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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.'''
 
'''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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'''
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6.b)'''

Revision as of 09:20, 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 \delta[n - t_0 -(k + 1)] = (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)

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Ruth Enoch, PhD Mathematics