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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"]] | ||
| + | ---- | ||
| + | |||
== The Formulas for Fourier Series == | == The Formulas for Fourier Series == | ||
<math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\,</math> | <math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\,</math> | ||
| Line 7: | Line 16: | ||
<math>x(t) = (5+3j)cos(4t) + (1+2j)sin(3t)</math> | <math>x(t) = (5+3j)cos(4t) + (1+2j)sin(3t)</math> | ||
| + | |||
| + | == Computation == | ||
| + | First we want to compute the period (T) for this function. The period of sin and cos is <math>2\pi</math>, therefore the combined period is also <math>2\pi</math>. | ||
| + | |||
| + | Next we compute the coefficients. Since we are using sin and cos, we can use their equivalent formulas. | ||
| + | |||
| + | <math>x(t) = (5+3j)cos(4t) + (1+2j)sin(3t)</math> | ||
| + | |||
| + | |||
| + | <math> = (5+3j)(\frac{e^{4jt} +e^{-4jt}}{2}) + (1+2j)(\frac{e^{j3t} - e^{-j3t}}{2j})</math> | ||
| + | |||
| + | <math> = \frac{5+3j}{2}e^{4jt} + \frac{5+3j}{2}e^{-4jt} + \frac{1+2j}{2j}e^{j3t} - \frac{1+2j}{2j}e^{-3jt}</math> | ||
| + | |||
| + | multiplying by complex conjugate we get: | ||
| + | |||
| + | <math> = \frac{5+3j}{2}e^{4jt} + \frac{5+3j}{2}e^{-4jt} + (1-\frac{j}{2})e^{j3t} - (1-\frac{j}{2})e^{-j3t}</math> | ||
| + | |||
| + | since <math>\omega_0=\frac{2\pi}{T}</math> and <math>T=2\pi</math> then <math>\omega_0=1</math> | ||
| + | |||
| + | Therefore for the first term k=4 and for the second term k=-4. Likewise, for the third and fourth terms k=3 and k=-3 respectively. Therefore our coefficients are: | ||
| + | |||
| + | <math>a_3 = a_{-3} = \frac{5+3j}{2}</math> | ||
| + | |||
| + | And <math>a_4 = a_{-4} = (1-\frac{j}{2})</math> | ||
| + | |||
| + | And <math>a_k = 0</math> else | ||
| + | ---- | ||
| + | Go back to [[Homework 4_ECE301Fall2008mboutin]] | ||
| + | |||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 09:57, 16 September 2013
Contents
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
The Formulas for Fourier Series
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $
where $ a_k=\frac{1}{T} \int_0^Tx(t)e^{-jk\omega_ot}dt $
Chosen Formula
$ x(t) = (5+3j)cos(4t) + (1+2j)sin(3t) $
Computation
First we want to compute the period (T) for this function. The period of sin and cos is $ 2\pi $, therefore the combined period is also $ 2\pi $.
Next we compute the coefficients. Since we are using sin and cos, we can use their equivalent formulas.
$ x(t) = (5+3j)cos(4t) + (1+2j)sin(3t) $
$ = (5+3j)(\frac{e^{4jt} +e^{-4jt}}{2}) + (1+2j)(\frac{e^{j3t} - e^{-j3t}}{2j}) $
$ = \frac{5+3j}{2}e^{4jt} + \frac{5+3j}{2}e^{-4jt} + \frac{1+2j}{2j}e^{j3t} - \frac{1+2j}{2j}e^{-3jt} $
multiplying by complex conjugate we get:
$ = \frac{5+3j}{2}e^{4jt} + \frac{5+3j}{2}e^{-4jt} + (1-\frac{j}{2})e^{j3t} - (1-\frac{j}{2})e^{-j3t} $
since $ \omega_0=\frac{2\pi}{T} $ and $ T=2\pi $ then $ \omega_0=1 $
Therefore for the first term k=4 and for the second term k=-4. Likewise, for the third and fourth terms k=3 and k=-3 respectively. Therefore our coefficients are:
$ a_3 = a_{-3} = \frac{5+3j}{2} $
And $ a_4 = a_{-4} = (1-\frac{j}{2}) $
And $ a_k = 0 $ else
Go back to Homework 4_ECE301Fall2008mboutin
