This document discusses Infinite Impulse Response (IIR) filters. IIR filters are more computationally efficient than FIR filters as they require fewer coefficients due to using feedback. The document covers IIR filter concepts, properties, design procedures including specification, coefficient calculation, structure selection, and implementation. It also provides examples of coefficient calculation methods like pole-zero placement and the bilinear transform method for converting analog filters to digital IIR filters.

1. Infinite Impulse Response (IIR) Filters
Vol-5

2. Introduction
 Infinite Impulse Response (IIR) filters are
the first choice when:
 Speed is paramount.
 Phase non-linearity is acceptable.
 IIR filters are computationally more
efficient than FIR filters as they require
fewer coefficients due to the fact that they
use feedback or poles.
 However feedback can result in the filter
becoming unstable if the coefficients
deviate from their true values.

3. Concept of Digital IIR filter
h(k), k=0,1,2,…,∞
x(n) (impulse response- y(n)
Input sequence infinite length) output sequence
y(n) h(k ) x(n k )
k 0

4. Properties of an IIR Filter
 The general equation of an IIR filter can
be expressed as follows:
b0 b1 z 1  bN z N
H z
1 a1 z 1  aM z M
N
k
bk z
k 0
M
k
1 ak z
k 1
 ak and bk are the filter coefficients.

5. Properties of an IIR Filter
 The transfer function can be factorised to
give:
z z1 z z 2  z z N Y z
H z k
z p1 z p2  z p N X z
 Where: z1, z2, …, zN are the zeros,
p1, p2, …, pN are the poles.

6. Properties of an IIR Filter .. continued
 Time domain theoretical equation of IIR
y(n) h(k ) x(n k )
k 0
 For the implementation of the above
equation we need the difference equation:
N M
y(n) b x(n k ) b y (n k )
k 0 k k 1k
b(0) x(n) b(1) x(n 1) b(2) x(n 2) 
a(1) y(n 1) a(2) y(n 2)

7. IIR difference Equation and Direct Implementation
Structure
y(n) b(0) x(n) b(1) x(n 1) b(2) x(n 2) 
a(1) y(n 1) a(2) y(n 2)
x(n) b0 y(n)
+ +
z-1 z-1
b1 + + a1
z-1 z-1
b2 a2
IIR structure for order=N = M = 2

8. IIR Direct Form I and Direct Form II
Structures
x[n] y[n]
+
Z-1 b0 Z-1
x[n-1] y[n-1] IIR
Z-1 b1
-a1 Z-1 Direct
x[n-2] Form I
y[n-2]
b2
-a2
x[n] + + y[n]
b0
Z-1 IIR Direct
-a1 b1 Form II
Z-1
(Canonic)
-a2 b2
y(n) b(0)x(n) b(1)x(n 1) b(2)x(n 2)  a(1) y(n 1) a(2) y(n 2)

10. Design Procedure
 To fully design and implement a filter five
steps are required:
(1) Filter specification.
(2) Coefficient calculation.
(3) Structure selection.
(4) Simulation (optional).
(5) Implementation.

13. Filter Specification - Step 1
|H(f )| pass-band stop-band
1
fc : cut-of f f requency fs/2 f (norm)
(a)
|H(f )| pass-band transition band stop-band |H(f )|
(dB) (linear)
p 1 p
0 1
1 p
pass-band
-3
ripple
stop-band
ripple
s s
fs/2 f (norm)
fsb : stop-band f requency
fc : cut-of f f requency
fpb : pass-band f requency
(b)

14. Coefficient Calculation - Step 2
 There are two different methods available
for calculating the coefficients:
 Direct placement of poles and zeros.
 Using analogue filter design.
 Impulse invariant
 Bilinear z-transform etc

15. Pole-zero Placement Method
 All that is required for this method is the
knowledge that:
 Placing a zero near or on the unit circle in
the z-plane will minimise the transfer
function at this point.
 Placing a pole near or on the unit circle in
the z-plane will maximise the transfer
function at this point.
 To obtain real coefficients the poles and
zeros must either be real or occur in
complex conjugate pairs.

