Difference between revisions of "HW4.1 Nicholas Gentry ECE301Fall2008mboutin" - Rhea

Revision as of 11:43, 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: $ f(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $ where $ a_k=\frac{1}{T}\int_0^Tf(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}\! $

Now, we can proceed to multiply everything out as follows:

$ f(t) = ((3+j)\frac{e^{2jt}}{2} + (3+j)\frac{e^{-2jt}}{2} + (10+j)\frac{e^{7jt}}{2j} - (10+j)\frac{e^{-7jt}}{2j}\! $

Therefore, we can conclude that the fourier series coefficients are:

$ a_{2} = \frac{3+j}{2}\! $

$ a_{-2} = \frac{3+j}{2}\! $

$ a_{7} = \frac{10+j}{2j}\! $

$ a_{-7} = -\frac{10+j}{2j}\! $