Line 31: Line 31:
 
<math>a_3 = \frac{-2-j}{2}\,</math>
 
<math>a_3 = \frac{-2-j}{2}\,</math>
  
<math>a_(-3) = \frac{-2-j}{2}\,</math>
+
<math>a_{-3} = \frac{-2-j}{2}\,</math>
  
 
<math>a_5 = 2\,</math>
 
<math>a_5 = 2\,</math>
  
<math>a_(-5) = -2\,</math>
+
<math>a_{-5} = -2\,</math>

Revision as of 18:30, 22 September 2008

CT signal:

$ x(t) = 4\sin(5 \pi t) - (2 + j)\cos(3 \pi t)\, $


$ x(t) = 4 * \frac{e^{j5\pi t} - e^{-j5\pi t}}{2} - (2+j)*\frac{e^{j3\pi t} + e^{-j3\pi t}}{2}\, $


$ x(t) = 2e^{j5\pi t} - 2e^{-j5\pi t} - \frac{2+j}{2}e^{j3\pi t} - \frac{2+j}{2}e^{j3\pi t}\, $


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


$ \omega_0\, $ ends up being $ \pi\, $ for this signal

So because of that, in my last step, i was able to determine what value of K's i had. (k = 3,-3,5,-5)

So then you just take the coefficients of those terms to get the $ a_k\, $


Therefore:


$ a_3 = \frac{-2-j}{2}\, $

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

$ a_5 = 2\, $

$ a_{-5} = -2\, $

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Followed her dream after having raised her family.

Ruth Enoch, PhD Mathematics