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CT Signal: | CT Signal: | ||
<br> | <br> | ||
| − | <math>X(t) = 6\cos(2\ | + | <math>X(t) = 6\cos(2\pi t) + 8\sin(4\pi t)\,</math><br> |
| − | <math>X(t) = 6\frac{e^{j2\ | + | <math>X(t) = 6\frac{e^{j2\pi t}+e^{-j2\pi t}}{2} + 8\frac{e^{j4\pi t}-e^{-j4\pi t}}{2}\,</math><br> |
| − | <math>X(t) = 3e^{j2\ | + | <math>X(t) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\,</math><br> |
With this expression we can conclude:<br> | With this expression we can conclude:<br> | ||
<math>a_1 = 3\,</math><br> | <math>a_1 = 3\,</math><br> | ||
Revision as of 06:41, 26 September 2008
CT Signal:
$ X(t) = 6\cos(2\pi t) + 8\sin(4\pi t)\, $
$ X(t) = 6\frac{e^{j2\pi t}+e^{-j2\pi t}}{2} + 8\frac{e^{j4\pi t}-e^{-j4\pi t}}{2}\, $
$ X(t) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\, $
With this expression we can conclude:
$ a_1 = 3\, $
$ a_{-1} = 3\, $
$ a_2 = 4\, $
$ a_{-2} = -4\, $
To calculate $ a_k\, $, we use this equation with $ \omega_0\, $ value as 2:
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}\, $
Then you will find the value of $ a_k \, $ is zero. -- For any k other than 1, -1, 2, -2
