View Full Version : Full width at half maximum
valesdn
Jul13-11, 01:50 PM
Hi. I need your help. What is the Full Width at Half Maximum (FWHM) of this function ( look at the picture) ? http://imageshack.us/photo/my-images/835/immaginezxa.png
The operating voltage of a PMT is in the order of 1710V. I've to find the PMT gain. The peak value
is near channel 630. The pedestal peak is near channel 66. Could you help me, please? Thanks anyway.
Redbelly98
Jul13-11, 09:03 PM
Welcome to Physics Forums.
It looks like the peak value near channel 630 is 50-60, so the half-max value is 25-30. This puts the half-max point at roughly channel 800 on the high side.
On the low side, we can't necessarily take where the value is 25-30, owing to the background. However, the background is falling off quite rapidly, roughly a factor of 10 every 100 channels. At channel 300 or so, it appears to be negligible (<5), so channel 300 appears to be a suitable half-max point as well.
If you want to get more accurate, you could model the background as an exponential decay -- seems to be approximately 800*10-x/100, but it's best to do a least-squares regression for a more accurate fit. Then subtract the background function from the data to get a more reasonable picture of the signal peak of interest.
Hope that helps.
valesdn
Jul14-11, 07:20 AM
Thanks Redbelly. I have Matlab but i can't use it =/ I have 3 sets of data ( at 1580V, at 1651V and at 1710V) and I need to fit a curve to them. Could it possible that the half max value is 37, at channel 728? How can i do a least-squares regression ( what is the equation in this example?). Thanks for your reply.
Redbelly98
Jul14-11, 07:33 AM
Thanks Redbelly. I have Matlab but i can't use it =/ I have 3 sets of data ( at 1580V, at 1651V and at 1710V) and I need to fit a curve to them. Could it possible that the half max value is 37, at channel 728?
Sounds reasonable. You have the actual data, I only have a graph (without gridlines) to make estimates.
How can i do a least-squares regression ( what is the equation in this example?).
Just to the right of the high, narrow peak, a portion of the data forms a straight line on the semilog graph. Take just that data, that makes the straight-line, and do the regression on that portion of the data only. You can either fit an exponential decay to the actual data, or do a linear fit to log(y) vs. x.
Thanks for your reply.
You're welcome.
EDIT: If you're not allowed to use a software package for regression (???), equations for linear regression are given here:
http://en.wikipedia.org/wiki/Simple_linear_regression#Fitting_the_regression_li ne
valesdn
Jul14-11, 08:45 AM
I have a text file. I've just imported data in Matlab and I've plotted data in that figure. But i've found only the peak value ( i can't add function for linear regression on the command window ). I plot log(y) vs. x. But i don't understand how can fit log(y) vs. x However, I'm using matlab for my first time. Do you know another simple programm? I need FWHM and a fitting curve vs. that distribution on that fig.
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