(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> ...) |
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| − | <math>x(t) = \frac{1}{j}e^{j2\pi t} - \frac{1}{j}e^{-j2\pi t} - \frac{1+3j}{2}e^{ | + | <math>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}\,</math> |
| − | <math>x(t) = \frac{ | + | <math>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}\,</math> |
| − | <math>\omega_0\,</math> | + | <math>\omega_0\,</math> = <math>\pi\,</math> therefore k = 2,-2,5,-5 |
| − | + | Applying the coefficients to get the <math>a_k\,</math> | |
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| + | <math>a_5 = \frac{-1-3j}{2}\,</math> <math>a_{-5} = \frac{-1-3j}{2}\,</math> | ||
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| + | <math>a_2 = \frac{1}{j}\,</math> <math>a_{-2} = \frac{-1}{j}\,</math> | ||
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| − | + | For K \neq [2,-2,-5,5], <math>a_k\, = 0</math> | |
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| − | <math>a_k\, = 0</math> | + | |
Revision as of 16:01, 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^{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 $
