Designing Bandpass Filters

We will design several filters for the following normalized frequencies and desired frequency response. First, specify

the desired frequency response point-wise, with 1.0 corresponding to half the sample rate. Plot the desired frequency response to make sure it is what we want

(unnormalize the frequency axis).

• f = [0 .4 .4 .6 .6 1];• H = [0 0 1 1 0 0];• fs = 1000; % assumed sampling rate• fhz = f*fs/2;• plot(fhz,H)• title('Desired Frequency Response')• xlabel('Frequency (Hz)')• ylabel('Magnitude')

• The YULEWALK function lets you to specify a piecewise shape for the desired frequency response magnitude. It then finds an infinite-impulse response filter of the desired order that fits the frequency response in a least-squares sense. Use YULEWALK to compute the coefficients of an 8th order filter that will approximate our desired response. Plot the frequency response magnitude and compare it to the desired response.

• N = 8; % Order of the filter (number of poles and zeros).

• [Bh,Ah] = yulewalk(N,f,H); % Working, please wait.....• n = 256;• hh = freqz(Bh,Ah,n); % compute complex frequency

response• hy = abs(hh); % compute magnitude• ff = fs/(2*n) * (0:n-1);• plot(fhz,H,ff,hy)• title('Actual vs. Desired Frequency Response')• xlabel('Frequency (Hz)')• ylabel('Magnitude')

• Now let's design Butterworth and Chebyshev bandpass filters with the same passband (defined between 0.0 and 1.0). Here we compare all three frequency responses.

• N = 4; passband = [.4 .6]; ripple = .1;• [Bb,Ab] = butter(N, passband);• [Bc,Ac] = cheby1(N, ripple, passband);• h = [abs(hh) abs(freqz(Bb,Ab,n))

abs(freqz(Bc,Ac,n))];• plot(ff,h)• legend({'YuleWalk','Butterworth','Chebyshev'});• title('Filter Performance')• xlabel('Frequency (Hz)')• ylabel('Magnitude')

• Finally, look at the frequency response on a logarithmic decibel (dB) scale. plot(ff(2:n),20*log10(h(2:n,:)))

• title('YuleWalk, Butterworth and Chebyshev filters')

• xlabel('Frequency (Hz)')

• ylabel('Magnitude in dB')

Filter Design

• This demonstration designs a filter with the Signal Processing Toolbox and applies it to a signal made up of harmonic components.

• Here's an example of filtering with the Signal Processing Toolbox. First make a signal with three sinusoidal components (at frequencies of 5, 15, and 30 Hz).

• Fs = 100;

• t = (1:100)/Fs;

• s1 = sin(2*pi*t*5); s2=sin(2*pi*t*15); s3=sin(2*pi*t*30);

• s = s1+s2+s3;

• plot(t,s);

• xlabel('Time (seconds)'); ylabel('Time waveform');

• To design a filter to keep the 15 Hz sinusoid and get rid of the 5 and 30 Hz sinusoids, we create an eighth order IIR filter with a passband from 10 to 20 Hz. Here is its frequency response. The filter was created with the ELLIP command.

• [b,a] = ellip(4,0.1,40,[10 20]*2/Fs);

• [H,w] = freqz(b,a,512);

• plot(w*Fs/(2*pi),abs(H));

• xlabel('Frequency (Hz)'); ylabel('Mag. of frequency response');

• grid;

• After filtering, we see the signal is a 15 Hz sinusoid, exactly as expected. sf = filter(b,a,s);

• plot(t,sf);

• xlabel('Time (seconds)');

• ylabel('Time waveform');

• axis([0 1 -1 1]);

• Finally, here is the frequency content of the signal before and after filtering. Notice the peaks at 5 and 30 Hz have been effectively eliminated. S = fft(s,512);

• SF = fft(sf,512);• w = (0:255)/256*(Fs/2);• plot(w,abs([S(1:256)' SF(1:256)']));• xlabel('Frequency (Hz)'); ylabel('Mag. of Fourier

transform');• grid;• legend({'before','after'})