Difference between revisions of "HW4.1 David Hartmann ECE301Fall2008mboutin" - Rhea

Latest revision as of 11:04, 16 September 2013

Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"

For the 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^{j5\pi t} - \frac{1+3j}{2}e^{j5\pi t}\, $

$ x(t) = \frac{1}{j}e^{2*j\pi t} - \frac{1}{j}e^{-2*j\pi t} - \frac{1+3j}{2}e^{5*j\pi t} - \frac{1+3j}{2}e^{-5*j\pi t}\, $

$ \omega_0\, $ = $ \pi\, $ therefore k = 2,-2,5,-5

Applying the coefficients to get the $ a_k\, $

$ a_5 = \frac{-1-3j}{2}\, $ $ a_{-5} = \frac{-1-3j}{2}\, $

$ a_2 = \frac{1}{j}\, $ $ a_{-2} = \frac{-1}{j}\, $

For K \neq [2,-2,-5,5], $ a_k\, = 0 $

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