Transcript
Infinite Impulse Response (IIR) FiltersVol-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,…,∞
(impulse response-infinite length)
x(n)Input sequence
y(n)output sequence
0
)()()(k
knxkhny
Properties of an IIR Filter
The general equation of an IIR filter can be expressed as follows:
ak and bk are the filter coefficients.
M
k
kk
N
k
kk
MM
NN
za
zb
zaza
zbzbbzH
1
0
11
110
1
1
Properties of an IIR Filter
The transfer function can be factorised to give:
zXzY
pzpzpz
zzzzzzkzH
N
N
21
21
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
For the implementation of the above equation we need the difference equation:
)2()2()1()1()2()2()1()1()()0(
0 1)()()(
nyanyanxbnxbnxb
N
k
M
kkny
kbknx
kbny
0
)()()(k
knxkhny
IIR difference Equation and Direct Implementation Structure
IIR structure for order=N = M = 2
x(n)+
+b1
b2
+
+ a1
a2
z-1
y(n)
z-1
z-1
z-1
b0
)2()2()1()1()2()2()1()1()()0()(
nyanyanxbnxbnxbny
IIR Direct Form I and Direct Form II Structures
Z-1
Z-1
+b0
b1
b2
x[n]
x[n-1]
y[n]
x[n-2]
Z-1
Z-1-a1
-a2
y[n-1]
y[n-2]
b0
b1
b2
x[n] y[n]
-a1
-a2
Z-1++
Z-1
)2()2()1()1()2()2()1()1()()0()( nyanyanxbnxbnxbny
IIR Direct Form I
IIR Direct Form II
(Canonic)
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
(a)
1
f(norm)fc : cut-off frequency
pass-band stop-band
pass-band stop-bandtransition band
1
s
pass-bandripple
stop-bandripple
fpb : pass-band frequency
fsb : stop-band frequency
f(norm)
(b)
p1
s
p0
-3
p1
fs/2
fc : cut-off frequency
fs/2
|H(f)|(dB)
|H(f)|(linear)
|H(f)|
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:
M
M
NN
M
k
kk
N
k
kk
zaza
zbzbb
za
zb
zX
zYzH
11
110
1
0
11
Difference equation:
M
kk
N
kk knyaknxbny
10
This leads to the following structure…
Realisation Structures - Step 3
Direct Form I:x(n)
+b0
+b1
+b2
+bN-1
+bN
+
+
+
+
+
a1
a2
aM-1
aM
z -1
y(n)
z -1
z -1z -1
z -1
z -1
Realisation Structures - Step 3
Direct Form II canonic realisation:
+b 0
+b 1
z -1
+b 2
+b N
-a 1
-a 2
-a N
+
+
+
+
y(n)x(n) P (n)
z -1
z -1
P (n-1)
P (n-2)
P (n-N)
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):
The digital filter is required to have: Cut-off frequency =100 Hz Sampling frequency of 1 kHz.
12
12
ss
sH
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
1221)(
sssH
1325.0
22
325.0
1)(
sssH
Now the transfer function of this Butterworth filter becomes (putting s=s/Ω)
H(s) for a Butterworth filter
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
1
1
1
1
1
z
z
z
zs
24127.011429.11
2067.01135.0067.0
111
11325.02
2
11
11
1)(
2325.01
zz
zz
z
z
z
z
zH
Design of a LP filter design using Bilinear Transformation ..continued
{bk}Coefficients {ak}coefficients
b0=0.067b1=0.0135b2=0.067
a1= -1.1429a2=0.4127
From the previous we get the filter coefficients-
)2(4127.0)1(1429.1)(67.0)1(135.0)(67.0
)2()2()1()1()2()2()1()1()()0()(
nynynxnxnxnyanyanxbnxbnxbny
The time domain difference equation is obtained
Direct realization of the design
x(n)
+
b0=0.067
+b1
b2
+
+ a1
a2
z-1
y(n)
z-1
z-1
z-1
b1=0.135
b2=0.067
a1=1.1429
a2= -0.4127
b0
Matlab Code for the BZT Methodclear 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 -
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