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<math>x(t) = (5+3j)cos(4t) + (1+2j)sin(3t)</math> | <math>x(t) = (5+3j)cos(4t) + (1+2j)sin(3t)</math> | ||
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<math> = (5+3j)(\frac{e^{4jt} +e^{-4jt}}{2}) + (1+2j)(\frac{e^{j3t} + e^{-j3t}}{2j})</math> | <math> = (5+3j)(\frac{e^{4jt} +e^{-4jt}}{2}) + (1+2j)(\frac{e^{j3t} + e^{-j3t}}{2j})</math> | ||
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| + | <math> = \frac{5+3j}{2}e^{4jt} + \frac{5+3j}{2}e^{-4jt} + \frac{1+2j}{2j}e^{j3t} + \frac{1+2j}{2j}e^{-3jt}</math> | ||
Revision as of 13:04, 23 September 2008
The Formulas for Fourier Series
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $
where $ a_k=\frac{1}{T} \int_0^Tx(t)e^{-jk\omega_ot}dt $
Chosen Formula
$ x(t) = (5+3j)cos(4t) + (1+2j)sin(3t) $
Computation
First we want to compute the period (T) for this function. The period of sin and cos is 2pi, therefore the combined period is also 2pi.
Next we compute the coefficients. Since we are using sin and cos, we can use their equivalent formulas.
$ x(t) = (5+3j)cos(4t) + (1+2j)sin(3t) $
$ = (5+3j)(\frac{e^{4jt} +e^{-4jt}}{2}) + (1+2j)(\frac{e^{j3t} + e^{-j3t}}{2j}) $
$ = \frac{5+3j}{2}e^{4jt} + \frac{5+3j}{2}e^{-4jt} + \frac{1+2j}{2j}e^{j3t} + \frac{1+2j}{2j}e^{-3jt} $
