Infinite Impulse Response (IIR) Filters Recursive Filters: with constant coefficients. Advantages: very selective filters with a few parameters; Disadvantages: a) in general nonlinear phase, b) can be unstable.

Design Techniques: discretization of analog filters s-plane z-plane analog digital Problem: we need to map the derivative operator “s” into the time shift operator “z”, and make sure that the resulting system is still stable.

approximation of “s” Two major techniques Euler Approximation (easiest), Bilinear Transformation (best). Euler Approximation of the differential operator: approximation of “s” take the z-Transform of both sides:

Example: take the analog filter with transfer function and discretize it with a sampling frequency . By Euler’s approximation The filter is implemented by the difference equation s-plane z-plane analog digital

if Re[s]<0. Problem with Euler Approximation: it maps the whole stable region of the s-plane into a subset of the stable region in the z-plane s-plane z-plane since if Re[s]<0.

Bilinear Transformation. It is based on the relationship nT-T nT A B C D Take the z-Transform of both sides: which yields the bilinear transformation:

Main Property of the Bilinear Transformation: it preserves the stability regions. s-plane z-plane since:

Mapping of Frequency with the Bilinear Transformation. Magnitude: Phase: where

See the meaning of this: it is a frequency mapping between analog frequency and digital freqiency.

Example: we want to design a digital low pass filter with a bandwith and a sampling frequency . Use the Bilinear Transformation. Solution: Step 1: specs in the digital freq. domain Step 2: specs of the analog filter to be digitized: or equivalently Step 3: design an analog low pass filter (more later) with a bandwith ; Step 4: apply Bilinear Transformation to obtain desired digital filter.

Design of Analog Filters Specifications: pass band transition band stop band

Two Major Techniques: Butterworth, Chebychev Specify from passband, determine N from stopband:

+ + + + Poles of Butterworth Filter: which yields the poles as solutions N=2 and choose the N poles in the stable region. + + poles + + s-plane

Example: design a low pass filter, Butterworth, with 3dB bandwith of 500Hz and 40dB attenuation at 1000Hz. Solution: solve for N from the expression poles at

Chebychev Filters. Based on Chebychev Polynomials: Property of Chenychev Polynomials: within the interval Chebychev polynomials have least maximum deviation from 0 compared to polynomials of the same degree and same highest order coefficient

Why? Suppose there exists with smaller deviation then root root root But: has degree 2 … … and it cannot have three roots!!! So: you cannot find a P(x) which does better (in terms of deviation from 0) then the Chebychev polynomial.

Chebychev Filter: Since (easy to show from the definition), then

Design of Chebychev Filters: Formulas are tedious to derive. Just give the results: Given: the passband, and which determines the ripple in the passband, compute the poles from the formulae s-plane

Example: design a Chebychev low pass filter with the following specs: passband with a 1dB ripple, stopband with attenuation of at least 40dB. Step 1: determine . The passband frequency For 1dB ripple, Step 2: determine the order N. Use the formula with , to obtain

Frequency Transformations We can design high pass, bandpass, bandstop filters from transformations of low pass filters. Low Pass to High Pass: same value at

Low Pass to Band Pass: The tranformation maps

Low Pass to Band Stop

How to make the transformation: Consider the transfer function then with we obtain with zeros and poles solutions of also n-m extra zeros at s where

IIR filter design using Matlab In Matlab there are three functions for each class of filters (Butterworth, Chebytchev1, Chebytchev2): BUTTAP CHEB1AP CHEB2AP Poles and Zeros of Analog Prototype Filter BUTTER CHEBY1 CHEBY2 Numerator and Denominator from N and BUTTORD CHEBY1ORD CHEBY2ORD N and from specifications Example. We want to design an IIR Digital Filter with the following specifications: Pass Band 0 to 4kHz, with 1dB ripple; Stop Band > 8kHz with at least 40 dB attenuation Sampling frequency 40kHz Type of Filter: Butterworth. Using Matlab: >> [N, fc]=butterord(fp, fs, Rp, Rs); % fp, fs=passband and stopband freq relative to Fs/2 >> [B, A]=butter(N, fc); % B, A vectors of numerator and denominator coefficients. In our case: [N, fc]=butterord(4/20, 8/20, 1, 40), would yield N=7, fc=0.2291; [B, A]=butter(7, 0.2291), would yield the transfer function B(z)/A(z).

Let’s verify these numbers: Step 1: specifications in the digital frequency domain: Step 2: specifications for analog filter from the transformation

Step 3: choose (say) Butterworh Filter with and from the ripple specification Step 4: determine order N from attenuation of 40dB with yields N=7

Step 5: finally the cutoff frequency, from the equation Which yields , corresponding to a digital frequency Step 6: the desired Filter is obtained by the function [num, den] = butter( 7 , 0.6889/)

Magnitude and Phase Plots: