Chapter 16 - Filter Circuits

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Chapter 16: Filter Circuits 
 
Exercises 
 
Ex. 16.3-1 
1( )
1
1 1250( ) 
1250 12501
1250
H sn s
sH s Hn s s
= +
⎛ ⎞= = =⎜ ⎟ +⎝ ⎠ +
 
Problems 
Section 16.3: Filters 
 
P16.3-1 
Equation 16-3.2 and Table 16-3.2 provide a third-order Butterworth low-pass filter having a 
cutoff frequency equal to 1 rad/s. 
2
1( )
( 1)( 1)n
H s
s s s
= + + + 
Frequency scaling so that = 2 100=628 rad/s:cω π 
 
3
L 2 2 22
1 628 247673152( ) 
( 628)( 628 628 ) ( 628)( 628 394384)
1 1
628 628 628
H s
s s s s s ss s s
= = =+ + + + + +⎛ ⎞⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
 
 
 
P16.3-2 
Equation 16-3.2 and Table 16-3.2 provide a third-order Butterworth low-pass filter having a 
cutoff frequency equal to 1 rad/s and a dc gain equal to 1. 
 
2
1( )
( 1)( 1
H sn s s s
=
)+ + + 
Multiplying by 5 to change the dc gain to 5 and frequency scaling to change the cutoff 
frequency to ωc = 100 rad/s: 
 
3
L 2 2 22
5 5 100 5000000( )
( 100)( 100 100 ) ( 100)( 100 10000)
1 1
100 100 100
H s
s s s s s ss s s
⋅= = =+ + + + + +⎛ ⎞⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
 
1 
P16.3-3 
Use Table 16-3.2 to obtain the transfer function of a third-order Butterworth high-pass filter 
having a cutoff frequency equal to 1 rad/s and a dc gain equal to 5. 
 
3
2
5( )
( 1)( 1
sH sn s s s
=
)+ + + 
Frequency scaling to change the cutoff frequency to 100 rad/s:cω = 
3
3 3
H 2 2 22
5
5 5100( )
( 100)( 100 100 ) ( 100)( 100 10000)
1 1
100 100 100
s
s sH s
s s s s s ss s s
⎛ ⎞⎜ ⎟⎝ ⎠= = =+ + + + + +⎛ ⎞⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
 
 
 
P16.3-4 
Use Table 16-3.2 to obtain the transfer function of a fourth-order Butterworth high-pass filter 
having a cutoff frequency equal to 1 rad/s and a dc gain equal to 5. 
 
( ) ( )( )
4
2 2
5
0.765 1 1.848 1n
sH s
s s s s
= + + + + 
 
Frequency scaling can be used to adjust the cutoff frequency 500 hertz = 3142 rad/s: 
 
( ) ( ) ( )
4
4
H 2 2 2 2 2
5
53142
2403.6 3142 5806.4 3142
0.765 1 1.848 1
3142 3142 3142 3142
s
sH s
s s s ss s s s
⎛ ⎞⋅⎜ ⎟ ⋅⎝ ⎠= =⎛ ⎞⎛ ⎞ + + + +⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠
2
 
 
 
P16.3-5 
First, obtain the transfer function of a second-order Butterworth low-pass filter having a dc gain 
equal to 2 and a cutoff frequency equal to 2000 rad/s: 
 
( )L 2 22 80000002828 4000000
1.414 1
2000 2000
H s
s ss s
= = + +⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 
 
Next, obtain the transfer function of a second-order Butterworth high-pass filter having a 
passband gain equal to 2 and a cutoff frequency equal to 100 rad/s: 
2 
( )
2
2
H 2 2
2
2100
141.4 10000
1.414 1
100 100
s
sH s
s ss s
⎛ ⎞⋅⎜ ⎟ ⋅⎝ ⎠= = + +⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 
 
