Chapter 14, Solution 1.
Vo R jωRC
H (ω) = = =
Vi R + 1 jωC 1 + jωRC
jω ω 0 1
H (ω) = , where ω 0 =
1 + jω ω 0 RC
ω ω0 π ω
H = H (ω) = φ = ∠H (ω) = − tan -1
1 + (ω ω0 ) 2 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
Chapter 14, Solution 2.
R 1 1 R
H (ω) = = = , where ω0 =
R + jωL 1 + jωL R 1 + jω ω 0 L
1 ω
H = H (ω) = φ = ∠H (ω) = - tan -1
1 + (ω ω0 ) 2 ω0
The frequency response is identical to the response in Example 14.1 except that
ω0 = R L . Hence the response is shown below.
H
1
0.7071
0 ω0 = R/L ω
φ
ω0 = R/L
0° ω
-45°
-90°
Chapter 14, Solution 3.
(a) The Thevenin impedance across the second capacitor where Vo is taken is
R
Z Th = R + R || 1 sC = R +
1 + sRC
1 sC Vi
VTh = Vi =
R + 1 sC 1 + sRC
ZTh
+
+ 1
VTh Vo
− sC
−
1 sC Vi
Vo = ⋅ VTh =
Z Th + 1 sC (1 + sRC)(1 + sCZ Th )
Vo 1 1
H (s) = = =
Vi (1 + sCZ Th )(1 + sRC) (1 + sRC)(1 + sRC + sRC (1 + sRC))
1
H (s) =
s R C + 3sRC + 1
2 2 2
(b) RC = (40 × 10 3 )(2 × 10 -6 ) = 80 × 10 -3 = 0.08
There are no zeros and the poles are at
- 0.383
s1 = = - 4.787
RC
- 2.617
s2 = = - 32.712
RC
Chapter 14, Solution 4.
1 R
(a) R || =
jωC 1 + jωRC
R
Vo 1 + jωRC R
H (ω) = = =
Vi R R + jωL (1 + jωRC)
jωL +
1 + jωRC
R
H (ω) =
- ω RLC + R + jωL
2
R + jωL jωC (R + jωL)
(b) H (ω) = =
R + jωL + 1 jωC 1 + jωC (R + jωL)
- ω 2 LC + jωRC
H (ω) =
1 − ω 2 LC + jωRC
Chapter 14, Solution 5.
Vo 1 jωC
(a) H (ω) = =
Vi R + jωL + 1 jωC
1
H (ω) =
1 + jωRC − ω 2 LC
1 R
(b) R || =
jωC 1 + jωRC
Vo jωL jωL (1 + jωRC)
H (ω) = = =
Vi jωL + R (1 + jωRC) R + jωL (1 + jωRC)
jωL − ω 2 RLC
H (ω) =
R + jωL − ω 2 RLC
Chapter 14, Solution 6.
(a) Using current division,
Io R
H (ω) = =
I i R + jωL + 1 jωC
jωRC jω (20)(0.25)
H (ω) = =
1 + jωRC − ω LC 1 + jω(20)(0.25) − ω2 (10)(0.25)
2
jω5
H (ω) =
1 + jω5 − 2.5 ω 2
(b) We apply nodal analysis to the circuit below.
0.5 Vx
Vx Io 1/jωC
+ −
Is R jωL
Vx Vx − 0.5Vx
Is = +
R jωL + 1 jωC
0.5 Vx
But Io =
→ Vx = 2 I o ( jωL + 1 jωC)
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
H (ω) = = =
I s 1 + 2 ( jωL + 1 jωC) R jωRC + 2 (1 − ω 2 LC)
jω
H (ω) =
jω + 2 (1 − ω2 0.25)
jω
H (ω) =
2 + jω − 0.5 ω 2
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, Solution 8.
(a) H = 0.05
H dB = 20 log10 0.05 = - 26.02 , φ = 0°
(b) H = 125
H dB = 20 log10 125 = 41.94 , φ = 0°
j10
(c) H(1) = = 4.472∠63.43°
2+ j
H dB = 20 log10 4.472 = 13.01 , φ = 63.43°
3 6
(d) H(1) = + = 3.9 − j1.7 = 4.254∠ - 23.55°
1+ j 2 + j
H dB = 20 log10 4.254 = 12.577 , φ = - 23.55°
Chapter 14, Solution 9.
1
H (ω) =
(1 + jω)(1 + jω 10)
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 10 100 ω
1
20 log 10
-20 1 + jω / 10
1
20 log10
-40 1 + jω
φ
0.1 1 10 100 ω
-45° 1
arg
1 + jω / 10
-90°
1
arg
1 + jω
-135°
-180°
Chapter 14, Solution 10.
