(44 intermediate revisions by the same user not shown)
Line 1: Line 1:
(a) Derive the condition for which the discrete time complex exponetial signal
+
(a) Derive the condition for which the discrete time complex exponetial signal x[n] is periodic.
  
    <math>x[n] = e^jwn</math><math>
+
  <math>x[n] = e^{jw_{o}n}</math>        
e^jwn
+
  <math>x[n] = x[n+N] = e^{jw_{o}(n+N)} = e^{jw_{o}n}e^{jw_{o}N}</math>
 +
  to be periodic
 +
  <math>e^{jw_{o}N} = 1 = e^{j2\pi k}</math>
 +
  <math>\therefore w_{o}N = 2\pi k</math>
 +
  <math>\Rightarrow \frac{w_{o}}{2\pi} = \frac{K}{N} \Rightarrow</math>Rational number
 +
  <math>\therefore \frac{w_{o}}{2\pi}</math> shold be a rational number
 +
 
 +
(b) Show that the system described by
 +
    <math>y[n] = x[n] + x[n+1] + x[n+2]</math> is a LTI system.
 +
 
 +
    <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>= ay_{1}[n]+by_{2}[n] \therefore</math>System is linear
 +
 
 +
 
 +
    <math>y_{1}[n-n_{0}] = x_{1}[n-n_{0}]+x_{1}[n-n_{0}+1]+x_{1}[n-n_{0}+2]</math>
 +
   
 +
    Let <math>x_{2}[n] = x_{1}[n-n_{0}]</math>
 +
   
 +
    <math>  x_{2}[n] \rightarrow x_{2}[n]+x_{2}[n+1]+x_{2}[n+2]</math>
 +
   
 +
            <math>= x_{1}[n-n_{0}]+x_{1}[n-n_{0}+1]+x_{1}[n-n_{0}+2] = y_{1}[n-n_{0}]\therefore</math>System is time-variant
 +
 
 +
(c) Prove that <math>x[n]*\delta[n] = x[n]</math>
 +
 
 +
    <math>x[n]*\delta[n] = \Sigma_{k=-\infty}^\infty x[k]\delta[n-k]</math>
 +
    <math>= \Sigma_{k=-\infty}^\infty x[n]\delta[n-k] = x[n]\Sigma_{k=-\infty}^\infty\delta[n-k] = x[n]</math>
 +
    <math>\therefore x[n]*\delta[n] = x[n]</math>
 +
 
 +
 
 +
Solved by Minwoong Kim

Latest revision as of 16:17, 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}[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

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett