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While studying Discrete-time Fourier Transform. I found that $X[\omega]$ is periodic with $2\pi$. So when did the proof: $$ X[\omega + 2\pi] = \sum_ {n= +\infty}^{-\infty}x[n]e^{-j \omega n}e^{-j 2\pi n}$$

In the books it was written $e^{-j 2\pi n}$ =1
But I was confused that how it will be equal to 1, I tried to use Euler Equation on that but it's not possible to use that.

Anyone help me to tell how it will be equal to 1?

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1 Answer 1

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$e^{i\theta} = \cos(\theta) + i \sin(\theta)$. With $\theta = 2\pi \cdot k$ for some integer k, $\sin(\theta)$ will always vanish, and $\cos(\theta)$ will equal 1. In your case, $k = -n\cdot j$. Since both $n$ and $j$ are integers, their product is an integer.

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  • $\begingroup$ Yes but there is also term $n$, how we can vanish that while using Euler equation $\endgroup$ Nov 9, 2015 at 8:03
  • $\begingroup$ I edited my anwer; use the fact that $n$ is an integer and then also $j\cdot n$. $\endgroup$ Nov 9, 2015 at 8:05

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