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From the above math, we can determine all the coefficients: | From the above math, we can determine all the coefficients: | ||
| − | <math> a_{-2} = 1; a_{-1} = -1; a_{0} = 1; a_{2} = -1 </math> | + | <math>\hspace a_{-2} = 1;\hspace a_{-1} = -1;\hspace a_{0} = 1;\hspace a_{2} = -1 </math> |
Revision as of 12:18, 25 September 2008
Homework 4 Ben Horst: 4.1 :: 4.2 :: 4.3:: 4.4:: 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} + 0 + 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: $ \hspace a_{-2} = 1;\hspace a_{-1} = -1;\hspace a_{0} = 1;\hspace a_{2} = -1 $
