Interpolation doesn't reproduce the character of original curve

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AR
AR on 10 Dec 2021
Commented: Matt J on 10 Dec 2021
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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Accepted Answer

Matt J
Matt J on 10 Dec 2021
Edited: Matt J on 10 Dec 2021
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')
  5 Comments
AR
AR on 10 Dec 2021
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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More Answers (1)

Matt J
Matt J on 10 Dec 2021
Edited: Matt J on 10 Dec 2021
I don't know how you are plotting, but it should look like this:
plot(x,v,xq,vq,'x--')
  7 Comments
Matt J
Matt J on 10 Dec 2021
interp1 will need to extrapolate outside the range of x. It will do that according to the extrapolation settings that you give to interp1:

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