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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"]] | ||
| + | ---- | ||
| + | |||
== Equations == | == Equations == | ||
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<br> | <br> | ||
<br> | <br> | ||
| − | The same method can be used to find each value of <math>a_k\!</math>. To compute the rest of the values I'll use complex exponential identities | + | The same method can be used to find each value of <math>a_k\!</math>. To compute the rest of the values I'll use complex exponential identities: |
<br> | <br> | ||
<br> | <br> | ||
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<math>x(t)=4sin(3t)+(1+2j)cos(2t)\!</math> | <math>x(t)=4sin(3t)+(1+2j)cos(2t)\!</math> | ||
<br> | <br> | ||
| − | <math>=\frac{ | + | <br> |
| + | <math>=\frac{2}{j}(e^{j3t}-e^{-j3t})+\frac{1+2j}{2}(e^{j2t}+e^{-j2t})</math> | ||
| + | <br> | ||
| + | <br> | ||
| + | <math>=\frac{2}{j}e^{j3t}-\frac{2}{j}e^{-j3t}+(\frac{1+2j}{2})e^{j2t}+(\frac{1+2j}{2})e^{-j2t}</math> | ||
| + | <br> | ||
| + | <br> | ||
| + | The coefficients can now easily be identified: | ||
| + | <br> | ||
| + | <br> | ||
| + | <math>a_{3}=\frac{2}{j}</math> | ||
| + | <br> | ||
| + | <math>a_{-3}=-\frac{2}{j}</math> | ||
| + | <br> | ||
| + | <math>a_{2}=\frac{1+2j}{2}</math> | ||
| + | <br> | ||
| + | <math>a_{-2}=\frac{1+2j}{2}</math> | ||
| + | <br> | ||
| + | <br> | ||
| + | For all other values of <math>_k\!</math>, <math>a_k=0\!</math> | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 10:55, 16 September 2013
Contents
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
Equations
Fourier series of x(t):
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
Signal Coefficients:
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
Defined Signal
$ x(t)=4sin(3t)+(1+2j)cos(2t)\! $
Solution
The fundamental period $ T\! $ is $ 2\pi\! $. Thus we use the equation $ \omega_0=\frac{2\pi}{T}\! $ to find $ \omega_0=1\! $
To find the value of $ a_0\! $ we simply plug and chug:
$ a_0=\frac{1}{2\pi}\int_0^{2\pi}[4sin(3t)+(1+2j)cos(2t)]e^{0}dt $
$ =\frac{2}{\pi}\int_0^{2\pi}sin(3t)dt+\frac{1+2j}{2\pi}\int_0^{2\pi}cos(2t)dt $
$ =\frac{-2}{3\pi}[cos(3t)]_0^{2\pi}+\frac{1+2j}{4\pi}[sin(2t)]_0^{2\pi} $
$ =\frac{-2}{3\pi}[cos(6\pi)-cos(0)]+\frac{1+2j}{4\pi}[(sin(4\pi)-sin(0)] $
$ =\frac{-2}{3\pi}[0]+\frac{1+2j}{4\pi}[0] $
$ =0\! $
The same method can be used to find each value of $ a_k\! $. To compute the rest of the values I'll use complex exponential identities:
$ x(t)=4sin(3t)+(1+2j)cos(2t)\! $
$ =\frac{2}{j}(e^{j3t}-e^{-j3t})+\frac{1+2j}{2}(e^{j2t}+e^{-j2t}) $
$ =\frac{2}{j}e^{j3t}-\frac{2}{j}e^{-j3t}+(\frac{1+2j}{2})e^{j2t}+(\frac{1+2j}{2})e^{-j2t} $
The coefficients can now easily be identified:
$ a_{3}=\frac{2}{j} $
$ a_{-3}=-\frac{2}{j} $
$ a_{2}=\frac{1+2j}{2} $
$ a_{-2}=\frac{1+2j}{2} $
For all other values of $ _k\! $, $ a_k=0\! $
