interp1 does not fail here. It merely fails to do what it was not programmed to do! Interp1 does as designed. That you try to use it in a way that is meaningless in context of what interp1 is written to do, then what could you possibly expect? This is NOT a question of linear versis cubic, or a spline! This is merely that interp1 is designed to interpolate a function that has only 1 value for any given value of x. Mathematically, interp1 assumes the mapping it will interpolate is a single valued one. So for ANY value of x, it will expect only one value for y, if that presumption fails, then you get garbage.
You have 31 points. You don't supply them, so I'll make some up.
t = linspace(1,2*pi,50);
x = cos(t);
y = sin(2*t);
plot(x,y,'o')
Clearly that is not a single valued mapping. You CANNOT just throw it at interp1 directly, not and hope interp1 will succeed. How can you interpolate it? The simplest way is to do it parametrically. That will mean to interpolate x and y SEPARATELY and independently.
I'll use a linear interpolant. Be careful if you try to use a spline interpolant. Then you would need to be more creative about how to build that parametric vector s. But you can be quite free in how you build the vector sinterp as long as it is merely linear interpolation that is desired, and the data I saw from you does not merit more than a linear interpolation.
s = 1:numel(x);
ninterp = 1000; % 1000 points along the curve
sinterp = linspace(1,numel(x),ninterp);
xinterp = interp1(s,x,sinterp,'linear');
yinterp = interp1(s,y,sinterp,'linear');
plot(x,y,'ro',xinterp,yinterp,'b-')
Again, if you want a smooth interpolant like a spline, a tool like cscvn is recommended. Or you could use my interparc utility (as found on the file exchange for free download.)
