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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"]] | ||
| + | ---- | ||
| + | |||
Lets define the signal | Lets define the signal | ||
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So, we get the coefficients: | So, we get the coefficients: | ||
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| + | <math>\ a_{0} = 0 </math> | ||
<math>\ a_{1} = \frac{1+2j}{2} </math> | <math>\ a_{1} = \frac{1+2j}{2} </math> | ||
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<math>\ a_{-2} = - \frac{5}{2j}</math> | <math>\ a_{-2} = - \frac{5}{2j}</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"
Lets define the signal
$ \ x(t) = (1+2j)cos(t)+5sin(4t) $
Knowing that its Fourier series is
$ \ x(t) = (1+2j)(\frac{e^{jt} + e^{-jt}}{2}) + 5(\frac{e^{j4t} - e^{-j4t}}{2j}) $
We simplify
$ \ x(t) = \frac{1+2j}{2} e^{jt} + \frac{1+2j}{2} e^{-jt} + \frac{5}{2j}e^{j4t} - \frac{5}{2j}e^{-j4t} $
So, we get the coefficients:
$ \ a_{0} = 0 $
$ \ a_{1} = \frac{1+2j}{2} $
$ \ a_{-1} = \frac{1+2j}{2} $
$ \ a_{2} = \frac{5}{2j} $
$ \ a_{-2} = - \frac{5}{2j} $
