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<math>\frac{1}{2j}(e^{j4\pi t}-e^{-j4\pi t}) + \frac{1}{2}(e^{j3\pi t} + e^{-j3\pi t}) + e^{j2\pi t}</math> | <math>\frac{1}{2j}(e^{j4\pi t}-e^{-j4\pi t}) + \frac{1}{2}(e^{j3\pi t} + e^{-j3\pi t}) + e^{j2\pi t}</math> | ||
| + | |||
| + | The terms come out to be | ||
| + | |||
| + | <math>4, -4, 3, -3, and 2</math> | ||
| + | |||
| + | <math>k^4 = \frac{1}{2j}</math> | ||
| + | <math>k^-4 = \frac{1}{2j}</math> | ||
| + | <math>k^3 = \frac{1}{2}</math> | ||
| + | <math>k^-3 = \frac{1}{2}</math> | ||
| + | <math>k^2 = 1</math> | ||
| + | |||
| + | <math>A_</math> | ||
Revision as of 18:27, 25 September 2008
The Signal
mmm lets randomly take...
$ \sin4\pi t + \cos3\pi t + e^{j2\pi t} $
The Coefficients
Remeber... $ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $
$ a_k=\frac{1}{T} \int_0^Tx(t)e^{-jk\omega_ot}dt $
Going to conver the equation into signal that is all in exponentials.
$ \frac{1}{2j}(e^{j4\pi t}-e^{-j4\pi t}) + \frac{1}{2}(e^{j3\pi t} + e^{-j3\pi t}) + e^{j2\pi t} $
The terms come out to be
$ 4, -4, 3, -3, and 2 $
$ k^4 = \frac{1}{2j} $ $ k^-4 = \frac{1}{2j} $ $ k^3 = \frac{1}{2} $ $ k^-3 = \frac{1}{2} $ $ k^2 = 1 $
$ A_ $
