(The Coefficients)
Line 19: Line 19:
 
<math>4, -4, 3, -3, and 2</math>
 
<math>4, -4, 3, -3, and 2</math>
  
<math>k^4 = \frac{1}{2j}</math>
+
<math>a_4 = \frac{1}{2j}</math>
<math>k^-4 = \frac{1}{2j}</math>
+
<math>a_-4 = \frac{1}{2j}</math>
<math>k^3 = \frac{1}{2}</math>
+
<math>a_3 = \frac{1}{2}</math>
<math>k^-3 = \frac{1}{2}</math>
+
<math>a_-3 = \frac{1}{2}</math>
<math>k^2 = 1</math>
+
<math>a_2 = 1</math>
  
 +
<math>a_k = 0, k \ne 4, -4, 3, -3, 2</math>
 
<math>A_</math>
 
<math>A_</math>

Revision as of 19:27, 25 September 2008

The Signal

mmm lets randomly take...

$ \sin4\pi t + \cos3\pi t + e^{j2\pi t} $


The Coefficients

Remeber... $ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $

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

Going to conver the equation into signal that is all in exponentials.

$ \frac{1}{2j}(e^{j4\pi t}-e^{-j4\pi t}) + \frac{1}{2}(e^{j3\pi t} + e^{-j3\pi t}) + e^{j2\pi t} $

The terms come out to be

$ 4, -4, 3, -3, and 2 $

$ a_4 = \frac{1}{2j} $ $ a_-4 = \frac{1}{2j} $ $ a_3 = \frac{1}{2} $ $ a_-3 = \frac{1}{2} $ $ a_2 = 1 $

$ a_k = 0, k \ne 4, -4, 3, -3, 2 $ $ A_ $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood