Effacer les filtres
Effacer les filtres

max and min in one cycle

3 vues (au cours des 30 derniers jours)
Mohanned Al Gharawi
Mohanned Al Gharawi le 16 Mai 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 commentaires
Walter Roberson
Walter Roberson le 16 Mai 2018
Perhaps run findpeaks twice, once on Signal and once on -Signal
Mohanned Al Gharawi
Mohanned Al Gharawi le 16 Mai 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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Réponse acceptée

Image Analyst
Image Analyst le 16 Mai 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 commentaire
Mohanned Al Gharawi
Mohanned Al Gharawi le 16 Mai 2018
Yes my mistake it must be 3.5, and the code worked. Thank you so much.

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Plus de réponses (1)

Walter Roberson
Walter Roberson le 16 Mai 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 commentaire
Mohanned Al Gharawi
Mohanned Al Gharawi le 16 Mai 2018
Thanks it works also, I appreciate your help.

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