Revision as of 08:18, 26 September 2008 by Nablock (Talk)

Let us take the periodic, CT signal: $ 3cos(4\pi t) + e^{j\frac{2\pi}{5}t} $


As we know, the Fourier Series for a CT signal is written as:

$ x(t) = \sum^{\infty}_{k = -\infty}a_k e^{j k w_o t} $

Where $ a_k $ is: $ a_k = \frac{1}{T} \int^{T}_{0} x(t) e^{-j k w_o t} $


Our signal, x(t), can also be written as:

$ x(t) = \frac{3}{2}(e^{j 4 \pi t} + e^{-j 4 \pi t}) + e^{j \frac{2 \pi}{5} t} $

Solving for a:

$ a_1 = \frac{3}{2} $

$ a_{-1} = \frac{3}{2} $

$ a_2 = a_{-2} = 1 $

$ a_k = 0 $ else

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood