How to find analytic expression for B-spline curve

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Darren Fletcher
Darren Fletcher 2017-8-16
回答： Yipeng Tang 2019-4-28
Suppose I have a set of points (x,y), I can use
sp=spap2(3,4,x,y);
to fit a smooth curve to my data, and that curve consists of 3 cubic polynomials joined at knots. Now the output of this gives 6 coefficients. I originally thought I ought to have 12 coefficients for the 3 cubic polynomials.
Q1: Why are there 6 and not 12? Is this because of conditions on the functions and derivatives at knots?
Q2: As the points are being estimated by a curve with 3 cubics I want to be able to write the curve as a linear combination of these 3 functions, but how do I do this if there are not enough coefficients? I want something like
y=a0+a1x+a2x^2+a3x^3 + b0+b1x+b2x^2+b3x^3 + c0+c1x+c2x^2+c3x^3
Any help would be greatly appreciated.

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回答（2 个）

Elizabeth Reese
Elizabeth Reese 2017-8-23
When you are executing the command
>> sp = spap2(3,4,x,y);
you are computing the B-form of a spline of order 4 with 3 polynomial pieces and knot locations chosen for you. The spline has 2 continuous derivatives.
The B-form of a spline is representing the spline in terms of a basis of B-splines. The formulation of these basis functions can be found at the following link.
https://en.wikipedia.org/wiki/B-spline#Definition
The coefficients that are returned from the "spap2" function are the coefficients for this basis. This is explained further at the following link.
https://www.mathworks.com/help/curvefit/types-of-splines-ppform-and-b-form.html#f9-19944
You can play with the B-spline basis functions using the "bspligui" GUI in MATLAB.
The definition of the spline still stands, independent of the form. They are piecewise polynomials, not a closed form for the entire interval.
Yipeng Tang
Yipeng Tang 2019-4-28
You can use fn2fm fuction to transform sp from 'bp' form to 'pp' form.
eg.
sp = spap2(augknt([x(1) x(length(x))],4),4,x,y)
pp = fnplt(sp);
...
z = pp.coefs(1)*(tt-x(length(x))).^3+ pp.coefs(2)*(tt-x(length(x))).^2+ pp.coefs(3)*(tt-x(length(x)))+pp.coefs(4);
z is what you want

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