(New page: == The signal used == In this example, I am using the signal <math>x(t)=cos(2t) + 2sin(2t)\!</math>. == Coefficient Breakdown == To get the coefficients of a simple signal like this, we ...) |
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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier series]] | ||
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| + | == Example of Computation of Fourier series of a CT SIGNAL == | ||
| + | A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]] | ||
| + | ---- | ||
== The signal used == | == The signal used == | ||
In this example, I am using the signal <math>x(t)=cos(2t) + 2sin(2t)\!</math>. | In this example, I am using the signal <math>x(t)=cos(2t) + 2sin(2t)\!</math>. | ||
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<math>a_{2,-2}=\frac{1}{j}\!</math> | <math>a_{2,-2}=\frac{1}{j}\!</math> | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Revision as of 09:46, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
The signal used
In this example, I am using the signal $ x(t)=cos(2t) + 2sin(2t)\! $.
Coefficient Breakdown
To get the coefficients of a simple signal like this, we have to expand it.
$ x(t)=\frac{e^{2jt}}{2}+ \frac{e^{-2jt}} {2}+ 2\frac{e^{2jt}} {2j}+ 2\frac{e^{-2jt}} {2j}! $
$ \omega_0\! $ is $ 2 \! $ in this example.
$ a_{1,-1}=\frac{1}{2}\! $
$ a_{2,-2}=\frac{1}{j}\! $
