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<math>X(t) = 6\frac{e^{j2t}+e^{-j2t}}{2} + 8\frac{e^{j4t}-e^{-j4t}}{2}\,</math><br> | <math>X(t) = 6\frac{e^{j2t}+e^{-j2t}}{2} + 8\frac{e^{j4t}-e^{-j4t}}{2}\,</math><br> | ||
<math>X(t) = 3e^{j2t}+3e^{-j2t} + 4e^{j4t}-4e^{-j4t}\,</math><br> | <math>X(t) = 3e^{j2t}+3e^{-j2t} + 4e^{j4t}-4e^{-j4t}\,</math><br> | ||
| − | With this expression we can conclude: | + | With this expression we can conclude:<br> |
<math>a_1 = 3\,</math><br> | <math>a_1 = 3\,</math><br> | ||
<math>a_{-1} = 3\,</math><br> | <math>a_{-1} = 3\,</math><br> | ||
Revision as of 20:08, 21 September 2008
CT Signal:
$ X(t) = 6\cos(2t) + 8\sin(4t)\, $
$ X(t) = 6\frac{e^{j2t}+e^{-j2t}}{2} + 8\frac{e^{j4t}-e^{-j4t}}{2}\, $
$ X(t) = 3e^{j2t}+3e^{-j2t} + 4e^{j4t}-4e^{-j4t}\, $
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.