16. Analogue to Digital Filter Conversion
 This is one of the simplest method.
 There is a rich collection of prototype
analogue filters with well-established
analysis methods.
 The method involves designing an
analogue filter and then transforming it to
a digital filter.
 The two principle methods are:
 Bilinear transform method
 Impulse invariant method.

17. Realisation Structures - Step 3
 Direct Form I:
N
k
bk z
Y z k 0 b0 b1 z 1  bN z N
H z M
X z k 1 a1 z 1  aM z M
1 ak z
k 1
 Difference equation:
N M
yn bk x n k ak y n k
k 0 k 1
 This leads to the following structure…

18. Realisation Structures - Step 3
 Direct Form I:
x(n) y(n)
b0 + +
z-1 z-1
b1 + + a1
z-1 z-1
b2 + + a2
bN-1 + + aM-1
z-1 z-1
bN + + aM

19. Realisation Structures - Step 3
 Direct Form II canonic realisation:
x(n) P(n) y(n)
+ b0 +
z-1
P(n-1)
+ -a1 b1 +
z-1
P(n-2)
+ -a2 b2 +
z-1
P(n-N)
+ -aN bN +

20. Bilinear Transform Method
 Practical example of the bilinear
transform method:
 The design of a digital filter to approximate a
second order low-pass analogue filter is
required.
 The transfer function that describes the
analogue filter is (This is an analog
BUTTERWORTH filter):
1
H s
s2 2s 1
 The digital filter is required to have:
 Cut-off frequency =100 Hz
 Sampling frequency of 1 kHz.

21. Example: Design of a LP filter design using
Bilinear Transformation
Design a digital equivalent of a 2nd order
Butterworth LP filter with a
cut-off freq fc=100 Hz
Sampling frequency fs=1000 Hz.
The normalized cut off frequency of the digital filter
ω=2πfc/fs=2πfc 100/1000=0.628
Now equivalent analog filter cut-off freq
Ω=ktan(ω/2)= 1.tan(0.628/2)=0.325 rads/sec

22. Design of a LP filter design using Bilinear
Transformation ..continued
H(s) for a Butterworth filter
1
H ( s)
s2 2s 1
Now the transfer function of this Butterworth filter
becomes (putting s=s/Ω)
1
H ( s)
s 2 s
2 1
0.325 0.325

23. Design of a LP filter design using Bilinear Transformation
..continued
Now convert the analog filter H(s) into equivalent
digital filter H(z) by applying the bilinear z-transform-
1
z 1 1 z
s 1
z 1 1 z
1
H ( z)
1 1 z 1 2 2 1 z 1
1
0.3252 1 z 1 0.325 1 z 1
0.067 0.135z 1 0.067 z 2
1 1.1429 z 1 0.4127 z 2

24. Design of a LP filter design using Bilinear Transformation
..continued
From the previous we get the filter coefficients-
{bk}Coefficients {ak}coefficients
b0=0.067 a1= -1.1429
b1=0.0135 a2=0.4127
b2=0.067
The time domain difference equation is obtained
y(n) b(0) x(n) b(1) x(n 1) b(2) x(n 2) 
a(1) y(n 1) a(2) y(n 2)
0.67 x(n) 0.135x(n 1) 0.67 x(n)
1.1429 y(n 1) 0.4127 y(n 2)

25. Direct realization of the design
x(n) y(n)
b0 + +
b0=0.067
z-1 z-1
b + + a
1 1
b1=0.135 a1=1.1429
z-1 z-1
b a
2 2
b2=0.067 a2= -0.4127

26. Matlab Code for the BZT Method
clear all
fc=100;fs=1000;
wp=2*pi*fc*(1/fs);
OmegaP=2*fs*tan(wp/2);
Omega=tan(wp/2);
Hs_den1=[(1/Omega)^2,((2^0.5)/Omega),1];
Hs_den=[(1/OmegaP)^2,((2^0.5)/OmegaP),1];
Hs_num=1;
Hs_denN=Hs_den/Hs_den(1);
Hs_numN=Hs_num/Hs_den(1);
[b,a]=bilinear(Hs_numN,Hs_denN,fs);
freqz(b,a,fs)

27. Infinite Impulse Response (IIR) Filters
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