Line 7: Line 7:
 
Using the property of linearity, the response is: <math>t\frac{1}{2}(e^{-j2t}+e^{j2t})</math> which is equal to <math>t*cos(2t)</math>
 
Using the property of linearity, the response is: <math>t\frac{1}{2}(e^{-j2t}+e^{j2t})</math> which is equal to <math>t*cos(2t)</math>
  
So if <math>x(t) = cos(2t) \rightarrow </math> Linear System <math> \rightarrow y(t) = tcos(2t) </math>
+
<math>x(t) = cos(2t) \rightarrow </math> Linear System <math> \rightarrow y(t) = tcos(2t) </math>

Latest revision as of 12:41, 18 September 2008

$ x_1(t) = e^{2jt} \rightarrow $ Linear System $ \rightarrow y_1(t) = te^{-2jt} $

$ x_2(t) = e^{-2jt} \rightarrow $ Linear System $ \rightarrow y_2(t) = te^{2jt} $

With an input of $ cos(2t) $, which is $ \frac{1}{2}(e^{j2t}+e^{-j2t}) $ according to Euler's Forumla.

Using the property of linearity, the response is: $ t\frac{1}{2}(e^{-j2t}+e^{j2t}) $ which is equal to $ t*cos(2t) $

$ x(t) = cos(2t) \rightarrow $ Linear System $ \rightarrow y(t) = tcos(2t) $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva