Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

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

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.