A simple Fourier Transformation I am a bit stuck with this small basic signal.

I have this

$$y(t)=\frac{\sin(200\pi\,t)}{\pi\,t}$$

and I want to take its Fourier Transformation. Obviously it looks like the sinc function. But that $200$ confuses me a lot.
In other words, the question is, how to rewrite that y(t), into a sinc(t) function , so i can use the FT table. –  RatzaJR


Elaborating on my comment, where I already answered this, we have that $y(t) = 200 \operatorname{sinc}(200t)$. Therefore, denoting the Fourier transform of a function $f$ by $\hat{f}$,
$$\hat{y}(\xi)=200 \frac{1}{200} \widehat{\operatorname{sinc}}\left( \frac{\xi}{200}\right) = \operatorname{rect} \left( \frac{\xi}{200}\right) = \begin{cases} 1 \qquad \text{if $|\xi|&lt; 100$} \\ 0 \qquad \text{if $|\xi| \geq 100$}\end{cases},$$
where the first equality follows from the basic fact that for real numbers $a,b$ where $a \neq 0$ and a function $f$, the Fourier transform of $a f(bx)$ is given by $\frac{a}{|b|} \hat{f}\left(\frac{\xi}{b} \right)$. You should really try to understand this step, otherwise you're toasted in your exam...
$$Y(\omega)=\int_{-\infty}^{\infty}y(t) \cos(\omega t) d\omega=2\int_0^{\infty}y(t) \cos(\omega t) d\omega=(sign(200\pi-\omega)+sign(200\pi+\omega))/2$$