Finally, the transfer function of the bandpass filter is 
 
( ) ( ) ( ) ( ) ( )
2
B L H 2 2
16000000
141.4 10000 2828 4000000
sH s H s H s
s s s s
×= × = + + + + 
 
 
P16.3-6 
( ) ( )
2
2
B 222
250
25000014 250 2 250 62500250
1
s sH s
s ss s
⎛ ⎞⎜ ⎟= =⎜ ⎟ + +⎜ ⎟+ +⎝ ⎠
 
 
P16.3-7 
First, obtain the transfer function of a second-order Butterworth high-pass filter having a dc gain 
equal to 2 and a cutoff frequency equal to 2000 rad/s: 
 
( )
2
2
L 2 2
2
22000 
2828 4000000
1.414 1
2000 2000
s
sH s
s ss s
⎛ ⎞⎜ ⎟⎝ ⎠= = + +⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 
 
Next, obtain the transfer function of a second-order Butterworth low-pass filter having a pass- 
band gain equal to 2 and a cutoff frequency equal to 100 rad/s: 
 
( )H 2 22 20000141.4 10000
1.414 1
100 100
H s
s ss s
= = + +⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 
 
Finally, the transfer function of the band-stop filter is 
 
( ) ( ) ( ) ( ) ( )( )( )
( )( )
2 2 2
N L H 2 2
4 3 2 10
2 2
2 141.4 10000 20000 2828 4000000
141.4 10000 2828 4000000
2 282.8 40000 56560000 8 10
141.4 10000 2828 4000000
s s s s s
H s H s H s
s s s s
s s s s
s s s s
+ + + + += + = + + + +
+ + + + ⋅= + + + +
 
 
 
 
3 
 
 
 
P16.3-8 
( ) ( )( )
2 2
N 222 2
250 24 62500
14 4 250 250 62500250
1
ss
H s
s ss s
⎛ ⎞ +⎜ ⎟= − =⎜ ⎟ + +⎜ ⎟+ +⎝ ⎠
 
 
 
P16.3-9 
( ) ( )
2
2 4
L 222 2
250 4 2504 250 250 62500250
1
H s
s ss s
⎛ ⎞⎜ ⎟ ⋅= =⎜ ⎟ + +⎜ ⎟+ +⎝ ⎠
 
 
 
P16.3-10 
( ) ( )
2
2 4
H 222 2
44 250 250 62500250
1
s sH s
s ss s
⎛ ⎞⎜ ⎟ ⋅= =⎜ ⎟ + +⎜ ⎟+ +⎝ ⎠
 
 
4 
 
Section 16.4: Second-Order Filters 
 
P16.4-1 
The transfer function is 
( ) ( )( )
2
o
2
2s
2
1
1
1
1 1
1
1
s L
C s
s L ss LV s s LC s s LC RCH s s L sV s s LCR s L Rs L R s
C s s LC RC LCR
s L
C s
×
+ += = = = =+ +× + +++
+
+
 
so 
2 0
0 0
1 11 , and CK Q
LC RC Q L
ωω ω= = = ⇒ = =RC R 
 
2
0
1Pick 1 F. Then 1 H and 1000 LC L R Q
C C
μ ω= = = = = Ω 
 
 
P16.4-2 
The transfer function is 
0
2
1
( )( ) 1( )s
I s LCT s sI s s
RC LC
= =
+ +
 
so 
2 0
0 0
1 11 , and Ck Q
LC RC Q L
ωω ω= = = ⇒ = =RC R 
 
2
0
1Pick 1 F then 25 H and 3535 LC L R Q
C C
μ ω= = = = = Ω 
 
 
 
 
 
 
 
 
 
1 
P16.4-3 
The transfer function is 
2
1
2
2 2
1
1
( ) 
1 12 
R RCT s
Rs s
RC R R C
−
= ⎛ ⎞+ + +⎜ ⎟⎝ ⎠
 
 
Pick 0.01 FC μ= , then 
0
0
1
1
1 2000 50000 50 
1 2 8333 8.33 
2
R k
RC
R RR k
Q RC R Q
ω
ω
= = ⇒ = = Ω
⎛ ⎞= + ⇒ = = =⎜ ⎟ −⎝ ⎠
Ω
 
