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} $
