How to create low pass filter for audio file?

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How do i create a low pass filter for an audio file?
I would like to keep 20 Hz and below Sampling rate 8000 Hz
Thanks

Accepted Answer

Star Strider
Star Strider on 24 Feb 2015
You gave enough information for the Signal Processing Toolbox (with a bit of help from me) to be able to design your filter:
Fs = 8000; % Sampling Frequency (Hz)
Fn = Fs/2; % Nyquist Frequency
Fco = 20; % Passband (Cutoff) Frequency
Fsb = 30; % Stopband Frequency
Rp = 1; % Passband Ripple (dB)
Rs = 10; % Stopband Ripple (dB)
[n,Wn] = buttord(Fco/Fn, Fsb/Fn, Rp, Rs); % Filter Order & Wco
[b,a] = butter(n,Wn); % Lowpass Is Default Design
[sos,g] = tf2sos(b,a); % Second-Order-Section For STability
figure(1)
freqz(sos, 2048, Fs) % Check Filter Performance
Then use the filtfilt function to filter your signal.
I chose a Butterworth design, but if you want a steeper cutoff, consider using a Chebychev design instead. You may want to experiment with the design to get the exact result you want with respect to your signal. See the relevant documentation for details.
  4 Comments
Daemian
Daemian on 24 Feb 2015
Hi, after i filter if i want to convert it to ftt, do i do it this way?
Fs = 8000; % Sampling Frequency (Hz)
Fn = Fs/2; % Nyquist Frequency
Fco = 20; % Passband (Cutoff) Frequency
Fsb = 30; % Stopband Frequency
Rp = 1; % Passband Ripple (dB)
Rs = 10; % Stopband Ripple (dB)
[n,Wn] = buttord(Fco/Fn, Fsb/Fn, Rp, Rs); % Filter Order & Wco
[b,a] = butter(n,Wn); % Lowpass Is Default Design
[sos,g] = tf2sos(b,a); % Second-Order-Section For STability
%figure(1) %freqz(sos, 2048, Fs) % Check Filter Performance yd = filtfilt(sos, g, myRecording); % Filter ‘xd’ To Get ‘y’
yd_len = length(yd); % Length Of ‘wav’ File Data
Ts = 1/Fs; % Sampling Time
Tv = linspace(0, yd_len-1, yd_len)*Ts; % Time Vector
y_fft = abs(fft(yd)); %Retain Magnitude
Nsamps = length(yd);
y_fft = y_fft(1:Nsamps/2); %Discard Half of Points
f = Fs*(0:Nsamps/2-1)/Nsamps; %Prepare freq data for plot
plot(handles.normalfft,f, y_fft)
xlim(handles.normalfft,[8 12])
Star Strider
Star Strider on 24 Feb 2015
I’m not certain what you’re doing. The usual procedure is to divide the output of the fft function by the length of the vector to normalise the amplitudes. I also don’t understand your frequency vector ‘f’ calculation (it doesn’t look correct to me), but if it works for you, go for it!
I follow the amplitude and frequency vector calculations in the documentation for the fft function, rather than creating my own calculations. It’s easier, and it works. I suggest you do that.

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More Answers (1)

Tengsim Momin
Tengsim Momin on 28 Mar 2016
how to filter eeg signal with(0.1db-80db) low pass filter using fdatool

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