MATLAB Answers

Low Pass filter not working

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Yipp Chun Munn
Yipp Chun Munn 2019 年 10 月 16 日
編集済み: Star Strider 2019 年 10 月 20 日
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.

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Daniel M
Daniel M 2019 年 10 月 16 日
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 2019 年 10 月 16 日
The frequency domain and Time domain. No matter what filter I apply, the Far right frequency components does'nt go away.

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Star Strider
Star Strider 2019 年 10 月 16 日
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:
freqz(sos, 2^14, Fs)

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Star Strider
Star Strider 2019 年 10 月 17 日
I always use sound, not soundsc, since soundsc scales the output to ±1. Using sound risks distorting the output, however it is possible to hear it. You can also multiply the argument matrix by a scalar (I usually use 10 to 100) to amplify it, then use sound (not soundsc) to hear it.
Otherwise, your filtered signal appears to have only one frequency at around 59 Hz. This is not a lung sound in my experience, since they cover a range of frequencies, and seems more like line-frequency noise. Since the 60 Hz hum could be overpowering everything else in your file, see Remove the 60 Hz Hum from a Signal to see if removing it allows you to hear the data in your file.
Yipp Chun Munn
Yipp Chun Munn 2019 年 10 月 20 日
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 2019 年 10 月 20 日
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
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.
Corrected typographical error.

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その他の回答 (1 件)

Daniel M
Daniel M 2019 年 10 月 16 日
編集済み: Daniel M 2019 年 10 月 16 日
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);
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.

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