You are working with regularly sampled data, so you want a digital filter, not an analog filter.

The Nyquist frequency is half the sampling rate.Plt.axvline(100, color='green') # cutoff frequencyĪnd got this which clearly does not cut-off at 23 rad/s: Plt.title('Butterworth filter frequency response')

I plotted the filter using: w, h = eqs(b, a) And why are you using the nyquist as a normalizer?Ĭan someone explain more about how to create filters with these functions? I know it's something about normalization but I thought the nyquist was 2 times the sampling requency, not one half. Secondly, what is the purpose of these two lines: nyq = 0.5 * fs I am very confused by this though because I am pretty sure the butter function takes in the cutoff and sampling frequency in rad/s but I seem to be getting a weird output. Y = butter_lowpass_filter(data, cutOff, fs, order) Here is the link the bandpass code, so I converted it to be this: from scipy.signal import butter, lfilterī, a = butter(order, normalCutoff, btype='low', analog = True)ĭef butter_lowpass_filter(data, cutOff, fs, order=4):ī, a = butter_lowpass(cutOff, fs, order=order)įs = 188.495559 #sampling frequency in rad/s The sampling frequency of the chip that takes data is 30Hz so I converted that to rad/s to get 188.495559 rad/s.Īfter looking up some stuff online I found some functions for a bandpass filter that I wanted to make into a lowpass. I converted 220/minute into 3.66666666 Hertz and then converted that Hertz to rad/s to get 23.0383461 rad/sec. Because heart rates should never be above about 220 beats per minute, I want to filter out all noise above 220 bpm. I am trying to filter a noisy heart rate signal with python.

Digital low pass filter equation