how to find out full width at half maximum from this graph

[John D'Errico]

This is not even a MATLAB question. Not even close to the point where you can consider writing code, because you need to make many decisions, answer many questions.

What value is the maximum here? You cannot compute a half-max until you know what is the max. And you would also seem to need to know the min, as the half-max would be midway.

How can you compute the max of a relationship that seems to be trending upwards at each end, with no upper limit?

Do you intend to do smoothing? It would seem that otherwise, any such computation would be highly inaccurate. Lots of ways to do smoothing.

Once you have a curve that you are happy is smooth (or not if you do no smoothing) and you have decided what value for y is the half-max, then you compute the locations of those points on the curves by interpolation. Linear interpolation will surely suffice here, given how noisy is the data. But you have a long way to go before you start worrying about that.

[parry sharma]

i am new to matlab .that graph wrongly generated by me in matlab. 1) previous chart was mistakely uploaded actual graph i am attached now, Is in this graph i have to first smooth data then how please tell me. 2) but smoothing of data result in average of data that part so this may make effect on my experiment or not. 3)after smotthing my experiment data is valid or not, 4) to find out FWHM, i am using formula is Dmax-Dmin/2 at gradient region but how to find that gradient region with help of matlab and then compute this

[Image Analyst]

OK, I've modified my answer below to take into account that your FWHM value is not based on the max but based on being between the min and the max.

Answer 1

[Image Analyst]

Use max():

% Find the half max value.
halfMax = (min(data) + max(data)) / 2;
% Find where the data first drops below half the max.
index1 = find(data >= halfMax, 1, 'first');
% Find where the data last rises above half the max.
index2 = find(data >= halfMax, 1, 'last');
fwhm = index2-index1 + 1; % FWHM in indexes.
% OR, if you have an x vector
fwhmx = x(index2) - x(index1);

[Tusharkumar Sorathiya]

Hello Sir,

I have been using this logic for my thesis algorithm. Thanks a lot!!

problem i am facing now is: what if I want to find bandwidth at 25% reflection level. I tried to solve it by taking halfMax = (min(data) + max(data)) / 4; which works well. But when I have spectra something like as shown in attached figure. it always takes the very last value which isnt something i am looking for. I wanted to find out the value as depicted in figure.

Thanks in advance

[Image Analyst]

Use a for loop

[maxValue, indexOfMax] = max(data);
value25 = 0.25 * maxValue
% Find 25% point on left
for index = indexOfMax : -1 : 1
    if data(index) < value25
        leftIndex = index;
        break;
    end
end
% Find 25% point on right
for index = indexOfMax : length(data)
    if data(index) < value25
        rightIndex = index;
        break;
    end
end

[Tusharkumar Sorathiya]

Thanks a lot, you saved my time and made it easy.

really appreciated. :)

[Image Analyst]

You're welcome. Then, can you please "Accept this answer"? Thanks in advance.

[Tusharkumar Sorathiya]

Sorry, but I do not see the option to accept. Since Main question is not asked by me, but I recommend your answer to readers.

[KALYAN ACHARJYA]

+1 Nice

[Eunan McShane]

Hey Image Analyst,

I am working with a 999x512 matrix, for which I can simply calculate tha half max value of each column, although when using the indexing to find the locations 1 and 2 in order to find the half max, the script spits out a huge logical array which I can't work out how to manipulate into two values at the locations of interest, per column which I can take away and find the effective FWHM value. I am ignoring the first row in every column as it is effectively not a part of the waveform, I have attached a mini version of my matrix for you if you get a chance to look at this, thank you!

[Image Analyst]

I don't understand. Try to describe in a different way, using a column of data to desmonstrate, or using images/diagrams. And describe what "Manipulate" means to you. Why can't you just do:

topRow = find(thisColumn, 1, 'first');
if ~isempty(topRow)
        bottomRow = find(thisColumn, 1, 'last');
end

[Eunan McShane]

Hey, I just want to find the two half max index locations as you do above, but for each of the 512 columns (whilst ignoring the first row in each column) by using your code above I can calculate the half max for each column but the index values will not calculate the FWHM for me.

Index 1 and 2 become 503049x512 logical arrays which I don't understand how they relate to the real index positions of 1 and 2, I hope this makes sense!

[Image Analyst]

If this

clc
clear
close all;
fontSize = 16;
fprintf('Working...\n');

yourMatrix = rand(40, 30);
[rows, columns] = size(yourMatrix);
subplot(1, 2, 1);
imshow(yourMatrix);
title('Original Matrix', 'FontSize', fontSize);
axis('on', 'image');
for col = 1 : columns
    thisColumn = yourMatrix(:, col);
    % Find the half max value.
    halfMax(col) = (min(thisColumn) + max(thisColumn)) / 2;
    % Find where the data first drops below half the max.
    index1(col) = find(thisColumn >= halfMax(col), 1, 'first');
    % Find where the data last rises above half the max.
    index2(col) = find(thisColumn >= halfMax(col), 1, 'last');
    fwhm(col) = index2(col) - index1(col) + 1; % FWHM in indexes.
end
subplot(1, 2, 2);
imshow(yourMatrix, []);
axis('on', 'image');
hold on;
x = 1 : columns;
plot(x, index1, 'r-', 'LineWidth', 3);
plot(x, index2, 'r-', 'LineWidth', 3);
title('With upper and lower boundaries shown', 'FontSize', fontSize);
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0, 0.04, 1, 0.96]);
msgbox('Done!');

doesn't work for you, I suggest you start a new question with your own data attached.

[Eunan McShane]

Thank you! This is working as I want it to, brilliant!

[SHANTANU KSHIRSAGAR]

Dear Image Analyst,

I wanted to calculate the FWHM and area under pulse ; for each pulse , in a data containing multiple pulses .

I have the data in the form of excel sheets.

Can you help me with that.

[Image Analyst]

Lots of us could. Post a new question, rather than here in parry's question.

[fitri zabudin]

@Image Analyst why I try use for fwhm it error. At command window it show ">> fwhm Error: File: fwhm.m Line: 4 Column: 1 The variable "fwhm" is also the name of the script "C:\Users\Admin PC\Pictures\clear.fi\fwhm.m". This is illegal, because it will be the name of a script and a variable in any context from which the script is called."

[Image Analyst]

Thanks for the notice. I didn't check to see if "fwhm" was the name of a built-in function or their custom function. If it is, then they should use a different variable name. How did you find out from parry, who posted this 5 years ago, that they had a file called "C:\Users\Admin PC\Pictures\clear.fi\fwhm.m"?

[fitri zabudin]

@Image Analyst thanks but I have other question. I use matlab to get the value of fwhm from my graph. My data is from picoscope graph that are detect the lightning waveform. How way to find the fwhm for the only maximum peak because the graph show many peak?

[Image Analyst]

Rather than hijack @parry sharma's question, I'll tell you in the new, separate question that you should post.

[fitri zabudin]

Sorry.. I will make a new question for this.

[SP]

Excuse me, sir. For the find function, can we limit the column (x-axis)? Such as find(data(:,50:100));

As I show the signal, because I want to calculate the full width at half maximum of this signal at the red arrow by using the 'first' and 'last' function. But the first and last are calculated at the red circle. So I want to limit the x-axis because I don't want to calculate at the red circle. Is it possible if I want to limit the x-axis in the range of the red arrow?

Now, I try to did such a find(data(:,50:100)); but it cannot to calculate correctly. I try to study the find function but I don't understand some sentence that is "If data is a multidimensional array, the find returns a column vector of the linear indices of the result". Maybe as this lesson, I cannot limit the data in x-axis.

If you have a good idea to resolve this problem, please suggest to me. If I'm not explaining well, I'm sorry.

of t

[Image Analyst]

@SP, yes you can limit the search range by cropping the data like you did with find(data(:,50:100));

Just be aware that the rows and columns you find will be relative to that cropped array, not the original array.

MATLAB is in column major order so an array goes first down rows in a column before it moves to the next column, so the linear indexes are like this:

1 6 11

2 7 12

3 8 13

4 9 14

5 10 15

So if you wanted to find elements above 40 and less than 60:

m = 100 * rand(5, 3)

m = 5×3

80.2771 71.7417 41.5551 66.5391 71.0821 59.4884 8.8467 53.8624 93.7960 52.8215 42.6820 8.3886 27.8592 6.0552 40.8630

% Find values above 40 and below 60
mask = m > 40 & m < 60

mask = 5×3 logical array
   0   0   1
   0   0   1
   0   1   0
   1   1   0
   0   0   1

[rows, column] = find(mask)

rows = 6×1

4 3 4 1 2 5

column = 6×1

1 2 2 3 3 3

linearIndexes = find(mask)

linearIndexes = 6×1

4 8 9 11 12 15

You can see that (row, column) = (4, 1) (the 1st elements) corresponds to 4.

(row, column) = (3, 2) (the 2nd elements) corresponds to 8.

(row, column) = (4, 2) (the 3rd elements) corresponds to 9.

(row, column) = (1, 3) (the 4th elements) corresponds to 11.

and so on

[SP]

@Image Analyst Thank you, sir. I understand about the concept of linear indexes from your explanation. But I did the code like this:

vq2 = resample(x_ccd,100,1);

for kkkkkk = 1:30

halfMax = 0.5; % Set FWHM at 0.30

vq1(kkkkkk,:) = resample(IFFT_hologram_bf_NM(kkkkkk,:),100,1);

% Find where the data first drops below half the max.

index1 = find(vq1(kkkkkk,20000:35000) >= halfMax, 1, 'first');

% Find where the data last rises above half the max.

index2 = find(vq1(kkkkkk,:) >= halfMax, 1, 'last');

xx1 = vq2(index1); %find the position in mm

xx2 = vq2(index2); %find the position in mm

fwhm(kkkkkk,:) = xx2-xx1;

end

I did for loop that the literation will be calculate from row-by-row. Even I difined kkkkkk as the literation of for loop then I limited the range of x axis array (20000:35000), but the values of index1 and index2 isn't calculated correctly. The reason that I limited (kkkkkk,20000:35000) because I don't want to calculate at the two red circles. But if I define like this:

index1 = find(vq1(kkkkkk,:) >= halfMax, 1, 'first');

or

index1 = find(vq1(kkkkkk,1:62900) >= halfMax, 1, 'first'); (the number array of vq1 = 62900 arrays)

it can calculated index1 and index2 correctly. May you suggest about this problem to me, sir?

[Image Analyst]

You either need to just add in the offset you searched from, like 20,000 to the index,

index1 = find(vq1(kkkkkk,20000:35000) >= halfMax, 1, 'first');  
index1 = index1 + 19999; % Get index relative to the original length of the vector, not the cropped length.

or just mask out the stuff you don't want so it's not found

vq1(kkkkkk, 1:19999) = -inf;
vq1(kkkkkk, 35001:end) = -inf;
index1 = find(vq1(kkkkkk,20000:35000) >= halfMax, 1, 'first');  

Answer 2

[Maks]

It should be "more or equal" for the first index: index1 = find(data >= halfMax, 1, 'first');

[Image Analyst]

You're right. Thanks for catching that. I've edited my answer to fix it.

[Muhammad Basit]

Hi, Could you please tell me whether the signs of >= should be on both indexes or for index2 it should be <=. Using your code, I'm not being able to find the correct answers.

[Image Analyst]

@Muhammad Basit, I don't understand what you're asking. What sign? The code is the way it should be. If it doesn't work for your data, then give us your data and we'll fix it.

Answer 3

[Xiaomeng Gao]

This would be limited by available sample points. For example, if major peak has only 3 sample points, but it's neighboring peak is above half of max, then the neighbouring peak can be mistaken as first point that cross half max, potentially widening the width.

[Image Analyst]

The data the poster showed has about 1200 sample points across it.