Question

6. MATLAB problem:

A filter has the following specifications:

Pass band edge = 0.3π.

Stop band edge = 0.5π. 20log10(?????????=?????) = - 40 dB.

Design this filter using various IIR and FIR methods using MATLAB. Make a "brief" statement comparing the designs.

Answer 1

FIR design

clc;close all;clear all;
wp=0.3*pi
ws=0.5*pi
wc=(wp+ws)/2
%using Hanning window
N=round(8*pi/(ws-wp));
M=N+1;
n=0:1:M-1;
%For Lpf, the desired impulse response is
alpha=(M-1)/2;
hd=(sin(wc*(n-alpha)))./(pi*(n-alpha))
hd(alpha+1)=wc/pi;
wh=hanning(M)';
h=hd.*wh;
%Magnitude response
w=0:0.01*pi:pi
[H,w]=freqz(h,[1],w)

figure;
subplot(221)
plot(w/pi,20*log10(abs(H)),'r')
title('Magnitude response |H(w)| -FiR Hanning window')
xlabel('w/pi')
ylabel('|H(exp(jw)|');grid;

subplot(222)
plot(w/pi,angle(H),'g')
title('phase response |H(w)|')
xlabel('w/pi')
ylabel('|H(exp(jw)|')

%Hamming window
wh=hamming(M)'
h=hd.*wh;
[H,w]=freqz(h,[1],w)

subplot(223)
plot(w/pi,20*log10(abs(H)),'m')
title('Magnitude response |H(w)| -FiR Hamming window')
xlabel('w/pi')
ylabel('|H(exp(jw)|');grid;

subplot(224)
plot(w/pi,angle(H),'g')
title('Phase response |H(w)|')
xlabel('w/pi')
ylabel('|H(exp(jw)|')

Magnitude response |H(w)| -FiR Hanning window phase response |H(w)|| (MI)dxəhl |H(exp(jw) my 0 0.2 0.8 1 0.4 0.6 w/pi 0 0.2 0.4 0.6 w/pi 0.8 1 Magnitude response (H(w)| -FiR Hamming window Phase response |H(w)|| |H(exp(jw) |H(exp(jw) 0 0.2 0.4 0.6 w/pi 0.8 1 0 0.2 0.4 0.6 w/pi 0.8 1

Observation:

For the same order M=41, the hamming window provides better attenuation than hanning window.

Both provides the same amount of transition width.

_____________________________________________________________________________

IIR design:

clc;
close all;
clear all;
%Specifications
Rs=40
wp=0.3
ws=0.5
Rp=0.5%assumption
%Butterworth filter
[N,wc]=buttord(wp,ws,Rp,Rs)
[b,a]=butter(N,wc)
figure;
freqz(b,a)
%chebyshev Filter 1
[N,wp] = cheb1ord(wp,ws,Rp,Rs)
[b,a]=cheby1(N,Rp,wp)
figure;
freqz(b,a)
%Elliptic Filter
[N,wp] = ellipord(wp,ws,Rp,Rs)
[b,a] = ellip(N,Rp,Rs,wp)
figure;
freqz(b,a)

Butterworth filter:

Magnitude (dB) 0.2 0.8 0.4 0.6 Normalized Frequency (xn rad/sample) Phase (degrees) -800 -1000 0.2 0.8 0.4 0.6 Normalized Frequency (xn rad/sample)

Chebyshev 1:

Magnitude (dB) 0.2 0.8 0.4 0.6 Normalized Frequency (xn rad/sample) Phase (degrees) -400 -500 R 0.2 0.8 0.4 0.6 Normalized Frequency (xn rad/sample)

elliptic:

Magnitude (dB) 0.2 0.8 0.4 0.6 Normalized Frequency (xn rad/sample) Phase (degrees) 0.8 0.2 0.4 0.6 Normalized Frequency (xn rad/sample)