Cut Audio Signals (.wav) in specific time.Create a repeat loop to reach the desired time

Hello friends. I want to make all the audio signals of a database the same in terms of time (for example 12.5 seconds). What loop and functions should be used?

Because some small signals must be repeated to reach 12.5 seconds.Some data cut and some are exactly the same size.So, apparently, we will be faced with 3 states, bigger, smaller and equal

% make a list of wave files in training_data folder
cd training_data\
folderInfo = dir('**/*.wav');
folderInfo1 = dir('**/*.tsv');
folderInfo2 = dir('**/*.txt');
cd ..\
addpath training_data\
% fir bandpass filter design with fc1 = 20/fs/2 and fc2 = 500/fs/2 with
% fs=4000
f = [0 0.005 0.015 0.245 0.255 1];
a = [0 0 1 1 0 0];
b = firpm(100,f,a);
% feature vector calculation
for i=1:length(folderInfo)
    x = audioread(folderInfo(i).name);
   
    % signal filtering
    xf = filter(b,1,x);

Answer 1

[Mathieu NOE]

hello

try this

I tested the code for mono and stereo wav files - should work even for more channels data

now please notice that I made sure to use the correct sampling rate associated with each file - in case they would differ from one file to another, the code will ensure that the total duration is correct - even though your filter would not have the same effect unless you include its coefficients computation inside the main loop (and take account of Fs);

folderInfo = dir('**/*.wav');
folderInfo1 = dir('**/*.tsv');
folderInfo2 = dir('**/*.txt');
cd ..\
addpath training_data\
% fir bandpass filter design with fc1 = 20/fs/2 and fc2 = 500/fs/2 with
% fs=4000
f = [0 0.005 0.015 0.245 0.255 1];
a = [0 0 1 1 0 0];
b = firpm(100,f,a);
% feature vector calculation
duration_target = 12.5; % seconds

for i=1:length(folderInfo)
    filename = folderInfo(i).name
    [x,Fs] = audioread(filename);
   
    % signal filtering
    xf = filter(b,1,x);
    
    % main code
    dt = 1/Fs;
    [samples,channels] = size(x);
    duration = (samples-1)*dt;
    
    duration_ratio = duration/duration_target;
    target_duration_samples = (1+duration_target*Fs);
    
    if duration_ratio>1  % truncate
        xf_out = xf(1:target_duration_samples,:);
        
    else                 % do nothing or expand(by repetition and trimming  
        
        repetitions = ceil(1/duration_ratio);
        xf_out = repmat(xf,repetitions,1); % duplicate (by excess)
        xf_out = xf_out(1:target_duration_samples,:); % exact number of samples
        
    end
        
    % export to wav 
    filename_out = [filename(1:length(filename)-4) '_out.wav']
    audiowrite(filename_out,xf_out,Fs);
     
    
end

[omid]

Thank you, dear Mathieu, the code worked correctly.Yes, the sampling frequency in all signals was 4000 Hz. Let me ask you two more questions here.

1- Do we have a 12-second signal without noise in the output (filename_out)?

2- If this is the case, I will save all the output signals in a new folder called training_data_out, and now I want to calculate the spectrum of all of them and save them in a new folder (for the next steps, which is the deep network).

I am sending the main code for converting to spectrum which was done on a sample. But I am still unable to write a loop to cover all the signals and save them.

a=audioread("2530_TV.wav");
[st,t,f]=st(a,25,350,1,1);
s=abs(st);
imagesc(s)

[Mathieu NOE]

hello

1/ yes , assuming you filter is appropriate for removing the noise. I didn't change anything to your filter design

the duration is 12.5 s and not 12 s to be exact (this was your initial request but you can easiky modify to 12 or another value in this line :

duration_target = 12.5; % seconds

2 / you want the spectrum or the spectrogram - according to these lines , it seems to me you want a spectrogram

[st,t,f]=st(a,25,350,1,1);

but the function st is unknown to me . I know specgram and the older spectrogram , but not st

or maybe you are using stft ?

