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I have a follow up to this question regarding the derivation of a sinc function from a complex exponential, because I haven't been able to figure out how the introduction of a scaling factor will affect the Euler's formula. Can anyone please help?

The window function is: \begin{align}g(t) = \begin{cases} 1, & \text{if |t| < $\frac{1}{2a}$} \\ 0, & \text{otherwise} \end{cases} \end{align}

So the Fourier transform is:

\begin{align} \int_{-1/2a}^{1/2a}e^{-j\omega t} dt &= \frac{1}{j\omega}[e^{j\omega/2a}-e^{-j\omega/2a}]\\ &= \frac{1}{j\omega}[\cos(\frac{\omega}{2a})+j\sin(\frac{\omega}{2a})-\cos(\frac{\omega}{2a})+j\sin(\frac{\omega}{2a})]\\ &= \frac{1}{j\omega}[2j\sin(\frac{\omega}{2a})]\\ &= \frac{\sin(\frac{\omega}{2a})}{\frac{w}{2}} \end{align}

How do I get to the sinc function?

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  • $\begingroup$ your question is not clear the linked question has already an accepted answer what do you want to know? $\endgroup$ – Deepesh Meena Aug 17 '18 at 21:46
  • $\begingroup$ $ \frac{\sin(\frac{\omega}{2a})}{\frac{w}{2}}= \frac{1}{a}\frac{\sin(\frac{\omega}{2a})}{\frac{w}{2a}} = \frac{1}{a}\text{sinc}\frac{\omega}{2a}$ $\endgroup$ – N74 Aug 17 '18 at 21:49
  • $\begingroup$ @James The linked question only had the answer to that specific problem. If I work it out in the exact same way I don't get to a step in which the $\frac{\sin(x)}{x}$ form is available. $\endgroup$ – Sai Vemula Aug 18 '18 at 0:56

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