(→Solution) |
(→Solution) |
||
| Line 43: | Line 43: | ||
| − | <math>a_0=\frac{1}{\pi}[sin(2t)]_0^{2\pi} | + | <math>a_0=\frac{1}{\pi}[sin(2t)]_0^{2\pi}-\frac{j}{2\pi}[cos(3t)]_0^{2\pi}</math> |
Revision as of 17:49, 24 September 2008
The signal
The signal I chose to use is as follows:
$ x(t) = 4cos(2t) + (3j)sin(3t)\! $
The fundamental period, denoted as $ T\! $, of this signal is $ 2\pi\! $. The fundamental frequency, denoted $ \omega_0\! $, is defined as:
$ \omega_0 = \frac{T}{2\pi}\! $
The value of this is $ \frac{2\pi}{2\pi}\! $, which coincidently, by no planning of mine, turns out to be $ 1\! $.
Solution
We know that the equation for signal coefficients is as follows:
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
And the equation for fourier series of a function is as follows:
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
We first put our signal into the first equation, and we get this monster:
$ a_0=\frac{1}{2\pi}\int_0^{2\pi}[4cos(2t) + (3j)sin(3t)]e^{0}dt $
We now solve.
$ a_0=\frac{2}{\pi}\int_0^{2\pi}cos(2t)dt+\frac{3j}{2\pi}\int_0^{2\pi}sin(3t)dt $
$ a_0=\frac{1}{\pi}[sin(2t)]_0^{2\pi}-\frac{j}{2\pi}[cos(3t)]_0^{2\pi} $