Short-time Fourier transform - MATLAB stft - MathWorks France

[omid]

hello again my friend.This is a stockwell transmission

function [st,t,f] = st(timeseries,minfreq,maxfreq,samplingrate,freqsamplingrate) 
% Returns the Stockwell Transform of the timeseries. 
% Code by Robert Glenn Stockwell. 
% DO NOT DISTRIBUTE 
% BETA TEST ONLY 
% Reference is "Localization of the Complex Spectrum: The S Transform" 
% from IEEE Transactions on Signal Processing, vol. 44., number 4, April 1996, pages 998-1001. 
% 
%-------Inputs Needed------------------------------------------------ 
%   
%   *****All frequencies in (cycles/(time unit))!****** 
%	"timeseries" - vector of data to be transformed 
%-------Optional Inputs ------------------------------------------------ 
% 
%"minfreq" is the minimum frequency in the ST result(Default=0) 
%"maxfreq" is the maximum frequency in the ST result (Default=Nyquist) 
%"samplingrate" is the time interval between samples (Default=1) 
%"freqsamplingrate" is the frequency-sampling interval you desire in the ST result (Default=1) 
%Passing a negative number will give the default ex.  [s,t,f] = st(data,-1,-1,2,2) 
%-------Outputs Returned------------------------------------------------ 
% 
% st     -a complex matrix containing the Stockwell transform.  
%			 The rows of STOutput are the frequencies and the  
%         columns are the time values ie each column is  
%         the "local spectrum" for that point in time 
%  t      - a vector containing the sampled times 
%  f      - a vector containing the sampled frequencies 
%--------Additional details----------------------- 
%   %  There are several parameters immediately below that 
%  the user may change. They are: 
%[verbose]    if true prints out informational messages throughout the function. 
%[removeedge] if true, removes a least squares fit parabola 
%                and puts a 5% hanning taper on the edges of the time series. 
%                This is usually a good idea. 
%[analytic_signal]  if the timeseries is real-valued 
%                      this takes the analytic signal and STs it. 
%                      This is almost always a good idea. 
%[factor]     the width factor of the localizing gaussian 
%                ie, a sinusoid of period 10 seconds has a  
%                gaussian window of width factor*10 seconds. 
%                I usually use factor=1, but sometimes factor = 3 
%                to get better frequency resolution. 
%   Copyright (c) by Bob Stockwell 
%   

After your help, I saved all the outputs in a new folder(training_data_out) and wrote a new code, unfortunately I don't know where my problem is.

clc
clear
close all
% make a list of wave files in training_data folder
cd training_data_out\
folderInfo = dir('**/*.wav');

cd ..\
addpath training_data_out\


% fir bandpass filter design with fc1 = 20/fs/2 and fc2 = 500/fs/2 with
% fs=4000
f = [0 0.005 0.015 0.245 0.255 1];
a = [0 0 1 1 0 0];
b = firpm(100,f,a);
% feature vector calculation
for i=1:length(folderInfo)
    x = audioread(folderInfo(i).name);
   
    % signal filtering
    xf = filter(b,1,x);
    
    [st,t,f]=st(xf,25,350,1,1);
    s=abs(st);
    imagesc(s)

    %readableDostIn(:,:,i) = ST.rearrangeDost(st,t,f);  ????
    %imagesc(readableDostIn(:,:,i))                     ????
     
     imshow(outpict)
imwrite(outpict,'myimage.png')
   
    
end

[Mathieu NOE]

seems to me the variable outpict is not defined

where do you generate it ?

your goal is simply to store the st result as an image ?

