Interpolation doesn't reproduce the character of original curve

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AR 2021-12-10

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评论： Matt J 2021-12-10

Using the Matlab function notation for interp1: x, v, xq, and vq {vq = interp1(x, v, xq, 'method')}, I am trying to create a 1-d vector vq as a function of 1-d vector xq to recreate the shape of the curve v(x). I am trying to attach images that show the x & xq on one plot, and v & the resultant vq on another plot.

I have tried various 'methods', and none reproduces the original character of the curve v(x). I don't understand why and what I need to do (other than laborious self-computation) to get there. Appreciate pointers.

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John D'Errico 2021-12-10

EXACTLY. Without your data, we cannot help you that much. A picture of your data might be interesting, but it does not tell us enough. Attach a .mat file to a comment, or to your original question by editting the question.

I will note that since xq is NOT monotonic, then I would expect vq to also not be monotonic.

Matt J 2021-12-10

Data posted by AR.

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采纳的回答

Matt J 2021-12-10

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编辑：Matt J 2021-12-10

在 MATLAB Online 中打开

Perhaps I misunderstood how this works. I expected v to be scaled (compressed as it were in this case) and be reproduced over the new range xq

The vector v is just a list of values. It doesn't have a unique domain x. If the goal is to compress the plot, you just need to redefine v's domain. It doesn't require interpolation,e.g.,

x=linspace(0,2*pi,100);

v=sin(x);

xc=linspace(0,pi,100);

plot(x,v,xc,v,'x'); legend('Original','Compressed')

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Matt J 2021-12-10

在 MATLAB Online 中打开

Easier to do,

xq=linspace(min(xq), max(xq), numel(v));
plot(x,v,xq,v,'x')

AR 2021-12-10

This may be workable, but here we are coming up with a new xq and keeping v. In my case, I don't think I can change xq. I would need to keep xq and come up with a new v. Anyway, we have a workable solution one way or another.

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