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$\textrm{DTFT}(\delta[n]) =1$, but $\textrm{IDTFT(1)} = \frac{\sin(\pi n)}{\pi n}$. Why it is not equal to the unit impulse $\delta[n]$?

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The IDTFT of $X(e^{j\omega})=1$ is indeed

$$x[n]=\frac{\sin(n\pi)}{n\pi}\tag{1}$$

Now, what happens for indices $n\neq 0$? As it turns out, you can safely rewrite $(1)$ as

$$x[n]=\delta[n]\tag{2}$$

where $\delta[n]$ is the discrete-time unit impulse. (HINT: think about where the zeros of $\sin(x)$ are).

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  • $\begingroup$ Thanks I have got it now $\endgroup$
    – user32834
    Commented Sep 13, 2019 at 1:37

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