How to calculate FWHM on the point

Question

mohd akmal masud 2023-6-18

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1984734-how-to-calculate-fwhm-on-the-point

评论： Star Strider 2023-6-18

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Dear all,

I have image1 as attached (image1.png). The statistics as attached. (stats.mat)

Then I plot the graph like picture attached. (graph.jpg)

Anyone can help me to calculate the FWHM?

I try using the function fwhm as attached, but got error

Error using max
Invalid data type. First argument must be numeric or logical.
Error in fwhm (line 13)
y = y / max(y);

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Star Strider 2023-6-18

在 MATLAB Online 中打开

How are you calling ‘fwhm’ and what data are you passing to it?

figure

imshow(imread('image1.png'))

figure

imshow(imread('graph.jpg'))

type('fwhm.m')

function width = fwhm(x,y) % function width = fwhm(x,y) % % Full-Width at Half-Maximum (FWHM) of the waveform y(x) % and its polarity. % The FWHM result in 'width' will be in units of 'x' % % % Rev 1.2, April 2006 (Patrick Egan) y = y / max(y); N = length(y); lev50 = 0.5; if y(1) < lev50 % find index of center (max or min) of pulse [garbage,centerindex]=max(y); Pol = +1; disp('Pulse Polarity = Positive') else [garbage,centerindex]=min(y); Pol = -1; disp('Pulse Polarity = Negative') end i = 2; while sign(y(i)-lev50) == sign(y(i-1)-lev50) i = i+1; end %first crossing is between v(i-1) & v(i) interp = (lev50-y(i-1)) / (y(i)-y(i-1)); tlead = x(i-1) + interp*(x(i)-x(i-1)); i = centerindex+1; %start search for next crossing at center while ((sign(y(i)-lev50) == sign(y(i-1)-lev50)) & (i <= N-1)) i = i+1; end if i ~= N Ptype = 1; disp('Pulse is Impulse or Rectangular with 2 edges') interp = (lev50-y(i-1)) / (y(i)-y(i-1)); ttrail = x(i-1) + interp*(x(i)-x(i-1)); width = ttrail - tlead; else Ptype = 2; disp('Step-Like Pulse, no second edge') ttrail = NaN; width = NaN; end

.

mohd akmal masud 2023-6-18

编辑：mohd akmal masud 2023-6-18

在 MATLAB Online 中打开

stats.mat

I tried this command,

h= fwhm (xstats, ystats); %xstats and ystats is from my data as attached

xstats and ystats is form my data as attached

But got error

dpb 2023-6-18

在 MATLAB Online 中打开

stats.mat

whos -file stats
  Name       Size            Bytes  Class     Attributes

  akmal      1x1              2184  struct              

Well, "Houston, we have a problem!". stats is a stuct, not a pair of x,y vectors.

load stats
fieldnames(akmal)
ans = 9×1 cell array
    {'distance'}
    {'mean'    }
    {'STD'     }
    {'min'     }
    {'max'     }
    {'profile' }
    {'cx'      }
    {'cy'      }
    {'position'}

OK, who inside there is the x,y for which to do the FWHM on?

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Answer 1

Star Strider 2023-6-18

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1984734-how-to-calculate-fwhm-on-the-point#answer_1258184

在 MATLAB Online 中打开

Those fields do not exist in ‘stats.mat’ so those are empty vectors and the error is obvious.

This works —

LD = load('stats.mat')

LD = struct with fields:

akmal: [1×1 struct]

akmal = LD.akmal

akmal = struct with fields:

distance: 23.5756 mean: 1.3427e+03 STD: 938.0745 min: 496 max: 3189 profile: [25×1 double] cx: [25×1 double] cy: [25×1 double] position: [2×2 double]

pf = LD.akmal.profile;

cx = LD.akmal.cx;

cy = LD.akmal.cy;

figure

plot3(cx, cy, pf)

grid

wx = fwhm(cx,pf)

Pulse Polarity = Positive Pulse is Impulse or Rectangular with 2 edges

wx = 8.0832

wy = fwhm(cy,pf)

Pulse Polarity = Positive Pulse is Impulse or Rectangular with 2 edges

wy = -0.0198

[wx,cxr] = myFWHM(cx,pf)

wx = 7.0372

cxr = 1×2

72.4166 79.4537

[wy,cyr] = myFWHM(cy,pf)

wy = -0.0172

cyr = 1×2

58.0415 58.0243

function [width,xr] = myFWHM(x, y)

p1 = polyfit(x([1 end]), y([1 end]), 1);

y_dtrnd = y - polyval(p1,x);

[ymax,yidx] = max(y_dtrnd);

[ymin,xidx] = min(y_dtrnd);

idxrng = {1:yidx; yidx:numel(y)};

xr(1) = interp1(y_dtrnd(idxrng{1})-ymin, x(idxrng{1}), (ymax-ymin)/2, 'linear');

xr(2) = interp1(y_dtrnd(idxrng{2})-ymin, x(idxrng{2}), (ymax-ymin)/2, 'linear');

width = xr(2)-xr(1);

end

My function returns slightly different values because it detrends the dependent variable first, and then interpolates to find the half-maximum values. (I coded it for fun, just to see if my values matched the others.)

.

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mohd akmal masud 2023-6-18

untitled.fig

Dear @Star Strider @dpb,

If I have the graph as attached, how to calculate the FWHM?

Star Strider 2023-6-18

在 MATLAB Online 中打开

That has different independent variable values.

Try this —

F = openfig('untitled.fig');

Lines = findobj(F, 'Type','Line');

x = Lines.XData;

y = Lines.YData;

wx = fwhm(x,y)

Pulse Polarity = Positive Pulse is Impulse or Rectangular with 2 edges

wx = 8.0080

[wm,xr,hm] = myFWHM(x,y)

wm = 6.9718

xr = 1×2

9.0430 16.0148

hm = 1.3460e+03

[ymax,idx] = max(y);

ymin = min(y);

figure

plot(x, y)

hold on

plot(xr, [1 1]*hm+ymin, '.-r')

hold off

grid

text(x(idx), hm+ymin, sprintf('FWHM = %.3f',wm), 'Horiz','center', 'Vert','bottom')

% ylim([min(y) max(y)])

function [width,xr,hm] = myFWHM(x, y)

p1 = polyfit(x([1 end]), y([1 end]), 1);

y_dtrnd = y - polyval(p1,x);

[ymax,yidx] = max(y_dtrnd);

[ymin,xidx] = min(y_dtrnd);

idxrng = {1:yidx; yidx:numel(y)};

hm = (ymax-ymin)/2;

xr(1) = interp1(y_dtrnd(idxrng{1})-ymin, x(idxrng{1}), hm, 'linear');

xr(2) = interp1(y_dtrnd(idxrng{2})-ymin, x(idxrng{2}), hm, 'linear');

width = xr(2)-xr(1);

end

.

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