[omid]

Yes, sir, it has not been defined yet. That is, I am not sure where and how to define it. And the second question is yes again

[Mathieu NOE]

hello again

so this code assumes that st function is working, but for the time neing I cannot run it (you did not provided the fully working function) , so I replaced the output of st function by some random arrays

but still you can see that the pictures are saved ith output diretory as you wanted :

folderInfo = dir('**/*.wav');
folderInfo1 = dir('**/*.tsv');
folderInfo2 = dir('**/*.txt');
cd ..\
addpath training_data\
% fir bandpass filter design with fc1 = 20/fs/2 and fc2 = 500/fs/2 with
% fs=4000
f = [0 0.005 0.015 0.245 0.255 1];
a = [0 0 1 1 0 0];
b = firpm(100,f,a);
% feature vector calculation
duration_target = 12.5; % seconds

% working (current) folder
current_folder = pwd;
output_folder = [current_folder '\training_data_out'];

for i=1:length(folderInfo)
    filename = folderInfo(i).name
    [x,Fs] = audioread(fullfile(current_folder,filename));
   
    % signal filtering
    xf = filter(b,1,x);
    
    % main code
    dt = 1/Fs;
    [samples,channels] = size(x);
    duration = (samples-1)*dt;
    
    duration_ratio = duration/duration_target;
    target_duration_samples = (1+duration_target*Fs);
    
    if duration_ratio>1  % truncate
        xf_out = xf(1:target_duration_samples,:);
        % debug
        [samples,channels] = size(xf_out);
        duration = (samples-1)*dt    
        
    else                 % do nothing or expand(by repetition and trimming  
        
        repetitions = ceil(1/duration_ratio);
        xf_out = repmat(xf,repetitions,1); % duplicate (by excess)
        xf_out = xf_out(1:target_duration_samples,:); % exact number of samples
         % debug
        [samples,channels] = size(xf_out);
        duration = (samples-1)*dt   
        
    end
        
    
%     % st transform
%     [st,t,f]=st(xf_out,25,350,1,1); % still need to bo worked out
%     st=abs(st);
    
    st = rand(100,100); % just for the sake of the demo until st is fixed (above)
    figure(i),
    imshow(st)
    filename_out = [filename(1:length(filename)-4) '_out.png']
    imwrite(st,fullfile(output_folder,filename_out));
    
    
     
    
end

[omid]

Thank you, dear friend. With a small change (removing the current folder), I run the code assuming st = rand(100,100), which works correctly. But why do I get this message when I run the main conversion?

"Insufficient number of outputs from right hand side of equal sign to satisfy assignment."

clc
clear
close all
cd training_data_out\
folderInfo = dir('**/*.wav');

cd ..\
addpath training_data_out\
% fir bandpass filter design with fc1 = 20/fs/2 and fc2 = 500/fs/2 with
% fs=4000
f = [0 0.005 0.015 0.245 0.255 1];
a = [0 0 1 1 0 0];
b = firpm(100,f,a);
% feature vector calculation
duration_target = 12.5; % seconds
% working (current) folder


for i=1:length(folderInfo)
    filename = folderInfo(i).name
    [x,Fs] = audioread(filename);
        % signal filtering
    xf = filter(b,1,x);
    
    % main code
    dt = 1/Fs;
    [samples,channels] = size(x);
    duration = (samples-1)*dt;
    
    duration_ratio = duration/duration_target;
    target_duration_samples = (1+duration_target*Fs);
    
    if duration_ratio>1  % truncate
        xf_out = xf(1:target_duration_samples,:);
        % debug
        [samples,channels] = size(xf_out);
        duration = (samples-1)*dt    
        
    else                 % do nothing or expand(by repetition and trimming  
        
        repetitions = ceil(1/duration_ratio);
        xf_out = repmat(xf,repetitions,1); % duplicate (by excess)
        xf_out = xf_out(1:target_duration_samples,:); % exact number of samples
         % debug
        [samples,channels] = size(xf_out);
        duration = (samples-1)*dt   
        
    end
        
    
%     % st transform
     [st,t,f]=st(xf_out,25,350,1,1); % still need to bo worked out
     st=abs(st);
    
  % st = rand(100,100); % just for the sake of the demo until st is fixed (above)
    figure(i),
    imshow(st)
    filename_out = [filename(1:length(filename)-4) 'o.png']
    imwrite(st,filename_out);
end

Insufficient number of outputs from right hand side of equal sign to satisfy assignment.

