(New page: <math>\cos{2t} = \frac{e^{2jt} - e^{-2jt}}{2}</math> Since we know, <math>e^{2jt} \Longrightarrow System \Longrightarrow te^{-2jt}</math> and <math>e^{-2jt} \Longrightarrow System \Lon...)
 
 
(One intermediate revision by the same user not shown)
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<math>\cos{2t} = \frac{e^{2jt} - e^{-2jt}}{2}</math>
+
<math>\cos{2t} = \frac{e^{2jt} + e^{-2jt}}{2}</math>
  
 
Since we know,
 
Since we know,
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then
 
then
  
<math>\cos{2t} \Longrightarrow System \Longrightarrow \frac{te^{-2jt} - te^{2jt}}{2} = t\cos{2t}</math>
+
<math>\cos{2t} \Longrightarrow System \Longrightarrow t(\frac{e^{-2jt} + e^{2jt}}{2}) = t\cos{2t}</math>

Latest revision as of 15:06, 18 September 2008

$ \cos{2t} = \frac{e^{2jt} + e^{-2jt}}{2} $

Since we know,

$ e^{2jt} \Longrightarrow System \Longrightarrow te^{-2jt} $

and

$ e^{-2jt} \Longrightarrow System \Longrightarrow te^{2jt} $

then

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

Alumni Liaison

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

Buyue Zhang