I've got this code, what filter, and how should i use it to only show me in the end the component with the 300Hz frequency? Using a fir filter
```
t = 0:1/8000:1023/8000;
w = 0:8000/1024:4000;
x = 0.5*sin(2*pi*50*t) + 0.5*square(2*pi*130*t) + 0.3*sawtooth(2*pi*250*pi) + 0.2*sin(2*pi*300*t);
h =
y = filter(h,1,x);
figure;
subplot(2,1,1), plot(t,x);
z = abs(fft(x)) / 512;
subplot(2,1,2), plot(w, z(1:513));
figure;
subplot(2,1,1), plot(t,y);
z = abs(fft(y)) / 512;
subplot(2,1,2), plot(w, z(1:513));
```

 Réponse acceptée

You could perhaps use a butterworth bandpass filter: (Note: Requires Signal Processing Toolbox)
t = 0:1/8000:1023/8000;
w = 0:8000/1024:4000;
x = 0.5*sin(2*pi*50*t) + 0.5*square(2*pi*130*t) + 0.3*sawtooth(2*pi*250*pi) + 0.2*sin(2*pi*300*t);
% h =
% y = filter(h,1,x);
% create bandpass butterworth filter @ 300 Hz
fCutoff1 = 250;
fCutoff2 = 350;
[b, a] = butter(6, [fCutoff1 fCutoff2]/(8000/2), 'bandpass');
% filter the signal
y = filter(b, a, x);
figure;
subplot(2,1,1), plot(t,x);
z = abs(fft(x)) / 512;
subplot(2,1,2), plot(w, z(1:513));
figure;
subplot(2,1,1), plot(t,y);
z = abs(fft(y)) / 512;
subplot(2,1,2), plot(w, z(1:513));

2 commentaires

Can i know why, or how did you know the order was 6?
Scott MacKenzie
Scott MacKenzie le 26 Mar 2022
Modifié(e) : Scott MacKenzie le 26 Mar 2022
I just used order six as it seemed to do a reasonable job in excluding neighboring frequencies. At order 3, for example, you get this:

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