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Discrete-time Fourier transform computation
Compute the discrete-time Fourier transform of the following signal:
$ x[n]= \cos \left( \frac{2 \pi }{500} n \right) $
(Write enough intermediate steps to fully justify your answer.)
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Answer 1
$ \mathcal{F}(x[n]) = \mathcal{F}(cos(\frac{2\pi}{500}n)) = \mathcal{F}(\frac{ e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2}) =\frac{1}{2}( \mathcal{F}(e^{j\frac{2\pi}{500}n})+\mathcal{F}(e^{-j\frac{2\pi}{500}n})) $
$ =\frac{1}{2}( \pi\sum_{l=-\infty}^{+\infty}\delta(w-\frac{2\pi}{500}-2\pi l) + \pi\sum_{l=- \infty}^{+\infty}\delta(w+\frac{2\pi}{500}-2\pi l) ) $
$ =\frac{\pi}{2} \sum_{l=-\infty}^{+\infty}[ \delta(w-\frac{2\pi}{500}-2\pi l)+\delta(w+\frac{2\pi}{500}-2\pi l) ] $
Answer 2
$ x[n] = \int_{-\pi}^{\pi} \mathcal{X} (w)e^{j\omega n} dw $
The input x[n] can can be written in the exponential form.
$ x[n] = cos(\frac{2\pi}{500}n) = \frac{e^{j\frac{2\pi}{500}n} + e^{-j\frac{2\pi}{500}n}}{2} $
In order for the input x[n] to have such a value,
$ \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500}) $
Answer 3
$ x[n] = \frac{2\pi}{500}n = \frac{e^{j\frac{2\pi}{500}n}}{2}+\frac{e^{-j\frac{2\pi}{500}n}}{2} $
$ \mathcal{X} (\omega) = \pi (\delta(\omega - \frac{2\pi}{500}) + \delta(\omega + \frac{2\pi}{500})) $
Answer 4
$ \begin{align} x[n]= \cos \left( \frac{2 \pi }{500} n \right) \\ = \frac{ e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2} \end{align} $
$ 0 < \frac{2\pi}{500}n < \pi $
$ -\pi < -\frac{2\pi}{500}n < \pi $
consider $ -\pi < \omega < \pi $
$ \begin{align}\mathcal{F}[x[n]] = 2\pi * \frac{1}{2} [\delta(\omega - \frac{2\pi}{500}n) + \delta(\omega + \frac{2\pi}{500}n)] \end{align} $
