(started math.) |
m (→Fourier Series) |
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By Euler's formula, we have: | By Euler's formula, we have: | ||
| + | <math> | ||
x(t)=2({ e^{j 6t} + -e^{-j6t} \over 2j}) + 4({ e^{j 6t} + e^{-j6t} \over 2}) | x(t)=2({ e^{j 6t} + -e^{-j6t} \over 2j}) + 4({ e^{j 6t} + e^{-j6t} \over 2}) | ||
| + | </math> | ||
Revision as of 11:45, 25 September 2008
Homework 4 Ben Horst: 4.1 :: 4.2 :: 4.3:: 4.4:: 4.5
Homework 4 Ben Horst: 4.1 :: 4.2 :: 4.3:: 4.4:: 4.5
Signal
x(t) = 2sin(6t) + 4cos(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^{j 6t} + e^{-j6t} \over 2}) $
