I have to take the fourier transform of the following:
$$h(t)= \begin{cases} \frac{2\sin(300\pi t)}{\pi t}, & t
> \ne 0 \\ 600, & t=0\end{cases}$$
So I used the two properties :
-
$\mathrm{rect}\left(\frac{t}{ \tau}\right) \xrightarrow{F.T.} \tau \mathrm{sinc}(\frac{\tau w}{2})$
-
$X(t) \xrightarrow{F.T.} 2\pi x(-w)$ (duality)
Here $F.T.$ means Fourier transform and $x(.)$ and $X(.)$ are fourier transform pairs and $\mathrm{rect}$ is :
$$\mathrm{rect}\left(\frac{t}{ \tau}\right) =\begin{cases}1, & |t| \leq \frac {\tau}{2}\\ 0, & otherwise \end{cases}$$
using them I got the $H(w)=2\pi\mathrm{rect}\left(\frac{w}{600\pi}\right)$
But the given answer is:
.
That is $H(w)=2\mathrm{rect}\left(\frac{w}{600\pi}\right)$
Can someone please tell me what is my mistake?
Best Answer
$$\mathrm{F.T.}\ \frac1\pi\mathrm{rect}(\frac{t}{600\pi}) = \frac1\pi600\pi\frac{\sin 300\pi w}{300\pi w}=2\frac{\sin 300\pi w}{\pi w}$$
$$\mathrm{F.T.}\ 2\frac{\sin 300\pi w}{\pi w}=2\pi\times\frac1\pi\mathrm{rect}(\frac{-w}{600\pi}) = 2\mathrm{rect}(\frac{w}{600\pi})$$