Re: FWHM, InterpolationFunction & Solve

• To: mathgroup at smc.vnet.net
• Subject: [mg80653] Re: FWHM, InterpolationFunction & Solve
• From: Mathieu G <ellocomateo at free.fr>
• Date: Tue, 28 Aug 2007 02:09:59 -0400 (EDT)
• References: <200708220839.EAA08527@smc.vnet.net> <faj4md\$8p2\$1@smc.vnet.net>

```Sseziwa Mukasa a =E9crit :
> 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
>
Hi!
Even though I read the documentation I still do not understand the
difference between FindRoot and Solve, can you help?
Regards,
Mathieu

```

• Prev by Date: Re: Fw: BarChart inside a Frame. How to remove Labels from the top of the frame? How to put x-axis label below and not on the side?
• Next by Date: Re: Drawing tick labels without ticks
• Previous by thread: RE: FWHM, InterpolationFunction & Solve
• Next by thread: Re: Re: FWHM, InterpolationFunction & Solve