% M07-filt.diary % Design a 3-pt filter to remove one of two sine tones mixed together % Dan Ellis dpwe@ee.columbia.edu % 2007-10-11 % filter is [alpha beta alpha] % analysis gives mag. resp. as beta + 2 alpha cos(w) % so to make it zero at w = 0.4, we have % alpha = -beta/(2 cos(0.4)) % and to make it one at w = 0.1, we have % beta (1 + 2(-1/(2 cos(0.4)))cos(0.1)) = 1 % hence... beta = 1/(1 + 2*(-1/(2*cos(0.4)))*cos(0.1)) % beta = % -12.4563 alpha = -beta/(2*cos(0.4)) % al = % 6.7619 % Plot results... % First, the impulse response h = [alpha beta alpha]; subplot(311) stem(h) axis([0 50 -15 15]) grid % Then the magnitude response (DTFT) subplot(312) [H,W] = freqz(h); % Plot the magnitude in dB (log scaling) plot(W, 20*log10(abs(H))) axis([0 1.2 -20 20]); grid % gain of zero at w = 0.4 corresponds -Inf dB % notice that gain goes through 0 dB (=1x) just around w = 0.1 % Finally, compare signal before and after filtering nn = 1:8000; w1 = 0.4; w2 = 0.1; x = 0.5*(sin(w1*nn) + sin(w2*nn)); y = conv(h,x); subplot(313) plot(1:200,x(1:200),1:200,y(1:200),'r') axis([0 200 -1 1]) grid % Check out the sound sound(x,16000); sound(y,16000); % Higher component (w1) clearly removed