finding missing value using interpolation

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S.Sil
S.Sil on 11 Nov 2015
Edited: S.Sil on 5 Dec 2016
Here is the data for the diameter and power of a hydraulic impulse

Answers (2)

Brandon
Brandon on 11 Nov 2015
The interp1 function does not allow for extrapolation. 2.01 is outside your domain for 0 to 1.92

dpb
dpb on 11 Nov 2015
Yeah, your second desired location is outside the range of the input and interp1 won't extrapolate unless it's ordered to--use
>> interp1(Diameter,Power,Given_Diameter,'linear','extrap')
ans =
2.0000 0.2367
>>
  2 Comments
dpb
dpb on 11 Nov 2015
>> plot(Diameter,Power,'x-') % original data showing points
>> hold on
>> pInterp1=interp1(Diameter,Power,Given_Diameter,'linear','extrap');
>> plot(Given_Diameter,pInterp1,'*') % linear interpolation
>> pSpln=interp1(Diameter,Power,[0:.1:2],'spline');
>> plot([0:.1:2],pSpln,'r-') % spline interpolation of original
>> pSplExt=interp1(Diameter,Power,Given_Diameter,'spline','extrap');
>> plot(Given_Diameter,pSplExt,'or') % and those interp/extrap values
>> [pInterp1;pSplExt] % compare the two choices...
ans =
2.0000 0.2367
2.0000 0.2465
>>
NB: the advantage of the spline over the linear is likely that it shows some effect of the curvature apparent in the data while the linear continues at the same slope as the last two points thus predicting lower value.
A quadratic looks like a reasonable try but if you fit it and add to the above you'll see it has a minimum before the last data point as it tries to minimize the whole curve equally.
Perhaps, depending on what it actually is, there may be reason to use a decaying exponential instead?

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