Problem 9.9 Circuit (b) in Fig. P9.9 is a scaled version of circuit (a). The scaling
process may have involved magnitude or frequency scaling, or both simultaneously.
If R1 = 1 kΩ gets scaled to R′1 = 10 kΩ, supply the impedance values of the other
elements in the scaled circuit.
j10 Ω
L1
j20 Ω
−j5 Ω L2
C1
R1
C2
R2
1 kΩ
−j50 Ω
2 kΩ
(a) Original circuit
L1
L2
C1
R1
10 kΩ
C2
R2
(b) Scaled circuit
Figure P9.9: Circuits for Problem 9.9.
Solution:
Km =
From the original circuit,
R′1
= 10.
R1
ZL1 = jω L1 = j10 Ω.
For the scaled circuit,
ZL′1 = jω ′ L′1 = jKf ω ·
Km
L1 = jKm ω L1 = j10 × 10 = j100 Ω,
Kf
ZL′2 = j200 Ω,
ZC1′ =
1
− jKm Kf − j10
=
=
= − j50 Ω,
jω ′C1′
Kf ωC1
ωC1
ZC2′ = − j500 Ω,
R′2 = Km R2 = 10 × 2 kΩ = 20 kΩ.
j100 Ω
j200 Ω
−j50 Ω
10 kΩ
−j500 Ω
20 kΩ
Scaled circuit
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Problem 9.12 Convert the following dB values to voltage ratios:
(a) 46 dB
(b) 0.4 dB
(c) −12 dB
(d) −66 dB
Solution:
(a) 10(46/20) = 102.3 = 199.5 ≃ 200.
(b) 10(0.4/20) = 100.02 = 1.047.
(c) 10(−12/20) = 10−0.6 = 0.25.
(d) 10(−66/20) = 10−3.3 = 5 × 10−4 .
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Problem 9.13 Generate Bode magnitude and phase plots for the following voltage
transfer functions:
j100ω
(a) H(ω ) =
10 + jω
0.4(50 + jω )2
(b) H(ω ) =
( jω )2
(40 + j80ω )
(c) H(ω ) =
(10 + j50ω )
(20 + j5ω )(20 + jω )
(d) H(ω ) =
jω
30(10 + jω )
(e) H(ω ) =
(200 + j2ω )(1000 + j2ω )
j100ω
(f) H(ω ) =
(100 + j5ω )(100 + jω )2
(200 + j2ω )
(g) H(ω ) =
(50 + j5ω )(1000 + jω )
Solution:
j100ω
j100ω
j10ω
(a) H(ω ) =
=
=
.
10 + jω
10(1 + jω /10) 1 + jω /10
• Constant factor 10
=⇒
+20 dB
• Zero @ origin
• Simple pole with ωc = 10 rad/s
M [dB] = 20 log |H|
= 20 log 10 + 20 log ω − 20 log |1 + jω /10|
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dB
M [dB]
40
20 log 10 = 20 dB
20
20 log ω
0
ω (rad/s)
0.1
1
10
100
20 log |1 + jω/10|
−20
Magnitude
φ
j
90o
φ(ω)
0o
ω (rad/s)
0.1
1
10
100
1
1 + jω/10
−90o
Phase
(b)
H(ω ) =
0.4(50 + jω )2
( jω )2
0.4 × 2500(1 + jω /50)2
−1000(1 + jω /50)2
=
.
ω2
ω2
M [dB] = 20 log |H(ω )| = 20 log 1000 + 40 log |1 + jω /50| − 40 log ω .
=
• Line starts at 60 dB at ω = 1 rad/s, and has slope of −40 dB/decade
– Constant factor 1000
=⇒
60 dB
– Pole @ origin of order 2
• Simple zero with ωc = 50 rad/s, of order 2
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dB
20 log 1000 = 60 dB
60
40 log |1 + jω/50|
40
M [dB]
20
0
ω (rad/s)
1
50
10
100
1000
−8 dB
−20
−40
−40 log ω
−60
Magnitude
φ
−180o
(1 + jω/50)
0
ω (rad/s)
1
10
100
1000
φ(ω)
−180o
−180o
Phase
(c) H(ω ) =
40 + j80ω
40(1 + j2ω ) 4(1 + jω /0.5)
=
=
.
