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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 :

  1. $\mathrm{rect}\left(\frac{t}{ \tau}\right) \xrightarrow{F.T.} \tau \mathrm{sinc}(\frac{\tau w}{2})$

  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:

enter image description here .

That is $H(w)=2\mathrm{rect}\left(\frac{w}{600\pi}\right)$

Can someone please tell me what is my mistake?

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1 Answer 1

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$$\mathrm{F.T.}\ \frac1\pi\mathrm{rect}(\frac{t}{600\pi}) = \frac1\pi600\pi\frac{\sin 300\pi w}{300\pi w}=2\frac{\sin 300\pi w}{\pi w}$$

$$\mathrm{F.T.}\ 2\frac{\sin 300\pi w}{\pi w}=2\pi\times\frac1\pi\mathrm{rect}(\frac{-w}{600\pi}) = 2\mathrm{rect}(\frac{w}{600\pi})$$

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  • $\begingroup$ Ohh yeah, I missed the $\pi$ while converting it back. Thanks a lot. I took $\tau = 600$ instead of $600\pi$ in the amplitude of sinc and that was the problem. $\endgroup$
    – paulplusx
    Commented Oct 25, 2018 at 15:03

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