📂Fourier Analysis

Fourier Transform of B-Splines

The Equation¹

The Fourier transform of a B-spline of order $m \in \mathbb{N}$ is given as follows.

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

Here, the definition of the Fourier transform of $f$ is as follows.

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

Explanation

Using the properties of B-splines, Fourier transforms, and convolutions, the calculation can be done without much difficulty.

Proof

First, computing the Fourier transform of $N_{1}$ results in:

$$\begin{align*} \mathcal{F}N_{1}(\gamma) =&\ \int _{-\infty} ^{\infty} N_{1}(x)e^{-2\pi i x \gamma }dx \\ =&\ \int_{0}^{1} e^{-2\pi i x \gamma}dx \\ =&\ \left[\frac{e^{-2\pi i x\gamma} }{-2\pi i \gamma} \right]_{x=0}^{1} \\ =&\ \frac{1-e^{2\pi i \gamma}}{2\pi i \gamma} \end{align*}$$

Since a B-spline is defined by $N_{m}=\overbrace{N_{1} * N_{1} * \cdots * N_{1}}^{m}$ and due to the properties of Fourier transforms,

$$\mathcal{F}\left[ f_{1} * f_{2}*\cdots * f_{n} \right]=\hat{f_{1}} \hat{f_{2}} \cdots \hat{f_{n}}$$

It follows that,

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

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