Low Pass filter not working

43 views (last 30 days)
Yipp Chun Munn
Yipp Chun Munn on 16 Oct 2019
Edited: Star Strider on 20 Oct 2019
I audioread() a signal and tried to apply low-pass filtering but it does not seem to have any change at all. The signal is a recording of lung sound and I wish to filter out the noise component.
[y,Fs] = audioread('mysound.wav')
Fs = 44100; % Sampling Frequency (Hz)
Fn = Fs/2; % Nyquist Frequency
Fco = 70; % Passband (Cutoff) Frequency
Fsb = 100; % 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);
filt_sig = filtfilt(sos,g,y)
Neither the plot(), FFT() or soundsc() shows anything different. I've tried cheby filters as well. Am I doing anything wrong? Thanks for the help.
  2 Comments
Daniel M
Daniel M on 16 Oct 2019
Edited: Daniel M on 16 Oct 2019
You would have to provide your data, or at least show the timeseries before and after, and the fft before and after filtering. As for soundsc, I suspect either your speakers do not play low-end frequencies very well, or the SNR of the signal is low below your cutoff frequency.
What does your noise consist of? This changes how you approach the filter.
Is it evenly distributed, white noise?
Is the noise separated from your signal spectrally?
Is the noise from a particular, characteristic external source?
Yipp Chun Munn
Yipp Chun Munn on 16 Oct 2019
Capture2.PNG
Capture1.PNG
The frequency domain and Time domain. No matter what filter I apply, the Far right frequency components does'nt go away.

Sign in to comment.

Accepted Answer

Star Strider
Star Strider on 16 Oct 2019
Your stopband attenuation is likely not sufficient to produce any difference.
Try this instead:
Rs = 60; % Stopband Ripple (dB)
Increase ‘Rs’ (to increase the stopband attenuation) as necessary to get the result you want.
Also, always use freqz (or similar functions) to see what your filter is doing:
figure
freqz(sos, 2^14, Fs)
  7 Comments
Yipp Chun Munn
Yipp Chun Munn on 20 Oct 2019
Dear Star Strider, I've tried the Remove the 60 Hz hum from a signal solution but still to no avail. I've continued to try various filters still and the best possible outcome I've gotten is Captur7.PNG
This is was obtained using a cheby2 bandpass filter at Passband 850-900hz and stopband at 600 1000Hz. Not perfect at all but at least I can make out the 3 waveform above are the 3 breathing sounds. Also when I sound() this signal, the breathing is more audible. Here's the confusion, when I abs(fft()) the filtered signal, I get components at > 5khz. How should this be as it is clearly way out of my bandpass. Very sorry for asking so many question, thanks.Capture8.PNG
Star Strider
Star Strider on 20 Oct 2019
Edited: Star Strider on 20 Oct 2019
The last image you posted appears to show your filter working as it should. No filter will completely eliminate the frequencies outside the passband and stopband limits, however they will be significantly attenuated. That is what you want.
A better and more efficient filter solution would likely be an elliptic filter.
Prototype code for that would be:
Fs = 2250;
Fn = Fs/2;
Wp = [59 61]/Fn; % Stopband Frequency (Normalised)
Ws = [58 62]/Fn; % Passband Frequency (Normalised)
Rp = 1; % Passband Ripple
Rs = 90; % Passband Ripple (Attenuation)
[n,Wp] = ellipord(Wp,Ws,Rp,Rs); % Elliptic Order Calculation
[z,p,k] = ellip(n,Rp,Rs,Wp); % Elliptic Filter Design: Zero-Pole-Gain
[sos,g] = zp2sos(z,p,k); % Second-Order Section For Stability
figure
freqz(sos, 2^18, Fs) % Filter Bode Plot
set(subplot(2,1,1), 'XLim',[0 100]) % Frequency Axis Limits
set(subplot(2,1,2), 'XLim',[0 100]) % Frequency Axis Limits
% signal_filt = filtfilt(sos, g, signal); % Filter Signal
Make appropriate changes for your sampling frequency, filter passbands and stopbands, and signal. Elliptic filters can handle very narrow transition regions, so:
Wp = [850 900]/Fn; % Stopband Frequency (Normalised)
Ws = [800 950]/Fn; % Passband Frequency (Normalised)
will produce a stable filter, likely with better charactersitics.
EDIT —
Corrected typographical error.

Sign in to comment.

More Answers (1)

Daniel M
Daniel M on 16 Oct 2019
Edited: Daniel M on 16 Oct 2019
You show a figure that essentially looks like the following (also attached):
fs = 1000;
t = 0:1/fs:10;
x = sin(2*pi*10*t);
X = fft(x);
figure
plot(abs(X))
This is completely normal. It is the two sided amplitude spectrum.
You should read more about fftshift, and follow the example for the single sided amplitude spectrum on page for fft.
"No matter what filter I apply, the Far right frequency components does'nt go away."
That's because they're not supposed to.

Products


Release

R2018a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!