This document discusses the design of infinite impulse response (IIR) digital filters. It begins by defining IIR filters and noting their advantages of high selectivity with few parameters, but also potential disadvantages of nonlinear phase and instability. It then covers two major techniques for designing IIR filters: the bilinear transformation and Euler approximation. The bilinear transformation maps the analog s-plane to the digital z-plane in a way that preserves stability. Examples are provided to illustrate the design process of mapping analog low-pass filters to digital filters using the bilinear transformation. The document also discusses analog filter design techniques, including Butterworth and Chebyshev filters characterized by their transfer functions.

1. Design of Infinite Impulse
Response Digital Filters
Mr. HIMANSHU DIWAKAR
Assistant Professor
JETGI
Mr. HIMANSHU DIWAKAR JETGI 1

2. Mr. HIMANSHU DIWAKAR JETGI 2

3. Infinite Impulse Response (IIR) Filters
Recursive Filters:
y n a y n a y n N b x n b x n NN N( ) ( ) ... ( ) ( ) ... ( )        1 01
with constant coefficients.a bi j, ... ,
Advantages: very selective filters with a few parameters;
Disadvantages: a) in general nonlinear phase,
b) can be unstable.
Mr. HIMANSHU DIWAKAR JETGI 3

4. Design Techniques: discretization of analog filters
s-plane z-plane







analog digital
H sa ( ) H zd ( )
Problem: we need to map
the derivative operator “s” into the time shift operator “z”,
and make sure that the resulting system is still stable.
Mr. HIMANSHU DIWAKAR JETGI 4

5. Two major techniques
• Euler Approximation (easiest),
• Bilinear Transformation (best).
 Z y nT
Y z z Y z
T
z
T
Y z( )
( ) ( )
( )








 1 1
1
Euler Approximation of the differential operator:
take the z-Transform of both sides:
( )
( ) ( )
y nT
y nT y nT T
T

 
approximation of “s”
s
z
T

 
1 1
Mr. HIMANSHU DIWAKAR JETGI 5

6. Example:
take the analog filter with transfer function and discretize it
with a sampling frequency .
By Euler’s approximation
H s
sa ( ) 

2
1
F Hzs  5
H z
z
z
zd ( )
.
.
.






 


2
1
02
1
0333
08331
The filter is implemented by the difference equation
y n y n x n( ) . ( ) . ( )  0833 1 0333
s-plane z-plane

analog digital

Mr. HIMANSHU DIWAKAR JETGI 6

7. Problem with Euler Approximation:
it maps the whole stable region of the s-plane into a subset of the stable region in the z-plane
s-plane z-plane
since

















T
s
T
s
sT
z
1
1
2
1
2
1
1
1
2
1
1
T
1
T

s
z  
1
2
1
2 if Re[s]<0.
Mr. HIMANSHU DIWAKAR JETGI 7

8. Bilinear Transformation. It is based on the relationship
 y nT y nT T y t dt y nT y nT T
T
nT T
nT
area ABCD
( ) ( ) ( ) ( ) ( )     
 2      
nT-T nT
A
B
C
D
Take the z-Transform of both sides:
 Y z z Y z
T
Y z z Y z( ) ( ) ( ) ( )   1
1
1
1
2
which yields the bilinear transformation:
s
T
z
z
z
s
T
s
T



  




2 1
1
2
2
1
1
Mr. HIMANSHU DIWAKAR JETGI 8

9. Main Property of the Bilinear Transformation: it preserves the
stability regions.
s-plane z-plane
| |z
s
T
s
T




2
2
1 2
T
2
T


ssince:
Mr. HIMANSHU DIWAKAR JETGI 9

10. Mapping of Frequency with the Bilinear Transformation.
| |e
j
T
j
T
j






2
2
1
2
T
2
T
j
2
phase ( ) 2
2
j
T
j
T
   
 
  
     
   
 
j
 j
2
T


Magnitude:
Phase:
where  






tg
T1
2

 






2
2T
tg

 






2
2
1
tg
T

Mr. HIMANSHU DIWAKAR JETGI 10

11. See the meaning of this:
 






2
2T
tg
 






2
2
1
tg
T

it is a frequency mapping between analog frequency and digital freqiency.
c
0
  0
  
c
cc
Mr. HIMANSHU DIWAKAR JETGI 11

12. Example: we want to design a digital low pass filter with a bandwith
and a sampling frequency . Use the Bilinear Transformation.
F kHzo  8
F kHzs  24
• Step 1: specs in the digital freq. domain
• Step 2: specs of the analog filter to be digitized:
  c rad ( )(8/ ) /2 24 2 3
Solution:
c s
c
F tg tg rad





    2
2
2 24 10 3 8313 103 3
( )( ) ( / ) . / sec
or equivalently F kHzc  1323.
• Step 3: design an analog low pass filter (more later) with a bandwith ;
• Step 4: apply Bilinear Transformation to obtain desired digital filter.
F kHzc  1323.
Mr. HIMANSHU DIWAKAR JETGI 12

13. Design of Analog Filters
Specifications:
p c
s
1
1 2
 
1
2
2
2
| ( )|H  2
pass
band
transition
band
stop
band
Mr. HIMANSHU DIWAKAR JETGI 13

14. Two Major Techniques: Butterworth, Chebychev
Butterworth:
| ( )|H
c
N
p
N




2
2
2
2
1
1
1
1
















Specify from passband, determine N from stopband: , p
| ( )|H s
s
p
N


2
2
2 2
21
1










  N
s
p













log
log
1
2
2
2
2
2
2

 


