angular data spline interpolation

Hi, I have the following query using spline interpolation for angular data. The code attached below is from matlab documentation:

x = pi*[0:.5:2];

y = [0 1 0 -1 0 1 0;

1 0 1 0 -1 0 1];

pp = spline(x,y);

yy = ppval(pp, linspace(0,2*pi,101));

plot(yy(1,:),yy(2,:),'-b',y(1,2:5),y(2,2:5),'or')

axis equal

This code gives the closed piecewise polynomial for circle of radius 1unit. Now if I calculate the value of plynomial for a given query point, it doesnt give me the correct answer. why??

For ex: xq = 0; yq = ppval(pp,xq);

It gives yq = [1;0] but it should be yq = [1,-1];

[Julius Muschaweck]

If you think my answer is correct, please make it the "accepted answer".Thanks

Answer 1

[Julius Muschaweck]

Your x has five values, your y has seven values. This is a mismatch.

I think you should use

y = [0 1 0 -1 0;

1 0 -1 0 1];

and when you do that,

xq = 0; yq = ppval(pp,xq);

gives you [0;1] as it should. Your "circle" looks more triangular, but that's ok since the sine is badly approximated by the spline with only 5 support points.

[mukesh bisht]

Here also it gives yq = [0;1] for xq = 0;

But in actual at x= 0, the y values are [-1,1]. This is the problem, it is not giving correct answers

[Julius Muschaweck]

Why do you think it should be [1, -1] (a row vector) and not [0;1] == [sin(0); cos(0)] (a column vector)?

[mukesh bisht]

Because, the piecewise polynomial is representing the circle of radius 1 with centre (0,0).

Hence, when my x = 0; y = radius i.e. y = -1,1. See, the attached figure, red dots corresponds to the query points at x = 0. Clearly, the point x= 0 & y = 0 doesn'nt lie on circle.

Here, my aim is to back calculate the y values using the piecewise polynomial obtained using spline interpolation

[Julius Muschaweck]

Isn't a radius a scalar number -- how can it be -1,1?

Your code, with

x = pi*[0:.5:2];

y = [0 1 0 -1 0;

1 0 -1 0 1];

uses x as the angle, and the first and second row of y are the first and second coordinate of unit circle points in a plane, i.e.

y = [sin(x);

cos(x)];

Then, at xq = 0, you get yq = [0; 1], which is the first column of y, and a point on the unit circle.

Your expected value [-1, 1] is not a point on the unit circle.

[mukesh bisht]

ok. Thanks. I got it

[darova]