[PDF] IIR Filter Design Example

39592_3IIRFilterDesign.pdf

IIR Filter Design Example

These notes summarize the design procedure for IIR lters as discussed in class on March 26.

Introduction:

We discuss in this lecture how to design a discrete-time lowpass lter using both the impulse invariance method and the bilinear transform method. The specs for the lter to be designed are Passband cuto frequency:!p= 0:15 Stopband cuto frequency:!s= 0:35 Passband ripple1:3 dBH(ej!)0 dB,j!j !p Stopband attenuation:H(ej!) 20 dB,!s j!j  In terms of the parameterspandsdiscussed in OSB, this means that

1p= 103=20= 0:7079

 s= 1020=20= 0:1 For convenience we dene two new parametersk1andk2that are closely related topandsas follows: k

11(1p)21 = 0:9953

k 21

2s1 = 991

Note that the values of passband and stopband attenuation are given in decibels (dB). Ifxhas the dimensions

of amplitude or magnitude (as opposed to energy or power), the value ofxexpressed in dB would be 20log10(x).

2 Finally, we are designing our lter by transforming the frequency response of a continuous-time Butterworth lter. As you will recall, the Butterworth lter has the transfer function jH(j )j2=11 +  c 2N The parameterNspecies the number of poles in the prototype lter, and the parameter c species the critical cuto frequency. Note that the magnitude response of the Butterworth lter decreases monotonically with increasing , that the fallo in response as we transit from passband to stopband becomes sharper asNincreases, and thatjH(j )j2equals 1:0 when = 0 and 1=2 when = cfor all values ofN. As you will recall from class, a Butterworth lter with parametersNand chasNpoles that fall in the left half of thes-plane, in complex conjugate locations along a circle of radius cseparated by angles of=N.

Design procedure using impulse invariance

Determining design parameters

Using the impulse invariance design procedure, we have noted that the relation between frequency in the continuous-time and discrete-time domains is!=

T, whereTis merely a design parameter

(and not necessarily the sampling period). LeavingTas an arbitrary constant for now, we obtain p=!p=T= 0:47124=T s=!s=T= 1:0996=T

At the passband edge frequency,

jH(j )j2=11 +  p c

2N= (1p)2

With a few algebraic manipulations we obtain

 p c 2N =1(1p)21 =k1

Similarly, at the stopband edge frequency we have

jH(j )j2=11 +  s c

2N=2s

3 which produces s p! 2N =1

2s1 =k2

Dividing, we obtain

( p= c)2N( s= c)2N=k1k 2or N=12 log(k2=k1)log( s= p)= 2:7144

Note that in using impulse invariance,

s p=!s=T! p=T=!s! p so the design parameterThas no eect at all on the value ofNthat is needed. Furthermore, as expected, the value ofNspecied increases as eitherporsdecreases in magnitude (which implies decreasing the ripple in the passband and/or stopband) or as the ratio s= pdecreases (which implies a decrease in the width of the transition band). Since the number of poles must be an integer, we round up toN= 3. Matching the frequency response exactly at passband produces  p c 2N =k1or c= pk 1=2N

1=0:4716T

If we were instead to match the frequency response at stopband, we would obtain  s c 2N =k2or c= sk 1=2N

2=0:5112T

In principle, any value of the critical frequency that satises

0:4716T

 c0:5112T would be valid. As discussed in class, we will choose the value that meets the passband spec exactly, c= 0:4716=T because this value leaves the greatest margin for error at the stopband edge. Since the impulse invariance procedure always incurs a certain degree of aliasing, it will be expected that theactual response of the lter at the stopband edge will be larger than designed for. Hence, having the stop-

band edge response be overdesigned will leave more room for error to allow for the eects of aliasing.

4

Prototype lter design

As noted above, the poles of a Butterworth lter lie in the left half of thes-plane on a circle of radius cin complex conjugate pairs separated by angles of=Nradians. Since there are three poles, this produces a lter with the system function

H(s) =

3c(s cej)(s cej(+=3))(s cej(=3))

Plugging in the value of

c= 0:4716=T, we obtain (with the help of MATLAB) H(s) =(0:4716=T)3(s+ 0:4716=T)(s(0:236 + 0:408j)=T)(s(0:2360:408j)=T)

Conversion to discrete-time form

Using partial fractions we can rewrite the system function of the continuous-time prototype lter as

H(s) =NX

k=1A k(ssk) where the parametersAkare the continuous-time residues of the polessk. Using the MATLABroutineresiduewe nd the residues for the three poles, producing the transfer function H(s) =0:4716s+ 0:4716=T+0:23160:136js(0:236 + 0:408j)=T+0:2316 + 0:136js(0:2360:408j)=T The corresponding discrete-time lter has the transfer function

H(z) =NX

k=1A k1eskTz1 Note that the poles of the continuous-time lter,sk, are all of the form of a (generally complex) constant divided by the parameterT. Sinceskis multiplied byTwherever it appears in the equation forH(z) above, the specic value chosen forThas no eect at all on the discrete-time lter that results from design process. Hence we frequently letT= 1 for simplicity. This produces the transfer function H(z) =0:47161e0:4716z1+0:23160:136j1e0:236+0:408jz1+0:2316 + 0:136j1e0:2360:408jz1 or H(z) =0:471610:624z1+0:23160:136j1(0:7215 + 0:313j)z1+0:2316 + 0:136j1(0:72150:313j)z1

Combining the second and third terms we obtain

H(z) =0:471610:624z1+0:472 + 0:341z111:45z1+ 0:624z2 This expression is easily implemented using any of the techniques discussed in OSB Chapter 6. 5

Design procedure using bilinear transformation

Determining design parameters

Here we will convert from continuous-time to discrete-time form using the bilinear transform s=2T

1z11 +z1

This produces the nonlinear relationship between continuous-time frequency and discrete-time fre- quency = 2T tan(!=2) Converting the critical frequencies and to their continuous-time counterparts produces p=2T tan(!p=2) = 0:4802=T s=2T tan(!s=2) = 1:226=T

We again use the design equation

N=12 log(k2=k1)log( s= p)= 2:4546 Note that although we still need to have 3 poles, the actual fractional value ofNsmaller because of the greater ratio of stopband to passband edge frequencies provided by the nonlinear frequency warping of the bilinear transform. We will match the specs exactly in the passband, although there is no specic need to do so when we use the bilinear transform. The equation for the critical frequency then becomes c= pk 1=2N

1=0:4806T

Prototype lter design

As before, the transfer function for the Butterworth lter is

H(s) =

3c(s cej)(s cej(+=3))(s cej(=3)) Using the MATLABroutinepolyand the value for that we had obtained we nd that the transfer function of the prototype lter is

H(s) =0:110=T3s

3+0:9612s2T

+0:4620sT

2+0:110T

3 6

Conversion to discrete-time form

We can now substitute directly the bilinear transform relationship s= 2T

1z11 +z1!

This produces

H(s) =0:110=T3

2T

1z11+z1

3+0:9612

2T

1z11+z1

2T +0:4620 2T

1z11+z1T

2+0:110T

3 Note that the design parameterTappears in cubic form in the denominator of every term in numerator and denominator. Hence its specic value does not matter and we can cancel it from the design equation. The value of the parameterTis frequently set to equal 2 to simplify the expression for the bilinear transform. We also note that simplifying the expression above is an algebraic nightmare. Using the MATLAB routinebilinearwith the poles and (null) zeros ofH(s) as input parameters (and a value of 1 for the parameterT), we obtain H(z) =0:00854(1 + 3z1+ 3z2+z3)12:064z1+ 1:519z20:386z3