(→Solution) |
|||
| Line 1: | Line 1: | ||
| + | [[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"]] | ||
| + | ---- | ||
| + | |||
==Fourier Transform== | ==Fourier Transform== | ||
| Line 19: | Line 28: | ||
else <math>a_k</math> equals 0 | else <math>a_k</math> equals 0 | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Revision as of 10:50, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
Fourier Transform
Let $ x(t)=sin(\pi t) + cos(2\pi t) $
Remember that the formula for CT Fourier Series are:
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
Solution
$ x(t)= \frac{e^{\pi jt}+e^{-\pi jt}}{2j} + \frac{e^{2\pi jt}+e^{-2\pi jt}}{2} $
$ \omega_0 = \pi $
$ a_1=\frac{1}{2j} $
$ a_2=\frac{1}{2} $
else $ a_k $ equals 0