Mr. HIMANSHU DIWAKAR JETGI 14

15. Poles of Butterworth Filter:
| ( )| ( ) ( )H H s H s s j
 
2
   H s H s
c
N
s
j
( ) ( ) 








1
1
2











1
1
2
2
s
c
N

which yields the poles as solutions
 

 
s
c
N
2
2
1
1
  s j e e ec
j
k
N
c
j j k
N
 


 
   2
2 2
2 1
2
( )
, ,...,k N 0 2 1
and choose the N poles in the stable region.
s-plane
poles
++
+ +
N=2
Mr. HIMANSHU DIWAKAR JETGI 15

16. Example: design a low pass filter, Butterworth, with 3dB bandwith of 500Hz and 40dB
attenuation at 1000Hz.
Solution:
 c s 1000 2000 ; ,
solve for N from the expression
1
1
1
1 2
0012 2
2












s
c
N N .
N N

  
log( )
log( )
.
10 1
2 2
6 64 7
4
poles at
s e kk
j k
  
1000 0 62 2 1 14
  ( / ( ) / )
, ,...,
| ( )|H dB

Mr. HIMANSHU DIWAKAR JETGI 16

17. Chebychev Filters.
Based on Chebychev Polynomials: T x NtN x t
( ) cos( ) cos( )
 
T x
T x t x
T x t t x
T x xT x T x
x t
x t
N N N
0
1
2
2 2
1 1
0 1
2 2 1 2 1
2
( ) cos( ) ,
( ) cos( ) ,
( ) cos( ) cos ( ) ,
( ) ( ) ( )
cos( )
cos( )
 
 
    
 


 

T x3 ( )
x
Property of Chenychev Polynomials:
within the interval Chebychev polynomials have least maximum deviation
from 0 compared to polynomials of the same degree and same highest order coefficient
   1 1x
T x x x3
3
4 3( )  
P x x ax bx c( )    4 3 2
x
Mr. HIMANSHU DIWAKAR JETGI 17

18. Why? Suppose there exists with smaller deviation thenP x x ax bx c( )    4 3 2
T x3( )
T x3 ( )
P x( )
P x T x( ) ( ) 3
P T  3 1
P T  3 1
P T  3 1
A
B
P T  3 1
D
A
B
C
C
D




rootroot root
But: has degree 2 …
… and it cannot have three roots!!!
P x T x x ax bx c x x( ) ( )      3
3 2 3
4 4 3
So: you cannot find a P(x) which does better (in terms of deviation from 0) then the
Chebychev polynomial.
Mr. HIMANSHU DIWAKAR JETGI 18

19. Chebychev Filter:
| ( )|H
TN
p



2
2 2
1
1








Since (easy to show from the definition), thenTN ( )1 1 | ( )|H p 2
2
1
1

 

 p
| ( )|H  2 1
1 2
 
Mr. HIMANSHU DIWAKAR JETGI 19

20. Design of Chebychev Filters:
Formulas are tedious to derive. Just give the results:
Given: the passband, and
which determines the ripple in the passband,
compute the poles from the formulae
p





 







1 12
1
N
r1 2
2
1
2, 


s-plane

 
k k
N
k N    
2
2 1
2
0 1( ) , ,...,
Mr. HIMANSHU DIWAKAR JETGI 20

21. Example: design a Chebychev low pass filter with the following specs:
• passband with a 1dB ripple,
• stopband with attenuation of at least 40dB.
F Hzp  500 ,
F Hzs  1000 ,
Step 1: determine . The passband frequency
For 1dB ripple,
 , p
20
1
1
10 1 12
1
2
2
log log( )






     

 dB    05088.
p  1000
Step 2: determine the order N. Use the formula
  
 
N
s p s p

   
 
log ( /
log ( / ) ( / )
1 1 1
1
2
2
2
2 2
2
2
   
   
with , to obtain2 0 01 . N  4
| ( )|H 

 p
Mr. HIMANSHU DIWAKAR JETGI 21

22. Frequency Transformations
We can design high pass, bandpass, bandstop filters from transformations of low pass filters.
Low Pass to High Pass:

| ( )|H 
H
j
c

2






c
c c
 c
same value at c
H
j
H
j
H jc
c
c
c




2





 





  ( )
s
s
c

 2
Mr. HIMANSHU DIWAKAR JETGI 22

23. Low Pass to Band Pass:
c c
ul l u
s
s
sc
l u
u l




 
 
2
( )
The tranformation
maps
s j
j
j
s j
j
j
l c
l l u
l u l
c
u c
u l u
u u l
c
 
 

 
 
 

 
 
  
  

 
  
  

2
2
( )
( )
Mr. HIMANSHU DIWAKAR JETGI 23

24. Low Pass to Band Stop
c c
l u
 l u
s
s
sc
u l
c l




 
 
( )
2
Mr. HIMANSHU DIWAKAR JETGI 24

25. How to make the transformation:
Consider the transfer function
H s
K s z s z s z
s p s p s p
m nm
n
( )
( )( )...( )
( )( )...( )
,
  
  
1 2
1 2
then with we obtains F s ( )  
   
   
H s H F s
K F s z F s z
F s p F s p
T
m
n
( ) ( )
( ) ... ( )
( ) ... ( )
 
 
 
1
1
with zeros and poles solutions of
F s z k m
F s p j n
k
j
( ) , ,...,
( ) , ,...,
 
 
1
1






also n-m extra zeros at s where F s( )  
Mr. HIMANSHU DIWAKAR JETGI 25

26. Mr. HIMANSHU DIWAKAR JETGI 26
THANK YOU