how can i find zero crossing irregularity of the sound data?

I have an sound file of an asthma patient. My aim here is to find the zero crossing irregularity of this sound file. Can you help me?

Answer 1

[Star Strider]

Find the zero-crossings using something like this approach:

[y,Fs] = audioread('Asthma.wav');                           % Import Sound File (Use The Correct File Name)
zxi = find(diff(sign(y(:,1))));                             % Approximate Indices Of Zero-Crossinga

ylen = numel(y(:,1));
tv = linspace(0, ylen, ylen).'/Fs;                          % Time Vector (Column Vector)

for k = 1:numel(zxi)
    idxrng = max(1,zxi(k)-2):min(ylen,zxi(k)+2);            % Index Range
    B = [t(idxrng) ones(size(idxrng))] \ y(idxrng);         % Parameters
    txz(k) = B(2)/B(1);                                     % Interpolated Precise Time Of Zero-Crossing
end

However, if you are doing time-frequency analysis of the wheezes, the Signal Processing Toolbox pspectrum function with the 'spectrogram' option is likely most appropriate. (The spectrogram function may also be useful, however the outputs are not as straightforward to interpret.)

[furkan akhan]

Thanks for your answer but i have a problem. You wrote it as [y, Fs] but I have a 5x1 structural file audio file. So while I was defining that sound file '' sound = structWheeze (2,1) .SoundData''. In the code you wrote, I could not convert 'y' and 'Fs' as I wanted. How can I find the zero crossing irregularity by typing '' sound = structWheeze (2,1) .SoundData; '' like this instead of y and Fs.

[Star Strider]

I wrote it as I did because MATLAB can only handle two-column sound files. (I believe the Audio Toolbox can handle more columns, however I do not have it so I do not have access to its functions.)

I am not certain what ‘a 5x1 structural file audio file’ actually refers to. If it is a structure with 5 sound files, it would be necessary to access each sound file from the structure and analyse it separately. I suspect you could do that with audioread, however since I have no idea what the elements of the structure array are, I can only guess.

[furkan akhan]

This is how I define the sound data. For example, this is the sound file of the second patient and as you said, I have to do it separately for each patient. I hope I was able to explain my problem well. How can I find the zero crossing irregularity by defining the sound data like this?

[Star Strider]

The image is not helpful.

See the documentation section on Access Data in Nested Structures to understand how tto retrieve the data and then import it using audioread.

[Star Strider]

First, I meant audioplayer not audioread in my earlier Comment.

I would have to have the structure array to experiment with to advise you on that.

[Star Strider]

I would do something like this to do the time-frequency analysis:

W = load('sampleWheezeData.mat');
% LungSounds = W.structWheeze;
% SoundData = LungSounds.SoundData;
% Properties = LungSounds.Properties;

C1 = struct2cell(W.structWheeze);

Fs = 44100;                                                     % Not Provided, Wild Guess

for k = 1:size(C1,2)
    figure(k)
    pspectrum(C1{2,k}, Fs, 'spectrogram')
end

See the documentation on pspectrum for details on how to extract necessary information from it. I am not sure what sort of analysis you want to do with these data, however this should get you started.

This routine will extract the zero-crossing data:

for k1 = 1:size(C1,2)
    y = C1{2,k1};
    zxi = find(diff(sign(y(:,1))));                             % Approximate Indices Of Zero-Crossinga
    
    ylen = numel(y(:,1));
    tv = linspace(0, ylen, ylen).'/Fs;                          % Time Vector (Column Vector)
    
    for k2 = 1:numel(zxi)
        idxrng = max(1,zxi(k)-2):min(ylen,zxi(k)+2);            % Index Range
        B = [tv(idxrng) ones(size(idxrng)).'] \ y(idxrng);      % Parameters
        txz(k1,k2) = -B(2)/B(1);                                % Interpolated Precise Time Of Zero-Crossing
    end
end

figure
contourf(txz)
set(gca, 'YTick',1:5)

however I did not find the result to be very informative when I plotted it.