Problem 9.36 Design an active lowpass filter with a gain of 4, a corner frequency
of 1 kHz, and a gain roll-off rate of −60 dB/decade.
Solution: The roll-off rate of −60 dB requires a three-stage LP filter, similar in
design to that in Fig. 9-26. To secure positive gain, we need an additional fourth stage.
Arbitrarily, we choose all resistors of the first three stages to be 10-kΩ resistors, and
we realize the overall gain through the last stage.
Cf
Cf
10 kΩ
10 kΩ
+
_
Vs
Cf
10 kΩ
_
10 kΩ
_
+
10 kΩ
10 kΩ
+
_
40 kΩ
10 kΩ
+
_
+
+
Vo
_
Figure P9.36(a)
The transfer function is given by
!
"3
Vo
1
H(ω ) =
=4
,
Vs
1 + jω /ωc1
with
ωc1 =
1
.
RfCf
The problem states that the corner frequency of the overall filter, which we will call
ωc3 , should be
ωc3 = 2π fc = 2π × 103 rad/s.
According to Exercise 9.14,
ωc3 = 0.51ωc1 .
Hence,
ωc1 =
ωc3
2π × 103
=
= 12.32 krad/s,
0.51
0.51
and since Rf = 10 kΩ,
Cf =
1
= 8.12 nF.
Rf ωc1
The spectral response of the magnitude of H(ω ) is shown in Fig. P9.36(b).
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©2009 National Technology and Science Press
20
0
−20
M (dB)
−40
−60
−80
−100
−120
2
10
10
3
10
4
10
5
10
6
ω
Figure P9.36(b)
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©2009 National Technology and Science Press
Problem 9.37 Design an active highpass filter with a gain of 10, a corner frequency
of 2 kHz, and a gain roll-off rate of 40 dB/decade.
Solution: To secure a roll-off rate of 40 dB/decade we need to use two stages of the
circuit in Fig. 9-24.
Rf1
Rs
Vs
Rf2
Cs
_
+
+
_
Rs
Cs
_
+
+
V
_o
The two stages have the same input impedances (Rs and Cs ). We choose
Rf1 = Rs = 10 kΩ,
Consequently,
G1 = −
Rf1
= −1,
Rs
Rf2 = 100 kΩ.
G2 = −
Rf2
= −10.
Rs
The overall response is:
!
"2
Vo
jω /ωHP
H(ω ) =
= G1 G2
Vs
1 + jω /ωHP
!
"2
jω /ωHP
= 10
,
1 + jω /ωHP
with
ωHP =
1
.
RsCs
The problem statement specifies a corner frequency fc = 2 kHz with a corresponding
angular frequency ωc given by
ωc = 2π fc = 4π × 103 rad/s.
By definition, ωc is the angular frequency at which the magnitude of H(ω ) is equal
to 0.707 of its maximum value. Thus, at ω = ωc ,
#!
"2 ##
#
j
ω
/
ω
#
#
c
HP
|H(ωc )| = 10 #
# = 7.07,
# 1 + jωc /ωHP #
which leads to
x2
= 0.707
1 + x2
with x = ωc /ωHP .
Solution of the above equation gives
x = 1.55.
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©2009 National Technology and Science Press
Hence
ωHP =
and
Cs =
1
= 1.55ωc = 1.55 × 4π × 103 = 1.95 × 104 rad/s,
RsCs
1
1
= 4
= 5.1 × 10−9 F = 5.1 nF.
Rs ωHP 10 × 1.95 × 104
A plot of M [dB] is shown in Fig. P9.37(b).
20
10
0
−10
M (dB)
−20
−30
−40
−50
−60
−70
−80
2
10
10
3
10
4
10
5
10
6
ω
Figure P9.37(b)
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©2009 National Technology and Science Press
Problem 9.38 The element values in the circuit of the second-order bandpass filter
shown in Fig. P9.38 are: Rf1 = 100 kΩ, Rs1 = 10 kΩ, Rf2 = 100 kΩ, Rs2 = 10 kΩ,
Cf1 = 3.98 × 10−11 F, Cs2 = 7.96 × 10−11 F. Generate a spectral plot for the
magnitude of H(ω ) = Vo /Vs . Determine the frequency locations of the maximum
value of M [dB] and its half-power points.
Cf1
Cf1
Rf1
Rs1
Vs
+
_
_
Rf1
Rs1
+
_
Rs2
Cs2
+
Rf2
_
Rs2
+
Cs2
Rf2
_
+
+
Vo
_
Figure P9.38: Circuit for Problem 9.38.
Solution: The overall transfer function is given by
Vout
H(ω ) =
= G2LP G2HP
Vs
!
1
1 + jω /ωLP
"2 !
jω /ωHP
1 + jω /ωHP
"2
,
(1)
with
Rf1
= −10,
R s1
Rf
GHP = − 2 = −10,
R s2
1
1
ωLP =
= 5
= 251.26 krad/s,
Rf1 Cf1
10 × 3.98 × 10−11
1
1
ωHP =
= 4
= 125.63 krad/s.
Rs2 Cs2
10 × 7.96 × 10−11
GLP = −
Figure P9.38(a) displays the calculated plot of M [dB].
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©2009 National Technology and Science Press
80
70
60
M (dB)
50
40
30
20
10
4
10
10
5
10
6
10
7
ω
Figure P9.38(b)
From the plot we determine that:
ω0 = 178 krad/s
(M(ω0 ) = 72.9563 dB),
ω1 (−3 dB) = 94 krad/s,
ω2 (−3 dB) = 336 krad/s.
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©2009 National Technology and Science Press
2. (a) The given system H ( s ) , has:
•=
Zeros: f z1 10
=
[ Hz ] , f z2 107 [ Hz ]
3
•=
Poles: f p1 10
=
[ Hz ] , f p2 105 [ Hz ]
The slopes are – 20  dB


