Design an FIR lowpass filter of order 255 with a transition region between $$0.25\pi$$ and $$0.3\pi$$. Use fvtool to display the magnitude and phase responses of the filter.

b = firls(255,[0 0.25 0.3 1],[1 1 0 0]);
fvtool(b,1,'OverlayedAnalysis','phase')

An ideal differentiator has a frequency response given by $$D(\omega )=j\omega$$. Design a differentiator of order 30 that attenuates frequencies above $$0.9\pi$$. Include a factor of $$\pi$$ in the amplitude because the frequencies are normalized by $$\pi$$. Display the zero-phase response of the filter.

b = firls(30,[0 0.9],[0 0.9*pi],'differentiator');

fvtool(b,1,'MagnitudeDisplay','zero-phase')

Design a 24th-order antisymmetric filter with piecewise linear passbands.

F = [0 0.3 0.4 0.6 0.7 0.9]; 
A = [0 1.0 0.0 0.0 0.5 0.5];
b = firls(24,F,A,'hilbert');

Plot the desired and actual frequency responses.

[H,f] = freqz(b,1,512,2);
plot(f,abs(H))
hold on
for i = 1:2:6, 
   plot([F(i) F(i+1)],[A(i) A(i+1)],'r--')
end
legend('firls design','Ideal')
grid on
xlabel('Normalized Frequency (\times\pi rad/sample)')
ylabel('Magnitude')

Design an FIR lowpass filter. The passband ranges from DC to $$0.45\pi$$ rad/sample. The stopband ranges from $$0.55\pi$$ rad/sample to the Nyquist frequency. Produce three different designs, changing the weights of the bands in the least-squares fit.

In the first design, make the stopband weight higher than the passband weight by a factor of 100. Use this specification when it is critical that the magnitude response in the stopband is flat and close to 0. The passband ripple is about 100 times higher than the stopband ripple.

bhi = firls(18,[0 0.45 0.55 1],[1 1 0 0],[1 100]);

In the second design, reverse the weights so that the passband weight is 100 times the stopband weight. Use this specification when it is critical that the magnitude response in the passband is flat and close to 1. The stopband ripple is about 100 times higher than the passband ripple.

blo = firls(18,[0 0.45 0.55 1],[1 1 0 0],[100 1]);

In the third design, give the same weight to both bands. The result is a filter with similar ripple in the passband and the stopband.

b = firls(18,[0 0.45 0.55 1],[1 1 0 0],[1 1]);

Visualize the magnitude responses of the three filters.

hfvt = fvtool(bhi,1,blo,1,b,1,'MagnitudeDisplay','Zero-phase');
legend(hfvt,'bhi: w = [1 100]','blo: w = [100 1]','b: w = [1 1]')