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<math>\begin{align}
 
<math>\begin{align}
 
x[n]= \cos \left( \frac{2 \pi }{500} n  \right)
 
x[n]= \cos \left( \frac{2 \pi }{500} n  \right)
\\ = \frac{e^{j\frac{2\pi}{500}n}}{2}+\frac{e^{-j\frac{2\pi}{500}n}}{2}
+
\\ = \frac{e^{j\frac{2\pi}{500}n + e^{-j\frac{2\pi}{500}n}}{2}
 
\end{align}</math>
 
\end{align}</math>
 
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Revision as of 18:21, 6 September 2011

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} $


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