Reproducing Polynomials with B-Splines: Unterschied zwischen den Versionen

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A B-Spline of order <math>N</math> is known to be able to reproduce any polynomial up to order <math>N</math><ref>I.J. Schoenberg: "Cardinal interpolation and spline functions", ''J. Approx. Theory volume 2'', pp. 167-206, 1969</ref>:
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<section begin="head"/>
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[[Datei:bspline_family.png|right|150px|Family of B-splines up to N=4]]
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A B-Spline of order <math>N</math> is known to be able to reproduce any polynomial up to order <math>N</math>:
  
 
<math>
 
<math>
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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.
 
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.
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<section end="head"/>
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Starting from
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<math>
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\sum_{n \in \mathbb{Z}} c_{m,n} \varphi(t - n) = t^m
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</math>
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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):
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<math>
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c_{m,n} = \int_{-\infty}^{\infty} t^m \tilde{\varphi}(t - n)\,dt
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</math>
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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!
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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):
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<math>
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c_{m,n} = \int t^m \tilde{\varphi}(-(n-t))\,dt = \int t^m \tilde{\varphi}(n-t)\,dt = (t^m * \tilde{\varphi})(n)
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</math>
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Now, this can be transformed to fourier domain:
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<math>
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(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\}
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</math>
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Writing the inverse of this expression yields:
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<math>
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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
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</math>
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It is known that<ref>http://en.wikipedia.org/wiki/Dirac_delta_function</ref>:
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<math>
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\int \delta^{(n)}(x) f(x)\,dx = (-1)^n f^{(n)}(0)
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</math>
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so that
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<math>
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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}
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</math>
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Now the whole procedure has been reduced to calculating the derivative of <math>f(\omega)</math> and set the result to zero.
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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>:
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<math>
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\tilde{\Phi}(\omega) = \frac{\Phi(\omega)}{\sum_{k \in \mathbb{Z}} |\Phi(\omega + 2\pi k)|^2}
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</math>
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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>):
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<math>
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\Beta_N(\omega) = \Phi(\omega) = \left( \frac{\sin(\omega/2)}{\omega/2} \right)^{N+1} =
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\mathrm{sinc}^{N+1}(\omega/2)
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</math>
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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:
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<math>
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\sum_{k \in \mathbb{Z}} |\Phi(\omega + 2\pi k)|^2 =
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\sum_{k \in \mathbb{Z}} \left|\mathrm{sinc}\left(\frac{1}{2}(\omega + 2\pi k)\right)^L \right|^2 =
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\sum_{k \in \mathbb{Z}} \left|\mathrm{sinc}\left(\frac{1}{2}(\omega + 2\pi k)\right) \right|^{2L}
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</math>
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and because <math>2L</math> is always even:
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<math>
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= \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}}
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= \sum_{k \in \mathbb{Z}}\frac{\sin^{2L}(\frac{\omega}{2} + \pi k))}{(\frac{\omega}{2} + \pi k)^{2L}}
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</math>
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Because of the periodicity it is known that
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<math>
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\sin^{2L}(x + \pi k) = \sin^{2L}(x)
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</math>
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such that
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<math>
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= \sin^{2L}\left(\frac{\omega}{2}\right) \sum_{k \in \mathbb{Z}}\frac{1}{(\frac{\omega}{2} + \pi k)^{2L}}
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</math>
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And finally the following relation is used<ref>L.V. Ahlfors: "Complex Analysis", ''McGraw-Hill'', 1979</ref>:
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<math>
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\sum_k \frac{1}{(x + \pi k)^{2L}} = -\frac{1}{(2L-1)!} \frac{d^{2L-1}}{dx^{2L-1}} \cot{x}
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</math>
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in order to finally obtain:
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<math>
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\sum_{k \in \mathbb{Z}} \left|\mathrm{sinc}\left(\frac{1}{2}(\omega + 2\pi k)\right)^L \right|^2 =
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-\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)}
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</math>
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and with <math>L = N+1</math>:
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<math>
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\sum_{k \in \mathbb{Z}} |\Phi(\omega + 2\pi k)|^2 =
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-\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)}
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</math>
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Therefore, together with <math>\Phi(\omega)</math> this yields:
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<math>
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\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)}}
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</math>
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and finally substituting for <math>t(\omega)</math>:
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<math>
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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}
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</math>
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As this function is not well defined it is better to use the limit:
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<math>
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c_{m,n} = j^m \lim_{\omega \rightarrow 0} f(\omega)
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</math>
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= Examples for a cubic spline =
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For a cubic spline (N=3) the coefficients are:
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<math>
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\begin{array}{lcl}
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c_{0,n}  & = & 1 \\
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c_{1,n}  & = & n \\
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c_{2,n}  & = & \frac{1}{3}\left( -1 + 3n^2 \right) \\
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c_{3,n}  & = & -n + n^3
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\end{array}
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</math>
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[[Datei:poly_repro_quad.png|center|cubic spline reproducing polynomial of order 2]]
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[[Datei:poly_repro_cubic.png|center|cubic spline reproducing polynomial of order 3]]
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= References =
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<ref name="schoenberg">I.J. Schoenberg: "Cardinal interpolation and spline functions", ''J. Approx. Theory volume 2'', pp. 167-206, 1969</ref>
  
 
<references/>
 
<references/>
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= Comments =
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<comments />{{:{{TALKSPACE}}:{{PAGENAME}}}}
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[[Kategorie:Weblog]]

Aktuelle Version vom 19. Juli 2010, 16:45 Uhr

Family of B-splines up to N=4

A B-Spline of order is known to be able to reproduce any polynomial up to order :

In words, a proper linear combination of shifted versions of a B-Spline can reproduce any polynomial up to order . This is needed for different applications, for example, for the Sampling at Finite Rate of Innovation (FRI) framework. In this case any kernel 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 for the reproduction-formula. In this small article, I describe one way.


Starting from

the coefficients can be obtained using the dual of , [1] (I set for consistency with my notes):

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 is symmetric which is the case):

Now, this can be transformed to fourier domain:

Writing the inverse of this expression yields:

It is known that[2]:

so that

Now the whole procedure has been reduced to calculating the derivative of and set the result to zero.

An open question is how to obtain the dual of . As the reproduction formula spans a vector space, must be at least bi-orthogonal to . This translates in fourier domain to[3]:

The fourier transform of a B-Spline of order is (e.g. [4]):

The following derivation of the sum is borrowed from [5]. For this derivation to work, I set temporarily:

and because is always even:

Because of the periodicity it is known that

such that

And finally the following relation is used[6]:

in order to finally obtain:

and with :

Therefore, together with this yields:

and finally substituting for :

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

Examples for a cubic spline

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

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

References

[7]

  1. 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
  2. http://en.wikipedia.org/wiki/Dirac_delta_function
  3. S. Mallat: "A Wavelet Tour of Signal Processing", Academic Press 1999
  4. M.Unser: "Splines - A Perfect Fit for Signal and Imaging Processing", IEEE Signal Processing Magazine Nov. 1999
  5. 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
  6. L.V. Ahlfors: "Complex Analysis", McGraw-Hill, 1979
  7. I.J. Schoenberg: "Cardinal interpolation and spline functions", J. Approx. Theory volume 2, pp. 167-206, 1969

Comments

<comments />

Manu said ...

Bussi

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