(→DT signal & its Fourier Coefficients) |
(→DT signal & its Fourier Coefficients) |
||
| Line 29: | Line 29: | ||
<math>\ a_1 = \frac{5}{j2\sqrt{2}}+ \frac{5}{2\sqrt{2}} </math> | <math>\ a_1 = \frac{5}{j2\sqrt{2}}+ \frac{5}{2\sqrt{2}} </math> | ||
| − | <math>\ | + | <math>\ a_-1 = \frac{5}{2\sqrt{2}} - \frac{5}{j2\sqrt{2}} </math> |
Revision as of 17:03, 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})}) $
We then proceed to compute:
$ \ 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}}) $
Knowing the following,
$ \ 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}} $
We substitute:
$ \ 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}}) ) $
And simplify:
$ \ 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)} $
$ \ x[n] = (\frac{5}{j2\sqrt{2}}+ \frac{5}{2\sqrt{2}}) e^{j (3\pi n)} + (\frac{5}{2\sqrt{2}} - \frac{5}{j2\sqrt{2}})e^{-j (3\pi n)} $
$ \ a_1 = \frac{5}{j2\sqrt{2}}+ \frac{5}{2\sqrt{2}} $
$ \ a_-1 = \frac{5}{2\sqrt{2}} - \frac{5}{j2\sqrt{2}} $
