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| − | CT signal: | + | [[Category:problem solving]] |
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier series]] | ||
| + | [[Category:signals and systems]] | ||
| + | |||
| + | == Example of Computation of Fourier series of a CT SIGNAL == | ||
| + | A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]] | ||
| + | ---- | ||
| + | For the CT signal: | ||
<math>x(t) = 2\sin(2 \pi t) - (1 + 3j)\cos(5 \pi t)\,</math> | <math>x(t) = 2\sin(2 \pi t) - (1 + 3j)\cos(5 \pi t)\,</math> | ||
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For K \neq [2,-2,-5,5], <math>a_k\, = 0</math> | For K \neq [2,-2,-5,5], <math>a_k\, = 0</math> | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 10:04, 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 $
