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max and min in one cycle

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Mohanned Al Gharawi
Mohanned Al Gharawi el 16 de Mayo de 2018
Comentada: Mohanned Al Gharawi el 16 de Mayo de 2018
Hi everyone
I have a signal, I need to find max and min values for each cycle in this signal.
Let's say we have the signal A which has two cycles:
Signal=[0 -2 -3 -3.5 -3.25 -2.5 -1.75 -1 0 1 1.8 2.6 3.25 3.5 3.25 2.6 1.7 0 -2 -3 -3.5 -3.25 -2.5 -1.75 -1 0 1 1.8 2.6 3.25 3.5 3.25 2.6 1.7 0]
as shown in the attached picture.
I should get for the first cycle -3.5 and 3.25 and the same values for the second cycles.
Thank you in advance
  2 comentarios
Walter Roberson
Walter Roberson el 16 de Mayo de 2018
Perhaps run findpeaks twice, once on Signal and once on -Signal
Mohanned Al Gharawi
Mohanned Al Gharawi el 16 de Mayo de 2018
Thank you for your responding, But I think findpeaks gives us only one value for one signal, while what I want finding max and min values (two values) for the first cycle then the max and min (also two values) for the second cycle and keep going to the last cycle for one signal. Thanks

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Respuesta aceptada

Image Analyst
Image Analyst el 16 de Mayo de 2018
If you have the Image Processing Toolbox, you can do this:
props = regionprops(Signal < 0, Signal, 'MinIntensity');
minIntensities = [props.MinIntensity]
props = regionprops(Signal > 0, Signal, 'MaxIntensity');
maxIntensities = [props.MaxIntensity]
but I get 3.5 for the maxima. Why are you saying 3.25?
  1 comentario
Mohanned Al Gharawi
Mohanned Al Gharawi el 16 de Mayo de 2018
Yes my mistake it must be 3.5, and the code worked. Thank you so much.

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Más respuestas (1)

Walter Roberson
Walter Roberson el 16 de Mayo de 2018
[maxpk, maxloc] = findpeaks(Signal);
[minpk, minloc] = findpeaks(-Signal);
extremevals = [maxpk.', -minpk.'];
extremlocs = [maxloc.', minloc.'];
>> extremevals
extremevals =
3.5 -3.5
3.5 -3.5
>> extremlocs
extremlocs =
14 4
31 21
  1 comentario
Mohanned Al Gharawi
Mohanned Al Gharawi el 16 de Mayo de 2018
Thanks it works also, I appreciate your help.

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