(DT signal & its Fourier Coefficients)
(DT signal & its Fourier Coefficients)
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<math>\ x[n] = \frac{5}{j2\sqrt{2}} e^{j (3\pi n)} + \frac{5}{2\sqrt{2}}e^{j (3\pi n)}  -    \frac{5}{j2\sqrt{2}}e^{-j (3\pi n)}  +  \frac{5}{2\sqrt{2}}e^{-j (3\pi n)}
 
<math>\ x[n] = \frac{5}{j2\sqrt{2}} e^{j (3\pi n)} + \frac{5}{2\sqrt{2}}e^{j (3\pi n)}  -    \frac{5}{j2\sqrt{2}}e^{-j (3\pi n)}  +  \frac{5}{2\sqrt{2}}e^{-j (3\pi n)}
 +
 +
<math>\ x[n] = (\frac{5}{j2\sqrt{2}}+ \frac{5}{2\sqrt{2}}) e^{j (3\pi n)}

Revision as of 16:57, 26 September 2008

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)} ({1 \over \sqrt{2}} + j{1 \over \sqrt{2}}) -e^{-j (3\pi n)} ({1 \over \sqrt{2}} - j{1 \over \sqrt{2}}) ) $

$ \ x[n] = \frac{5}{j2\sqrt{2}} e^{j (3\pi n)} + \frac{5}{2\sqrt{2}}e^{j (3\pi n)} - \frac{5}{j2\sqrt{2}}e^{-j (3\pi n)} + \frac{5}{2\sqrt{2}}e^{-j (3\pi n)} <math>\ x[n] = (\frac{5}{j2\sqrt{2}}+ \frac{5}{2\sqrt{2}}) e^{j (3\pi n)} $

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