Revision as of 15:43, 26 September 2008 by Cdleon (Talk)

DT signal & its Fourier Coefficients

$ \ x[n] = 5sin(3 \pi n + \frac{\pi}{4}) $

Knowing its Fourier series is

$ \ x[n] = \frac{5}{2j} (e^{j (3\pi n + \frac{\pi}{4})}-e^{-j (3\pi n + \frac{\pi}{4})}) $

$ \ x[n] = \frac{5}{2j} (e^{j (3\pi n)} e^{j\frac{\pi}{4}}-e^{-j (3\pi n)}e^{j\frac{\pi}{4}}) $

$ \ e^{j {\pi \over 4}} = {1 \over \sqrt{2}} + j{1 \over \sqrt{2}} $

$ \ e^{-j{\pi \over 4}} = {1 \over \sqrt{2}} - j{1 \over \sqrt{2}} $

$ \ x[n] = \frac{5}{2j} (e^{j (3\pi n)} () -e^{-j (3\pi n)} () ) $

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