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Find the transfer function V o / V i of the RC circuit in Fig. 14.68. Express it using ω o =
1/ RC.
Figure 14.
For Prob. 14.1.
Chapter 14, Solution 1.
1 j RC
j RC
R 1 j C
R
i
o
ω
ω = = V
V
H
H (ω) = 0
0
1 j
j
ω ω , where RC
ω 0 =
2 0
0
H ( )
ωω = H ω = ⎟ ⎠
ω
ω −
π φ =∠ ω = 0
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.
ω 0 = 1/RC (^) ω
φ
ω 0 = 1/RC ω
H
Obtain the transfer function V (^) o ( s )/ V (^) i of the circuit in Fig. 14.69.
Figure 14.
For Prob. 14.2.
Chapter 14, Solution 2.
s 0. 6667
s 4
12 8 /s
2 8 /s
s/ 8
s/ 8
V
V
H(s) i
o
Find the transfer function H ( ω ) = V (^) O /V (^) i of the circuits shown in Fig. 14.71.
Figure 14.
For Prob. 14.4.
Chapter 14, Solution 4.
(a) 1 j RC
R
j C
R ||
ω
R j L( 1 j RC )
R
1 j RC
R
j L
1 j RC
R
i
o
ω +
V
H
H (ω) =
- RLC R j L
R
2 ω + + ω
(b) 1 j C(R j L)
j C(R j L)
R j L 1 j C
R j L ( )
ω + ω
H (ω) = 1 LC j RC
- LC j RC 2
2
−ω + ω
ω + ω
For each of the circuits shown in Fig. 14.72, find H ( s ) = V (^) o ( s )/ V (^) s ( s ).
Figure 14.
For Prob. 14.5.
Chapter 14, Solution 5.
(a) Let //
sRL Z R sL R sL
o s s
Z
V V
Z R
o
s s s s s
sRL
V Z (^) R sL sRL H s V Z R sRL RR s R R L R R sL
(b) Let
Rx sC R Z R sC sRC R sC
o s
Z
V V
Z sL
s LRC sL R
R
1 sRC
R
sL
1 sRC
R
Z sL
Z
V
V
H( s) 2 i
o
Calculate H ( ω) if H dB equals
(a) 0.05dB (b) -6.2 dB (c) 104.7 dB
Chapter 14, Solution 7.
(a) 0. 05 = 20 log 10 H
- 5 10 log 10 H
× =
- 5 × 10 - H 10 1. 005773
(b) - 6.2= 20 log 10 H
-0. H 10 0. 4898
(c) 104. 7 = 20 log 10 H
- 235 =log 10 H
- 235 H 10
**5
- 718** × 10
Chapter 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)
j
j
(d)
1 + j ω
2 + j ω
Chapter 14, Solution 8.
(a) H = 0. 05
H (^) dB = 20 log 100. 05 = - 26.02 , φ = 0 °
(b) H = 125
H (^) dB = 20 log 10125 = 41.94 , φ = 0 °
(c) = ∠ °
2 j
j 10 H( 1 )
H (^) dB = 20 log 104. 472 = 13. 01 , φ = 63. 43 °
(d) = − = ∠ °
= 3. 9 j 2. 7 4. 743 - 34. 2 j
1 j
H( 1 )
= , φ = –34.7˚
A ladder network has a voltage gain of
H ( ω ) =
+ j ω + j ω
Sketch the Bode plots for the gain.
Chapter 14, Solution 9.
( 1 j )( 1 j 10 )
H ω =
H (^) dB =- 20 log 101 +jω − 20 log 101 +jω/ 10
-1 - φ= ω − ω
The magnitude and phase plots are shown below.
HdB
(^1 10100) ω
-20 (^1) j / 10
1 20 log 10
1 + j ω
1 20 log (^10)
φ
0.1 (^1 10100) ω
1 j / 10
1 arg
1 + j ω
1 arg
j j
j
Chapter 14, Solution 11.
j ( 1 j 2 )
5 ( 1 j 10 ) ( ) ω + ω
H (^) dB = 20 log 105 + 20 log 101 +jω 10 − 20 log 10 jω − 20 log 101 +jω 2
-1 - φ= °+ ω − ω
The magnitude and phase plots are shown below.
φ
0.1 (^1 10100) ω
HdB
0.1 (^1 10100) ω
s s
s
Sketch the magnitude and phase Bode plots.
Construct the Bode plots for
G ( s ) = ( 10 )
2
s s
s
, s = j ω
Chapter 14, Solution 13.
(j ) ( 1 j 10 )
( 110 )( 1 j )
(j ) ( 10 j )
1 j ( ) 2 2 ω + ω
G (^) dB =- 20 + 20 log 101 +jω − 40 log 10 jω − 20 log 101 +jω 10
-180 tan tan 10
-1 - φ= °+ ω− ω
The magnitude and phase plots are shown below.
φ
0.1 (^1 10100) ω
GdB
0.1 (^1 10100) ω