 
 
 
P16.4-4 
1 2 3Pick 0.02 F. Then 40 k , 400 k and R =3.252 k .C R Rμ= = Ω = Ω Ω
 
 
 
 
 
P16.4-5 
1 2Pick 1 FC C C μ= = = . Then 
6
0
1 2
10 
R R
ω= 
and 
1 10
2 2
1 2
1 
R R
Q R
R C Q R Q
ω= ⇒ = ⇒ = 
 
6
2 1 1
10In this case, since 1, we have and 1000 1 k
1000
Q R R R= = = = = Ω 
 
 
 
 
 
 
 
 
 
 
 
2 
P16.4-6 
 
 
The node equations are 
( ) ( )
( ) ( ) ( ) ( )( )
2
0 a
2
2
0 a
1 a i
1
1
0
R
V s V s
R
C s
V s V s
C s V s V s
R
=
+
− − − =
 
Doing a little algebra 
( ) ( ) ( )o a 1 1
1 1
1V s V s sC sC V s
R R
⎛ ⎞− + = −⎜ ⎟⎜ ⎟⎝ ⎠ i
 
Substituting for ( )aV s
 
( ) ( ) ( )o 2 2 1 1o 1
1 2 2 1
1 1V s R C s R C s
V s sC V s
R R C s R
⎛ ⎞+ +− − + =⎜ ⎟⎜ ⎟⎝ ⎠ i
 
( )
( )
( )
( )( )
2
1 1 2 2o 1 2 1 2
2
i 1 2 1 2 1 12 2 2 2 1 1 11 1
C s R R C sV s R R C C s
V s R R C C s R C sR C s R C s R C s
= = + +− + + + 
 
The transfer function is: 
( ) ( )( )
2
0
2i
2 2 1 2 1 2
1
V s sH s sV s s
R C R R C C
= =
+ +
 
 
0 2 2
1 2 1 20
2 11 2
1 1Pick 1 F. Then and 
R
C C C Q R Q R
R C Q RC R R
ωμ ω= = = = = ⇒ = ⇒ = . 
1 2 0
1In this case and 1000 R R R R
C R
ω= = = ⇒ = Ω . 
 
3 
P16.4-7 
 
( ) 0
2i
1 1
( )
1 1( )
V s C s LCT s RV s L s R s s
C s L LC
= = =
+ + + +
 
 
2 6When 25 , 10 H and 4 10 F, thenR L C− −= Ω = = × the transfer function is 
 
( ) 62 625 102500 25 10H s s s
×= + + × 
so 
6
old 25 10 5000ω = × = 
and 
new
f
old
250 0.05
5000
k ωω= = = 
The scaled circuit is 
 
 
 
 
 
 
 
 
P16.4-8 
 
 
The transfer function of this circuit is 
 
4 
( ) ( )( ) ( )( )
2
2 2 2 1 1 20
i 1 1 2 2 21
1 1 1 2 2 1 2 1 2
1
1
1 1 1 1 1 1 
R
s
R C s R C s R CV s
H s
V s R C s R C sR s sC s R C R C R R C C
+= = − = − = − ⎛ ⎞+ ++ + + +⎜ ⎟⎜ ⎟⎝ ⎠
 
 
( )
m
1 1 2 2
100 F 500 FPick 1000 so that the scaled capacitances will be 0.1 F and 0.5 F.
1000 1000
Before scaling 20 , 100 F, = 10 and 500 F
k
R C R C
μ μμ μ
μ μ
= =
= Ω = Ω=
=
 
 
( ) 2 5-100700 10
sH s
s s
= + + 
 ( )1 1 2 2After scaling 20000 20 k , 0.1 F, 10000 10 k , 0.5 FR C R Cμ μ= Ω= Ω = = Ω= Ω = 
 
