L R j LC R R -
+ ω + ω
− ω
−
+ ω
− ω
− ω
− ω
+ ω
Trang 43PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
2 1
1 1 2
1 2
1
C L R L R LC R R
1 3 2
1 2 2 2
2 1
ω
=
LC 1
C R
0
C R LC
1
−
= ω
2 6 - 2
6 - 0
) 10 9 ( ) 1 0 ( ) 10 9 )(
02 0 (
=
ω0 2 357 krad / s
(b) At s ω = ω0 = 2 357 krad / ,
14 47 j ) 10 20 )(
10 357 2 ( j L
0212 0 j 9996 0 14 47 j 1
14 47 j L j
||
+
= ω
14 47 j 1 0 ) 10 9 )(
10 357 2 ( j
1 1
0 C j
1
×
× +
= ω +
) C j 1 R (
||
) L j
||
R ( )
Z
) 14 47 j 1 0 ( ) 0212 0 j 9996 0 (
) 14 47 j 1 0 )(
0212 0 j 9996 0 ( ) ( 0
− +
= ω
Z
=
ω ) ( 0
in
Trang 44* For the circuit in Fig 14.83, find:
(a) the resonant frequency ωo
1 j
2 j
3 1
ω + + ω + ω
jC C -j0.5 jC
1
jC j -j1.5
1
+
+ +
= + + +
1
C -0.5
+ +
2
1 C 1
C 1
2
2
Z Z
20 ) 10 )(
2
=
= I Z V
V ) t sin(
20 ) t (
Trang 45PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
For the circuit shown in Fig 14.84, find ωo, B, and Q, as seen by the voltage across the
Trang 46For the network illustrated in Fig 14.85, find
(a) the transfer function H( ω ) = Vo( ω )/I(ω ),
(b) the magnitude of H at ωo = 1 rad/s
x 10 x ) 10 x 15 x 2 (
1 L
1 C LC
1
32
3o
2
π
= ω
Q = ωo = π 3 −3 = π =
Trang 47PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
Show that a series LR circuit is a lowpass filter if the output is taken across the resistor Calculate the corner frequency fc if L = 2 mH and R = 10 k Ω
Chapter 14, Solution 47
R L j 1
1 L
j R
R )
(
i
o
ω +
= ω +
1 ) 0 (
H = and H ( ∞ ) = 0 showing that this circuit is a lowpass filter
At the corner frequency,
2
1 ) (
H ωc = , i.e
R
L 1 R
L 1
1 2
2 c
L
R
c = ω
=
= ω
=
×
×
⋅ π
=
⋅ π
10 10 2
1 L
R 2 1
Trang 48Find the transfer function Vo/Vs of the circuit in Fig 14.86 Show that the circuit is a lowpass filter
Figure 14.86
For Prob 14.48.
Chapter 14, Solution 48
C j
1
||
R L j
C j
1
||
R )
(
ω +
ω
ω
= ω
H
C j 1 R
C j R L j
C j 1 R
C j R )
(
ω +
ω +
ω
ω +
ω
= ω
H
= ω) (
H
RLC L
j R
R
2
ω
− ω + 1 ) 0 (
H = and H ( ∞ ) = 0 showing that this circuit is a lowpass filter
Trang 49PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part
Determine the cutoff frequency of the lowpass filter described by
H( ω ) =
10 2
Hence,
2
2 ) 0 ( H 2
1 ) (
2 c
100 4
4 2
2
ω +
=
2 0 8
100
ω +
10 j 1
2 20 j 2
4 ) 2 ( H
+
= +
=
199 0 101
2 ) 2 (
In dB, 20 log10 H ( 2 ) = - 14.023
=
= -tan 10 )
2 ( H
Trang 50Determine what type of filter is in Fig 14.87 Calculate the corner frequency fc
Figure 14.87
For Prob 14.50.
Chapter 14, Solution 50
L j R
L j )
(
i
o
ω +
0 )
R 1
1 2
1 )
(
c 2
=
= ω
H
L
R π
=
= ω
=
⋅ π
=
⋅ π
=
1 0
200 2
1 L
R 2
1