Difference between revisions of "HW4.1 Virgil Hsieh ECE301Fall2008mboutin" - Rhea

Revision as of 14:53, 26 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} $

$ a_0=\frac{1}{\pi}[sin(4\pi)-sin(0)]+\frac{j}{2\pi}[(cos(6\pi)-cos(0)] $

$ a_0=\frac{1}{\pi}[0]+\frac{j}{2\pi}[0] $

$ a_0=0\! $

The exact same method can be used to compute any $ a_k\! $. However, that would take quite a hefty sum of time, and I, unlike most double E students, have a life. Therefore, I will use complex exponential identities to do the rest of them. Here we go, home slice:

$ x(t) = 4cos(2t) + (3j)sin(3t)\! $ turns into the following:

$ x(t) = \frac{4}{2}(e^{j2t} + e^{-j2t}) + \frac{3j}{2j}(e^{j3t} - e^{-j3t}) $

$ x(t) = 2(e^{j2t} + e^{-j2t}) + \frac{3}{2}(e^{j3t} - e^{-j3t}) $

$ x(t) = 2e^{j2t} + 2e^{-j2t} + \frac{3}{2}e^{j3t} - \frac{3}{2}e^{-j3t} $

$ a_2 = 2, a_{-2} = 2, a_3 = \frac{3}{2}, a_{-3} = -\frac{3}{2}\! $

Hey Virgil it's Joe.. I am not sure if its period is $ 2\pi $. I would appreciate your explanation. (Jungu -Joe- Choi)