find how many times a signal becomes above a certain signal

Dear all,

I have to count the times when a signal goed under a certain value and gets above a different value.

For example: if y < 80 and goes above 85 count as 1, so the next time it does this the output has to be 2.

Thanks

Answer 1

[Adam Danz]

I think what you mean is that you want to count the number of times a continuous signal goes below 80 and then later goes above 85. Here's a method that identifies all samples that fall below 80 ("low") and all samples that rise above 85 ("high"). Then it loops through all samples and keeps track of when a "high" follows a recent "low". There's probably a more clever method that doesn't involve loops but this is fast and it works.

It starts by creating fake data so it can be verified.

% Create fake data
data = randi(30, 1, 20) + 70; 
% Which samples dip below 80?
isLow = data < 80; 
% Which samples rise above 85?
isHigh = data > 85; 

% put two index vectors together
isThreshold = [isLow', isHigh'];

% Loop through each sample; for each 'low' detected, 
% look for the nest 'high' and increment the counter. There
% might be a better, non-loop way of doing this. 
count = 0;   
lowFound = false; 
highFound = false; 
for i = 1 : length(isLow)
    % search next row
    if lowFound
        highFound = isThreshold(i,2); 
    else
        lowFound = isThreshold(i,1); 
    end
    % if a high was found after finding a low, increment counter; reset flags
    if lowFound && highFound
        count = count + 1; 
        lowFound = false; 
        highFound = false; 
    end
    
end

Answer 2

[Image Analyst]

Assuming each element above 80 is one count, you can do this

count = sum(y > 80);

Assuming that if it's above 80 for multiple indexes and count that as one region, use bwlabel(), which counts the regions:

y = [1 2 81 83 0 84 86 0]
[L, count] = bwlabel(y > 80)

You'll see the labeled regions and the total region count (2 in this example):

y =
     1     2    81    83     0    84    86     0
L =
     0     0     1     1     0     2     2     0
count =
     2