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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"]] | ||
| + | ---- | ||
Let the input signal be | Let the input signal be | ||
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<math>a_-1 = \frac{j}{2}</math> | <math>a_-1 = \frac{j}{2}</math> | ||
| − | < | + | <math>a_2 = 1 </math> |
| + | |||
| + | <math>a_-2 = 1 </math> | ||
| − | < | + | <math>a_k = 0</math> |
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 09:59, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
Let the input signal be
$ x(t)=sint +2cos2t $
= $ \frac{e^{jt}-e^{-jt}}{2i} $+$ 2*\frac{e^{2jt}+{e^{-2jt}}}{2} $
= $ \frac{-j{e^{jt}}}{2}+\frac{j{e^{-jt}}}{2}+{e^{2jt}}+{e^{-2jt}} $
The coefficients are as follows:
$ a_1 = \frac{-j}{2} $
$ a_-1 = \frac{j}{2} $
$ a_2 = 1 $
$ a_-2 = 1 $
$ a_k = 0 $