Error in OMG (line 52)
     [st,t,f]=st(xf_out,25,350,1,1); % still need to bo worked out

[Walter Roberson]

Are you getting st from https://www.mathworks.com/matlabcentral/fileexchange/51808-time-frequency-distribution-of-a-signal-using-s-transform-stockwell-transform ?

Answer 2

[Walter Roberson]

     [st,t,f]=st(xf_out,25,350,1,1); % still need to bo worked out

The first time you execute that statement, there is not variable named st so MATLAB looks for an st function. The function is run, and you assign the outputs to several variables, including st

The second time you execute that statement, there is now a variable named st so that is an indexing request. But indexing can never have multiple outputs, so MATLAB complains.

[omid]

Thank you dear Walter. Yes I got the st from https://www.mathworks.com/matlabcentral/fileexchange/51808-time-frequency-distribution-of-a-signal-using-s-transform-stockwell-transform.

I understand your point, but what should be done to solve this problem?

[Mathieu NOE]

@omid

avoid using function names as variable name

this

[st,t,f]=st(xf_out,25,350,1,1);

should be replaced by something like

[st_out,t,f]=st(xf_out,25,350,1,1);

so matlab is not confused between the function st and a variable

NB : i haven't checked yet that st(xf_out,25,350,1,1) actually works (btw) !

[omid]

Yes dear Mathieu, you are right. Friends, let me express my last problem here.

This is my current modified and final code

clc
clear
close all
cd training_data_out\
folderInfo = dir('**/*.wav');

cd ..\
addpath training_data_out\
% fir bandpass filter design with fc1 = 20/fs/2 and fc2 = 500/fs/2 with
% fs=4000
f = [0 0.005 0.015 0.245 0.255 1];
a = [0 0 1 1 0 0];
b = firpm(100,f,a);
% feature vector calculation
duration_target = 12.5; % seconds
% working (current) folder


for i=1:length(folderInfo)
    filename = folderInfo(i).name
    [x,Fs] = audioread(filename);
        % signal filtering
    xf = filter(b,1,x);
    
    % main code
    dt = 1/Fs;
    [samples,channels] = size(x);
    duration = (samples-1)*dt;
    
    duration_ratio = duration/duration_target;
    target_duration_samples = (1+duration_target*Fs);
    
    if duration_ratio>1  % truncate
        xf_out = xf(1:target_duration_samples,:);
        % debug
        [samples,channels] = size(xf_out);
        duration = (samples-1)*dt    
        
    else                 % do nothing or expand(by repetition and trimming  
        
        repetitions = ceil(1/duration_ratio);
        xf_out = repmat(xf,repetitions,1); % duplicate (by excess)
        xf_out = xf_out(1:target_duration_samples,:); % exact number of samples
         % debug
        [samples,channels] = size(xf_out);
        duration = (samples-1)*dt   
        
    end
        
    
%     % st transform
     [st_out,t,f]=st(xf_out,25,350,1,1); % still need to bo worked out
     zz=abs(st_out);
     
     
    
    %st = rand(100,100); % just for the sake of the demo until st is fixed (above)
    figure(i),
    %imshow(zz)
    imagesc(zz)
    filename_out = [filename(1:length(filename)-4) '_out.png']
    imwrite(zz,filename_out);
end

I want the shape to appear in the loaded image (z1) in the output (line 62).But in the saved outputs, I face the z2 image

[Mathieu NOE]

You are not using imwrite the right way

imwrite(zz) works with zz = an image object not a numerical array

see the doc

imwrite(A,FILENAME,FMT) writes the image A to the file specified by

FILENAME in the format specified by FMT.

as an alternative you can do this

imagesc(zz)       
%export figure (use print or savefig)
print('-dpng','-r300',filename_out); % choose your format and resolution

read the doc for print

[Walter Roberson]

@Mathieu NOE

No,

https://www.mathworks.com/help/matlab/ref/imwrite.html#btv3cny-1-A

A

Image data, specified as a full (nonsparse) matrix.

