(New page: As mentioned in the problem, the response of :<math>e^{2jt}\,</math> is :<math>te^{-2jt}\,</math> Suppose we let <math>y(t)</math> be the response of <math>x(t)</math>, in order to...)
 
 
Line 27: Line 27:
  
  
Therefore, the system's response of <math>cos(2t)</math> is:
+
Therefore, using the Euler formula:
  
  
:<math>tcos(-2t)\,</math>
+
:<math>cos(2t) = \frac{1}{2}(e^{-2jt}+e^{2jt})\,</math>
 +
 
 +
 
 +
and the two responses mentioned above, the response of <math>cos(2t)</math> is:
 +
 
 +
 
 +
:<math>\frac{1}{2}(te^{2jt}+te^{-2jt}) = tcos(2t)\,</math>

Latest revision as of 19:39, 18 September 2008

As mentioned in the problem, the response of


$ e^{2jt}\, $

is

$ te^{-2jt}\, $


Suppose we let $ y(t) $ be the response of $ x(t) $, in order to make $ x(t) $ produce the output corresponding to $ y(t) $, we need to multiply the input by $ t $ and make the $ t $ of $ x $ negative. ie.


$ y(t) = tx(-t)\, $


This can be confirmed by the second condition, which is


$ te^{2jt}\, $


is the response of


$ e^{-2jt}\, $


Therefore, using the Euler formula:


$ cos(2t) = \frac{1}{2}(e^{-2jt}+e^{2jt})\, $


and the two responses mentioned above, the response of $ cos(2t) $ is:


$ \frac{1}{2}(te^{2jt}+te^{-2jt}) = tcos(2t)\, $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang