(New page: == Signal == <math> x(t) = 5cos(2t) + 3sin(4t)\!</math> == Fourier Series == <math> x(t) = 5(\frac{e^{j2t} + e^{-j2t}}{2}) + 3(\frac{e^{j4t} - e^{-j4t}}{2j}) \!</math> We take <math>...) |
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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"]] | ||
| + | ---- | ||
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== Signal == | == Signal == | ||
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According to the Formula, <math> a_k = 0\!</math> whenever <math> K \neq \pm2, \pm 4\!</math> | According to the Formula, <math> a_k = 0\!</math> whenever <math> K \neq \pm2, \pm 4\!</math> | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 09:58, 16 September 2013
Contents
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
Signal
$ x(t) = 5cos(2t) + 3sin(4t)\! $
Fourier Series
$ x(t) = 5(\frac{e^{j2t} + e^{-j2t}}{2}) + 3(\frac{e^{j4t} - e^{-j4t}}{2j}) \! $
We take $ \omega_0 = 2\! $
$ x(t) = \frac{5}{2}e^{j2t} + \frac{5}{2}e^{-j2t} + \frac{3}{2j}e^{2j2t} - \frac{3}{2j}e^{-2j2t}\! $
Fourier Coefficients
From the Fourier Series, we determine the coefficients to be:
$ a_1 = a_{-1} = \frac{5}{2}\! $
$ a_2 = a_{-2} = \frac{3}{2j}\! $
Other Coefficients
$ w_0 = 2\! $
$ x(t) = \sum^{\infty}_{k = -\infty} a_ke^{jKw_0t}\! $
where $ a_k = \frac{1}{T}\int{0}^{T} x(t)e^{-jKw_0t}dt\! $
According to the Formula, $ a_k = 0\! $ whenever $ K \neq \pm2, \pm 4\! $
