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| − | [[Homework 4_ECE301Fall2008mboutin|<< Back to Homework 3]] | + | [[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"]] | ||
| + | ---- | ||
| + | |||
| + | |||
| + | [[Homework 4_ECE301Fall2008mboutin|<< Back to Homework 4]] | ||
| + | |||
| + | Homework 4 Ben Horst: [[HW4.1 Ben Horst _ECE301Fall2008mboutin| 4.1]] :: [[HW4.3 Ben Horst _ECE301Fall2008mboutin| 4.3]] :: [[HW4.5 Ben Horst _ECE301Fall2008mboutin| 4.5]] | ||
| + | ---- | ||
| + | |||
| + | ==Signal== | ||
| + | <math>x(t) = 2\sin(6t) + 4\cos(3t)</math> | ||
| + | |||
| + | ==Fourier Series== | ||
| + | <math> | ||
| + | x(t) = \sum_{k=- \infty }^ \infty a_ke^{jk\omega_0t} | ||
| + | </math> | ||
| + | |||
| + | By Euler's formula, we have: | ||
| + | <math> | ||
| + | x(t)=2({ e^{j 6t} + -e^{-j6t} \over 2j}) + 4({ e^{j3t} + e^{-j3t} \over 2 }) | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | x(t) = ({ e^{j 6t} + -e^{-j6t} \over j}) + 2e^{j3t} + 2e^{-j3t} | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | x(t) = -e^{2 j3t} + e^{-2 j3t} + 2e^{1 j3t} + 2e^{-1 j3t} | ||
| + | </math> | ||
| + | |||
| + | Ordering our k's to form a proper series: | ||
| + | |||
| + | <math> | ||
| + | x(t)= e^{(-2) j3t} + 2e^{(-1)j3t} + 2e^{(1) j3t} - e^{(2) j3t} | ||
| + | </math> | ||
| + | |||
| + | And making sure we don't forget about <math>a_0</math>: | ||
| + | |||
| + | <math> | ||
| + | x(t)= (1)e^{(-2) j3t} + (2)e^{(-1)j3t} + (0)e^{(0)j3t} + (2)e^{(1) j3t} + (-1)e^{(2) j3t} | ||
| + | </math> | ||
| + | |||
| + | ==Summary== | ||
| + | |||
| + | From the above math, we can determine all the coefficients: | ||
| + | <math> \ \ a_{-2} = 1; \ \ a_{-1} = 2; \ \ a_{0} = 0; \ \ a_1 = 2;\ \ a_{2} = -1 </math> | ||
| + | |||
| + | The fundamental period of the function is found from: <math>e^{j\omega_0}</math> where he period T = <math>{2\pi \over \omega_o}</math> | ||
| − | + | Thus, the fundamental period = <math> {2\pi \over 3} </math> | |
---- | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 09:54, 16 September 2013
Contents
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
Homework 4 Ben Horst: 4.1 :: 4.3 :: 4.5
Signal
$ x(t) = 2\sin(6t) + 4\cos(3t) $
Fourier Series
$ x(t) = \sum_{k=- \infty }^ \infty a_ke^{jk\omega_0t} $
By Euler's formula, we have: $ x(t)=2({ e^{j 6t} + -e^{-j6t} \over 2j}) + 4({ e^{j3t} + e^{-j3t} \over 2 }) $
$ x(t) = ({ e^{j 6t} + -e^{-j6t} \over j}) + 2e^{j3t} + 2e^{-j3t} $
$ x(t) = -e^{2 j3t} + e^{-2 j3t} + 2e^{1 j3t} + 2e^{-1 j3t} $
Ordering our k's to form a proper series:
$ x(t)= e^{(-2) j3t} + 2e^{(-1)j3t} + 2e^{(1) j3t} - e^{(2) j3t} $
And making sure we don't forget about $ a_0 $:
$ x(t)= (1)e^{(-2) j3t} + (2)e^{(-1)j3t} + (0)e^{(0)j3t} + (2)e^{(1) j3t} + (-1)e^{(2) j3t} $
Summary
From the above math, we can determine all the coefficients: $ \ \ a_{-2} = 1; \ \ a_{-1} = 2; \ \ a_{0} = 0; \ \ a_1 = 2;\ \ a_{2} = -1 $
The fundamental period of the function is found from: $ e^{j\omega_0} $ where he period T = $ {2\pi \over \omega_o} $
Thus, the fundamental period = $ {2\pi \over 3} $
