ELG4135: Tutorial on Chapter 12
Q1 (12.5)
A low-pass filter is specified to have Amax = 1 dB and Amin = 10 dB. It is found that its specifications can
be just met with a single-time-constant RC circuit having a time constant of 1 s and a dc transmission of
unity. What must ω p and ω s of this filter be? What is the selectivity factor?
Solution:
The frequency response of the low-pass filter can be plotted as
T ( jω )
dB
0
-1
-10
ω
ωp
ωs
An R-C low-pass filter can be plotted as
R
Vi
Vo
C
The voltage transfer function
T ( jω ) is then
1
V
1
jωC
=
T ( jω ) = o =
1
Vi
1 + jωRC
R+
jωC
Given τ = RC = 1 s,
1
j ωC
T ( jω ) =
1
1 + jω
T ( jω ) is then
The magnitude of
1
| T ( jω ) |=
1+ ω 2
At the pass-band edge:
−1
−1
| T ( jω p ) |= 10 20 . (Note: 10 20 is the linear scale of -1 dB loss.)
ie.
1
1+ ωp
= 10
2
−1
20
=> ω p = 0.5088 rad/s
At the stop-band edge:
| T ( jω s ) |= 10
ie.
1
1 + ωs
2
−10
20
. (Note: 10
= 10
−10
20
is the linear scale of -10 dB loss.)
−10
20
=> ω s = 3 rad/s
=>Selectivity=
ωs
ωp
=5.9
#
Q2 (12.11)
Analyze the RLC network of Fig. P11.11 to determine its transfer function T ( s ) =
Vo ( s )
and hence its
Vi ( s )
poles and zeros.
(1)
1 Ω
2 H
(2)
(3)
Vi
Vo
1 F
1 F
1 Ω
Solution:
The easiest way to solve the circuit is to use nodal analysis at nodes (1), (2) and (3).
At node (3), ΣI = 0
Vo Vo Vo − V1
+
+
=0,
1 1/ s
2s
where V1 is the voltage at node (2).
-> V1 = Vo (2 s + 2 s + 1)
2
(a)
At node (2), ΣI = 0
V1 − Vi V1 V1 − Vo
+
+
=0,
1
1/ s
2s
->
V1 (2s 2 + 2s + 1) = Vo + 2sVi
Substitute (a) to (b)
Vo (2s 2 + 2s + 1) 2 = Vo + 2sVi
=> T ( s ) =
Vo ( s)
0.5
= 3
Vi ( s ) s + 2s 2 + 2s + 1
=>There is no zero.
=>Poles are given by.
s 3 + 2s 2 + 2s + 1 = 0
( s + 1)( s 2 + s + 1) = 0
1
3
s = −1 , s = − ± j
2
2
#
(b)
Q3 (12.12)
Determine the order N of the Butterworth filter for which Amaz = 1 dB, Amin ≥ 20 dB, and the selectivity
ratio
ωs
= 1 .3 .
ωp
What is the actual value of minimum stopband attenuation realized? If Amin is to be
exactly 20 dB, to what value can Amax be reduced?
Solution:
From equation (12.15)
A(ω s ) = 10 log[1 + ε 2 (
ωs 2 N
) ] = Amin
ωp
A
ωs 2N
1 + ε ( ) = 10 10
ωp
min
2
Amin
ω
10 10 − 1
( s )2N =
ωp
ε2
Amin
ω
10 10 − 1
log[( s ) 2 N ] = log[
]
ωp
ε2
N=
10
log[
Amin
10
−1
ε
ωs 2
log[( ) ]
ωp
2
]
(a)
From equation (11.14), as Amaz = 1 dB,
ε = 10
Amax
10
1
10
− 1 = 10 − 1 = 0.5088
Substitute Amin ≥ 20 dB,
ε , and the selectivity ratio
ωs
= 1 .3
ωp
into (a)
N ≈ 11.3
=> Choose N=12
The actual value of stopband attenuation can be calculated using N=12 as
A(ω s ) = 10 log[1 + ε 2 (
ωs 2N
) ] = 27.35 dB
ωp
If Amin is to be exactly 20 dB, from equation (12.15)
ε2 =
10
Amin
10
−1
ω
( s )2N
ωp
= 0.1824 , with Amin = 20 dB and N=12
From equation (11.13),
Amax = 10 log(1 + ε 2 ) = 0.73 dB
#
Q4 (12.14)
3
Find the natural modes of a Butterworth filter with a 1-dB bandwidth of 10 rad/s and N=5.
Solution:
From equation (11.14), as Amaz = 1 dB,
Amax
10
1
10
ε = 10
− 1 = 10 − 1 = 0.5088
3
And ω p = 10 rad/s, N=5, the solutions can be found graphically. (The method is shown in Fig. 11.10)
The radius of the circle is
P1 = ω 0 e
π
π
π
π
± j( +
)
2 2N
P2 = ω0 e
P3 = ω0 e
= −269.96 ± j830.84
π
± j( +
+ )
2 2N N
± j (π )
= −706.75 ± j 513.49
= −873.59
π
π
N
π
N
1
ω 0 = ω p ( )1 / N = 873.59
ε
2N
Q5 (12.22)
By cascading a first-order op amp-RC low-pass circuit with a first-order op amp-RC high-pass circuit one
can design a wideband bandpass filter. Provide such a design for the case the midband gain is 12 dB and
the 3-dB bandwidth extends from 100 Hz to 10 kHz. Select appropriate component values under the
constraint that no resistors higher than 100 kohms are to be used, and the input resistance is to be as high as
possible.
Solution:
C1
R4
R1
R2
Vo
R3
12 / 20
Gain= 10
C2
= 3.98 ≈ 4
Want Ri=R1 large, so R1=100 kOhm
Total gain=Alp*Ahp=4
Alp=-R2/R1 => R2=-Alp*R1 and R2<100 kOhm, so
Alp=-1, R2=100 kOhm
From Fig. 11.14, R2 C1 =
C1 =
1
R2ω0,lp
=
1
ω 0,lp
,
1
= 0.159 nF
2π (10 ×10 3 )100 ×10 3
Ahp=-4, R4=100 kOhm, R3=25 kOhm
From Fig. 11.14, R3C 2 =
C1 =
1
R2ω0,lp
=
1
ω 0,hp
,
1
= 63.7 nF
2π × (100) × 25 ×10 3