( ) 2 5100700 10
sH s
s s
−= + + 
 
 
 
 
 
P16.4-9 
This is the frequency response of a bandpass filter, so 
 
( )
0
02 2
0
k s
QH s
s s
Q
ω
ω ω
=
+ +
 
From peak of the frequency response 
 
6 6
0 2 10 10 62.8 10 rad/s and =10 dB 3.16kω π= × × = × = 
Next 
6 6 6 60 BW (10.1 10 9.9 10 ) 2 (0.2 10 )2 1.26 10 rad/s
Q
ω π π= = × − × = × = × 
 
So the transfer function is 
 
6 6
2 6 2 12 2 6
3.16(1.26)10 (3.98)10( ) 
(1.26)10 62.8 10 (1.26)10 3.944 10
s sH s
s s s s
= =+ + × + + × 15 
 
5 
P16.4-10 
 
(a) ( ) ( )( )
2
220
 1 1 2
s 1 1
2
2
1
 = where and 1
R
j CjR
C R
j C
ωωω ω ω
ω
×
= − = − =
+
ZV
H Z Z
V Z
 
( ) 2 1 1 2
1 1 2 2
1 2
 1 1 = where , 
1+ 1+
j R C
R C Rj j
ωω ω
ω ω
ω ω
∴ − = =⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
H
C
ω 
 
 
(b) 1 2
1 1 2 2
1 1, 
R C R
ω ω= =
C
 
 
(c) 
( )
( )
2 1 22 1
1 2 1
1
11
 
 = 
0+ 1+0
R C Rj R C
R C
Rj
ωωω ωωω ωω
− = =⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
H = 
 
 
P16.4-11 
 
Voltage division: 
( ) ( ) ( ) ( )0 1 1
1
s , (1 )1
C sV s V V s s RC V s
R
C s
= ⇒ = +
+ 0
 
KCL: 
( )1 s 1 0 1 0+ + 0V V V V V V nC smR R
− − − = 
Combining these equations gives: 
 
2 2
0
1 sVsCV sC s n RC
mR m mR
⎡ ⎤+ + + =⎢ ⎥⎣ ⎦ 
6 
Therefore 
( )
( ) ( )0 22 2 2
0 0
V 1 1 ( ) =
V 1 1 s
1 
s m RC nm R C s
j
Q
ωω ω ω ω
ω ω
= =+ + + ⎛ ⎞− +⎜ ⎟⎜ ⎟⎝ ⎠
H 
0
1where and 
1 
mn
Q
mmn RC
ω = = + 
 
 
 
P16.4-12 
 
 
2
2
2 2 2 1 20
1 1S 21
11 1 1 2 2 1 2 1 2
1 1
1( )( ) = 1 1( ) 1 1 1 
RR s
C s R C s R CV sH s R C sV s R s sC s C s R C R C R R C C
−+= − = =+ ⎛ ⎞+ + + +⎜ ⎟⎜ ⎟⎝ ⎠
 
 
Substituting the element values , 1 100 R = Ω 2 200 R = Ω , 1 0.02 FC μ= and we 
determine the center frequency and bandwidth of the filter to be 
2 50 pFC =
 
( )
( )
0
1 2 1 2
0
1 1 2 2
1 70.7 k rad sec = 2 11.25 kHz
1 1 150 k rad =2 23.9 kHz
R R C C
BW s
Q R C R C
ω π
ω π
= =
= = + =
 
 
 
 
 
 
 
7 
 
 
 
P16.4-13 
 
 
( ) a 01 a s
1 2 20
2a s
2 0
1 1 1 2 1 22
1 = 0 s
( ) ( )= 1 1( ) ss 0
V VC s V V
R R CV sH s
V V s sC V R C R R C CR
− ⎫− + −⎪⎪ ⇒ =⎬⎪ + +− − = ⎪⎭
 
