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==Problem 3==
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[[Category: ECE]]
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[[Category: ECE 301]]
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[[Category: Summer]]
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[[Category: 2008]]
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[[Category: asan]]
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[[Category: Exams]]
 
(a) Derive the condition for which the discrete time complex exponetial signal x[n] is periodic.
 
(a) Derive the condition for which the discrete time complex exponetial signal x[n] is periodic.
  

Revision as of 09:14, 21 November 2008

(a) Derive the condition for which the discrete time complex exponetial signal x[n] is periodic.

 $ x[n] = e^{jw_{o}n} $         
 $ x[n] = x[n+N] = e^{jw_{o}(n+N)} = e^{jw_{o}n}e^{jw_{o}N} $
 to be periodic 
 $ e^{jw_{o}N} = 1 = e^{j2\pi k} $
 $ \therefore w_{o}N = 2\pi k $
 $ \Rightarrow \frac{w_{o}}{2\pi} = \frac{K}{N} \Rightarrow $Rational number
 $ \therefore \frac{w_{o}}{2\pi} $ shold be a rational number

(b) Show that the system described by

   $ y[n] = x[n] + x[n+1] + x[n+2] $ is a LTI system.
   $      ax_{1}[n]+bx_{2}[n] \rightarrow ax_{1}[n]+bx_{2}[n]+ax_{1}[n+1]+bx_{2}[n+1]+ax_{1}[n+2]+bx_{2}[n+2]      $
                  $ = a(x_{1}[n]+x_{1}[n+1]+x_{1}[n+2])+b(x_{2}[n]+x_{2}[n+1]+x_{2}[n+2]) $
                  
                  $ = ay_{1}[n]+by_{2}[n] \therefore $System is linear


   $ y_{1}[n-n_{0}] = x_{1}[n-n_{0}]+x_{1}[n-n_{0}+1]+x_{1}[n-n_{0}+2] $
   
   Let $ x_{2}[n] = x_{1}[n-n_{0}] $
   
   $   x_{2}[n] \rightarrow x_{2}[n]+x_{2}[n+1]+x_{2}[n+2] $
   
       $ = x_{1}[n-n_{0}]+x_{1}[n-n_{0}+1]+x_{1}[n-n_{0}+2] = y_{1}[n-n_{0}]\therefore $System is time-variant

(c) Prove that $ x[n]*\delta[n] = x[n] $

   $ x[n]*\delta[n] = \Sigma_{k=-\infty}^\infty x[k]\delta[n-k] $
   $ = \Sigma_{k=-\infty}^\infty x[n]\delta[n-k] = x[n]\Sigma_{k=-\infty}^\infty\delta[n-k] = x[n] $
   $ \therefore x[n]*\delta[n] = x[n] $


Solved by Minwoong Kim

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood