Filtfilt returns NaN matrix

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Aboltabol
Aboltabol 2023-5-10
评论： Star Strider 2023-5-11
I am running the following operation (please download the attached file to your working directory):
load('Channel_Sim.mat');
[b,a] = butter(4, [3 20] ./ (1000/2)); % Sampling frequency is 1000 Hz.
chan_filtered = filtfilt(b, a, chan_data)
However, chan_filtered is a NaN matrix. Why?
I have checked that chan_data does not contain any NaN or Inf. I also tried resetting the butterworth filter range (3-20) over a wide range of values but to no avail. A PSD plot (see attached jpeg file) shows that chan_data encompasses a wide range of frequencies inluding the target frequency (3-20).
  2 个评论
Walter Roberson
Walter Roberson 2023-5-10
If you have even 1 nan or inf in your data the filtered results will likely be nan
Aboltabol
Aboltabol 2023-5-10
@Walter Roberson : I have checked that.
>> sum(isnan(chan_data))
ans =
0
>> sum(isinf(chan_data))
ans =
0

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Star Strider
Star Strider 2023-5-10
Ran in:
The transfer function implementation of the filter is unstable, however I cannot determine the reasson for the instability, although it may simply be the relative magnitudes of the numerator and denominator coefficients.
The easiest way to solve that problem is to use second-order-section implementation of the same filter —
load('Channel_Sim.mat')
% chan_data
Fs = 1000;
[b,a] = butter(4, [3 20] ./ (1000/2));
[z,p,k] = butter(4, [3 20] ./ (1000/2));
[sos,g] = zp2sos(z,p,k);
sos = 4×6
1.0000 2.0000 1.0000 1.0000 -1.8460 0.8546 1.0000 2.0000 1.0000 1.0000 -1.9175 0.9319 1.0000 -2.0000 1.0000 1.0000 -1.9598 0.9604 1.0000 -2.0000 1.0000 1.0000 -1.9885 0.9889
g = 7.1024e-06
figure
freqz(b,a, 2^18, 1000)
set(subplot(2,1,1), 'XLim',[0 50])
set(subplot(2,1,2), 'XLim',[0 50])
Check = nnz(isnan(chan_data))
Check = 0
L = numel(chan_data);
L = 7000
t = linspace(0, L-1, L)/Fs;
figure
plot(t, chan_data)
grid
Fn = Fs/2;
NFFT = 2^nextpow2(L)
NFFT = 8192
FTcd = fft((chan_data-mean(chan_data)).*hann(L), NFFT)/L;
Fv = linspace(0, 1, NFFT/2+1)*Fn;
Iv = 1:numel(Fv);
figure
plot(Fv, abs(FTcd(Iv))*2)
grid
chan_filtered = filtfilt(b, a, chan_data)
chan_filtered = 7000×1
NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
chan_filtered = filtfilt(sos, g, chan_data);
figure
plot(t, chan_filtered)
grid
chan_filtered = bandpass(chan_data, [3 20], Fs, 'ImpulseResponse','iir');
figure
plot(t, chan_filtered)
grid
The second-order-section implementation of the same Butterworth filter then works, and produces essentially the same result as an elliptic bandpass filter with the same essential parameters.
.
  2 个评论
Aboltabol
Aboltabol 2023-5-11
编辑：Aboltabol 2023-5-11
Okay, so, on running this code-block:
load('Channel_Sim.mat') % Sampling frequency is 1000 Hz.
[b,a] = butter(4, [3 20] ./ (1000/2));
[z,p,k] = butter(4, [3 20] ./ (1000/2));
[sos,g] = zp2sos(z,p,k);
chan_filtered_1 = filtfilt(b, a, chan_data);
chan_filtered_2 = filtfilt(sos, g, chan_data);
chan_filtered_1 is NaN but chan_filtered_2 is all right. This is just weird, because they are equivalent representations?! Why does using a SOS implementations skirts around the issue?
Star Strider
Star Strider 2023-5-11
The second-order-section implementation of a filter is always stable (at least in my experience).
The reason has to do with the way the filters are implemented and how filtfilt handles them. The relevant discussion is in the sos documentation section for filtfilt, and a more extensive discussion is in the sosfilt documentation. (I cannot determine from reading the documentation if sosfilt does the same forward-backward filtering that filtfilt does, so filtfilt still appears to be the preferred method of doing the actual filtering.)
.

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