[Walter Roberson]

@omid

    imwrite(zz,filename_out);

That line would write out the original data in zz. If zz is double precision data (as it appears to be) then imwrite() would assume that 0 represents the first color in the colormap, and 1 represents the last color in the color map, and that less than 0 should be treated as 0 and that greater than 1 should be treated as 1.

imwrite() does not scale the image according to its range of data, and imwrite() does not get told to do so based on the fact that you recently did imagesc(zz) .

In order to write out the scaled image you should do

  zzs = rescale(zz, 0, 1);
  imwrite(zzs, filename_out);

[Mathieu NOE]

@Walter Roberson

thanks for the explanations; I almost never used imwrite so I'll try to remember this for next time !

all the best

[omid]

@Mathieu NOE @Walter Roberson Thank you. What command should be added to not display the numbers in the row and column.

[Walter Roberson]

change

duration = (samples-1)*dt

to add a semi-colon

duration = (samples-1)*dt;

Answer 3

[omid]

Hello friends. I executed this code and saved it with .mat extension to save time.Due to the large number of samples, it took 4 hours to perform the calculations, and the size of the saved file is 360 MB, which seems reasonable (that is, the signal-to-image conversions are apparently done correctly).

clc
clear
close all
cd training_data_out\
folderInfo = dir('**/*.wav');

cd ..\
addpath training_data_out\


% working (current) folder


for i=1:length(folderInfo)
    filename = folderInfo(i).name
    [x,Fs] = audioread(filename);
      
        
    
%     % st transform
     [st_out,t,f]=st(x,25,350,1,1); % still need to bo worked out
     zz=abs(st_out);
end

1= Now I can't have the output of the photos(how should I write this new ring?)

2- As it is clear in the uploaded photo (a1), the number of rows and columns of the photo is 326*5001, but with the command print('-dpng','-r300',filename_out) the rows and columns are messed up.

3=Look at the next photo (a2). How can I prevent these numbers from being printed in the output?

[Walter Roberson]

(3) -- you can use xticks() and yticks() to set the tick positions to be empty.

(2) -- so don't display a array as an image and then use print(). Instead, rescale to 0 to 255, uint8(), and then ind2rgb() to convert the relative intensities into an RGB array. Then imwrite() that array. Full resolution.

[omid]

Excuse me, dear Walter, can you put what you said (items 2 and 3) into my code.Because I didn't get an answer, I must have misunderstood.

[Walter Roberson]

projectdir = 'training_data_out';
imgext = '.png';

folderInfo = dir( fullfile(projectdir, '**', '*.wav');

cmap = parula(256);
% working (current) folder


for i=1:length(folderInfo)
    filename = fullfile(folderInfo(i).folder, folderInfo(i).name);
    [folder, basename] = fileparts(filename);
    outfilename = fullfile(folder, [basename imgext]);
    [x,Fs] = audioread(filename);
      
    %     % st transform
    [st_out,t,f]=st(x,25,350,1,1); % still need to bo worked out
    zz=abs(st_out);
    
    zzind = uint8(rescale(zz, 0, 255));
    imwrite(zzind, cmap, outfilename);
end

This creates .png with the same file name in the same folder as the original wav file.

Note that there is no colorbar and no labels or title. Also note that the data is scaled relative to itself, so the colors have no absolute meaning -- if one zz result ranged up to +738 then that +738 would be mapped to the last colour in cmap, and if a different zz result ranged up to +1192 then that +1192 would be mapped to the last colour in cmap for that image. This is the same behaviour you would get with imagesc()