How to calculate FWHM on the point

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);

[Star Strider]

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]

Dear @Star Strider, here attached is my statistic

[mohd akmal masud]

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]

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?

Answer 1

[Star Strider]

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.)

.

[mohd akmal masud]

Dear @Star Strider @dpb,

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

[Star Strider]

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

.