10 + j50ω
10(1 + j5ω )
(1 + jω /0.2)
• Constant factor 4
=⇒
12 dB
• Simple pole with ωc = 0.2 rad/s
• Simple zero with ωc = 0.5 rad/s
M [dB] = 20 log |H(ω )| = 20 log 4 + 20 log |1 + jω /0.5| − 20 log |1 + jω /0.2|
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dB
20
20 log |1 + jω/0.5|
15
12 dB
12
10
M [dB]
5
0
0.01
0.1
1
10
ω (rad/s)
−5
−20 log |1 + jω/0.2|
−10
Magnitude
90o
(1 + jω/0.5)
45o
0
0.01
−45o
0.1
1
φ(ω)
10
ω (rad/s)
1
(1 + jω/0.2)
−90o
Phase
(d)
H(ω ) =
=
(20 + j5ω )(20 + jω )
jω
− j20(1 + jω /4)20(1 + jω /20) − j400(1 + jω /4)(1 + jω /20)
=
.
ω
ω
• Constant term 400
=⇒
52 dB
• Pole @ origin
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• Simple zero with ωc = 4 rad/s
• Simple zero with ωc = 20 rad/s
dB
60
52 dB
52
40
20 log |1 + jω/4|
M [dB]
20 log |1 + jω/20|
20
0
0.1
1
10
100
1000
ω (rad/s)
−20
−20 log ω
−40
−60
Magnitude
90o
(1 + jω/20)
(1 + jω/4)
45o
φ(ω)
0
0.1
1
10
100
1000
ω (rad/s)
−45o
−j
−90o
Phase
(e)
H( ω ) =
300(1 + jω /10)
30(10 + jω )
=
(200 + j2ω )(1000 + j2ω ) 200 × 1000(1 + jω /100)(1 + jω /500)
=
1.5 × 10−3 (1 + jω /10)
(1 + jω /100)(1 + jω /500)
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• Constant term 1.5 × 10−3
=⇒
−56.5 dB
• Simple zero with ωc = 10 rad/s
• Simple pole with ωc = 100 rad/s
• Simple pole with ωc = 500 rad/s
dB
40
20 log |1 + jω/10|
20
0
1
10
1000
100
10000
−20
ω (rad/s)
−20 log |1 + jω/500|
−36.5
−40
−20 log |1 + jω/100|
−56.5 dB
−56.5
−60
Magnitude
φ
90o
(1 + jω/10)
45o
φ(ω)
0
1
100
10
−45o
1
(1 + jω/100)
1000
10000
ω (rad/s)
1
(1 + jω/500)
−90o
Phase
(f) H(ω ) =
j100ω
j10−4 ω
=
2
(100 + j5ω )(100 + jω )
(1 + jω /20)(1 + jω /100)2
• Constant term 10−4
=⇒
−80 dB
• Zero @ origin
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• Simple pole with ωc = 20 rad/s
• Simple pole with ωc = 100 rad/s, of order 2
dB
20
20 log ω
0
1
10
100
1000
−40 log |1 + jω/100|
10000
ω (rad/s)
−20
−20 log |1 + jω/20|
−40
M [dB]
−54
−60
−80
−80 dB
−94
−100
Magnitude
φ
90o
j
45o
0
−45o
φ(ω)
1
10
100
1000
ω (rad/s)
1
(1 + jω/100)
1
(1 + jω/20)
−90o
−180o
Phase
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(g) H(ω ) =
(200 + j2ω )
(1 + jω /100)
=
(50 + j5ω )(1000 + jω ) 250(1 + jω /10)(1 + jω /1000)
• Constant term 1/250
=⇒
−48 dB
• Simple pole with ωc = 10 rad/s
• Simple zero with ωc = 100 rad/s
• Simple pole with ωc = 1000 rad/s
dB
20
20 log |1 + jω/100|
0
1
10
100
1000
10,000
ω (rad/s)
−20 log |1 + jω/1000|
−20
−20 log |1 + jω/10|
−40
−48
−48 dB
−60
−68
M [dB]
−80
−88
−100
Magnitude
φ
90o
(1 + jω/100)
0
1
10
100
1000
φ(ω)
1
1 + jω/1000
10,000
ω (rad/s)
1
1 + jω/10
−90o
Phase
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Problem 9.16 Determine the voltage transfer function H(ω ) corresponding to the
Bode magnitude plot shown in Fig. P9.16. The phase of H(ω ) is 90◦ at ω = 0.
