# How to interpolate values given a starting and end point

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Bradley 2024 年 6 月 14 日

I have a voltmeter that I was using to measure voltage on a battery, unfortunatly it wasnt working properly and Im not able to do the test again. Before and after I ran the test I tested the voltage, so I know the real starting (50.6V) and ending (50.3V). My voltmeter recorded the power draw correctly but the values are off, they hover around 18V, how do I interpolate the bad values with a known starting and end point?
Example:
I know its 50.6 to 50.3 and my recordings are 18.2 to 17.9
I want the readings to be 50.6, 50.5, 50.4, 50.3 instead of 18.2, 18.1, 18.0, 17.9. I cant just add the difference because power draw wasnt exactly linear so Im trying to use the interp1 function but it isnt working for me. Thanks!
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Voss 2024 年 6 月 14 日

xl = [18.2, 17.9];
yl = [50.6, 50.3];
real_values = (recordings-xl(1))./(xl(2)-xl(1)).*(yl(2)-yl(1))+yl(1);

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### 採用された回答

Star Strider 2024 年 6 月 14 日
Use the rescale function —
Vrec = [18.2, 18.1, 18.0, 17.9];
Vrange = [50.6 50.3];
Vres = rescale(Vrec, Vrange(2), Vrange(1))
Vres = 1x4
50.6000 50.5000 50.4000 50.3000
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Bradley 2024 年 6 月 14 日
This worked great, thank you!
Star Strider 2024 年 6 月 14 日
As always, my pleasure!

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### その他の回答 (1 件)

John D'Errico 2024 年 6 月 14 日
If you ONLY have two data points, the only possible interpolation is LINEAR. There is no fancy method to interpolate in a nonlinear way. Interp1 cannot help you.
Caveat: You could perform a sort of exponential interpolation, so a nonlinear interpolant. Effectively, that means you might take the log of your data, then use linear interpolation, then use exp to back out the logs.
As far as why interp1 was not working for you, we cannot guess, because we do not see what you did with interp1, how you called it. We cannot see into your computer, or read your mind.

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