Infinite Impulse Response (IIR) FiltersRecursive 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 filtersProblem: we need to map the derivative operator s into the time shift operator z,and make sure that the resulting system is still stable.

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

Example:take the analog filter with transfer function and discretize itwith a sampling frequency .By Eulers approximationThe filter is implemented by the difference equation

Problem with Euler Approximation: it maps the whole stable region of the s-plane into a subset of the stable region in the z-planes-planez-planesinceif Re[s]

Bilinear Transformation. It is based on the relationshipTake the z-Transform of both sides:which yields the bilinear transformation:

Main Property of the Bilinear Transformation: it preserves the stability regions.s-planez-planesince:

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. Step 1: specs in the digital freq. domain Step 2: specs of the analog filter to be digitized:Solution: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 FiltersSpecifications:passbandtransitionbandstopband

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

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

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 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 formulaes-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 formulawith , to obtain

Frequency TransformationsWe 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 tranformationmaps

Low Pass to Band Stop

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

IIR filter design using MatlabIn Matlab there are three functions for each class of filters (Butterworth, Chebytchev1, Chebytchev2):BUTTAPCHEB1APCHEB2APPoles and Zeros of Analog Prototype FilterBUTTERCHEBY1CHEBY2Numerator and Denominator from N and BUTTORDCHEBY1ORDCHEBY2ORDN and from specifications

Example. We want to design an IIR Digital Filter with the following specifications:Pass Band0 to 4kHz, with 1dB ripple;Stop Band> 8kHz with at least 40 dB attenuationSampling frequency 40kHzType 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).

Lets 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 Filterwith and from the ripple specification Step 4: determine order N from attenuation of 40dBwith yields N=7

Step 5: finally the cutoff frequency, from the equationWhich 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: