# How to find the width of a peak where there is no actual data, just individual points?

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Alfredo Scigliani il 27 Giu 2023
I am trying to obtain the width at the half maximum location of the peak displayed in the figure. I have already found the location of the half max and have depicted it with dotted lines. Now, I would like to calculate the width inside that peak at the specified height. I would just get the intercept of the blue line and the dotted line and calculate it but since there is no actual data in the blue line, just a straight line connecting the individual points, I am not sure how to do it.
clear; clc; clf; close all;
xdata = [0 0.012 0.03 0.051 0.08 0.11 0.15];
ydata = [0.060583 0.098032 0.1407 0.40137 0.1407 0.098032 0.060583];
halfMax_y = 0.20069;
maxLocation_x = 0.051;
figure;
plot(xdata, ydata, 'o-');
hold on;
yline(halfMax_y, 'k--');
xline(maxLocation_x, 'k--');
I would like to find this distance, you can assume that the units of the x axis are distance in meters:
If somebody has an idea on how to do it I would really appreciate it. And if it can also be kept as simple as possible because I am a beginner I appreciate it. I have seen a couple peak processing toolbox online but I am not experienced with those options.
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### Risposta accettata

David Goodmanson il 27 Giu 2023
Hi Alfredo,
Usually your ydata would be the frequency domain version of a linear quantity such as voltage or displacement. Half maximum refers to half of a squared quantity such as power, so it would be down from the peak by a factor of sqrt(2) for the linear quantity. I will arbitrarily assume that here, but if you really intend to drop by a factor of 2 in ydata you can make the appropriate change below.
With this small number of points it seems best to use some spline-type interpolation, which I did with pchip.
xdata = [0 0.012 0.03 0.051 0.08 0.11 0.15];
ydata = [0.060583 0.098032 0.1407 0.40137 0.1407 0.098032 0.060583];
npts = 1000;
xp = linspace(0,.15,npts);
yp = pchip(xdata,ydata,xp);
[ymax,indmax] = max(yp)
halfh = ymax/sqrt(2); % down by sqrt(2)
x1 = xp(1:indmax);
x2 = xp(indmax:end);
y1 = yp(1:indmax);
y2 = yp(indmax:end);
w1 = interp1(y1,x1,halfh);
w2 = interp1(y2,x2,halfh);
fwhm = w2-w1 % full width half max
figure(1);
plot(xdata, ydata, 'o-',xp,yp);
hold on
plot([w1 w2],[halfh halfh],'og-')
hold off
grid minor
ymax = 0.4014
indmax = 341
fwhm = 0.0246
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Alfredo Scigliani il 27 Giu 2023
Thank you!

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