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

Revision as of 16:52, 30 June 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} = 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

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Ryne Rayburn