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max and min in one cycle

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Mohanned Al Gharawi
Mohanned Al Gharawi 2018 年 5 月 16 日
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 件のコメント
Walter Roberson
Walter Roberson 2018 年 5 月 16 日
Perhaps run findpeaks twice, once on Signal and once on -Signal
Mohanned Al Gharawi
Mohanned Al Gharawi 2018 年 5 月 16 日
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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採用された回答

Image Analyst
Image Analyst 2018 年 5 月 16 日
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 件のコメント
Mohanned Al Gharawi
Mohanned Al Gharawi 2018 年 5 月 16 日
Yes my mistake it must be 3.5, and the code worked. Thank you so much.

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その他の回答 (1 件)

Walter Roberson
Walter Roberson 2018 年 5 月 16 日
[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 件のコメント
Mohanned Al Gharawi
Mohanned Al Gharawi 2018 年 5 月 16 日
Thanks it works also, I appreciate your help.

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