Signal and System IIR Filter Filbert H. Juwono Wireless and Signal Processing (WaSP) Research Group Department of Electrical Engineering University of Indonesia www.ee.ui.ac.id/wasp

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

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 :

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

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

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

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

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

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

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

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 +- 360o 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 :

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

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 = 0.877969 b2 = -1

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

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

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

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

We can write So

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

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 :

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

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

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

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

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

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

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 :

The cutoff frequency may be expressed as ωc= 2π x 150 = 942.4778 rad/s . The transfer function of the denormalized analog filter is obtained by repla- cing s by s/ωc : Find poles by abc formula

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

Then, prT = -0.52065 cos (piT) = 0.8675 piT = +0.52065 e prT = 0.5941 The transfer function become :

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

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

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)

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

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

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 :

lowpass to lowpass lowpass to highpass lowpass to bandpass lowpass to bandstop

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

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

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

Denormalized analogue filter The difference equation

When it transfers into highpass filter Simplifying

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

Butterworth Filter Here is sketch of frequency response on Butterworth filter

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

Butterworth filter contains zero at infinity and poles:

Butterworth polynomials, normalized 1 2 3 4 5 6 7 8

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

The poles of the normalized Chebyshev

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

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

Here is sketch of frequency response on Elliptics

The critical frequencies: 0, 1,

Example Obtain the transfer function of a lowpass digital filter meeting the following specifications: Passband 0 – 60 Hz Stopband > 85 Hz Stopband attenuation > 15 dB Assume a sampling frequency of 256 Hz and a Butterworth characteristic

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

The prewarped equivalent analogue frequencies We need to obtain H(s) with Butterworth characteristics, a 3 dB cutoff of 0.906347 and a response at 85 Hz that is down by 15 dB

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 = 2.68. We use N = 3. A normalized third order filter

Performing transform Combining