Question

(a) The impulse response hfn of an FIR filter satisfies the following property: h[n]- otherwise where M is an even integer. Derive the filters frequency response and show that it has a linear phase. Why is linear phase a desired property ? (b) You are asked to design a linear-phase FIR filter. The required pass-band is from 1,000 Hz to 3,000 Hz. The input signals sampling frequency is 16, 000Hz e the pass-band in the w domain 1. GlV n ii. Derive the ideal impulse response (which is non-causal); iii. By preserving the most significant filter coefficients, give the impulse response of a linear-phase FIR filter of order 8; iv. Plot the magnitude response of the FIR filter. (c) Use the MATLAB function firl to re-do (b). Provide the MATLAB codes and list the filter coefficients obtained

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Answer #1
load chirp
y = y + 0.25*(rand(size(y))-0.5);
f = [0 0.48 0.48 1];
mhi = [0 0 1 1];
bhi = fir2(34,f,mhi);

freqz(bhi,1)
outhi = filter(bhi,1,y);
t = (0:length(y)-1)/Fs;

subplot(2,1,1)
plot(t,y)
title('Original Signal')
ylim([-1.2 1.2])

subplot(2,1,2)
plot(t,outhi)
title('Higpass Filtered Signal')
xlabel('Time (s)')
ylim([-1.2 1.2])
mlo = [1 1 0 0];
blo = fir2(34,f,mlo);
outlo = filter(blo,1,y);

subplot(2,1,1)
plot(t,y)
title('Original Signal')
ylim([-1.2 1.2])

subplot(2,1,2)
plot(t,outlo)
title('Lowpass Filtered Signal')
xlabel('Time (s)')
ylim([-1.2 1.2])
f = [0 0.6 0.6 1];
m = [1 1 0 0];

b1 = fir2(30,f,m);
[h1,w] = freqz(b1,1);

plot(f,m,w/pi,abs(h1))
xlabel('\omega / \pi')
lgs = {'Ideal','fir2 default'};
legend(lgs)
b2 = fir2(30,f,m,64);
h2 = freqz(b2,1);

hold on
plot(w/pi,abs(h2))
lgs{3} = 'npt = 64';
legend(lgs)
b3 = fir2(30,f,m,64,13);
h3 = freqz(b3,1);

plot(w/pi,abs(h3))
lgs{4} = 'lap = 13';
legend(lgs)-50 100 0 0.1 0.2 0.3 0.40.5 0.6 0.7 0.8 0.9 1 Normalized Frequency (× π rad/sample) Ф -500 9-1000 -1500 2000 0 0.1 0.2 0.3 0.40.5 0.6 0.7 0.8 0.9 1 Normalized Frequency (x rad/sample)
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