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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier series]] | ||
| + | |||
| + | == Example of Computation of Fourier series of a CT SIGNAL == | ||
| + | A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]] | ||
| + | ---- | ||
CT Signal: | CT Signal: | ||
<br> | <br> | ||
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<br><math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}\,</math><br> | <br><math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}\,</math><br> | ||
Then you will find the value of <math>a_k \,</math> is zero. -- For any k other than 1, -1, 2, -2 | Then you will find the value of <math>a_k \,</math> is zero. -- For any k other than 1, -1, 2, -2 | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Revision as of 09:27, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
CT Signal:
$ X(t) = 6\cos(2\pi t) + 8\sin(4\pi t)\, $
$ X(t) = 6\frac{e^{j2\pi t}+e^{-j2\pi t}}{2} + 8\frac{e^{j4\pi t}-e^{-j4\pi t}}{2}\, $
$ X(t) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\, $
With this expression we can conclude:
$ a_1 = 3\, $
$ a_{-1} = 3\, $
$ a_2 = 4\, $
$ a_{-2} = -4\, $
To calculate $ a_k\, $, we use this equation with $ \omega_0\, $ value as 2:
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}\, $
Then you will find the value of $ a_k \, $ is zero. -- For any k other than 1, -1, 2, -2
