📂Fourier Analysis

Central B-spline

Definition¹

For $m\in \mathbb{N}$, the centered B-spline $B_{m}$ is defined as follows:

$$B_{m}(x):= T_{-\frac{m}{2}}N_{m}(x)=N_{m}(x+{\textstyle \frac{1}{2}})$$

Here, $T$ is the translation of the space $L^{2}$ space.

Description

It can also be defined as follows:

$$B_{1}:= \chi_{[-1/2,1/2]},\quad B_{m+1}:=B_{m}*B_{1},\ m\in\mathbb{N}$$

Both definitions actually mean the same function. The key point here is that $B_{m}$ is defined to be an even function. As with the B-spline, it is easy to see that the following equation holds:

$$B_{m+1}(x)=\int _{-\infty} ^{\infty} B_{m}(x-t)B_{1}(t)dt=\int_{-{\textstyle \frac{1}{2}}}^{{\textstyle \frac{1}{2}}}B_{m}(x-t)dt$$

The centered B-spline is merely a translation of the B-spline, and therefore has the same properties of the B-spline.

Properties

(a) $\mathrm{supp} B_{m}=[-\frac{m}{2},\frac{m}{2}]$

(b) $\displaystyle \int _{-\infty} ^{\infty} B_{m}(x)dx=1$

(c) For $m\ge 2$,

$$\sum \limits_{k\in \mathbb{Z}}B_{m}(x-k)=1,\quad \forall x\in R$$

(c’) When $m=1$, the above equation holds for $x\in \mathbb{R}\setminus \left\{ \pm\frac{1}{2},\pm \frac{3}{2},\dots \right\}$.

(d) The Fourier transform of the centered B-spline is as follows:

$$\widehat{B_{m}}(\gamma)=\left( \frac{e^{\pi i \gamma}-e^{-\pi i \gamma}}{2\pi i \gamma} \right)^{m}=\left( \frac{\sin (\pi\gamma)}{\pi \gamma} \right)^{m}$$

Here, the definition of the Fourier transform $\widehat{f}$ of $f$ is as follows:

$$\widehat{f}(\gamma):=\int _{-\infty} ^{\infty} f(x)e^{-2\pi i x\gamma}dx$$

Proof

(a)

Since $\mathrm{supp}N_{m}=\left[ -\frac{m}{2}, \frac{m}{2} \right]$ and $B_{m}=T_{- \frac{m}{2}}N_{m}$,

$$\mathrm{supp} B_{m} = \left[0-\frac{m}{2},m-\frac{m}{2} \right] = \left[ -\frac{m}{2},\frac{m}{2} \right]$$

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(b)

$$\int _{-\infty} ^{\infty} N_{m}(x)dx=1$$

and since $B_{m}=T_{- \frac{m}{2}}N_{m}$,

$$\int _{-\infty} ^{\infty} B_{m}(x)dx=\int _{-\infty} ^{\infty} T_{-\frac{m}{2}}B_{m}(x)dx=1$$

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(c)

$$\sum \limits_{k \in \mathbb{Z}} N_{m}(x-k)=1,\quad \forall x\in \mathbb{R}$$

thus,

$$\sum \limits_{k\in \mathbb{Z}}B_{m}(x-k)=\sum \limits_{k \in \mathbb{Z}} T_{-\frac{m}{2}}N_{m}(x-k)=1,\quad \forall x\in \mathbb{R}$$

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(c')

When $m=1$, if $N_{m}$ and $x\in \mathbb{R}\setminus \mathbb{Z}$ holds and $B_{m}=T_{- \frac{m}{2}}N_{m}$, then it holds for $x\in \mathbb{R}\setminus \left\{ \pm\frac{1}{2},\pm \frac{3}{2},\dots \right\}$.

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(d)

The Fourier transform of the B-spline is as follows:

$$\widehat{N_{m}}(\gamma)=\left( \frac{1-e^{-2\pi i\gamma}}{2\pi i \gamma} \right)^{m}$$

Then, by the properties of the Fourier transform,

$$\mathcal{F}\left[ N_{m}(x+\frac{m}{2}) \right] (\gamma)=e^{2\pi i \frac{ m}{2} \gamma}\widehat{N_{m}}(\gamma)$$

thus,

$$\begin{align*} \widehat{B_{m}}(\gamma) =&\ \mathcal{F}\left[ B_{m}(x) \right] (\gamma) =\mathcal{F}\left[ N_{m}(x+\textstyle \frac{m}{2}) \right] (\gamma) \\ =&\ e^{2\pi i \frac{m}{2} \gamma}\widehat{N_{m}}(\gamma) \\ =&\ \left(e^{\pi i \gamma}\right)^{m}\left( \frac{1-e^{-2\pi i\gamma}}{2\pi i \gamma} \right)^{m} \\ =&\ \left( \frac{e^{\pi i \gamma}-e^{-\pi i \gamma}}{2\pi i \gamma} \right)^{m} \\ =&\ \left( \frac{\sin (\pi\gamma)}{\pi \gamma} \right)^{m} \end{align*}$$

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