Quickest way to do a bandpass filter?

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I have a huge dataset of .wav files that I would like to bandpass filter.
Is the following the most time efficient way to do it..?
This filter takes about 2 minutes to run so it's quite time consuming with such a huge dataset.
[xbit, fs]=audioread(filename, [1*fs,120*fs]); %read in file from 1-120 secs
xbit=detrend(xbit);%removes DC offset
tic
xbit_filt=bandpass(xbit,[50 24000],fs);
toc

Accepted Answer

Mathieu NOE
Mathieu NOE on 9 Jan 2023
Hello Louise
welcome back and happy new year !
I made some speed comparison with your code and two other options using a butterworth digital filter , then using the function filter or filtfilt (filtering twice , forward and backward in time, for phase distorsion cancellation)
see my suggestion are much faster than using the bandpass function as in your code
I wonder why you put the low pass frequency at 24 kHz as this is already above the human hearing upper frequency limit...
at what sampling frequency are your wav files ?
with the my small audio file sampled at 11025 Hz , the differences are significant :
  • your code : Elapsed time is 1.574109 seconds.
  • my code with filter : Elapsed time is 0.016102 seconds.
  • my code with filtfilt : Elapsed time is 0.042606 seconds.
filename = 'test_voice_mono.wav' ;
% NB this file is sampled at 11025 Hz, so the bandpass filter cut off
% frequencies are modified (LPF : 24kHz => 2.4kHz)
[xbit, fs]=audioread(filename); %we don't know yet what if fs so I cannot define range = [1*fs,120*fs] as in your original code
samples = length(xbit);
start = 1*fs;
stop = min(120*fs,samples);
xbit=xbit(start:stop); % extract data for t = 1-120 secs
xbit=detrend(xbit);%removes DC offset
% your code
tic
xbit_filt=bandpass(xbit,[50 2400],fs);
toc
% good old butter bandpass filter (applied once)
tic
N = 2;
[B,A] = butter(N,[50 2400]/(fs/2));
xbit_filt = filter(B,A,xbit); % NB filter applies the filter once !
toc
% good old butter bandpass filter (applied twice)
tic
N = 2;
[B,A] = butter(N,[50 2400]/(fs/2));
xbit_filt = filtfilt(B,A,xbit); % NB filtfilt applies the filter twice !
toc
  6 Comments
Louise Wilson
Louise Wilson on 9 Jan 2023
Thanks Mathieu for your brilliant and detailed answer.
I am just using the filter to get the 5th and 95th percentiles, median, and rms level of the file, then store those in an output. Got the loop sorted :-)
Mathieu NOE
Mathieu NOE on 9 Jan 2023
ok
glad I could help !
all the best

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