finding several values close to the set point

Hello!

I would like to calculate the period in a curve using the mean. However, my attempts to date have not been successful, and I think there is a simple way to do it.

I put random data - eventually I will have a lot of different data and this is to be the method for determining it.

I managed to find the closest value, but unfortunately 1, and I need a few of them to determine the curve's symmetry axis.

Edit: It doesn't search for peaks, because the data on which I work has a few of them on the main peaks, and filtration changes the shape of the plot very much.

Thank you for every advice.

A=[1.5 2.3 3.7 4.6 5.3 4.2 3.8 2.7 1.3 2.6 3.9 4.8 6.1 5.3 4.3 3.6 2.8 1.6 2.4 3.6 4.7 5.4 4.0 2.8 1.5 2.7 3.8 4.7];

A=A';

s = size(A);

t = 1:1:s(1);

plot (t,A)

a = mean(A);

Edit2:

In the attachment I add the real data I work on. From each cell I choose 1 column as X and 3 columns as Y in plot (X, Y).

Answer 1

[Ameer Hamza]

The following code works by detecting the peak of each cycle and then calculate the average distance between each peak

A=[1.5 2.3 3.7 4.6 5.3 4.2 3.8 2.7 1.3 2.6 3.9 4.8 6.1 5.3 4.3 3.6 2.8 1.6 2.4 3.6 4.7 5.4 4.0 2.8 1.5 2.7 3.8 4.7];
A=A';
t = 1:1:size(A,1);
plot(t,A);
[~, index_peaks] = findpeaks(A);
t_peaks = t(index_peaks);
period = mean(diff(t_peaks));

[Caroline]

Unfortunately, the real data has a few peaks on the main peak ... oh I won't work :(

[Ameer Hamza]

In that case, please attach a sample of real data. It will be easy to suggest a working solution using your actual data.

[Caroline]

Okay, I added in the attachment.

[Ameer Hamza]

Check this code

load('test.mat');

a = sila{1};
s = a(:,3);
s = detrend(s);
s_ = s-mean(s);
[~, idx] = findpeaks(s, 'MinPeakProminence', max(s_));
period = mean(diff(idx));

t = 1:numel(s);
plot(t, s, '-', t(idx), s(idx), '+');

It reads the third column and then find the peak of each cycle. It then finds the average difference between peaks to calculate the period.

[Caroline]

Thank you very much, this is the first way I was thinking but I couldn't deal with these peaks, Thank you very much !!

[Ameer Hamza]

Glad to be of help.

[Caroline]

Analyzing the data, I noticed that the solution has 1 drawback. Peaks occur in different places, therefore the periods are not equal :(.

[Ameer Hamza]

Yes, that can be issue, but since you are taking the average of each interval so this difference is somewhat compensated.