(Solution)
(Solution)
Line 14: Line 14:
 
:<math> a_2 = \frac{1}{2} </math>
 
:<math> a_2 = \frac{1}{2} </math>
  
 +
:<math> \omega_0 = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1 </math>
  
:<math> \omega_0 = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1
+
:else
:
+
:<math> a_k = 0 </math>

Revision as of 08:10, 25 September 2008

Useful Info

$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $

$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.

Let
$ x(t) = 2sin(2\pi t) + cos(\pi t). $

Solution

$ x(t) = 2\frac{e^{2 \pi jt}+e^{-2 \pi jt}}{2j} + \frac{e^{\pi jt}+e^{-\pi jt}}{2} $
$ x(t) = \frac{1}{j}(e^{2 \pi jt} + e^{-2 \pi jt}) + \frac{1}{2}(e^{\pi jt}+e^{-\pi jt}) $
$ a_1 = \frac{1}{j} $
$ a_2 = \frac{1}{2} $
$ \omega_0 = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1 $
else
$ a_k = 0 $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood