Difference between revisions of "HW4-1 Steve Anderson ECE301Fall2008mboutin" - Rhea

Revision as of 09:28, 16 September 2013

Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"

CT signal:

$ x(t) = 4\sin(5 \pi t) - (2 + j)\cos(3 \pi t)\, $

$ x(t) = 4 * \frac{e^{j5\pi t} - e^{-j5\pi t}}{2j} - (2+j)*\frac{e^{j3\pi t} + e^{-j3\pi t}}{2}\, $

$ x(t) = \frac{2}{j}e^{j5\pi t} - \frac{2}{j}e^{-j5\pi t} - \frac{2+j}{2}e^{j3\pi t} - \frac{2+j}{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 $

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