Dsp lecture vol 5 design of iir

Infinite Impulse Response (IIR) Filters
Vol-5

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.

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

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.

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.

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)

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

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)

Dsp lecture vol 5 design of iir

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.

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)

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

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.

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.

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…

 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

 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 +

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.

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

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

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

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)

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

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)

Infinite Impulse Response (IIR) Filters
- End -

Dsp lecture vol 5 design of iir

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