 
Comparing this transfer function to the standard form of the transfer function of a second order 
bandpass filter and substituting the element values 1 1 kR = Ω , 2 100 R = Ω , 1 1 FC μ= and 
2 0.1 FC μ= gives: 
4
0
1 2 1 2
3
 
1 1
0
1 10 rad sec
1BW = 10 rad/sec
10
BW
R R C C
R C
Q
ω
ω
= =
=
= =
 
 
 
 
8 
P16.4-14 
 
 
Node equations: 
 
( )( ) ( ) ( )
c s c 0
c
s c
s 0
( ) ( ) ( ) ( )( ) 0
( ) 0 
V s V s V s V saC sV s
R R
V s V sC s V s V s
a R
 
− −+ + =
−− + =
 
 
Solving these equations yields the transfer function: 
 
( )
( )
( )
( )
2
2
0
2s
2
2 1 1 
( )
2 1 1 
s a s
a RCV s RC
H s
V s s s
a RC RC
⎛ ⎞+ + +⎜ ⎟⎝ ⎠= = ⎛ ⎞+ +⎜ ⎟⎝ ⎠
 
 
We require 5 110 
RC
= . Pick 0.01 F then 1000 C Rμ= = Ω . Next at 0s jω= 
( ) 20
2
12 2
a aa
a
ω
+
= = +H 
The specifications require 
( ) 20201 1 202
a aω= = + ⇒ =H 
 
9 
 
P16.4-15 
 
Node equations: 
 
( )
a a 0
2 1
a b
a
b s b a
b 0
=0
 0
+ + 
V V V
R R
V VC sV
R
V V V VC s V V 0
R R
−+
−+ =
− −− =
 
 
 
Solving the node equation yields: 
( )
( )
1
2 2
20
s 12
2 2
2
11
1 12 
R
R R CV s
V s R
s s
R RC R C
⎛ ⎞+⎜ ⎟⎜ ⎟⎝ ⎠= ⎛ ⎞+ − +⎜ ⎟⎜ ⎟⎝ ⎠
 
( ) ( )0 3 91 1 41.67 k rad sec1.2 10 20 10RCω −= = =× × 
 
 
10 
Section 16.5: High-Order Filters 
 
P16.5-1 
This filter is designed as a cascade connection of a Sallen-Key low-pass filter designed as 
described in Table 16.4-2 and a first-order low-pass filter designed as described in Table 16.5-2. 
 
Sallen-Key Low-Pass Filter: 
 
 
 
MathCad Spreadsheet (p16_5_1_sklp.mcd) 
A 2=Calculate the dc gain.
R A 1−( )⋅ 1.592 104×=R 1.592 104×=A 3 1
Q
−:=R 1
C ω0⋅
:=Calculate resistance values:
C 0.1 10 6−⋅:=Pick a convenient value for the capacitance:
Q 1=ω0 628=Q
ω0
b
:=ω0 a:=Determine the Filter Specifications:
b 628:=a 6282:=Enter the transfer function coefficitents:
 c
 ----------------- .
 s^2 + bs + a
 
The transfer function is of the form H(s) =
 
1 
First-Order Low-Pass Filter: 
 
 
MathCad Spreadsheet (p16_5_1_1stlp.mcd) 
 
The transfer function is of the form H(s) =
 -k
 -------- .
 s + p
Enter the transfer function coefficitents: p 628:= k 0.5p:=
Pick a convenient value for the capacitance: C 0.1 10 6−⋅:=
Calculate resistance values: R2
1
C p⋅:= R1
1
C k⋅:= R1 3.185 10
4×= R2 1.592 104×=
 
 
P16.5-2 
This filter is designed as a cascade connection of a Sallen-key high-pass filter, designed as 
described in Table 16.4-2, and a first-order high-pass filter, designed as described in Table 16.5-
2. 
 
