(→DT signal & its Fourier Coefficients) |
(→DT signal & its Fourier Coefficients) |
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<math>\ x[n] = \frac{5}{2j} (e^{j (3\pi n + \frac{\pi}{4})}-e^{-j (3\pi n + \frac{\pi}{4})}) </math> | <math>\ x[n] = \frac{5}{2j} (e^{j (3\pi n + \frac{\pi}{4})}-e^{-j (3\pi n + \frac{\pi}{4})}) </math> | ||
| + | |||
| + | <math>\ x[n] = \frac{5}{2j} (e^{j (3\pi n)} e^{\frac{\pi}{4})}-e^{-j (3\pi n + \frac{\pi}{4})}) </math> | ||
:<math>e^{j {\pi \over 4}} = {1 \over \sqrt{2}} + j{1 \over \sqrt{2}} </math><br> | :<math>e^{j {\pi \over 4}} = {1 \over \sqrt{2}} + j{1 \over \sqrt{2}} </math><br> | ||
:<math>e^{-j{\pi \over 4}} = {1 \over \sqrt{2}} - j{1 \over \sqrt{2}} </math> | :<math>e^{-j{\pi \over 4}} = {1 \over \sqrt{2}} - j{1 \over \sqrt{2}} </math> | ||
Revision as of 16:36, 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^{\frac{\pi}{4})}-e^{-j (3\pi n + \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}} $
