(CT SIGNAL & ITS FOURIER COEFFICIENTS)
(CT SIGNAL & ITS FOURIER COEFFICIENTS)
Line 5: Line 5:
 
<math>x(t)=sint +2cos2t</math>
 
<math>x(t)=sint +2cos2t</math>
  
= <math>\frac{e^{jt}-e^{-jt}}{2i}</math>+<math>2*\frac{e^{2jt}+{e^{-2jt}}}{2}</math>
+
=   <math>\frac{e^{jt}-e^{-jt}}{2i}</math>+<math>2*\frac{e^{2jt}+{e^{-2jt}}}{2}</math>
 +
 
 +
=    <math>\frac{-j {e^{jt}}{2}</math>

Revision as of 14:49, 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} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood