Synthesis of Butterworth low-pass filter
(short version)
ωp - passband frequency
ωs - stopband frequency
αp - attenuation in passband
αs - attenuation in stopband
filter order can be calculated as
 10S

 10  1 
log   P

 10 10  1 


n
 
2 log  S 
 P 
S


 10S



10
10

1
 10  1 
log


log   P
P


 10 10  1 
 10 10  1 




n

2
   
 
log  S 
log   S  
  P  
 P 


or
0
it needs to be roundup n to the integer value
90
180
 1

 k  1
 901  2k  1
n
n
 n

 1

or  k  1  2k  1
2 n

for 0 = 1
 k  90 
ak = - cos  k 
Qk = -
0
2a


0
bk =  sin  k 
 sin  
 cos 
1
2 cos  
Frequency-attenuation relationship:
1
    2n 
  10 log 1    
  0  


S
0 
1
S
2
 10
 n
10  1





  10  2 n
 10  1
0 

0 
P
P
 10

10  1




P
1
2n
0 dB
P
1
  P   S
 4 n
10 10  110 10  1



P
0 dB
P
0
S
0 dB
P
S
S
0
 P S
S
P
S
0 
S
0
1
Synthesis of Chebyshev low-pass filter
ωp - passband frequency
ωs - stopband frequency
αp - attenuation in passband
αs - attenuation in stopband
1
1
cosh 1
10
S
10
1  2
1
P
10
10  1
 
cosh 1  S 
 P 
n
x    ln x  x  1
sinh 1 x   ln x  x 2  1
cosh
1
0
2
Location of poles of Chebyshev filter
90
180
 1

 k  1
 901  2k  1
n
n
 n

 1

 k  1  2k  1
2 n

 k  90 
or
  10
for
P
10
1
0 = 1




1
1
1 1
1
  sinh    sinh 

P
n
  n


10
 10  1 
 ak  cos k sinh  
0k  a  b
2
k
Qk 
 bk  sin  k cosh  
cosh  
sinh  
cosh  sin  

sinh   cos 
2
k
k
2 ak
2
Design of Low Pass Inverse Chebyshev filter
S
cosh 1
10 10  1
P
10 10  1
 
cosh 1  S 
 P 
n
S


1
1 1
1 
  sinh    sinh
10 10  1 


n
  n


1

S
10 10  1
notice that for Chebyshev was
k 
  10
P
10
1




1
1
1 1
1
  sinh    sinh 

P
n
  n


 10 10  1 

1

1  2k  1

2 n

pole locations:
0 = 1
ak  cos k sinh  
for
pIC 
0 k 
cosh  
 bk  sin  k cosh  
1
a  jbk
 k2
ak  jbk
ak  bk2
1
a b
2
k
zero locations:
2
k
Qk 
k
2 ak
1
 i 
i 
 sec 
 i 
 2n 
cos 
 2n 
i  1, 3, 5, ..... i=0 to i<np number of poles
i  2j - 1
j=1 to nc number of conjugate poles
sinh  
cosh  sin  

sinh   cos 
3
% Butterwoth filter
wp=1000;
ws=1725;
ap=0.5;
as=20;
n=0.5*log10((10^(0.1*as)-1)/(10^(0.1*ap)-1))/log10(ws/wp);
n
n=ceil(n)
for k=1:n,
th(k)=0.5*pi*(1+(2*k-1)/n);
b(k)=sin(th(k));
a(k)=-cos(th(k));
s(k)=complex(a(k),b(k));
Q(k)=abs(1/(2*a(k)));
end;
% Chebyshev filter
wp=1000;
ws=1725;
ap=0.5;
as=20;
%x1=sqrt((10^(as/10)-1)/(10^(ap/10)-1))
%x2=ws/wp
n=acosh(sqrt((10^(as/10)-1)/(10^(ap/10)-1)))/ acosh(ws/wp)
%n=log(x1-sqrt(x1*x1-1))/log(x2-sqrt(x2*x2-1)); % arcosh(x1)/arcosh(x2)
n=ceil(n)
eps=sqrt(10^(ap/10)-1)
gama=1/n*asinh(1/eps)
for k=1:n,
th(k)=0.5*pi*(1+(2*k-1)/n);
b(k)=cosh(gama)*sin(th(k));
a(k)=-sinh(gama)*cos(th(k));
w(k)=sqrt(a(k)*a(k)+b(k)*b(k));
s(k)=complex(a(k),b(k));
Q(k)=abs(w(k)/(2*a(k)));
end;
% Inverse Chebyshev filter
ws=1500
wp=1000
ap=0.2
as=40
n=acosh(sqrt( (10^(as/10)-1)/(10^(ap/10)-1) ))/ acosh(ws/wp)
n=ceil(n);
eps=1/sqrt(10^(as/10)-1)
gama=1/n*asinh(1/eps)
wscal=ws/wp
if mod(n,2)~=0, odd=1; else odd=0; end;
for k=1:n, %find poles
th(k)=0.5*pi*(1+(2*k-1)/n);
pole(k)=complex(sinh(gama)*cos(th(k)),cosh(gama)*sin(th(k)));
pole(k)=wscal/pole(k);
end;
nc=(n-odd)/2;
for k=1:nc, %find zeros
i=2*k -1;
theta=0.5*pi*i/n;
zero(k)=wscal*complex(0,1/cos(theta));
zero(k+nc)=wscal*complex(0,-1/cos(theta));
end;
for k=1:ceil(n/2), %find q and w
w(k)=abs(pole(k));
Q(k)=w(k)/(-2*real(pole(k)));
end;
w=w*wp
4
Frequency transformations Low-pass to High-pass
S
1
s
1
j
j 
T (S ) 
S 2


02
0
Q
1
j
2

1

02
0 1
02 s 2
0
s2
T ( s) 


1
1
1
2
2 2



1

s


s

s  s2
0
0
2
2
s
Q s
Q
0 0Q
S  02
s2
T ( s) 
1
1
s2 
s 2
0 Q
0
Transformation of poles
T (S ) 
1
S  S1
T ( s) 
1
1
 S1
s

s
S1
s
s


1  sS1 1  sS1 1  s
S1
sa  jb
s
sa  jb
2
2
02
a  jb
T (s) 
 a b

a  jb  s s  a  jb
1
s
a  jb
a 2  b2
02
0
0

please notice that if conjugate poles are transformed complex numerator will simplify
s a  jb
sa  jb
02
02
s
s 202
a  jb  s  a  jb  

2
0

2
0
s2
04
a  jb    s  a  jb 
s 



2
0
 
 

2
0



02

a  jb    s  a  jb 
 s 
02  
02 

5
Frequency transformations Low-pass to High-pass
1
S
s
T (S ) 
1
1


S  S1 S  a  jb
T (s) 
1  a  jbs  0
1
a  jb
a  jb
s
 2

2
a  jb
a b
02
1
s

1
 a  jb 1  a  jbs
s
solving
0
0

Q=12.8 and Ωo = 1.15
a = 0.0449 and b = 1.1491
low pass pole => -0.0449 + j1.1491
Transformation of poles
T (S ) 
s a  jb
s
s a  jb
2
2
1
02
a  jb
T ( s) 


 a b

a  jb  s s  a  jb
1
1
1
 S1
s
s
s
S1
a  jb
a 2  b2
02
s
S1
1
S  S1
a  jb
High-pass pole:
For
 02
0  1.15
low pass pole =
high pass pole =
-0.0449 + j1.1491
-0.0340 - j0.8689
Short cut: The same Q and 0 
1
0
6
Frequency transformations Low-pass to Band-pass
S
 s  
s 2  02 0 s 2  02

 Q  0 
Bs
B 0 s
s 
 0
S
for normalized frequency:
S 2
T ( s) 
B

 0
02
0
Q

0
1

S  Q s  
s

T (S ) 
s 2  02
Bs
1
S  02
2
02
0  s 2  02 
02 s 2

2
 s 2  02 
 s 2  02  0 s  s 2  02  2 2

 

  02 
 

  s 0
Q  Bs 
Q  Bs 
 Bs 
 B 
1
1
Bs
T (S ) 
T ( s)  2
 2
2
s  0
S  S1
s  02  sBS1
 S1
Bs
Bs
T (s)  2
 ...... has one zero and two complex poles
s  sB a  jb  02
C
2
 ....
2
2
 b  b 2  4ac  Ba  jb  B a  jb  40
s12 

 ....
2a
2
2
0
C

7
low-pass prototype:
T ( s) 
 02
 02
 02
 02


 2
2
2

s 2  s   02 s  s1 s  s2  s  a  jbs  a  jb s  2as  a  b
Q
0
 2a
Q

a 0
2Q
0  a 2  b 2
b   02  a 2
Q=12.8 and Ωo = 1.15
a = 0.0449 and b = 1.1491
low pass pole => -0.0449 + j1.1491
T ( s) 
1
s 
 S1
Bs
2
2
C

Bs
s  C2  sBS1
2
T ( s) 
Bs
s  sB a  jb  C2
2
Assume B=1 and ωC = 1
band pass poles
 B a  jb  B 2 a  jb  4C2  B a  jb  B 2 a 2  b 2  j 2ab   4C2
s12 

2
2
2
s1 = -0.0337 + j0.0040
s2 = -0.0561 - j2.3023
Frequency transformations Low-pass to Band-stop
S
1
s
2
H s   P21 
8
Foster Reactance functions
SH


s s 2   Z21
s 2   P21 s 2   P2 2




0

 0
 Z 0  P1
Z 1
P 2
Foster Functions
S
s
H
low pass
S
H
s
high pass
SH
s
s   P21 
2






double band pass (low pass)
s   
s s   
double band pass (high pass)
S
1 s 2   Z21
H
s
S
1 s s 2  Z21
H s 2   P21
SH
band stop
2
2
2
Z1
2
P1
band pass
9