(New page: CT signal: <math>x(t) = 2\sin(2 \pi t) - (1 + 3j)\cos(5 \pi t)\,</math> <math>x(t) = 2 * \frac{e^{j2\pi t} - e^{-j2\pi t}}{2j} - (1 + 3j)*\frac{e^{j5\pi t} + e^{-j5\pi t}}{2}\,</math> ...) |
(No difference)
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Revision as of 15:51, 26 September 2008
CT signal:
$ x(t) = 2\sin(2 \pi t) - (1 + 3j)\cos(5 \pi t)\, $
$ x(t) = 2 * \frac{e^{j2\pi t} - e^{-j2\pi t}}{2j} - (1 + 3j)*\frac{e^{j5\pi t} + e^{-j5\pi t}}{2}\, $
$ x(t) = \frac{1}{j}e^{j2\pi t} - \frac{1}{j}e^{-j2\pi t} - \frac{1+3j}{2}e^{j3\pi t} - \frac{1+3j}{2}e^{j3\pi t}\, $
$ x(t) = \frac{2}{j}e^{5*j\pi t} - \frac{2}{j}e^{-5*j\pi t} - \frac{2+j}{2}e^{3*j\pi t} - \frac{2+j}{2}e^{-3*j\pi t}\, $
$ \omega_0\, $ ends up being $ \pi\, $ for this signal
So because of that, in my last step, i was able to determine what value of K's i had. (k = 3,-3,5,-5)
So then you just take the coefficients of those terms to get the $ a_k\, $
Therefore:
$ a_3 = \frac{-2-j}{2}\, $
$ a_{-3} = \frac{-2-j}{2}\, $
$ a_5 = \frac{2}{j}\, $
$ a_{-5} = \frac{-2}{j}\, $
For every other k:
$ a_k\, = 0 $