50 10
H( jω) = =
jω(5 + jω) jω
1 jω1 +
5
HdB
40
20 log1
20
10
0.1 1 100 ω
1
-20 20 log 1
jω 20 log
jω
1+
-40 5
φ 1
0.1 10 100 ω
-45° 1
arg
1 + jω / 5
-90°
1
arg
-135° jω
-180°
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 1 10 100 ω
-20
-40
φ
90°
45°
0.1 1 10 100 ω
-45°
-90°
Chapter 14, Solution 12.
0.1(1 + jω )
T ( w) = , 20 log 0.1 = −20
jω (1 + jω / 10)
The plots are shown below.
|T| (db)
20
0 ω
0.1 1 10 100
-20
-40
arg T
90o
0 ω
0.1 1 10 100
-90o
Chapter 14, Solution 13.
1 + jω (1 10)(1 + jω)
G (ω) = =
( jω) (10 + jω) ( jω) 2 (1 + jω 10)
2
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 1 10 100 ω
-20
-40
φ
90°
0.1 1 10 100 ω
-90°
-180°
Chapter 14, Solution 14.
50 1 + jω
H (ω) =
25 jω10 jω 2
jω1 +
+
25 5
H dB = 20 log10 2 + 20 log10 1 + jω − 20 log10 jω
− 20 log10 1 + jω2 5 + ( jω 5) 2
ω10 25
φ = -90° + tan -1 ω − tan -1
1 − ω2 5
The magnitude and phase plots are shown below.
HdB
40
26
20
6
0.1 1 10 100 ω
-20
-40
φ
90°
0.1 1 10 100 ω
-90°
-180°
Chapter 14, Solution 15.
40 (1 + jω) 2 (1 + jω)
H (ω) = =
(2 + jω)(10 + jω) (1 + jω 2)(1 + jω 10)
H dB = 20 log10 2 + 20 log10 1 + jω − 20 log10 1 + jω 2 − 20 log10 1 + jω 10
φ = tan -1 ω − tan -1 ω 2 − tan -1 ω 10
The magnitude and phase plots are shown below.
HdB
40
20
6
0.1 1 10 100 ω
-20
-40
φ
90°
45°
0.1 1 10 100 ω
-45°
-90°
Chapter 14, Solution 16.
jω
G (ω) =
2
jω
100(1 + jω)1 +
10
GdB
20 log jω
20
0.1 1 10 100
jω ω
− 40 log
-20 10
-40
20 log(1/100)
-60
φ
arg(jω)
90°
ω
0.1 1 10 100
1
-90° 1 arg
arg 1 + jω
2
jω
-180° 1 +
10
Chapter 14, Solution 17.
(1 4) jω
G (ω) =
(1 + jω)(1 + jω 2) 2
G dB = -20log10 4 + 20 log10 jω − 20 log10 1 + jω − 40 log10 1 + jω 2
φ = -90° - tan -1ω − 2 tan -1 ω 2
The magnitude and phase plots are shown below.
GdB
20
0.1 1 10 100
-12 ω
-20
-40
φ
90°
0.1 1 10 100 ω
-90°
-180°
Chapter 14, Solution 18.
4 (1 + jω 2) 2
G (ω) =
50 jω (1 + jω 5)(1 + jω 10)
G dB = 20 log10 4 50 + 40 log10 1 + jω 2 − 20 log10 jω
− 20 log10 1 + jω 5 − 20 log10 1 + jω 10
where 20 log10 4 50 = -21.94
φ = -90° + 2 tan -1 ω 2 − tan -1 ω 5 − tan -1 ω 10
The magnitude and phase plots are shown below.
GdB
20
0.1 1 10 100 ω
-20
-40
-60
φ
180°
90°
0.1 1 10 100 ω
-90°
Chapter 14, Solution 19.
jω
H (ω) =
100 (1 + jω 10 − ω2 100)
H dB = 20 log10 jω − 20 log10 100 − 20 log10 1 + jω 10 − ω2 100
ω 10
φ = 90° − tan -1
1 − ω2 100
The magnitude and phase plots are shown below.
HdB
40
20
0.1 1 10 100 ω
-20
-40
-60
φ
90°
0.1 1 10 100 ω
-90°
-180°
Chapter 14, Solution 20.
10 (1 + jω − ω2 )
N(ω) =
(1 + jω)(1 + jω 10)
N dB = 20 − 20 log10 1 + jω − 20 log10 1 + jω 10 + 20 log10 1 + jω − ω2
ω
φ = tan -1 − tan -1 ω − tan -1 ω 10
1 − ω2
The magnitude and phase plots are shown below.
