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By Euler's formula, we have: | By Euler's formula, we have: | ||
<math> | <math> | ||
| − | x(t)=2({ e^{j 6t} + -e^{-j6t} \over 2j}) + 4({ e^{j3t} + e^{-j3t} \over 2}) | + | x(t)=2({ e^{j 6t} + -e^{-j6t} \over 2j}) + 4({ e^{j3t} + e^{-j3t} \over 2 }) |
</math> | </math> | ||
<math> | <math> | ||
| − | x(t)=({ e^{j 6t} + -e^{-j6t} \over j}) + 2e^{j3t} + 2e^{-j3t} | + | x(t) = ({ e^{j 6t} + -e^{-j6t} \over j}) + 2e^{j3t} + 2e^{-j3t} |
</math> | </math> | ||
<math> | <math> | ||
| − | x(t)= -e^{2 j3t} + e^{-2 j3t} | + | x(t) = -e^{2 j3t} + e^{-2 j3t} + 2e^{1 j3t} + 2e^{-1 j3t} |
</math> | </math> | ||
Revision as of 12:33, 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: $ \ \ a_{-2} = 1; \ \ a_{-1} = -1; \ \ a_{0} = 1; \ \ a_{2} = -1 $
