Difference between revisions of "HW4.1 Sean Ray ECE301Fall2008mboutin" - Rhea

Revision as of 09:49, 16 September 2013

Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"

Problem

Find the Fourier series coefficients of a CT signal.

I chose the signal $ x(t) = 3cos(2t) $.

Fourier Series Coefficients

The Fourier series of a CT signal can be found by:

$ x(t) = \sum_{n=-\infty}^\infty a_k e^{jkw_0t} $

Fourier series coefficients can be found with the equation:

$ a_k = \frac{1}{T} \int_{0}^{T} x(t)e^{-jkw_0t} \,\ dt $

However, in the case of sinusoidal waves, the coefficients can be found more simply by applying Euler's formula to solve for the Fourier series and take the coefficients from that.

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

$ = \frac{3}{2}e^{j2t}+\frac{3}{2}e^{-j2t} $

And so, we know that our coefficients are both $ \frac{3}{2} $. Now, we need to find which $ a_k $s these belong to.

Knowing $ T $ is $ 2\pi $, we can find $ w_0 $.

$ w_0 = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1 $

Finding the corresponding $ k $s should be easy.

We know

$ a_k e^{jkw_0t} $ corresponds to $ \frac{3}{2}e^{j2t} $ and $ \frac{3}{2}e^{-j2t} $

and since $ w_0 = 1 $, then we can conclude

$ a_2 = \frac{3}{2} $

and

$ a_{-2} = \frac{3}{2} $

and

$ a_k = 0 \! $ for $ K \neq 2, -2 $

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