Average normalised Shannon energy for phonocardigraphy signals

I have a code that has to calculate average normalised Shannon energy for phonocardigraphy signals but have a problem with calculating standard deviation. mayve something wrong with my code above this expression. Can somedody help me. please.

function segmentation_Shennon
close all 
clear all 

% Загружаем файл с данными
[F,Fs] = audioread('1.wav'); 

pcg = F(:,2); 
t=0:1/Fs:(length(F)-1)/Fs; 
%%Нормируем и фильтруем сигнал фильтром Чебышева 1 8го порядка НЧ с частотой среда 882 Гц 
pcg_filter = chebushev1(pcg,Fs);
pcg_norm = pcg_filter./abs(max(pcg_filter));
%рассчитываем энергию Шеннона
shennon_energy= -((pcg_norm.^2).*log(pcg_norm.^2));
figure(1)
plot(t,shennon_energy)

%рассчитываем усредненную энергию Шеннона в сегменте длительностью 0,02с
%с перекрытием 0.01 с
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

% нормируем усреднeнyую энергию Шеннона
M_Es = (mean(Es)); %среднее значение энергии сегмента
S_Es = (std(Es(k-1)));  %стандартное отклонение энергии сегмента
P(k) = (Es(k-1)-M_Es)/S_Es; % Нормированная усредненная энергия Шеннона,
plot(t,P)

[Brian Sang;Sang]

Hi, I understand what question you are trying to solve as it refers to Heart Sound Segmentation Algorithm Based on Heart Sound Envellogram and this is how I solved it.

N=0.02*Fs;

win = N;

k=1;

i = 1;

Es=[];

Es_t=[];

P=[];

m = length(filter)/N;

t=0:1/Fs:(length(norm)-1)/Fs;

for k = 1:m

for i = win*(k-1)+1:win*(k)

F1(i) = (norm(i).^2 * log(norm(i).^2));

end

Es(k) = -1/N* sum(F1);

F1 = [];

end

for k = 1:m

M_Es = (mean(Es));

S_Es = (std(Es));

P(k) = (Es(k)-M_Es)/S_Es;

end

tk = 0:60/(m-1):60;

[yupper,ylower] = envelope(P);

plot(tk,yupper)

[Arthur Guilherme Santos Fernandes]

Does the "Es" have the same length as the normalized signal ?

I am trying to reproduce this thecnique based on a paper, and they have this picture:

As if the Average Shannon Energy has the same length of the original signal.

This got me confused, i want to reproduce this picture but i can't because in my case i'm getting "Es" as a shorter array, since we do the splits of 0.02s.

Answer 1

[Star Strider]

I am guessing here because I cannot run your code. You are taking the standard deviation of the second-to-last value of ‘Es’ in your loop. See if changing the standard deviation calculation to:

S_Es = (std(Es));  %стандартное отклонение энергии сегмента

gives you the correct result.

[Stefaniya Pirogova]

yes, you're right, thank you. but still have problem with P(k) here is an archive where all the files that you need for running, can you, please, have a look at them?

[Star Strider]

My pleasure.

I do not download and open .zip files. If your .zip file contains your data, please save your data as a .mat file instead. (See the documentation for the save function for all the details.)

I understand phonocardiography (I am an Internal Medicine physician), but I have no idea what analysis you are doing. Do you have a PDF reference for it (in English), since I do not recognise what you are doing from your code. I cannot promise any solution even then, but at least I might have some idea.