M [dB]
60 dB
40 dB
20 dB
0
0.5
5
50
500
ω (rad/s)
Figure P9.16: Bode magnitude plot for Problem 9.16.
Solution: H(ω ) consists of:
(1) A constant term K whose dB value is 60 dB, or
K = 1060/20 = 1000.
(2) A simple pole of order 3 with ωc = 5 rad/s (slope = −60 dB/decade)
(3) A simple zero of order 6 with ωc = 50 rad/s (slope reverses from −60
dB/decade to +60 dB/decade)
(4) A simple pole of order 3 with ωc = 500 rad/s (slope changes to 0 dB at
ωc = 500 rad/s).
Hence,
H(ω ) =
( j)N 1000(1 + jω /50)6
j1000(50 + jω )6
=
.
(1 + jω /5)3 (1 + jω /500)3
(5 + jω )3 (500 + jω )3
Given that the phase of H(ω ) is 90◦ at ω = 0, it follows that N = 1.
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Problem 9.26 For the circuit shown in Fig. P9.26:
(a) Obtain an expression for H(ω ) = Vo /Vi in standard form.
(b) Generate spectral plots for the magnitude and phase of H(ω ), given that
R1 = 1 Ω, R2 = 2 Ω, L1 = 1 mH, and L2 = 2 mH.
(c) Determine the cutoff frequency ωc and the slope of the magnitude (in dB) when
ω /ωc ≪ 1 and when ω /ωc ≫ 1.
R1
L2
V1
+
+
L1
Vi
_
R2
Vo
_
Figure P9.26: Circuit for Problem 9.26.
Solution:
(a) At node V1 , KCL gives:
V1 − Vi
V1
V1
+
+
= 0.
R1
j ω L1 R2 + j ω L2
Also, voltage division gives
Vo =
R2 V1
.
R2 + j ω L2
Solution is
H(ω ) =
Vo
j ω L1 R2
=
.
Vi
(R1 + jω L1 )(R2 + jω L2 ) + jω L1 R1
To express denominator in standard form, we factor out R1 R2 and expand terms:
H(ω ) = jω L /R
1 1 L1
L2
L1
1 + jω
1 + jω
+ jω
R1
R2
R2
jω L1 /R1
r
2
L1 L2 L1
L1 L2
1 + jω
+
+
+ jω
R1 R2 R2
R1 R2
jω K
=
,
1 + j2ξ ω /ωc + ( jω /ωc )2
=
where
r
L1
R1 R2
K=
,
ωc =
,
R1
L1 L2
ωc L1 L2 L1
ξ=
+
+
.
2 R1 R2 R2
(b) For R1 = 1 Ω, R2 = 2 Ω, L1 = 1 mH, and L2 = 2 mH,
K = 10−3 ,
ωc = 103 rad/s,
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ξ = 1.25.
©2009 National Technology and Science Press
Hence,
H(ω ) =
with ωc = 103 rad/s.
j × 10−3 ω
,
1 + j2.5ω /ωc + ( jω /ωc )2
M [dB] = 20 log |H(ω )|
Spectral plots of M [dB] are φ (ω ) are shown in Figs. P9.26(a) and (b).
0
M (dB)
−20
−40
−60
−80
1
10
10
2
10
3
10
4
10
5
10
6
ω
ϕ (deg)
100
0
−100
10
1
10
2
10
3
10
4
10
5
10
6
ω
Figures P9.26(a) and (b)
(c) Low-frequency asymptote (ω /ωc ≪ 1):
H(ω ) ≃ j10−3 ω
=⇒
M [dB] has slope of +20 dB/decade.
High-frequency asymptote (ω /ωc ≫ 1):
H(ω ) ≃ − j10−3
ωc2
103
= −j
ω
ω
=⇒
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M [dB] has slope of −20 dB/decade.
©2009 National Technology and Science Press