-2

I am confused about the fourier transform of the $\operatorname{sinc}$ function. First I don't know if

$$\operatorname{sinc} (x) = \frac{\sin(\pi x)}{(\pi x)}$$

or

$$\operatorname{sinc} (\pi x) = \frac{\sin(\pi x)}{(\pi x)}$$

Also, is the Fourier transform of $\operatorname{sinc} (a \pi f)$

$$\left( \frac{1}{\pi |a|} \right ) \text{rect} \left( \frac{f}{a \pi} \right )$$

or

$$\left( \frac{1}{a} \right) \text{rect} \left( \frac{f}{a} \right)$$?

Could someone help me understand which equation I need to use.

Sean Lake
  • 1,757
  • 10
  • 23
J.ba
  • 163
  • 2
    $\text{sinc}{n}(x) = \sin( \pi x)/(\pi x)$ is called the normalised sinc function. If you start with the un-normalised sinc function $$\text{sinc}{u}(x) = \sin(x)/x$$ then mapping $x \mapsto \pi x \implies \text{sinc}{u}( \pi x) = \sin( \pi x)/(\pi x) = \text{sinc}{n}(x)$ gives the normalised sinc function. In response to your question about the Fourier transform, which convention of the FT are you using? – Matthew Cassell Sep 17 '16 at 03:12

2 Answers2

1

This is one of those times where there are different definitions of what is meant by $\operatorname{sinc}(x)$ used by different people, analogous to mathematicians using $i=\sqrt{-1}$ and electrical engineers using $j=\sqrt{-1}$. When mathematicians say $\operatorname{sinc}(x)$ they usually mean $\sin(x)/x$. When information or signal processing people say it, they usually mean $\sin(\pi x)/(\pi x)$. So you're going to need to check which context you're working in, be explicit about what you mean, and ask when you're unsure.

Sean Lake
  • 1,757
  • 10
  • 23
0

As indicated by Sean Lake, the definition of $\mathrm{sinc}$ can depend on the context. Moreover, the definition of the Fourier transform also depends on the context.

To avoid any doubt, the best is as always to know how to prove the formula quickly. One can for example proceed by the Fourier inversion theorem. For $a\geq 0$, $$ \int_{\Bbb R} \mathbf{1}_{[-a/2,a/2]}(y) \,e^{2i\pi\,x\,y}\,\mathrm d y = \int_{-a/2}^{a/2} \,e^{2i\pi\,x\,y}\,\mathrm d y = \frac{e^{i\pi\, a\,x} - e^{-i\pi\,a\,x}}{2i\pi\,x} = a\,\frac{\mathrm{sin}(\pi\,a\,x)}{\pi\,a\,x} $$ one deduces by the Fourier inversion theorem that $$ \int_{\Bbb R} \frac{\mathrm{sin}(\pi\,a\,x)}{\pi\,a\,x} \,e^{-2i\pi\,x\,y}\,\mathrm d x = \frac{1}{a}\,\mathbf{1}_{[-a/2,a/2]}(y). $$

LL 3.14
  • 13,938