How to find the locations or indices of lower and upper ends of FWHM of a peak

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Vinay Killamsetty
Vinay Killamsetty 2021년 7월 14일
댓글: Simon Chan 2021년 7월 14일
How to find the locations and Index of lower and upper limit of FWHM of the largest peak shown in the pic below
Whn I use findpeaks function I am able to get the FWHM but I find it hard to get the location/index of the upper and lower limits of the FWHM
since there exists another peak in the range having a peak value above the half maximum of the main peak.

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Simon Chan
Simon Chan 2021년 7월 14일
편집: Simon Chan 2021년 7월 14일
Suppose y is the data, try this:
You may change the interpolation method, mine is only an example.
The location of the limits on the left side is between idx1 & idx1+1,
while on the right side, it is between idx2-1 & idx2.
[~,location] = max(y);
yn = y/max(y); % Normalized data
idx1 = location - find(diff(yn(location:-1:1)>0.5)~=0,1,'first');
idx2 = location + find(diff(yn(location:1:end)>0.5)~=0,1,'first');
xaxis_left = interp1([yn(idx1), yn(idx1+1)], [x(idx1), x(idx1+1)], [yn(idx1), 0.5, yn(idx1+1)], 'linear');
xaxis_right = interp1([yn(idx2-1), yn(idx2)], [x(idx2-1), x(idx2)], [yn(idx2-1), 0.5, yn(idx2)], 'linear');
FWHM = xaxis_right(2) - xaxis_left(2)
  댓글 수: 2
Simon Chan
Simon Chan 2021년 7월 14일
Try this:
idx3 and idx4 are the minimum points on the left and right side respectively.
Hope it helps
idx3 = location - find(diff(yn(location:-1:1))>0,1,'first') + 1;
idx4 = location + find(diff(yn(location:1:end))>0,1,'first') - 1;

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추가 답변(1개)

KSSV
KSSV 2021년 7월 14일
  댓글 수: 1
Vinay Killamsetty
Vinay Killamsetty 2021년 7월 14일
Deat KSSV,
Thank you very much for your suggestion.
I have used the below command as you have suggested
knnsearch(x,y,'k',2)
But this function has helped me in finding the location of higher end of the FWHM but it is giving an inappropriate value for the location oflower end FWHM.
Instead of loacting the lower end of FWHM it is giving me some location from 1st peak

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