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FIltering for multiple band of frequncies

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Ramya Raman
Ramya Raman am 6 Feb. 2020
Kommentiert: Star Strider am 24 Okt. 2022
I have below two questions.
  1. I want to filter a signal with sampling 100Hz in range of 4-7Hz and 10-13Hz and 30-35Hz. I saw the bandpass function that says to filter seperating and concatenate the filtered signal. But I want to know if we could filter these 3 range of frequncies with any specific filter?
  2. What is bandpass or bandstop ripple and how to we choose the ripple ?

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Star Strider
Star Strider am 6 Feb. 2020
As much as I like elliptic filters, creating three of them and filtering them in parallel is simply not efficient. Multiband filters are much easier to implement as FIR filters.
Try a FIR filter such as this one:
Fs = 100; % Sampling Frequency (Hz)
fcuts = [3.5 4 7 7.5 9.5 10 13 13.5 29 30 35 36]; % Frequencies
mags = [0 1 0 1 0 1 0 ]; % Passbands & Stopbands
devs = [0.05 0.01 0.05 0.01 0.05 0.01 0.05]; % Tolerances
[n,Wn,beta,ftype] = kaiserord(fcuts,mags,devs,Fs); % Kaiser Window FIR Specification
n = n + rem(n,2);
hh = fir1(n,Wn,ftype,kaiser(n+1,beta),'noscale'); % Filter Realisation
figure
freqz(hh,1,2^14,Fs)
set(subplot(2,1,1), 'XLim',[0 50]); % Zoom Frequency Axis
set(subplot(2,1,2), 'XLim',[0 50]); % Zoom Frequency Axis
Then use the filtfilt function to do the actual filtering:
signal_filtered = filtfilt(hh, 1, signal);
The Bode plot for this filter:
FIltering for multiple band of frequncies - 2020 02 06.png
  10 Kommentare
Ameen
Ameen am 24 Okt. 2022
So the stopbands are [55 59], [61,64], [61 64]+60, [61 64]+120 and passbands are [59,61], [55 59]+60, [55 59]+120 in the first example?
Star Strider
Star Strider am 24 Okt. 2022
No, they’re [59 61], [119 121], and [179 181] (or [59 61], [59 61]+60 and [59 61]+120) in this illustration.
They’re the centre two frequencies in those vectors.

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