The passband gain of the Sallen key stage is 2 and the passband gain of the first-order stage is 
2.5 So the overall passband gain is 2 × 2.5 = 5 
 
Sallen-Key High-Pass Filter: 
 
 
 
2 
MathCad Spreadsheet (p16_5_2_skhp.mcd) 
A 2=Calculate the passband gain.
R A 1−( )⋅ 1 105×=R 1 105×=A 3 1
Q
−:=R 1
C ω0⋅
:=Calculate resistance values:
C 0.1 10 6−⋅:=Pick a convenient value for the capacitance:
Q 1=ω0 100=Q
ω0
b
:=ω0 a:=Determine the Filter Specifications:
b 100:=a 10000:=Enter the transfer function coefficitents:
 A s^2
 ----------------- .
 s^2 + bs + a
 
The transfer function is of the form H(s) =
 
 
 
First-Order High-Pass Filter: 
 
 
 
MathCad Spreadsheet (p16_5_2_1sthp.mcd) 
 
The transfer function is of the form H(s) =
 -ks
 -------- .
 s + p
Enter the transfer function coefficitents: p 100:= k 2.5:=
Pick a convenient value for the capacitance: C 0.1 10 6−⋅:=
Calculate resistance values: R1
1
C p⋅:= R2 k R1⋅:= R1 1 10
5×= R2 2.5 105×=
 
 
 
3 
P16.5-3 
This filter is designed as a cascade connection of a Sallen-key low-pass filter, a Sallen-key high-
pass filter and an inverting amplifier. 
 
Sallen-Key Low-Pass Filter: 
 
 
 
MathCad Spreadsheet (p16_5_3_sklp.mcd) 
A 1.586=Calculate the dc gain.
R A 1−( )⋅ 2.93 103×=R 5 103×=A 3 1
Q
−:=R 1
C ω0⋅
:=Calculate resistance values:
C 0.1 10 6−⋅:=Pick a convenient value for the capacitance:
Q 0.707=ω0 2 103×=Q
ω0
b
:=ω0 a:=Determine the Filter Specifications:
b 2828:=a 4000000:=Enter the transfer function coefficitents:
 c
 ----------------- .
 s^2 + bs + a
 
The transfer function is of the form H(s) =
 
 
 
 
4 
Sallen-Key High-Pass Filter: 
 
 
 
MathCad Spreadsheet (p16_5_3_skhp.mcd) 
A 1.586=Calculate the passband gain.
R A 1−( )⋅ 5.86 104×=R 1 105×=A 3 1
Q
−:=R 1
C ω0⋅
:=Calculate resistance values:
C 0.1 10 6−⋅:=Pick a convenient value for the capacitance:
Q 0.707=ω0 100=Q
ω0
b
:=ω0 a:=Determine the Filter Specifications:
b 141.4:=a 10000:=Enter the transfer function coefficitents:
 c s^2
 ----------------- .
 s^2 + bs + a
 
The transfer function is of the form H(s) =
 
 
 
Amplifier: The required passband gain is 
61.6 10 4.00
141.4 2828
× =× . An amplifier with a gain equal to 
4.0 1.59
2.515
= is needed to achieve the specified gain.5 
 
P16.5-4 
This filter is designed as the cascade connection of two identical Sallen-key bandpass filters: 
 
Sallen-Key BandPass Filter: 
 
 
 
MathCad Spreadsheet (p16_5_4_skbp.mcd) 
A Q⋅ 2=Calculate the pass-band gain.
R A 1−( )⋅ 4 104×=2 R⋅ 8 104×=R 4 104×=
A 3
1
Q
−:=R 1
C ω0⋅
:=Calculate resistance values:
C 0.1 10 6−⋅:=Pick a convenient value for the capacitance:
Q 1=ω0 250=Q
ω0
b
:=ω0 a:=Determine the Filter Specifications:
b 250:=a 62500:=Enter the transfer function coefficitents:
 cs
 ----------------- .
 s^2 + bs + a
 
The transfer function is of the form H(s) =
 
 
 
 
 
 
 
 
 
 
 
