Reproducing Polynomials with B-Splines

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Family of B-splines up to N=4

A B-Spline of order <math>N</math> is known to be able to reproduce any polynomial up to order <math>N</math>:

<math> \sum_{n \in \mathbb{Z}} c_{m,n} \beta_N (t - n) = t^m </math>

In words, a proper linear combination of shifted versions of a B-Spline can reproduce any polynomial up to order <math>N</math>. This is needed for different applications, for example, for the Sampling at Finite Rate of Innovation (FRI) framework. In this case any kernel <math>\varphi</math> reproducing polynomials (that is, satisfying the Strang-Fix conditions) can be used. However, among all possible kernels, the B-Splines have the smallest possible support.

An important question is how to obtain the coefficients <math>c_{m,n}</math> for the reproduction-formula. In this small article, I describe one way.


Starting from

<math> \sum_{n \in \mathbb{Z}} c_{m,n} \varphi(t - n) = t^m </math>

the coefficients can be obtained using the dual of <math>\varphi</math>, <math>\tilde{\varphi}</math><ref>P.L. Dragotti, M. Vetterli, T.Blu: "Sampling Moments and Reconstructing Signals of Finite Rate of Innovation: Shannon Meets Strang-Fix", IEEE Transactions on Signal Processing, vol. 55, No. 5, May 2007</ref> (I set <math>\beta_N = \varphi</math> for consistency with my notes):

<math> c_{m,n} = \int_{-\infty}^{\infty} t^m \tilde{\varphi}(t - n)\,dt </math>

However, even if the dual would be known, solving the infinite integral is only feasible when the dual has finite support. This is the case with the B-Spline itself but not with its dual!

A closer look at the formula tells that this is nothing more than a convolution (under the assumption that <math>\tilde{\varphi}</math> is symmetric which is the case):

<math> c_{m,n} = \int t^m \tilde{\varphi}(-(n-t))\,dt = \int t^m \tilde{\varphi}(n-t)\,dt = (t^m * \tilde{\varphi})(n) </math>

Now, this can be transformed to fourier domain:

<math> (t^m * \tilde{\varphi})(n) = \mathcal{F}^{-1}\left\{ \mathcal{F}\left\{t^m\right\} \tilde{\Phi}(\omega)\right\} = \mathcal{F}^{-1}\left\{ j^m \sqrt{2\pi} \delta^{(n)}(\omega) \tilde{\Phi}(\omega) \right\} = j^m \sqrt{2\pi} \mathcal{F}^{-1}\left\{\delta^{(n)}(\omega) \tilde{\Phi}(\omega) \right\} </math>

Writing the inverse of this expression yields:

<math> j^m \sqrt{2\pi} \frac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} \delta^{(n)}(\omega) \tilde{\Phi}(\omega) e^{j\omega n}\,d\omega = j^m \int_{-\infty}^{\infty} \delta^{(n)}(\omega) \underbrace{\tilde{\Phi}(\omega) e^{j\omega n}}_{f(\omega)}\,d\omega </math>

It is known that<ref>http://en.wikipedia.org/wiki/Dirac_delta_function</ref>:

<math> \int \delta^{(n)}(x) f(x)\,dx = (-1)^n f^{(n)}(0) </math>

so that

<math> j^m \int_{-\infty}^{\infty} \delta^{(n)}(\omega) f(\omega)\,d\omega = j^m (-1)^m \left. \frac{\partial^m}{\partial \omega^m} f(\omega) \right|_{\omega = 0} </math>

Now the whole procedure has been reduced to calculating the derivative of <math>f(\omega)</math> and set the result to zero.

An open question is how to obtain the dual of <math>\varphi</math>. As the reproduction formula spans a vector space, <math>\varphi</math> must be at least bi-orthogonal to <math>\tilde{\varphi}</math>. This translates in fourier domain to<ref>S. Mallat: "A Wavelet Tour of Signal Processing", Academic Press 1999</ref>:

<math> \tilde{\Phi}(\omega) = \frac{\Phi(\omega)}{\sum_{k \in \mathbb{Z}} |\Phi(\omega + 2\pi k)|^2} </math>

The fourier transform of a B-Spline of order <math>N</math> is (e.g. <ref>M.Unser: "Splines - A Perfect Fit for Signal and Imaging Processing", IEEE Signal Processing Magazine Nov. 1999</ref>):

<math> \Beta_N(\omega) = \Phi(\omega) = \left( \frac{\sin(\omega/2)}{\omega/2} \right)^{N+1} = \mathrm{sinc}^{N+1}(\omega/2) </math>

The following derivation of the sum is borrowed from <ref>M.J.C.S. Reis, P.J.S.G. Ferreira, S.F.S.P. Soares: "Linear combinations of B-splines as generating functions for signal approximation", Elsevier Digital Signal Processing 15, 2005</ref>. For this derivation to work, I set <math>L=N+1</math> temporarily:

<math> \sum_{k \in \mathbb{Z}} |\Phi(\omega + 2\pi k)|^2 = \sum_{k \in \mathbb{Z}} \left|\mathrm{sinc}\left(\frac{1}{2}(\omega + 2\pi k)\right)^L \right|^2 = \sum_{k \in \mathbb{Z}} \left|\mathrm{sinc}\left(\frac{1}{2}(\omega + 2\pi k)\right) \right|^{2L} </math>

and because <math>2L</math> is always even:

<math> = \sum_{k \in \mathbb{Z}}\frac{\sin^{2L}(\frac{1}{2}(\omega + 2\pi k))}{\left(\frac{1}{2}(\omega + 2\pi k)\right)^{2L}} = \sum_{k \in \mathbb{Z}}\frac{\sin^{2L}(\frac{\omega}{2} + \pi k))}{(\frac{\omega}{2} + \pi k)^{2L}} </math>

Because of the periodicity it is known that

<math> \sin^{2L}(x + \pi k) = \sin^{2L}(x) </math>

such that

<math> = \sin^{2L}\left(\frac{\omega}{2}\right) \sum_{k \in \mathbb{Z}}\frac{1}{(\frac{\omega}{2} + \pi k)^{2L}} </math>

And finally the following relation is used<ref>L.V. Ahlfors: "Complex Analysis", McGraw-Hill, 1979</ref>:

<math> \sum_k \frac{1}{(x + \pi k)^{2L}} = -\frac{1}{(2L-1)!} \frac{d^{2L-1}}{dx^{2L-1}} \cot{x} </math>

in order to finally obtain:

<math> \sum_{k \in \mathbb{Z}} \left|\mathrm{sinc}\left(\frac{1}{2}(\omega + 2\pi k)\right)^L \right|^2 = -\sin^{2L}\left(\frac{\omega}{2}\right) \frac{1}{(2L-1)!} \frac{d^{2L-1}}{d\left(\frac{\omega}{2}\right)^{2L-1}} \cot{\left(\frac{\omega}{2}\right)} </math>

and with <math>L = N+1</math>:

<math> \sum_{k \in \mathbb{Z}} |\Phi(\omega + 2\pi k)|^2 = -\sin^{2(N+1)}\left(\frac{\omega}{2}\right) \frac{1}{(2N+1)!} \frac{d^{2N+1}}{d\left(\frac{\omega}{2}\right)^{2N+1}} \cot{\left(\frac{\omega}{2}\right)} </math>

Therefore, together with <math>\Phi(\omega)</math> this yields:

<math> \tilde{\Phi}(\omega) = \frac{(2N+1)!}{\left(\frac{\omega}{2}\right) \sin\left(\frac{\omega}{2}\right)^{N+1} \frac{d^{2N+1}}{d\left(\frac{\omega}{2}\right)^{2N+1}} \cot{\left(\frac{\omega}{2}\right)}} </math>

and finally substituting for <math>t(\omega)</math>:

<math> f(\omega) = \frac{(2N+1)!}{\left(\frac{\omega}{2}\right) \sin\left(\frac{\omega}{2}\right)^{N+1} \frac{d^{2N+1}}{d\left(\frac{\omega}{2}\right)^{2N+1}} \cot{\left(\frac{\omega}{2}\right)}} e^{j \omega n} </math>

As this function is not well defined it is better to use the limit:

<math> c_{m,n} = j^m \lim_{\omega \rightarrow 0} f(\omega) </math>

Examples for a cubic spline

For a cubic spline (N=3) the coefficients are:

<math> \begin{array}{lcl} c_{0,n} & = & 1 \\ c_{1,n} & = & n \\ c_{2,n} & = & \frac{1}{3}\left( -1 + 3n^2 \right) \\ c_{3,n} & = & -n + n^3 \end{array} </math>

cubic spline reproducing polynomial of order 2
cubic spline reproducing polynomial of order 3

References

<ref name="schoenberg">I.J. Schoenberg: "Cardinal interpolation and spline functions", J. Approx. Theory volume 2, pp. 167-206, 1969</ref>

<references/>

Comments

Manu said ...

Bussi

--Manu 19:47, 19. Jul. 2010 (MSD)