dec 
20*log|H(s)|
45
40
35
30
25
|H(s)| [dB]
20
15
10
5
0
H
f [Hz]
The actual values are calculated from the following formula –
 







ω2
ω2
ω2
ω2
H ( s ) = 20 ⋅ log 
+
+
+
−
+
−
+ 1  =
1
log
1
log
1
log






2
2
2
14
2
6
2
10

 4π ⋅10

 4π ⋅10

 4π ⋅10

  4π ⋅10







  f2

 f2

 f2

 f2

+
+
+
−
+
−
+
20 ⋅ log 
1
log
1
log
1
log
1







2

 1014

 106

 1010

  10







Values can be determined by plugging the frequency into the above formula, e.g. (marked on actual
plot below):
  1012

 1012

 1012

 1012

=
+
+
+
−
+
−
+
H ( s = j 2π ⋅1MHz ) = 20 ⋅ log 
1
log
1
log
1
log
1







2

 1014

 106

 1010

  10







100 [ dB ] + 0.04 [ dB ] − 60 [ dB ] − 20.04 [ dB ] =
20 [ dB ]
Magnitude [dB] vs. frequency [Hz])
Actual plot
(b) H ( s =
j 2π ⋅10 Hz ) =
3.01[ dB ]
H (s =
j 2π ⋅10 KHz ) =
39.91[ dB ]
H (s =
j 2π ⋅10 MHz ) =
3.01[ dB ]
3.
(a) Notice that the given input impedance is real, therefore it either consists of resistors only, or
resistors in addition to a combination of capacitors and inductors which cancel each other at the
center frequency.
R ' = Km ⋅ R ⇒ Km =
20 K Ω
= 20
1K Ω
The center frequency is shifted, therefore –
ω ' = K f ⋅ω ⇒ K f =
100 KHz
= 20
5 KHz
The net scaling factors for the components of the circuit are as following –
R ' = K m ⋅ R = 20 ⋅ R
C ='
1
1
⋅ C=
⋅C
400
Km ⋅ K f
L '=
Km
⋅ L= L
Kf
In order for the quality factor to remain the same we require –
ω0 ' ω0
2π ⋅100 KHz 2π ⋅ 5 KHz
Q' =
Q⇒
=
=
=
500 Hz
B'
B
B'
10 KHz
⇒ B' =
, ωc2 ' 105 KHz
=
⇒ ωc1 ' 95 KHz=
These values may be applied to different circuits, e.g. the example shown on page 441 in the
book or others band-pass filters.
(b) Op-amp should be able to operate on high frequencies.
Op-amp should be as ideal as possible (High input resistance, low output resistance, high gain).
4. The general form of a transfer function with two zeros and two poles –
H (s) =
(
)( s − s )
)( s − s )
K ⋅ s − sz1
(s − s
p1
z2
p2
In the given problem we have the following –
•
Zeros: sz1 = 0, sz2 = −3
•
Poles: s p1 =−1 + 4 j , s p2 =−1 − 4 j
Therefore –
H (s)
K ⋅ ( s − 0 )( s + 3)
=
( s − ( −1 + 4 j ) ) ( s − ( −1 − 4 j ) )
K ⋅ s ⋅ ( s + 3)
( s − ( −1 + 4 j ) ) ( s − ( −1 − 4 j ) )
Applying the given assumption –
K ⋅s⋅s
=1 ⇒ K =1
lim H ( s ) =
s →∞
s⋅s
s ⋅ ( s + 3)
⇒ H (s) =
( s − ( −1 + 4 j ) ) ( s − ( −1 − 4 j ) )