M2-3S Active Filter (Part II)
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•
•
•
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Biquadratic function filters
Positive feedback active filter: VCVS
Negative feedback filter: IGMF
Butterworth Response
Chebyshev Response
Active Filter (Part 2)
1
Biquadratic function filters
s 
2
Z
s 
2
Z
s  cs  d
QZ
H ( s)  2
K
P
s  as  b
2
2
s 
s  P
QP
2
Realised by:
(I) Positive feedback
(II) Negative feedback
Active Filter (Part 2)
(III) Band Pass
Biquadratic functions
s
s

K

s 2  as  b
s 2  P s  P2
QP
(IV) Band Stop
(I) Low Pass
H ( s)  K
H ( s)  K
1
1

K

s 2  as  b
s 2  P s  P2
QP
s2  b
s 2  Z2
H ( s)  K 2
K

s  as  b
s 2  P s  P2
QP
(V) All Pass
(II) High Pass
2
2
s
s
H ( s)  K 2
K

s  as  b
s 2  P s  P2
QP
s2 
Z
s  Z2
s 2  as  b
QZ
H ( s)  K 2
K

s  as  b
s 2  P s  P2
QP
Active Filter (Part 2)
Low-Pass Filter
Voltage Gain (dB)
Voltage Gain
Qp = 1.5
1.5
Qp = 1.5
Qp = 1
0
K = 1, p = 1
Qp = 1
1
Qp =
Qp =
1
2
-10
2
Qp =
0.5
-20
p
Q
=
Qp =
0 .5
0.5
Qp
0
1
0
=0
-30
K = 1, p = 1
.1
1
2
3
4
0 .1
0
1
2
Frequency
Frequency
H s  
1
s
2
s 
1
Qp
Active Filter (Part 2)
3
4
High-Pass Filter
Voltage Gain
Voltage Gain (dB)
5
Qp = 1.5
1.5
Qp = 1.5
K = 1, p = 1
0
Qp = 1
1
Qp =
Qp = 1
1
Qp = 0.5
-5
2
Qp =
Qp = 0
.5
2
-10
0.5
Qp =
Qp = 0
0
1
0
1
-15
.1
2
0 .1
K = 1, p = 1
3
4
-20
0
Frequency
1
2
Frequency
s2
H s  
s
s2 
1
Qp
Active Filter (Part 2)
3
4
Band-Pass Filter
Voltage Gain (dB)
Voltage Gain
5
1.5
Qp = 1.5
K = 1, p = 1
0
-5
Qp = 1
1
Qp =
1
-10
2
-15
0.5
-30
2
Qp = 0.1
-25
Qp = 0.1
0
Qp = 0.5
-20
Qp = 0.5
0
Qp = 1.5
Qp = 1
1
Qp =
2
4
Frequency
6
8
K = 1, p = 1
0
2
4
Frequency
H s  
s
s
s2 
1
Qp
Active Filter (Part 2)
6
8
Band-Stop Filter
Voltage Gain
Voltage Gain (dB)
5
10
Qp = 1.5
K = 1, p = 1, z = 2
Qp = 1.5
4
0
3
Qp =
-10
2
0 .1
0.5
Qp =
-20
1
2
Qp
1
1
Qp =
Qp =
Voltage Gain
Qp = 1
2
K = 1, p = 1, z = 2
.1
=0
0
0
2
4
Frequency
6
-30
8
0
2
s 2
H s  
s
2
s 
1
Qp
2
2
Active Filter (Part 2)
4
Frequency
6
8
Voltage Controlled Votage Source (VCVS)
Positive Feedback Active Filter (Sallen-Key)
Z3
if
ii ia
Z1
Vi
By KCL at Va:
where,
Z4
Va
ii  i f  ia  0
V  Va
ii  i
Z1
if 
Vo  Va
Z3
Va
ia 
Z2  Z4
K
Z2
Vo
Therefore, we get
V i  V a Vo  V a
Va


0
Z1
Z3
Z2  Z4
Re-arrange into voltage group gives:
 1
Vi Vo
1
1 

 Va  

 0
Z1 Z3
 Z1 Z3 Z2  Z4 
Active Filter (Part 2)
(1)
But,
Vo  Ki a Z 4  K
Z 4Va
Z2  Z4
(2)
Substitute (2) into (1) gives
Vi Vo Vo Z2  Z4   1
1
1 



 
 0
Z1 Z3
KZ4
 Z1 Z3 Z2  Z4 
or
H
Vo
K

ZZ
1
Vi Z1
1  K   1 2  1  Z1  Z 2 
Z3
Z3Z 4
Z4
(3)
In admittance form:
H
K
 1
1  Y3
Y3Y4


1  Y4    
1 K 
Y1Y2
 Y1 Y2  Y1
(4)
* This configuration is often used as a low-pass filter, so a
specific example will be considered.
Active Filter (Part 2)
VCVS Low Pass Filter
1
H (s )  K 2
s  as  b
In order to obtain the above response, we let:
Z 1  R1
Z 2  R2
Z3 
R1
R2
1
1

jC 3 sC 3
Z4 
1
1

jC 4 sC 4
C3
C4
Then the transfer function (3) becomes:
H ( s) 
K
K'

1  sC4 R1  R2   sR1C3 1  K   s 2 R1 R2C3C4 s 2  P s   2
P
QP
Active Filter (Part 2)
(5)
we continue from equation (5),
K
H ( s)  2
s R1 R2 C3C4  sC4 R1  R2   sR1C3 1  K   1
1
R1 R2 C3C 4
H ( s) 
 C 4 R1  R2   R1C 3 1  K  

1
2
s  s



R
R
C
C
R
R
C
C

  1 2 3 4
1
2
3
4
K
Equating the coefficient from equations (6) and (5), it gives:
P 
1
R1 R2 C3C4

1
1
R1C3
R2 C4
QP 
1
RC
R1C4
R2 C4

 1  K  1 3
R2 C3
R1C3
R2 C4
Now, K=1, equation (5) will then become,
H (s ) 
1
1  sC 4 R1  R 2   s 2 R 1R 2C 3C 4
Active Filter (Part 2)
Simplified Design (VCVS filter)
Z1  mR
H ( s) 
Z2  R
Z3 
1
1

j (nC ) snC
1
1  sRC m  1  s 2 nmR 2C 2
mR
Comparing with the low-pass response:
1
H (s)  K

s 2  P s   P2
QP
It gives the following:
p 
1
RC nm
mn
Qp 
m 1
Active Filter (Part 2)
Z4 
1
1

jC sC
R
nC
C
Example (VCVS low pass filter)
To design a low-pass filter with f O  512Hz and Q 
1
2
nC
Let m = 1
mn
1 n
n
1
QP 



m 1 11
2
2
mR
R
+
vin
-
C
 n=2
1
1
P 


 2 (512Hz )
RC mn RC 1  2 RC 2
vo
1
Choose C  100nF
2k2
200nF
2k2
+
Then
R  2,198 ~ 2.2k
vin
What happen if n = 1?
Active Filter (Part 2)
100nF
-
vo
VCVS High Pass Filter
C1
C2
R3
+
vin
R4
-
vo
KS 2
K'S2
H ( s) 

P
2
2
 1

1
1
1
2
s

s


1  K  
s  s


P
Q
P
 R4C1 R4C2 R3C1
 R3 R4C1C2
Active Filter (Part 2)
VCVS Band Pass Filter
R1
vin
C2
+
R4
C5
K
H ( s) 
R3
-
s
R1C5
s 1
1
1
C 
R1  R3
s     1  K   5  
C5  R1 R4 R3
R4C2  R1 R3 R4C2C5
2
Active Filter (Part 2)
vo
K'S

s2 
P
QP
s  P2
Infinite-Gain Multiple-Feedback (IGMF)
Negative Feedback Active Filter
Z
4
Z1
V
i
vi  0
vi  0 
By KCL at Vx ,
Z
5
Z
3
V
x
Z
2
Vo  
Z5
Z
Vx  Vx   3 Vo
Z3
Z5
Vi  Vx Vx Vx Vx  Vo

 
Z1
Z 2 Z3
Z4
V
o
Note: because no
current flows into v+,
v- terminals of op-amp.
Therefore, from KCL at
node v- :
Vo/Z5+ Vx/Z3 = 0
(1)
(2)
substitute (1) into (2) gives
Vi
Z3
Z3
Vo
Z3
Vo

V 
V 

V 
Z1 Z1Z5 o
Z 5Z 2 o Z 5 Z 4 Z 5 o Z 4
Active Filter (Part 2)
(3)
rearranging equation (3), it gives,
1
V
Z1Z 3
H o 
Vi
1  1
1
1
1
1

 
 
Z5  Z1 Z 2 Z 3 Z 4  Z 3Z 4
Or in admittance form:
H
Vo
Y1Y3

Vi
Y5 Y1  Y2  Y3  Y4   Y3Y4
Filter Value
LP
Z1
Z2
Z3
Z4
Z5
R1
C2
R3
R4
C5
HP
C1
R2
C3
C4
R5
BP
R1
R2
C3
C4
R5
Active Filter (Part 2)
IGMF Band-Pass Filter
s
Band-pass: H ( s )  K 2
s  as  b
To obtain the band-pass response, we let
Z1  R1
Z 2  R2
Z3 
1
1

jC3 sC3
Z4 
sC 3
R1
H (s )  
C  C4
1  1
1
s 2C 3C 4  s 3

  
R5
R 5  R1 R 2 
*This filter prototype has a very low
sensitivity to component tolerance when
compared with other prototypes.
Active Filter (Part 2)
1
1

jC4 sC 4
R1
R2
C4
C3
Z 5  R5
R5
Simplified design (IGMF filter)
sC
R1
H (s)  
1
2C
s
 s 2C 2
R1 R5
R5
R1
C
R5
C
Comparing with the band-pass response
s
H ( s)  K
s2 
P
QP
s  P2
Its gives,
p 
1
C R1 R5
Qp 
1 R5
2 R1
K-
1
CR1
Active Filter (Part 2)
H  j p   2Q 2
Example (IGMF band pass filter)
To design a band-pass filter with
P 
f O  512Hz and Q  10
1
 2 (512Hz )
C R1 R5
C  100 nF 
R1R5  9,662,741 2
1 R5
QP 
 10
2 R1
 R1  155.4
100nF
62,170
155.4
R5  62,170
vin
100nF
With similar analysis, we can choose the following values:
C  10 nF R1  1,554  and R5  621,700 
Active Filter (Part 2)
+
vo
Butterworth Response (Maximally flat)
H  j  
1

1   
 o 
2n
where n is the order
Normalize to o = 1rad/s
Hˆ ( j ) 
1
1   2n
Butterworth polynomials:
1
Hˆ ( j ) 
Bn ( j )
Butterworth polynomials
B1 s   s  1
B2 s   s 2  2 s  1
B3 s   s 3  2s 2  2s  1


 s  1 s 2  s  1
B4 s   s 4  2.61s 3  3.41s 2  2.61s  1



 s 2  0.77 s  1 s 2  1.85s  1
B5 s   s 5  3.24 s 4  5.24s 3  5.24s 2  3.24s  1



 s  1 s 2  0.62s  1 s 2  1.62s  1
Active Filter (Part 2)
Butterworth Response
Active Filter (Part 2)
Second order Butterworth response
Started from the low-pass biquadratic function H ( s )  K
For
p  1
K 1
Q
1
2
1
s2 
P
QP
1
(second order butterwoth polynomial )
2
s  2s  1
1
H ( j ) 
  2  2 j  1
1
H ( j ) 
2
2
1   2  2
H (s) 
 

H ( j ) 
H ( j ) 
H ( j ) 

1
1  2 2   4  2 2
1
1  4
1
1   
22

1
1   
2n
Active Filter (Part 2)
s  P2
Bode plot (n-th order Butterworth)
x

1
1 
2n
In dB form :
Hˆ ( j )  20 log
1 st
-20
1
-60
d
or
de
o rd
r
er
r
h
or
-80
de
r
de
-100
or
r
suppose   1
3r
5t h
Hˆ ( j )  20 log( 1   2 n )
2nd
-40
o rd e
4t
1 
2n
voltage gain (dB)
Hˆ ( j ) 
x
Hˆ ( j )  20n log(  )
Butterworth response
For decade condition,   10 x
The Butterwort h filter wou ld have (-20n)dB/d ecade
For 2nd order : - 40 dB/decade
For 3rd order : - 60 dB/decade
For n - th order : - 20n dB/.decade
Active Filter (Part 2)
Second order Butterworth filter
K
K'

