(→FOURIER SERIES) |
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When we apply these identities (above) to our original function, we obtain the following equation: | When we apply these identities (above) to our original function, we obtain the following equation: | ||
| − | <math>f(t) = (3+j)cos(2t) + (10+j)sin(7t) = (3+j)\frac{e^{2jt}+e^{-2jt}}{2} + (10+j)\frac{e^{7jt}-e^{-7jt}}{2j}\!</math> | + | <math>f(t) = (3+j)cos(2t) + (10+j)sin(7t)\!</math> |
| + | <br> | ||
| + | <math>f(t) = (3+j)\frac{e^{2jt}+e^{-2jt}}{2} + (10+j)\frac{e^{7jt}-e^{-7jt}}{2j}\!</math> | ||
<br> | <br> | ||
STILL UNDER CONSTRUCTION | STILL UNDER CONSTRUCTION | ||
Revision as of 11:29, 25 September 2008
CT SIGNAL
I chose the signal: f(t) = (3+j)cos(2t) + (10+j)sin(7t)
FOURIER SERIES
In order to find the fourier series coefficients, we must first understand the operations associated with taking the fourier transform of a signal. The fundamental period of the signal (above) is 2$ \pi\! $. We know that $ \omega_0\! $ = 2$ \pi / T\! $ (where T is the fundamental period). Therefore, the fundamental frequency is $ 1\! $.
We know that:
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $ where $ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
$ a_0=\frac{1}{2\pi}\int_0^{2\pi}[(3+j)cos(2t) + (10+j)sin(7t)]e^{0}dt $
$ a_0=\frac{1}{2\pi}[\frac{(3+j)sin(2t)}{2} + \frac{-(10+j)cos(7t)}{7}]_0^{2\pi} $
$ a_0=\frac{1}{2\pi}[\frac{-(10+j)}{7} - \frac{-(10+j)}{7}] $
$ a_0 = 0\! $
Now, we can use the same process to find $ a_k\! $ However, the most efficient way to solve for the coefficients is to use complex identities. First, we must consider the following complex identities.
$ sin(t) = \frac{e^{jt}-e^{-jt}}{2j}\! $
$ cos(t) = \frac{e^{jt}+e^{-jt}}{2}\! $
When we apply these identities (above) to our original function, we obtain the following equation:
$ f(t) = (3+j)cos(2t) + (10+j)sin(7t)\! $
$ f(t) = (3+j)\frac{e^{2jt}+e^{-2jt}}{2} + (10+j)\frac{e^{7jt}-e^{-7jt}}{2j}\! $
STILL UNDER CONSTRUCTION
