How do I find the (x,y) coordinates of the peaks and valleys of a graph?
78 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Dave Phillips
el 1 de Nov. de 2015
Respondida: Sergio Yanez-Pagans
el 28 de Mzo. de 2021
I am new to Matlab and I am not sure how to find the coordinates of the peaks or valleys of my graph. After looking online, I tried using findpeaks() which did give me the y-values of the local maxima of my function ((e^(−at))*cos(2πft), where t is time and a and f are constants). However, I have not been able to find the corresponding x-values (and findpeaks() omits a local max at the y-intercept). I am also not at all sure how to find the coordinates for the valleys (minima or "troughs").
0 comentarios
Respuesta aceptada
Image Analyst
el 1 de Nov. de 2015
The x values are the second return argument of findpeaks(). It's the index number. Your formula does not have an x by name so you have to go with the index number. If you have a second array for t, then to get the t values you'd do
[peakValues, indexes] = findpeaks(y);
tValues = t(indexes);
To get valleys, you invert the signal, so that now what used to be valleys are now peaks, and use findpeaks() again
invertedY = max(y) - y;
[peakValues, indexes] = findpeaks(invertedY);
tValues = t(indexes);
5 comentarios
Walter Roberson
el 1 de Nov. de 2015
You could try appending -inf to the data so that the adjacent value is seen as a local maximum.
Markus Wahl
el 24 de Nov. de 2018
An alternative could be to take the absolute value of your vector to find only the indexes of both peaks and values.. Would need to go back and evaluate the original function to get the y-values, though.
Más respuestas (1)
Sergio Yanez-Pagans
el 28 de Mzo. de 2021
You can use my MATLAB file exchange function, it's really easy to implement and use:
Hope you find this useful!
0 comentarios
Ver también
Categorías
Más información sobre Annotations en Help Center y File Exchange.
Productos
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!