Butterworth filter: Number of coefficients?

12 vues (au cours des 30 derniers jours)
Max Behr
Max Behr le 19 Mai 2020
Commenté : Star Strider le 19 Mai 2020
Hello
I am trying to create a low pass filter. My device has a sample rate of 100 and the cutoff value should be 10 Hz.
But according to the manual the filter should also have a setting "number of coefficients". And this should be 40 in my case...
How can I implement the number of coefficients into the filter?
Thanks.
My Code:
Fs = 100;
Fco = 10;
Fn = Fs/2;
Wn = Fco/Fn;
[b,a]=butter(1,Wn,'low')
m_filtered=filter(b,a,m)

Réponse acceptée

Star Strider
Star Strider le 19 Mai 2020
The first argument to butter is the filter order, and the number of coefficients in the transfer function vectors will be 1 less than that, so:
Fs = 100;
Fco = 10;
Fn = Fs/2;
Wn = Fco/Fn;
n = 40;
[b,a]=butter(n-1,Wn,'low');
figure
freqz(b,a,2^14,Fs)
should do what you want.
Note that this creates an unstable filter.
Are you absolutely certain that you want to use a IIR filter here? A filter order of 40 is more typical of a FIR filter, especially for a lowpass application such as this. (If you want a FIR filter instead, use the kaiserord and fir1 functions to create it.)
  8 commentaires
Max Behr
Max Behr le 19 Mai 2020
Thanks again :)
I did it!
Star Strider
Star Strider le 19 Mai 2020
As always, my pleasure!
Thank you!

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