Signals and Systems - Electrical Engineering

11.4 IIR Filter Design 663

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% Example 11.7

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% LP Butterworth

alphamax = 2; alphamin = 9; % loss specifications

wp = 0.47;ws = 0.6; % passband and stopband frequencies

[N,wh] = buttord(wp,ws,alphamax,alphamin) % minimal order, half-power frequency

[b,a] = butter(N,wh); % coefficients of designed filter

[H,w] = freqz(b,a);w = w/pi;N = length(H); % frequency response

spec1 = alphamax∗ones(1,N); spec2 = alphamin∗ones(1,N); % specification lines

hpf = 3.01∗ones(1,N); % half-power frequency line

disp(‘poles’) % display poles

roots(a)

disp(‘zeros’) % display zeros

roots(b)

alpha = -20∗log10(abs(H)); % loss in dB

The results of the design are shown in Figure 11.13. The order of the designed filter isN=3 and

the half-power frequency isωhp=0.499π. The poles are along the imaginary axis of thez-plane

(Kb= 1 )and there are three zeros atz=−1. The transfer function of the designed filter is

H(z)=

0.166+0.497z−^1 +0.497z−^2 +0.166z−^3

1 −0.006z−^1 +0.333z−^2 −0.001z−^3

Finally, to verify that the specifications are satisfied we plot the loss functionα(ejω)along with three

horizontal lines corresponding toαmax=2 dB, 3 dB for the half-power frequency, andαmin=9 dB.

− 5 0 5

− 1

0

1

3

Real part

Imaginary

part

0 0.1 0.2 0. 30. 40. 50. 60. 70. 80. 9

0

2

4

6

8

10

12

α(

ω

)dB

ω/π

(b)

0 0. 1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 0. 9

0

0. 5

1

|H

(e

jω

)|

0 0. 1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 0. 9 1

− 6

− 4

− 2

0

<H

(e

jω

)

ω/π

(a)

FIGURE 11.13
Design of low-pass Butterworth filter using MATLAB: (a) poles and zeros and magnitude and phase responses,
and (b) verification of specifications using loss functionα(ω).