6 
 
P16.5-5 
This filter is designed using this structure: 
 
 
 
Sallen-Key Low-Pass Filter: 
 
 
 
MathCad Spreadsheet (p16_5_5_sklp.mcd) 
A 1.586=Calculate the dc gain.
R A 1−( )⋅ 5.86 104×=R 1 105×=A 3 1
Q
−:=R 1
C ω0⋅
:=Calculate resistance values:
C 0.1 10 6−⋅:=Pick a convenient value for the capacitance:
Q 0.707=ω0 100=Q
ω0
b
:=ω0 a:=Determine the Filter Specifications:
b 141.4:=a 10000:=Enter the transfer function coefficitents:
 c
 ----------------- .
 s^2 + bs + a
 
The transfer function is of the form H(s) =
 
7 
Sallen-Key High-Pass Filter: 
 
 
 
MathCad Spreadsheet (p16_5_5_skhp.mcd) 
A 1.586=Calculate the passband gain.
R A 1−( )⋅ 2.93 103×=R 5 103×=A 3 1
Q
−:=R 1
C ω0⋅
:=Calculate resistance values:
C 0.1 10 6−⋅:=Pick a convenient value for the capacitance:
Q 0.707=ω0 2 103×=Q
ω0
b
:=ω0 a:=Determine the Filter Specifications:
b 2828:=a 4000000:=Enter the transfer function coefficitents:
 c s^2
 ----------------- .
 s^2 + bs + a
 
The transfer function is of the form H(s) =
 
 
 
Amplifier: The required gain is 2, but both Sallen-Key filters have passband gains equal to 1.586. 
The amplifier has a gain of 2 1.26
1.586
= to make the passband gain of the entire filter equal to 2. 
 
 
 
 
 
 
8 
P16.5-6 
This filter is designed as the cascade connection of two identical Sallen-key notch filters. 
 
Sallen-Key Notch Filter: 
 
 
MathCad Spreadsheet (p16_5_6_skn.mcd) 
A 1.5=Calculate the pass-band gain.
R A 1−( )⋅ 2 104×=R
2
2 104×=R 4 104×=
A 2
1
2 Q⋅−:=R
1
C ω0⋅
:=Calculate resistance values:
2 C⋅ 2 10 7−×=C 0.1 10 6−⋅:=Pick a convenient value for the capacitance:
Q 1=ω0 250=Q
ω0
b
:=ω0 a:=Determine the Filter Specifications:
b 250:=a 62500:=Enter the transfer function coefficitents:
 c(s^2 + a)
 ----------------- .
 s^2 + bs + a
 
The transfer function is of the form H(s) =
 
 
Amplifier: The required passband gain is 4. An amplifier having gain equal to 4 1.78
(1.5)(1.5)
= 
is needed to achieve the required gain. 
 
 
 
 
 
 
 
9 
P16.5-7 
 
(a) 
 
Voltage division gives: ( ) ( )( )
1 1 1
a
s 1
1
1 1
V s R R C s
H s
V s R C sR
C s
= = = ++
 
 
(b) 
 
Voltage division gives: ( ) ( )( )
2
b
1 2
V s L sH s
V s R L s
= = + 
 
(c) 
 
Voltage division gives: 
 
 
 
 
Doing some algebra: 
( ) ( )( )
( )
( )
1 22
c
s 2
1 2
||
1 ||
R R L sV s L sH s
V s R L sR R L s
C s
+= = × ++ +
 
( ) ( )( )
( )
( )
( )
( )
( )
( )
( )
1 2
1 22
c
s 21 2
1 2
2
1 2 1
2
1 2 1 1 2 2
1 2
2
21 1 2 1 2
2
1
2
1 1 2 1 2
1
R R L s
R R L sV s L sH s
V s R L sR R L s
C s R R L s
R R C s R LC s L s
R R C s R LC s R R L s R L s
R C s R L s L s
R L sR LC s R R C L s R R
R LC s
R LC s R R C L s R R
× +
+ += = × +× ++ + +
+= ×+ + + + +
+= × ++ + + +
= + + + +
 
(d) ( ) ( ) ( )c a bH s H s H s≠ × because the 2 ,R L s voltage divider loads the 11 , RC s voltage 
divider. 
 
 
 
 
 
 
 
 
 
 
10 
P16.5-8 
( ) 100 20
1 1 1 1
200 20,000 20 4000
2000
1 1 1 1
20 200 4000 20,000
H s
s s s s
π
s s s s
π π π
π π π
= ×⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞+ + + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
= ⎛ ⎞⎛ ⎞⎛ ⎞⎛+ + + +⎜ ⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠⎝ ⎠⎝ ⎠⎝ π
⎞⎟⎠
 
 
 
 
 
 
P16.5-9 
(a) The transfer function of each stage is 
( )
2
2 2
2 2
2 1
i
1 1 1
1
1 1|| 1
1
R
C s
R R
R R R C s RC s C sH s
2R R R R
×
+ += − = − = − = − + C s 
 
The specification that the dc gain is 0 dB = 1 requires 2 1R R= . 
The specification of a break frequency of 1000 rad/s requires 
2
1 1000
R C
= . 
Pick 0.1 FC μ= . Then 2 10 kR = Ω so 1 10 kR = Ω . 
 
(b) ( ) ( )
2
2
1 1 1 110,000 40.1 dB
1011 101 1
1000 1000
j j
ω ω ω
⎛ ⎞− −= × ⇒ = = = −⎜ ⎟+⎝ ⎠+ +
H H 
 
11 
Section 16.7 How Can We Check…? 
 
P16.7-1 
0
0
10010000 100 rad and 25 4 5
25
s Q
Q
ωω = = = ⇒ = = ≠ 
This filter does not satisfy the specifications. 
 
 
P16.7-2 
0
0
100 75 10000 100 rad , 25 4 and 3
25 25
s Q k
Q
ωω = = = ⇒ = = = = 
 
This filter does satisfy the specifications. 
 
 
P16.7-3 
0
0
20 600 400 20 rad s , 25 0.8 and 1.5
25 400
Q k
Q
ωω = = = ⇒ = = = = 
 
This filter does satisfy the specifications. 
 
 
P16.7-4 
0
0
25 750 625 25 rad s , 62.5 0.4 and 1.2
62.5 625
Q k
Q
ωω = = = ⇒ = = = = 
 
This filter does satisfy the specifications. 
 
 
P16.7-5 
0
0
12 144 12 rad/s and 30 0.4
30
Q
Q
ωω = = = ⇒ = = 
 
This filter does not satisfy the specifications. 
 
 
 
1 
	Ch16sec3
	Chapter 16: Filter Circuits 
	Exercises 
	Ex. 16.3-1 
	Problems 
	Section 16.3: Filters 
	P16.3-1 
	P16.3-2 
	P16.3-3 
	P16.3-4 
	P16.3-5 
	P16.3-6 
	P16.3-7 
	P16.3-8 
	P16.3-9 
	P16.3-10 
	Ch16sec4
	Section 16.4: Second-Order Filters 
	P16.4-1 
	P16.4-2 
	P16.4-3 
	P16.4-4 
	P16.4-5 
	P16.4-6 
	P16.4-7 
	P16.4-8 
	P16.4-9 
	P16.4-10 
	P16.4-11 
	P16.4-12 
	P16.4-13 
	P16.4-14 
	P16.4-15 
	Ch16sec5
	Section 16.5: High-Order Filters 
	P16.5-1 
	P16.5-2 
	P16.5-3 
	P16.5-4 
	P16.5-5 
	 P16.5-6 
	P16.5-7 
	P16.5-8 
	P16.5-9 
	Ch16sec7
	Section 16.7 How Can We Check…? 
	P16.7-1 
	P16.7-2 
	P16.7-3 
	P16.7-4 
	P16.7-5