How to Butterworth Filter with Bandpass [10 500] with sampling rate 1000?

Hi,

I want to butterworth filter with bandpass [10 500] with sampling frequency 1000 but there is an error stating "The cutoff frequencies must be within the interval of (0,1)." How do I solve this problem?

Answer 1

[Star Strider]

You have to divide your passband and stopband frequencies by the Nyquist frequency to normalise them:

Fs = 1000;                                          % Sampling Frequency
Fn = Fs/2;                                          % Nyquist Frequency
Wp = [10  495]/Fn;                                  % Norrmalised Passband Frequencies
Ws = [5   499]/Fn;                                  % Normalised Stopband Frequencies
Rp =  1;                                            % Passband Ripple (dB)
Rs = 25;                                            % Stopband Ripple (dB)
[n,Wn] = buttord(Wp, Ws, Rp, Rs);                   % Optimal Filter Order
[b,a] = butter(n, Wn);                              % Calculate Filter Coefficients
[sos,g] = tf2sos(b,a);                              % Convert To Second-Order Sections For Stability

figure(1)
freqz(sos, 4096, Fs)                                % Filter Bode Plot

You cannot have any frequency exactly at zero or exactly at the Nyquist frequency, so I designed your passband to be close enough to them to be stable. In your filter, you can also use a high-pass design with a lower cutoff of 10 Hz, since your upper passband is at the Nyquist frequency. The passband and stopband ripple values are necessary to specify in the design but irrelevant to a Butterworth filter. These are acceptable values for most filter designs (for example Chebyshev) that require them.

You would use the ‘sos’ and ‘g’ variables in your actual filter:

filtered_signal = filtfilt(sos, g, signal);

(This design works well in R2016a and should also work in earlier releases.)

[Hazwani Harun]

Thanks sir. I had tried and it works. I would like to ask , if there any effect on my result if the bandpass demand is 10-500 but i use 495/499?

[Star Strider]

My pleasure.

The ‘495,499’ will not produce a noticeable difference as an upper passband frequency in a band-pass filter as compared to a high-pass filter with a 500 Hz Nyquist frequency (the highest resolvable frequency in a sampled signal, in this instance with a sampling frequency, ‘Fs’ of 1 kHz). Those frequencies are necessary to (1) produce a stable filter, and (2) to avoid the upper passband frequency equaling the Nyquist frequency.