Usıng the impulse sampling method Prepared by: Dr. Erhan A. INCE Desıgn of IIR Fılters Usıng the impulse sampling method Prepared by: Dr. Erhan A. INCE

A digital filter, 𝐻 𝑒 𝑗𝑤 , with infinite impulse response (IIR) can be designed by first transforming it into a prototype analog filter 𝐻 𝑎 𝑗Ω and then design this analog filter using a standard procedure. Once the analog filter has been designed it is then mapped back to the discrete-time domain to obtain a digital Filter that meets the specifications. Commonly used analog filters are: 1. Buttorworth filters – no ripples in passband or stopband 2. Chebychev filters – ripples in the passband OR in the stopband 3. Elliptic filters – ripples in both the passband and stopbands A disadvantage of IIR filters is that they usually have non-linear phase. As a result some minor signal distortion is experienced. There are two mapping techniques : 1. The impulse sampling method 2. The bilinear transformation method.

Impulse Sampling Method: The impulse response of the digital filter (ℎ[𝑛]) is made(approximately) equal to the impulse response of an analog filter ( ℎ 𝑎 𝑡 ) by taking samples from filter ℎ 𝑎 𝑡 at 𝑡=𝑛 𝑇 𝑑 where 𝑇 𝑑 is an arbitrary sampling period. ℎ 𝑛 = 𝑇 𝑑 ∙ ℎ 𝑎 𝑛 𝑇 𝑑 Aliasing would occur if 𝐻 𝑎 𝑗Ω is not bandlimited to 𝜋 𝑇 𝑑 (rad/s) and 𝐻 𝑒 𝑗𝜔 = 𝑘=−∞ ∞ 𝐻 𝑎 𝑗 𝜔 𝑇 𝑑 +𝑗 2𝜋𝑘 𝑇 𝑑 If 𝐻 𝑎 𝑗Ω is bandlimited to 𝜋 𝑇 𝑑 then 𝐻 𝑒 𝑗𝜔 = 𝐻 𝑎 𝑗𝜔/ 𝑇 𝑑 In general all the commonly used prototype analog filters used in the impulse sampling method are indeed non-bandlimited. We can minimize the aliasing however by overdesigning the analog filter (especially in the stopband).

Given the digital filter specifications we map the digital frequency 𝜔 onto the analog frequency Ω= 𝜔 𝑇 𝑑 Let Ha 𝑠 = 𝑘=1 𝑁 𝑅 𝑘 𝑠− 𝑠 𝑘 After taking inverse laplace the corresponding impulse response is ℎ 𝑎 𝑡 = 𝑘=1 𝑁 𝑅 𝑘 𝑒 𝑠 𝑘 𝑡 0 , 𝑡≥0 𝑡<0 Impulse response of the discrete time filter is obtained by sampling 𝑇 𝑑 ℎ 𝑎 𝑡 and is h[n] = 𝑇 𝑑 ℎ 𝑐 𝑛 𝑇 𝑑 = 𝑘=1 𝑁 𝑇 𝑑 𝑅 𝑘 𝑒 𝑠 𝑘 𝑛 𝑇 𝑑 u[n] = 𝑛 𝑇 𝑑 = 𝑘=1 𝑁 𝑇 𝑑 𝑅 𝑘 𝑒 𝑠 𝑘 𝑇 𝑑 𝑛 u[n]

𝐻 𝑧 = 𝑛=−∞ ∞ ℎ[𝑛] 𝑧 −𝑛 = 𝑛=0 ∞ 𝑘=1 𝑁 𝑇 𝑑 𝑅 𝑘 𝑒 𝑠 𝑘 𝑇 𝑑 𝑛 𝑧 −1 = 𝑘=1 𝑁 𝑇 𝑑 𝑅 𝑘 𝑛=0 ∞ 𝑧 −1 𝑒 𝑠 𝑘 𝑇 𝑑 𝑛 = 𝑘=1 𝑁 𝑇 𝑑 𝑅 𝑘 1− 𝑧 −1 𝑒 𝑠 𝑘 𝑇 𝑑 The system function of the discrete-time filter is therefore H(z) = 𝑘=1 𝑁 𝑅 𝑘 𝑇 𝑑 1− 𝑒 𝑠 𝑘 𝑇 𝑑 𝑧 −1 Note: a pole at s = pk transform into a pole z= 𝑒 𝑝 𝑘 𝑇 𝑑 in the z-plane.

If all the 𝑠 𝑘 are on the left half of the s-plane, i If all the 𝑠 𝑘 are on the left half of the s-plane, i.e 𝜎 𝑘 <0, then all 𝑝 𝑘 s are within the unit-circle in the Z-plane. This means that a stable analog filter always yields a stable digital filter! with the impulse sampling method. In general 𝑧≠ 𝑒 𝑠 𝑇 𝑑

EXAMPLE Design a digital lowpass IIR filter with the following specifications: Passband: 0.89125 ≤ 𝐻 𝑒 𝑗𝜔 ≤ 1 0≤ 𝜔 ≤0.2𝜋 Stopband: 𝐻 𝑒 𝑗𝜔 ≤ 0.17783 0.3𝜋 ≤ 𝜔 ≤𝜋 Use the impulse sampling technique and an analog Butterworth filter. Assume Td = 1.

With 𝑇 𝑑 =1 this means Ω=𝜔 and 𝐻 𝑎 𝑗Ω =𝐻 𝑒 𝑗Ω Therefore the specifications for the analog filter are: Passband: 0.89125 ≤ 𝐻 𝑎 𝑗Ω ≤ 1 0≤ Ω ≤0.2𝜋 Stopband: 𝐻 𝑎 𝑗Ω ≤ 0.17783 0.3𝜋 ≤Ω Upper limit for the stopband frequency has not been specified. The magnitude square response of a Butterworth filter of order N is: 𝐻 𝑎 𝑗Ω 2 = 1 1+ Ω Ω 𝑐 2𝑁 Where Ω 𝑐 is the 3dB frequency of the filter. Larger the N value the closer is the Butterworth Filter’s frequency response to that of an ideal lowpass filter.

𝐻 𝑎 𝑗Ω Ω Ω 𝑐

The 3dB frequency and the filter order N are solutions of the two simultaneous equations below: 1+ 0.2𝜋 Ω 𝑐 2𝑁 = 1 0.89125 2 1+ 0.3𝜋 Ω 𝑐 2𝑁 = 1 0.17783 2 The values for N and Ω 𝑐 are: N = 5.8858 Ω 𝑐 =0.70474 Since filter order has to be integer we round and take N = 6 and recompure Ω 𝑐 and it is 0.7032. These results guarantee that passband requirement is met exactly at w = 0.2 Note that 𝐻 𝑎 𝑗0.3𝜋 =0.1700 ≤ 0.17783.

The next step is to find the poles of the Butterworth filter The next step is to find the poles of the Butterworth filter. Recall that we have 2N poles whose locations in the S-plane are depicted in the diagram below:

Mathematically these poles are: 𝑟 𝑘 =0.7032 𝑒 𝑗 𝜋 12 + 𝜋 2 +(𝑘−1) 2𝜋 12 k = 1,2,……, 2N Half of these poles are the poles of the Butterworth filter. Specifically we choose those 𝑟 𝑘 on the left-half s-plane to be the poles of 𝐻 𝑎 𝑗Ω . Consequently the poles of 𝐻 𝑎 𝑗Ω are:

The poles of the corresponding digital filter are:

The transfer function of the Butterworth Filter is:

The transfer function of the digital filter H(z) is (recall Td=1)