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I'm looking for the Fourier transformation of the (constant) uniform B-Spline $$N_0(x) = \begin{cases}1 & 0 \leqslant x < 1 \\ 0 & otherwise \end{cases}$$

If $N_0(x)$ would also attain value $1$ when $x=1$ (i.e. the $<$ would be $\leqslant$), it would just be a shifted version of the rectangular function R(x), which has (if I'm not mistaken) the following Fourier transform ($T=1$ and shifted $1/2$ to the right):

$$\hat{R}(\omega) = e^{-\tfrac{i\omega}{2}}\, \frac{\sin(\omega/2)}{\omega/2}$$

However, I'm not sure if this is helpful in finding $\hat{N_0}(\omega)$, the Fourier transform of $N_0(x)$...

Once I know $\hat{N_0}(\omega)$, it is easy to find the Fourier transforms of higher order uniform B-Splines, since they are (or can be) defined using convolution. Of course, the Fourier transform of a convolution is just a multiplication of the two Fourier transformed functions, therefore:

$$\hat{N_k}(\omega) = \left( \hat{N_0}(\omega) \right)^{k+1}$$

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up vote 2 down vote accepted

Since the Fourier transform is defined through an integral, the condition $f=g$ almost everywhere implies $\hat{f} = \hat{g}$. So in this case the Fourier transform of $N_0$ is indeed the same as that of $R$, and your formula is correct, assuming you use the convention $\hat{f}(\omega) = \int_{-\infty}^\infty f(x) e^{-i\omega x} \, dx$ for the Fourier transform.

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Thanks. Somewhere I found stated that $\hat{N_0}(\omega)=\dfrac{1-e^{-i \omega}}{i \omega}$. I'm not quite sure whether this is the same as the function I found -- I'll take a more detailed look at a later time and update my post / add a comment. –  Ailurus Oct 21 '12 at 20:11
Yes, that is the same, using the identity $\sin z = \frac{e^{iz} - e^{-iz}}{2i}$ with $z = \omega/2$. –  Lukas Geyer Oct 21 '12 at 20:27
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