1. Signal and System IIR Filter Filbert H. Juwono
Wireless and Signal Processing (WaSP)
Research Group
Department of Electrical Engineering
University of Indonesia

2. State of the art
The basic IIR filter is characterized by the following equation :
Where h(k) is the impulse response of the filter which is theoretically infinite in duration

3. bk and ak are the coefficients of the filter
x(n) and y(n) are the input and output to the filter
Transfer function for the IIR filter is :

4. Zeros are those values of z which H(z) becomes zero
Poles are values of z for which H(z) is infinite

5. The important thing is to find suitable values for the coefficients bk and ak
Note that the current output y(n) is a function of the past outputs y(n-k) . So that it show the feedback system of some sort
The strength of IIR filters comes from the flexibility the feedback arrangement provides
Remember that the transfer function of IIR filter can be shown as the pole and zero equations

6. For frequency selective filters, such as lowpass and bandpass filters, the frequency response specifications are often in the form of tolerance scheme
Here is an example tolerance scheme for an IIR bandpass filter

7. ε2 : passband ripple parameter
δp : passband deviation
δs : stopband deviation
fp1 and fp2 : passband edge frequency
fp1 and fp2 : stopband edge frequency
Ap : passband ripple = 10. log10(1+ ε2)
= 20. log10(1- δp)
As : stopband attenuation = -20. log10(δs)
The band edge frequencies are sometimes given in normalized form, that is a fraction of the sampling frequency

8. Coefficient calculation methods for IIR filters
There are 4 methods to calculate the coefficients :
Pole-zero placement
Impulse invariant
Matched z-transform
Bilinear z-transform
Learn carefully

9. Pole-zero placement Method
The idea is :
when a zero is placed at a given point on the z-plane, the frequency response will be zero at the corresponding point
while a pole produces a peak at the corresponding frequency point
Note that for the coefficient of the filter to be real, the poles and zeros must either be real

11. Example
A bandpass digital filter which is required to meet the following specifications :
- complete signal rejection at dc and 250 Hz
- a narrow passband centered at 125 Hz
- a 3dB bandwidth of 10 Hz
Assume the sampling frequency of 500 Hz

12. Solution Zeros are at angles of 0o and 360o x 250/500 = 180o
Fist we must determine where to place the poles and
zeros on the z-plane .
Watch the frequency on 250 Hz and 125 Hz
Zeros are at angles of 0o and 360o x 250/500 = 180o
and the place poles at o x 125/500 = +-90o
The radius, r, of the poles is determined by the desi-
red bandwidth. An approximate relationship between
r , for r > 0.9 and bandwidth bw is given by :
r = 1 – (bw/Fs).π
So that, by substituting the value of bw=10 Hz and
Fs=500 Hz , giving r = 0.937
After it, we can draw the pole-zero diagram below :

14. From the pole-zero diagram, the transfer function
can be written as follow :

15. So that, the difference equation is :
Look at again the transfer function. It shows filter
which is a second-order section, with coefficients :
b0 = 1 a1 = 0
b1 = 0 a2 =
b2 = -1

16. Converting analogue filters into equivalent digital filters
This represents the most successful approach of obtaining the coefficients of IIR filters
Two common approaches:
The impulse invariant
The matched z-transform
The bilinear z-transform

17. Impulse Invariant Method
First, consider these component :
- H (s) : a suitable analog transfer function
- h (t ) : the impulse response
- h (nT) : z transforming with T sampling interval
- H (z) : desired transfer function
Those component are useful and obtained by using Laplace Transform and also z-transformation

18. Illustrating The Impulse Invariant Method
Digitize, using the impulse invariant method, the simple analogue filter with the transfer function given by:

19. The impulse response of the equivalent digital filter
The transfer function of H(z) is obtained by z-transforming h(nT)

20. We can write So

21. For the case when M = 2
If the poles are conjugates, then

22. Here are the steps in Impulse Invariant Method :
Determine a normalized analog filter H(s) that satisfies the specifications for the desired digital filter
If necessary, expand H(s) using partial fraction to simplify the next step
Obtain the z-transform of each partial fraction to obtain :

23. 4. Obtain H(z) by combining the z-transforms of the partial fractions into second-order terms and possibly one-first-order term.
If the actual sampling frequency is used then multiply H(z) by T

24. Example
It is required to design a digital filter to approximate the following normalized analogue transfer function
Assumed a 3 dB cutoff frequency of 150 Hz and a sampling
frequency of 1.28 kHz

25. Solution We need to scale the normalized transfer function
This is achieved by replacing s by s/α, where
Thus

27. To keep the gain down and to avoid overflows when the filter is implemented, it is common to multiply H(z) by T or divide it by the sampling frequency

