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MathGroup Archive 2007

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Re: FWHM, InterpolationFunction & Solve

  • To: mathgroup at smc.vnet.net
  • Subject: [mg80468] Re: FWHM, InterpolationFunction & Solve
  • From: dh <dh at metrohm.ch>
  • Date: Thu, 23 Aug 2007 01:06:00 -0400 (EDT)
  • References: <fagstu$8du$1@smc.vnet.net>


Hi Mathieu,

if your data has noise you may have some troubles with Max and Solve. In 

this case a fit will be more adequate. If the peak has gaussian form I 

would fit a Gaussian (take log of data to make the fit linear!). If you 

have the Gaussian, you can get FWHM analytically.

hope this helps, Daniel



Mathieu G wrote:

> Hello,

> I have a set of discrete data, representing a peak.

> I would like to compute the Full Width at Half Maximum (FWHM) of this peak.

> For that I would like to know which points correspond to half the peak value.

> SO far I was considering using an interpolating function, but this does 

> not seem to work:

> 

> 

> DataFile = Import["FFT.dat"];

> CleanDataFile = Part[DataFile, 2 ;; Length[DataFile]];

> WorkingData = CleanDataFile[[All, {1, 3}]];

> 

> ListLinePlot[WorkingData, PlotRange -> All]

> 

> MaximumFFTY = Max[WorkingData[[All, 2]]];

> MaximumFFTPosition = Position[WorkingData, MaximumFFTY][[1, 1]];

> MaximumFFTX = WorkingData[[MaximumFFTPosition, 1]];

> 

> DataInterpolation = Interpolation[WorkingData];

> 

> Solve[DataInterpolation[x] == MaximumFFTValue/2, x]

> 

> 

> Can you help me please? How would you do that?

> I then want to compute the area under the peak:

> 

> Integrate[DataInterpolation[x], {x, BegFWHM, EndFWHM}]

> 

> Which works fine with the interpolating function.

> 

> Thank you for your help!

> Mathieu

> 




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