I have to create my own function to find peaks in a signal

Like it said in the title, I have to create my own function to find peaks. It should also output the locations of the peaks in the x and y coordinates. Can someone give me a hint? I am a beginner so I don't really have any solid ideas.

[Voss]

A peak is essentially a point where the signal changes from moving up to moving down. You can use diff(y) to find the difference between adjacent samples of the signal y, and sign(diff(y)) will tell you the sign of the difference (i.e., sign == 1 -> moving up, sign == -1 -> moving down, sign == 0 -> staying the same).

For reference: diff sign

Answer 1

[Image Analyst]

See What can be programmed "without any built-in functions" ? - MATLAB Answers - MATLAB Central

It should be pretty easy to write your own version of islocalmax() and islocalmin() from low level commands. Just a simple for loop to scan the vector, and inside an "if" to compare the current value to the prior and later values. If the current value is greater than each of the values on either side, then log that location. I can't do it for you because you said you need to create your own function. Really trivial.

If you want to get fancier you can try duplicating findpeaks() though that would be quite an undertaking because it's a sophisticated function with lots of options.

[Lukas Zizka]

I came up with this, but it doesn't entirely work.

function findpeaks
   
    y = [0, 4, 1, 3, 2, 0, 6, 9, 7, 3, 1, 3, 0];
    
for ii = 1:length(y)
    if y(ii) >ii-1 && y(ii)>ii+1
       v(ii)= true
    else
        v(ii)=false
    end 
end

   
    
    
    
    
    
end

[Voss]

You are comparing the value of y at index ii to the value of the index ii-1.

Your comparisons should be y(ii) > y(ii-1) && y(ii) > y(ii+1), and your for loop should go 2:length(y)-1 because otherwise you'll try to index off the end and get an error.

[Lukas Zizka]

Thanks for the tip, but I get this error:

Array indices must be positive integers or logical values.

Error in findpeaks (line 8)

if y(ii) > y(ii-1) && y(ii) > y(ii+1)

[Image Analyst]

You need to compare y values to y values, NOT y values to indexes. There are also special cases for the first and last element since for those there is only one neighbor to compare the value to.

% function itsAPeak = findpeaks(y)
y = [0, 4, 1, 3, 2, 0, 6, 9, 7, 3, 1, 3, 0];
itsAPeak = false(1, length(y));
for ii = 1:length(y)
    if ii == 1
        % Check first element.  Can compare to only the second element.
        if y(ii) >= y(ii+1)
            % First location is greater than second location.
            itsAPeak(ii)= true
        end
    elseif ii == length(y)
        % Check last element.  Can compare to only the next-to-the-last element.
        if y(ii) >= y(ii-1)
            % Last position is greater than prior location.
            itsAPeak(ii)= true;
        end
    else
        if y(ii) >= y(ii-1) && y(ii) >= y(ii+1)
            % Current position is greater than prior or next location.
            itsAPeak(ii)= true;
        end
    end
end
itsAPeak % Show in command window;
% end

[Lukas Zizka]

I managed to get this and it gives me the same outputs

function [peak_y,peak_x] = findpeaks(y)
  
    y = [0, 4, 1, 3, 2, 0, 6, 9, 7, 3, 1, 3, 0];
    x = 1:length(y);
    
for ii = 2:length(y)
    if y(ii) > y(ii-1) && y(ii) > y(ii+1)
     
       v(ii)= true
    else
        v(ii)=false
    end 
   
end

plot(x,y,x(v),y(v),'o')

    
   peak_y = y(y(v))
   peak_x = x(x(v))
    
    
end

 
    

[Image Analyst]

Uh, okay, but why did you reject the checking I did for the first and last elements? Do you not consider it a peak if a peak is located there?

[Walter Roberson]

ii becomes lenth(y) eventually. When it does, what is y(ii+1)?

[Image Analyst]

@Walter Roberson I also thought of that but apparently he lucked out because of the order of the tests on the "if" line. It runs without error because of the short circuit -- it worked for that particular example vector, but it would fail if the last element was higher than the prior one.

I suggest he use my code -- it's more robust.

[Lukas Zizka]

I did use your code and I really appreciate your help.