28. Matched z-transform (MZT) method
It provides a simple way to convert an analog filter into an equivalent digital filter
The idea is : each of the poles and zeros of the analog filter is mapped directly from the s-plane to the z-plane using the following equation :
It maps a pole or zero at the location s=a in the s-plane, onto a pole or zero in the z-plane at z=eaT

29. In case where M = N =2 and the poles and zeros are conjugates

30. Example
Having a filter with a 3 dB cutoff frequency of 150 Hz in sampling frequency of 1.28 kHz. The normalized of transfer function of an analog filter is given by :

31. The cutoff frequency may be expressed as ωc=
2π x 150 = rad/s . The transfer function of
the denormalized analog filter is obtained by repla-
cing s by s/ωc :
Find poles by
abc formula

32. Remember : so that :
We have the real and imaginary poles :

33. Then, prT = cos (piT) =
piT = e prT =
The transfer function become :

34. Bilinear z-transform (BZT) Method
It is the most important method
The idea is: to convert an analog filter H(s) into an equivalent digital filter is to replace s as follow:
That transformation maps the analog transfer function, H(s), from the s-plane into the discrete transfer function, H(z), in the z-plane

35. Look at the figure below. It shows the transforma-tion using BZT method
S-plane
Z-plane

36. Direct replacement of s in H(s) may lead to a digital filter with undesirable response
We find that the analogue frequency ω’ and the digital frequency ω are related as
We can see that ω’ and ω are almost linear for small values of ω, but becomes not linear for large values of ω, leading to distortion (warping)

37. This effect is compensated by prewarping tha analogue filter before applying the bilinear transformation

38. Here are the steps for using BZT
Use the digital filter specifications to find suitable normalized, prototype, analog low pass filter H(s)
Determine and prewarpe the bandedge or critical frequencies of the desired filter
when : ωp = specified cutoff frequency
ωp’ = prewarped cutoff frequency

39. Remember that in bandpass and bandstop filter, there are the lower and upper passband edge frequencies or we can say ωp1’ and ωp2’ .
3. Denormalize the analog prototype filter by replacing s in the transfer function, H(s), using these following transformation :

40. lowpass to lowpass
lowpass to highpass
lowpass to bandpass
lowpass to bandstop

41. 4. Apply the BZT to obtain the desired digital filter transfer function, H(z), by replacing s in the frequency-scaled (i.e. denormalized) transfer function, H’(s) as follows

42. Example
Determine the transfer function and the difference equation for the digital equivalent of the analogue lowpass filter transfer function
Assume a sampling frequency of 150 Hz and a cutoff frequency of 30 Hz

43. Solution The critical frequency The cutoff frequency after prewarping
With T = 1/150 Hz, so

44. Denormalized analogue filter
The difference equation

45. When it transfers into highpass filter
Simplifying

46. Classical Analog Filter
There are four types of Classical Analog filter :
Butterworth filter
Chebysev type I
Chebysev type II
Elliptic
All types of filter are derived from lowpass prototype filter

47. Butterworth Filter
Here is sketch of frequency response on Butterworth filter

48. The important equations on Butterworth filter are :
Magnitude square
Frequency response
Filter order

49. Butterworth filter contains zero at infinity and poles:

50. Butterworth polynomials, normalized1 2 3 4 5 6 7 8

51. Chebysev Filter Chebysev Type I : equal ripple in the passband,
monotonic in the stopband
Chebysev Type II : equal ripple in the stopband,
monotonic in the passband
Chebysev Type I
Chebyshev polynomial which exhibits equal ripple in the passband
Determine the passband ripple

52. The poles of the normalized Chebyshev

53. Here is sketch of frequency response on Chebysev
Type 1
Type 2

54. Elliptic Filter
The elliptic filter exhibits equiripple behavior in both the passband and the stopband
This is the following magnitude-squared response:
GN(ω’) is a Chebysev rational function

55. Here is sketch of frequency response on Elliptics

56. The critical frequencies: 0, 1,

62. Example
Obtain the transfer function of a lowpass digital filter meeting the following specifications:
Passband – 60 Hz
Stopband > 85 Hz
Stopband attenuation > 15 dB
Assume a sampling frequency of 256 Hz and a Butterworth characteristic

63. Solution
This illustrates how to combine steps 4 and 5 in the BZT process
The critical frequencies

64. The prewarped equivalent analogue frequencies
We need to obtain H(s) with Butterworth characteristics, a 3 dB cutoff of and a response at 85 Hz that is down by 15 dB

65. For attenuation 15 dB and a passband ripple of 3 dB we get N = 2. 68
For attenuation 15 dB and a passband ripple of 3 dB we get N = We use N = 3.
A normalized third order filter

66. Performing transform
Combining