1  sC4 R1  R2   sR1C3 1  K   s 2 R1 R2C3C4 s 2  P s   2
P
Q
1
P
R1C4
R2C4
RC

 1  K  1 3
R2C3
R1C3
R2C4
C3
R
R
H ( s) 
QP 
1
2
+
Setting R1= R2 and C1 = C2
1
1
1
QP 


1  1  1  K  1 2  1  K  3  K
C4
vin
-
RB
vo
RA
Now K = 1 + RB/ RA
QP 
1

3 K
1
 R 
3  1  B 
 RA 

1
R
2 B
RA
For Butterworth response:
1
QP 
2

1
1
QP 

2 2  RB
RA
Therefore, we have 2 
RB
 2  1.414
RA
We define Damping Factor (DF) as:
Active Filter (Part 2)
DF 
R
1
 2  B  1.414
QP
RA
Damping Factor (DF)
• The value of the damping factor required to produce desire response
characteristic depends on the order of the filter.
• The DF is determined by the negative feedback network of the filter
circuit.
• Because of its maximally flat response, the Butterworth characteristic
is the most widely used.
• We will limit our converge to the Butterworth response to illustrate
basic filter concepts.
Active Filter (Part 2)
Values for the Butterworth response
B1 s   s  1
B2 s   s 2  2 s  1


 2.61s  1  s
B3 s   s 3  2 s 2  2 s  1  s  1 s 2  s  1
B4 s   s 4  2.61s 3  3.41s 2
B5 s   s 5  3.24 s 4  5.24 s 3  5.24 s 2

 3.24 s  1  s  1s
2

 0.77 s  1 s 2  1.85s  1
Order
Roll-off
dB/decade
1st stage
poles
DF
1
-20
1
optional
2
-40
2
1.414
3
-60
2
4
-80
5
6
2
2nd stage
poles
DF
1.000
1
1.000
2
1.848
2
0.765
-100
2
1.000
2
-120
2
1.932
2
Active Filter (Part 2)


 0.62 s  1 s 2  1.62 s  1
3rd stage
poles
DF
1.618
1
0.618
1.414
2
0.518
Forth order Butterworth Filter
R3  R4 , C3  C 4
R1  R2 , C1  C2
C1 0.01 F
R1 8.2 k
C3 0.01 F
+15
V
R2 8.2 k
R3 8.2 k
+
R4 8.2 k
-741C
C2
0.01 F
-15 V
RB 1.5 k
Vout
RB 27 k
-15 V
RA 22 k
RB
 1.848
RA
RB
 0.152
RA
+
- 741C
C4
0.01 F
RA 10 k
DF  2 
+15
V
DF  2 
RB
 0.765
RA
RB
 1.235
RA
RB  0.152R A  0.15210k
RB  1.235R A  1.23522k
RB  1.52k
RB  27.17 k
Active Filter (Part 2)
Chebyshev Response (Equal-ripple)
H  j  
1
1   2Cn2  
Where  determines the ripple and
Cn2 is the Chebyshev cosine polynomial defined as
Active Filter (Part 2)
Chebyshev Cosine Polynomials
C0    1
C1    
C2    2 2  1
C3    4 3  3
C4    8 4  8 2  1
C5    16 5  20 3  5
Cn    2Cn 1    Cn  2  
Active Filter (Part 2)
Second order Chebychev Response
H  j  
1
1   2Cn2  
Example: 0.969dB ripple gives  = 0.5, C2    2 2  1
H 22   
1
1   2Cn2  
1
2
1  0.52 2  1
1
 4
   2  1.25

H 22 ( )
 s / f
Roots:
s2 
1 1 5
1
  j
2
2
H 2 s H 2  s  
1
 2 1
 2 1

 s   j  s   j 
2
2



 H 2 s H 2  s 

1
Note:  20 log 10 (
s  s  1.25
4
2
Active Filter (Part 2)
1
1 
2
)  0.969
Roots
Roots of first bracketed term
Roots of second bracketed term
1

0   4  j 
2

s
2
1
  j
2
1
0   4 
2
s
2
1
   j
2
 1 5
1 
1  5  1 

   
      j 
     
2  4  2 
 2  4  2  

 0.566  j 0.899
1

j

 1 5
1 
1  5  1 

   
      j 
     
2  4  2 
 2  4  2  

 0.566  j 0.899


 Ce j  C e j 2  C (cos( )  j sin( ))
2
2
1
1
H 2 s H 2  s  
．
s  0.556  j 0.899s  0.556  j 0.899 s  0.566  j 0.899s  0.556  j 0.899
Note:
or H 2 s  
a  bj 
a 2  b 2 e j tan
1
(b / a )
s  0.556  j 0.899s  0.556 
j 0.899

1
1

s  0.5562  0.899 2 s 2  0.112s  1.117
Active Filter (Part 2)