NdB
40
20
0.1 1 10 100 ω
-20
-40
φ
180°
90°
0.1 1 10 100 ω
-90°
Chapter 14, Solution 21.
jω (1 + jω)
T(ω) =
100 (1 + jω 10)(1 + jω 10 − ω2 100)
TdB = 20 log10 jω + 20 log10 1 + jω − 20 log10 100
− 20 log10 1 + jω 10 − 20 log10 1 + jω 10 − ω2 100
ω 10
φ = 90° + tan -1 ω − tan -1 ω 10 − tan -1
1 − ω2 100
The magnitude and phase plots are shown below.
TdB
20
0.1 1 10 100 ω
-20
-40
-60
φ
180°
90°
0.1 1 10 100 ω
-90°
-180°
Chapter 14, Solution 22.
20 = 20 log10 k
→ k = 10
A zero of slope + 20 dB / dec at ω = 2
→ 1 + jω 2
1
A pole of slope - 20 dB / dec at ω = 20
→
1 + jω 20
1
A pole of slope - 20 dB / dec at ω = 100
→
1 + jω 100
Hence,
10 (1 + jω 2)
H (ω) =
(1 + jω 20)(1 + jω 100)
10 4 ( 2 + jω)
H (ω) =
( 20 + jω)(100 + jω)
Chapter 14, Solution 23.
A zero of slope + 20 dB / dec at the origin
→ jω
1
A pole of slope - 20 dB / dec at ω = 1
→
1 + jω 1
1
A pole of slope - 40 dB / dec at ω = 10
→
(1 + jω 10) 2
Hence,
jω
H (ω) =
(1 + jω)(1 + jω 10) 2
100 jω
H (ω) =
(1 + jω)(10 + jω) 2
Chapter 14, Solution 24.
The phase plot is decomposed as shown below.
φ
90°
arg (1 + jω / 10)
45°
0.1 1 10 100 1000 ω
-45° arg ( jω) 1
arg
1 + jω / 100
-90°
k ′ (1 + jω 10) k ′ (10)(10 + jω)
G (ω) = =
jω (1 + jω 100) jω (100 + jω)
where k ′ is a constant since arg k ′ = 0 .
k (10 + jω)
Hence, G (ω) = , where k = 10k ′ is constant
jω (100 + jω)
Chapter 14, Solution 25.
1 1
ω0 = = = 5 krad / s
LC (40 × 10 -3 )(1 × 10 -6 )
Z(ω0 ) = R = 2 kΩ
ω0 4
Z(ω0 4) = R + j L −
4 ω0 C
5 × 10 3 4
Z(ω0 4) = 2000 + j ⋅ 40 × 10 -3 −
4 (5 × 10 3 )(1 × 10 -6 )
Z(ω0 4) = 2000 + j (50 − 4000 5)
Z(ω0 4) = 2 − j0.75 kΩ
ω0 2
Z(ω0 2) = R + j L −
2 ω0 C
(5 × 10 3 ) 2
Z(ω0 2) = 2000 + j (40 × 10 -3 ) −
2 (5 × 10 3 )(1 × 10 -6 )
Z(ω0 4) = 2000 + j (100 − 2000 5)
Z(ω0 2) = 2 − j0.3 kΩ
1
Z(2ω0 ) = R + j 2ω0 L −
2ω0 C
1
Z(2ω0 ) = 2000 + j (2)(5 × 10 3 )(40 × 10 -3 ) −
(2)(5 × 10 3 )(1 × 10 -6 )
Z(2ω0 ) = 2 + j0.3 kΩ
1
Z(4ω0 ) = R + j 4ω0 L −
4ω0 C
1
Z(4ω0 ) = 2000 + j (4)(5 × 10 3 )(40 × 10 -3 ) −
(4)(5 × 10 )(1 × 10 )
3 -6
Z(4ω0 ) = 2 + j0.75 kΩ
Chapter 14, Solution 26.
1 1
(a) fo = = = 22.51 kHz
2π LC 2π 5 x10 −9 x10 x10 −3
R 100
(b) B= = = 10 krad/s
L 10 x10 −3
ωo L 1 L 10 6 10 x10 −3
(c ) Q= = = 3
= 14.142
R LC R 50 0.1x10
Chapter 14, Solution 27.
At resonance,
1
Z = R = 10 Ω , ω0 =
LC
R ω 0 ω0 L
B= and Q= =
L B R
Hence,
RQ (10)(80)
L= = = 16 H
ω0 50
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