Hi,i am trying to run the following program,but it shows index out of bounds .length of my signal loaded is=10240 and when i am taking square for x ,it shows index out of bounds error and also one more error is undefined function chebyshev

function segmentation_Shennon
 close all 
 clear all 
  % Load the data file
 [F, Fs] = audioread ('1.wav'); 
  pcg = F (:, 2); 
 t = 0: 1 / Fs: (length (F) -1) / Fs; 
 %%We normalize and filter the signal with a Chebyshev filter of 1 8th order NP with a frequency of medium 882 Hz 
 pcg_filter = chebushev1 (pcg, Fs);
 pcg_norm = pcg_filter./abs(max(pcg_filter));
 % calculate the energy of Shannon
 shennon_energy = - ((pcg_norm. ^ 2). * log (pcg_norm. ^ 2));
 figure (1)
 plot (t, shennon_energy)
  % calculate the average Shannon energy in a segment lasting 0.02 s
 % with an overlap of 0.01 s
 win = 0.002 * Fs; 
 i = 1;
 k = 1;
 Es = [];
 Es_t = [];
 P = [];
  while i <length (pcg_norm)
  for i = i: i + win
 square = pcg_norm (i). ^ 2; 
 Es (k) = -1 / win * sum (square. * Log (square));
 end
 ES_t (k) = t (i); 
 i = i + round (win / 2);
 k = k + 1;
 end
  % we normalize Shannon's averaged energy
 M_Es = (mean (Es));  % mean energy of the segment
 S_Es = (std (Es (k-1)));  % standard deviation of the energy of the segment
 P (k) = (Es (k-1) -M_Es) / S_Es;  % Shannon's normalized averaged energy,
 plot (t, P)

[Greg]

Highlight your code and then click the {} code button to help format it so we can read it.

Answer 1

[Walter Roberson]

You appear to have a typing mistake: you used chebushev1 but the English version of the name is chebyshev . However there is no chebyshev1 function: there is cheby1 https://www.mathworks.com/help/signal/ref/cheb1ord.html and there is the MuPAD orthopoly:chebyshev1 which is not likely what you are wanting to call.

With respect to the out of range: you have

    pcg = F (:, 2);

which assumes that F has two or more columns, which would correspond to two or more channels. If your '1.wav' is mono then it would only have one column, F(:,1);

Answer 2

[poornima.a A]

Thank you sir and one more doubt sir,plz clarify me ,my signal length is 20324 sample,when I taking square of this signal.it shows me index out of bounds error.Give me code to do this sir

[Walter Roberson]

Please show your code and a complete copy of the error message.

Answer 3

[poornima.a A]

function segmentation_Shennon

   close all 
   clear all 
    % Load the data file
   [pcg, Fs] = audioread ('D:\phase 2 sets\audio signals\trim\normalcut.wav'); 
    %pcg = F (:, 1); 
   t = 0: 1 / Fs: (length (pcg) -1) / Fs; 
   %%We normalize and filter the signal with a Chebyshev filter of 1 8th order NP with a frequency of medium 882 Hz 
   pcg_filter =pcg;
   pcg_norm = pcg_filter./abs(max(pcg_filter));
   % calculate the energy of Shannon
   shennon_energy = - ((pcg_norm.^2).* log (pcg_norm.^2));
   figure (1)
   plot (t, shennon_energy)
    % calculate the average Shannon energy in a segment lasting 0.02 s
   % with an overlap of 0.01 s
   win = 0.002 * Fs; 
   i = 1;
   k = 1;
   Es = [];
   Es_t = [];
   P = [];
    while i <length (pcg_norm)
    for i = i: i + win
   square = pcg_norm (i).^2; 
   Es (k) = -1 / win * sum (square.*log (square));
   end
   ES_t (k) = t (i); 
   i = i + round (win / 2);
   k = k + 1;
   end
    % we normalize Shannon's averaged energy
   M_Es = (mean (Es));  % mean energy of the segment
   S_Es = (std (Es));  % standard deviation of the energy of the segment
   P (k) = (Es (k-1) -M_Es) / S_Es;  % Shannon's normalized averaged energy,
  plot (t, P)

[Walter Roberson]

Your line

    square = pcg_norm (i).^2;

assumes that the input is one channel. You should probably be selecting one channel, like you do in your commented out code.

You have

while i <length (pcg_norm)
  for i = i: i + win
    square = pcg_norm (i).^2; 
    Es (k) = -1 / win * sum (square.*log (square));
  end
  ES_t (k) = t (i); 
  i = i + round (win / 2);
  k = k + 1;
end

When you have a for loop, the loop variable is left at the last value it had, so after the "for i=i:i+win" loop, i is left at what was previously i+win . You then add round(win/2) to that. That causes half a window to be skipped in the input. However, your earlier code is trying to use overlapping windows, in which i would have to be advanced by round(win/2) further than where it had started, as opposed to your current code that advances it from where it had ended.

But either way, suppose you reach the point where i is one window or less away from the end. For example suppose that the buffer is 100 and window is 10 and you have reached 91 by starting at 1 then 6 then 11 then 16 and so on. 91 < 100 so the while loop continues. You then execute the for loop, for i = 91:91+10 which is for i = 91:101 . That is a problem as the last entry is 100 not 101. i:i+win is a window of length (win+1) not a win. But suppose you fix that with

for i = i : i + win - 1

now the 91 works processing 91 through 100. Now the buffer advances to 96. 96 < 100 so the while loop continues. Now you have for i = 96:96+10-1 which is for i = 96:105 . But elements 101 onward are after the end of the buffer...

[poornima.a A]

so this correction is enough to solve my error sir

[poornima.a A]

thank you so mach sir its worked out well.but one more error was arises .i attached that error message,sir plz view that and give me the correction according to that error

[Walter Roberson]

You plot(t,P) . t is your time points, so it has as many entries as you have times. P, though, has as many entries as you have overlapping windows.

Just before what you had posted before as

for i = i: i + win

insert

window_times(k) = t(i);

Make sure this is before that for loop, not inside the loop.

Then at the end

plot(window_times, P)

Answer 4

[poornima.a A]

sir while running my above code it shows value of p and Es are NaN .i thing error in the below code

   M_Es = (mean (Es));  % mean energy of the segment
   S_Es = (std (Es));  % standard deviation of the energy of the segment
   P (k) = (Es (k-1) -M_Es) / S_Es;  % Shannon's normalized averaged energy,
  plot (t, P)

[Walter Roberson]

I will need to see your updated code.

[poornima.a A]

 close all 

 clear all 
  % Load the data file
 [pcg, Fs] = audioread ('D:\phase 2 sets\audio signals\trim\normalcut.wav'); 
  %pcg = F (:, 1); 
 t = 0: 1 / Fs: (length (pcg) -1) / Fs; 
 %%We normalize and filter the signal with a Chebyshev filter of 1 8th order NP with a frequency of medium 882 Hz 
 pcg_filter =pcg;
 pcg_norm = pcg_filter./abs(max(pcg_filter));
 % calculate the energy of Shannon
 shennon_energy = - ((pcg_norm.^2).* log (pcg_norm.^2));
 figure (1)
 plot (t,shennon_energy)
% calculate the average Shannon energy in a segment lasting 0.02 s
   % with an overlap of 0.01 s
   win = 0.002 * Fs; 
   i = 1;
   k = 1;
   Es = [];
   Es_t = [];
   P = [];
    while i <length (pcg_norm)
    for i = i: i + win-1
   square = pcg_norm (i).^2; 
   Es (k) = -1 / win * sum (square.*log (square));
   end
   ES_t (k) = t (i); 
   i = i + round (win / 2);
   k = k + 1;
   end
    % we normalize Shannon's averaged energy
   M_Es = (mean (Es));  % mean energy of the segment
   S_Es = (std (Es));  % standard deviation of the energy of the segment
   P (k) = (Es (k-1) -M_Es) / S_Es;  % Shannon's normalized averaged energy,
   while i <length (pcg_norm)
    for i = i: i + win-1
     window_times(k) = t(i);
     plot(window_times, P)
    end
  end

[Walter Roberson]

P would come out nan if both (Es (k-1) -M_Es) and S_Es come out as 0. S_Es can come out as 0 if all of the values in Es are the same. (Es (k-1) - M_Es) would come out as 0 if all of the values in Es are the same.

Your code should expect P to be all 0 except for one location, as you compute P(k) once after all of the loops and you do that on the basis of the single Es(k-1) where k has been left over from all of the loops and is 1 more than the number of windows that were created. Perhaps you wanted

P = (Es - M_es) / S_Es;

The way you use k-1 after the end of the loop is confusing. I would suggest to you that instead of using

k = 1
for ...
   variable(k) = ...
   k = k + 1;
end

that you instead use

k = 0
for ...
   k = k + 1;
   variable(k) = ...
end

then k records the number of entries that have been created and does not need to have 1 subtracted from it at the end.

[poornima.a A]

Thank you so much sir.

[weii chieun lim]

@poornima.a A

Can i have your updated source code ?

[Faizul Hakim]

@poornima.a A Can I have your updated code