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and "conveniently":
 
and "conveniently":
  
<math>e^{2jt} \; + \; e^{-2jt} = \cos{2t} \; + \; j \sin{2t} \; + \; \cos{-2t} \; + \; j \sin{-2t} \; \; \; \; \;</math> (by Euler's Formula)
+
<math>e^{2jt} \; + \; e^{-2jt} = cos(2t) \; + \; jsin(2t) \; + \; \cos{-2t} \; + \; j \sin{-2t}</math> (by Euler's Formula)

Revision as of 20:18, 16 September 2008

Problem

A linear system’s response to $ e^{2jt} $ is $ te^{-2jt} $, and its response to $ e^{-2jt} $ is $ te^{2jt} $. What is the system’s response to $ cos(2t) $?

Solution

If the system is linear, then the following is true:

For any $ x_{1}(t) \; \rightarrow \; y_{1}(t) $ and $ x_{2}(t) \; \rightarrow \; y_{2}(t) $

and any complex constants $ a $ and $ b $


then


$ ax_{1}(t) \; + \; bx_{2}(t) \; \rightarrow \; ay_{1}(t) \; + \; by_{2}(t) $


and "conveniently":

$ e^{2jt} \; + \; e^{-2jt} = cos(2t) \; + \; jsin(2t) \; + \; \cos{-2t} \; + \; j \sin{-2t} $ (by Euler's Formula)

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood