Question

Can someone please explain to me how to find the Fourier Transform of sinc(2t)?

Answer 1

what is sinc(t) sine integral?

Answer 2

I don't really know :/

Answer 3

I know sinc(t) = \[\sin (\pi t)/\pi t\]

Answer 4

use Dirichlet integral

Answer 5

http://en.wikipedia.org/wiki/Dirichlet_integral

Answer 6

I still don't understand... the answer is apparently \[(\pi/2) rect(\omega/4)\]

Answer 7

I don't know how they obtained it, I tried using the formula.

Answer 8

http://www.wolframalpha.com/input/?i=Fourier+Transform+Sinc%5BPi+t%5D
\[
\begin{align*}
f(\omega) &= \frac{1}{\sqrt{2 \pi }} \int_{-\infty}^\infty \frac{\sin (\pi t)}{(\pi t)} e^{-i \omega t}dt \\
&= \frac{1}{\sqrt{2 \pi }} \left( \int_{-\infty}^0 \frac{\sin (\pi t)}{(\pi t)} e^{-i \omega t}dt + \int_0^{\infty} \frac{\sin (\pi t)}{(\pi t)} e^{-i \omega t}dt\right )\\
&= \frac{1}{\sqrt{2 \pi }} \left( \int_0^{\infty} \frac{\sin (\pi t)}{(\pi t)} e^{i \omega t}dt + \int_0^{\infty} \frac{\sin (\pi t)}{(\pi t)} e^{-i \omega t}dt\right )\\
&= \frac{1}{\sqrt{2 \pi }} \int_0^{\infty} \frac{\sin (\pi t)}{(\pi t)} (e^{i \omega t} +e^{i \omega t}) dt\\
&= \frac{1}{\sqrt{2 \pi }} \int_0^{\infty} \frac{\sin (\pi t)}{(\pi t)} (2 \cos(\omega t)) dt \\
&= \frac{2}{\sqrt{2 \pi }}\int_0^{\infty} \frac{\sin (\pi t)}{(\pi t)} (\cos(\omega t)) dt
\end{align*}
\]

You can plug this integral into WA and solve for it.
There are many ways you can evaluate this integral

Answer 9

\[ \begin{align*}
\int_0^{\infty} \frac{\sin (\pi t)}{(\pi t)} \cos(\omega t) dt &= \int_0^\infty \frac{\sin ((\pi + \omega)t) +\sin ((\pi - \omega)t) }{\pi t} dt\\
&= \int_0^\infty \frac{\sin ((\pi + \omega)t)}{ \pi t}dt + \int_0^\infty \frac{\sin ((\pi - \omega)t)}{ \pi t}dt\\
\end{align*} \]
\[ \int_0^\infty \frac{\sin (kx)}{x}dx = \text{sign}(k) \frac \pi 2 \]