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==Linearity and Time Invariance==
 
==Linearity and Time Invariance==
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'''6.a)'''
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'''the system is defined as'''
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<math>X_k[n] = \delta[n - k] \to sys \to  Y_k[n] = (k + 1)^2 \delta[n - (k + 1)]</math>
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'''let us check for time invariance'''
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'''System followed by time delay'''
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'''now,let us apply a time-delay of <math>t_0</math> to the system.'''
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<math>\delta[n - k] \to sys \to (k + 1)^2 \delta[n - (k + 1)] \to timedelay \to (k + 1)^2  = (k + 1)^2 \delta[n -(k + 1 +t_0)]  </math>
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'''Time-delay followed by system:'''
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<math>\delta[n - k] \to timedelay \to \delta[n-(k + t_0)] \to sys \to (k + t_0 + 1)^2 \delta[n - (k + t_0 + 1)]</math>
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'''For the system to be time invariant both the outputs should be same but they are not. so the system is not time variant it is rather a time variant system as the output varies with time.'''
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'''6.b)'''
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Assumed that the system given is linear, then '''X[n]=u[n]''' is the input to yield the output''' Y[n] = u[n-1].
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'''
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since u[n] is simply the summation of shifted delta functions we can say that
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<math> X_k[n]= \ u[n] = \delta[n]- \delta[n - N]</math>
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substituting k = 0 and N = 1 in the above equation will give the output Y'''[n] =u[n − 1]''' after the input X[n] goes through the system.

Latest revision as of 15:34, 12 September 2008

Linearity and Time Invariance

6.a)

the system is defined as

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

let us check for time invariance

System followed by time delay

now,let us apply a time-delay of $ t_0 $ to the system.


$ \delta[n - k] \to sys \to (k + 1)^2 \delta[n - (k + 1)] \to timedelay \to (k + 1)^2 = (k + 1)^2 \delta[n -(k + 1 +t_0)] $


Time-delay followed by system:

$ \delta[n - k] \to timedelay \to \delta[n-(k + t_0)] \to sys \to (k + t_0 + 1)^2 \delta[n - (k + t_0 + 1)] $

For the system to be time invariant both the outputs should be same but they are not. so the system is not time variant it is rather a time variant system as the output varies with time.

6.b)

Assumed that the system given is linear, then X[n]=u[n] is the input to yield the output Y[n] = u[n-1]. since u[n] is simply the summation of shifted delta functions we can say that

$ X_k[n]= \ u[n] = \delta[n]- \delta[n - N] $

substituting k = 0 and N = 1 in the above equation will give the output Y[n] =u[n − 1] after the input X[n] goes through the system.

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ECE462 Survivor

Seraj Dosenbach