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<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)} \; + \; j \sin{(2t)} \; + \; \cos{(-2t)} \; + \; j \sin{(-2t)}</math> (by Euler's Formula)
  
<math>=\cos{(2t)} \; + \; j \sin{(2t)} \; + \; \cos{(2t)} \; - \; j \sin{(2t)}</math> (<math>\cos{(-x)}=\cos{(x)}</math> and <math>\sin{(-x)}=-\sin{(x)}</math>
+
<math>=\cos{(2t)} \; + \; j \sin{(2t)} \; + \; \cos{(2t)} \; - \; j \sin{(2t)}</math> (<math>\cos{(-x)}=\cos{(x)}</math> and <math>\sin{(-x)}=-\sin{(x)}</math>)

Revision as of 21:22, 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)} \; + \; j \sin{(2t)} \; + \; \cos{(-2t)} \; + \; j \sin{(-2t)} $ (by Euler's Formula)

$ =\cos{(2t)} \; + \; j \sin{(2t)} \; + \; \cos{(2t)} \; - \; j \sin{(2t)} $ ($ \cos{(-x)}=\cos{(x)} $ and $ \sin{(-x)}=-\sin{(x)} $)

Alumni Liaison

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

Francisco Blanco-Silva