Re: FWHM, InterpolationFunction & Solve
- To: mathgroup at smc.vnet.net
- Subject: [mg80460] Re: [mg80418] FWHM, InterpolationFunction & Solve
- From: Sseziwa Mukasa <mukasa at jeol.com>
- Date: Thu, 23 Aug 2007 01:01:52 -0400 (EDT)
- References: <200708220839.EAA08527@smc.vnet.net>
On Aug 22, 2007, at 4:39 AM, 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?
Use FindRoot not Solve:
FindRoot[DataInterpolation[x]==MaximumFFTValue/2,{x,MaximumFFTX}]
But if you have a model for your signal you should fit that with
FindFit rather than doing interpolation. At any rate this will only
work if there is one extremum in your data.
Regards,
Ssezi
- References:
- FWHM, InterpolationFunction & Solve
- From: Mathieu G <ellocomateo@free.fr>
- FWHM, InterpolationFunction & Solve