(CT SIGNAL & ITS FOURIER COEFFICIENTS)
(CT SIGNAL & ITS FOURIER COEFFICIENTS)
Line 8: Line 8:
  
 
=    <math>\frac{-j{e^{jt}}}{2}+\frac{j{e^{-jt}}}{2}+{e^{2jt}}+{e^{-2jt}}</math>
 
=    <math>\frac{-j{e^{jt}}}{2}+\frac{j{e^{-jt}}}{2}+{e^{2jt}}+{e^{-2jt}}</math>
 +
 +
The coefficients are as follows:
 +
<math>a_1 = \frac{-j}{2}</math>

Revision as of 13:57, 23 September 2008

CT SIGNAL & ITS FOURIER COEFFICIENTS

Let the input signal be

$ x(t)=sint +2cos2t $

= $ \frac{e^{jt}-e^{-jt}}{2i} $+$ 2*\frac{e^{2jt}+{e^{-2jt}}}{2} $

= $ \frac{-j{e^{jt}}}{2}+\frac{j{e^{-jt}}}{2}+{e^{2jt}}+{e^{-2jt}} $

The coefficients are as follows: $ a_1 = \frac{-j}{2} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood