To find the max of the signal and what element it occurs at, you can do this:
[maxSignal, indexOfMax] = max(signal)
I don't know how many elements correspond to 12 to 15 seconds, but let's just assume here that it is 100 elements. So you can do
index1 = indexOfMax - 100;
if index1 < 1
index1 = 1; % Don't allow negative or zero indexes.
end
index2 = indexOfMax + 100;
if index2 > length(signal)
index2 = length(signal); % Don't allow negative or zero indexes.
end
Now you want the max in the range from index1 to index2 but obviously not the first max you found. So you can set the array before index1, and after index2, and at the first max to minus infinity. Then call max again. This will find " max value of the signal 's', but only within a range of 12 to 15 seconds" of the first max as you asked for
signal(1:index1) = -inf; % Make a copy of signal if you want to save the original.
signal(index2:end) = -inf;
signal(indexOfMax) = -inf;
[secondMax, indexOfSecondMax] = max(signal);
An alternate method is to sort the signal but then you have to check the sorted values to see if their indexes are within 12-15 seconds of the first index.
[sortedValues, sortIndices] = sort(signal, 'descend');
maxSignal = sortedValues(1);
indexOfMax = sortIndices(1);
% Now find second, but has to be within, say, 100 elements
k = 2;
while k < length(sortIndices)
indexesApart = sortIndices(k) - sortIndices(1);
if abs(indexesApart) < 100
break;
end
% If we get here, it's not within 12-15 seconds of the first max.
% So increment the index and try again.
k = k +1;
end
% Get the actual index
indexOfSecondMax = sortIndices(k)
% Get the signal value at that location.
secondMax = signal(indexOfSecondMax)
