(→‎Question 6a)
(→‎Question 6a)
 
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== Question 6a ==
 
== Question 6a ==
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The system is time variant because of the following example:
  
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I'm assuming <math>n\,</math> is the variable I will be applying the time shift to. I looked at some other peoples work and although they all thought <math>k\,</math> was the time variable, I think <math>k\,</math> is just the time step moving the function forward relative to some time position <math>n\,</math>. In other words , <math>k=2\,</math> doesn't mean time = 2 sec, it just means 2 time steps ahead of time <math>n\,</math>. Another reason I chose <math>n\,</math> to be the time variable is because when you discussed the sifting property in class you sifted by <math>n_0\,</math>, not <math>k\,</math>.
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Dealing with the following system
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+
  
 
<math> X_k[n]=Y_k[n] \,</math>
 
<math> X_k[n]=Y_k[n] \,</math>
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Consider the input and output of the system when k = 1
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Consider the input and output of the system when k = 0
  
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<math> X_1[n]=\delta[n-1]\,</math>     
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<math> X_0[n]=\delta[n]\,</math>     
  
  
 
and  
 
and  
  
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<math> Y_1[n]=4\delta[n-2] \,</math>
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<math> Y_0[n]=\delta[n-1] \,</math>
  
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If I time shift the input by <math> n_0\,</math> , then run it through the system I obtain:
 
  
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<math>X_1[n] \longrightarrow X_1[n-n_0-1] \longrightarrow Y_1[n]=4\delta[n-n_0-2]\,</math>
 
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If I run it through the system, then time shift the output by <math> n_0\,</math>  I obtain:
 
−
 
−
 
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<math>X_1[n] \longrightarrow X_1[n-n_0-1] \longrightarrow Y_1[n]=4\delta[n-n_0-2]\,</math>
 
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Consider the system: <math>y(t)=x(t^2-3) \,</math>
 
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If <math>x(t) \,</math> is first time shifted, then put into the system:
 
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<math>x(t) \longrightarrow x(t-t_0) \longrightarrow y(t)=x(t^2-3-t_0)\,</math>
 
  
 +
If I time shift the input by <math> n_0\,</math> , then run it through the system I obtain:
  
 +
<math>X_0[n] \longrightarrow X_0[n-n_0] \longrightarrow Y_1[n]=\delta[n-n_0-1],</math>
  
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If <math>x(t) \,</math> is first entered into the system, then time shifted:
 
  
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<math>x(t) \longrightarrow y(t)=x(t^2-3) \longrightarrow y(t-t_0)=x((t-t_0)^2-3)\,</math>
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but this isn't equal to running the input through the system, then time shifting the output by <math> n_0\,</math>
  
 +
<math>X_0[n] \longrightarrow Y_1[n]=4\delta[n-1] \longrightarrow Y_1[n-n_0]=4\delta[n-n_0-1]\,</math>
  
 +
== Question 6b ==
  
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Thus this system isn't T.I.
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Assuming this system is linear, an input
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<math> X_0[n]=u[n]\,</math> 
 +
would result in an output
 +
<math> Y_0[n]=u[n-1]\,</math>.

Latest revision as of 16:07, 12 September 2008

Question 6a

The system is time variant because of the following example:

Dealing with the following system

$ X_k[n]=Y_k[n] \, $


where

$ X_k[n]=\delta[n-k]\, $


and

$ Y_k[n]=(k+1)^2 \delta[n-(k+1)] \, $


Consider the input and output of the system when k = 0

$ X_0[n]=\delta[n]\, $


and

$ Y_0[n]=\delta[n-1] \, $


If I time shift the input by $ n_0\, $ , then run it through the system I obtain:

$ X_0[n] \longrightarrow X_0[n-n_0] \longrightarrow Y_1[n]=\delta[n-n_0-1], $


but this isn't equal to running the input through the system, then time shifting the output by $ n_0\, $

$ X_0[n] \longrightarrow Y_1[n]=4\delta[n-1] \longrightarrow Y_1[n-n_0]=4\delta[n-n_0-1]\, $

Question 6b

Assuming this system is linear, an input $ X_0[n]=u[n]\, $ would result in an output $ Y_0[n]=u[n-1]\, $.

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