(→Define a periodic CT signal and compute its Fourier series coefficients.) |
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| + | [[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"]] | ||
| + | ---- | ||
| + | |||
==Define a periodic CT signal and compute its Fourier series coefficients.== | ==Define a periodic CT signal and compute its Fourier series coefficients.== | ||
For CT, | For CT, | ||
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<math>x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}</math> | <math>x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}</math> | ||
| − | + | where | |
<math>a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt</math>. | <math>a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt</math>. | ||
| Line 10: | Line 19: | ||
Let the signal be | Let the signal be | ||
| − | y(t) = 2*sin(2t)+2*cos( | + | y(t) = 2*sin(2t)+2*cos(4t) |
| + | |||
| + | <math> y(t) = 2(\frac{e^{j2t} - e^{-j2t}}{2j}) + 2(\frac{e^{2j2t} + e^{-2j2t}}{2}) \!</math> | ||
| + | |||
| + | <math> a_1 = a_-1 = (\frac{1}{j})</math> | ||
| + | |||
| + | <math> a_2 = a_-2 = 1 </math> | ||
| + | |||
| + | a_k = 0 elsewhere | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 09:56, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
Define a periodic CT signal and compute its Fourier series coefficients.
For CT,
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
where
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
Let the signal be
y(t) = 2*sin(2t)+2*cos(4t)
$ y(t) = 2(\frac{e^{j2t} - e^{-j2t}}{2j}) + 2(\frac{e^{2j2t} + e^{-2j2t}}{2}) \! $
$ a_1 = a_-1 = (\frac{1}{j}) $
$ a_2 = a_-2 = 1 $
a_k = 0 elsewhere
