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Both <math>Y_k</math> yield the same output; therefore, the system is time invariant. | Both <math>Y_k</math> yield the same output; therefore, the system is time invariant. | ||
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Revision as of 07:27, 12 September 2008
6A - Is the system time invariant?
System: $ Y_k[n]=(k+1)^2\delta[n-(k+1)] $
Time-delay: $ n-n_0 $
Time-delay, then system:
$ T_k[n]=\delta[(n-n_0)-(k+1)] $
$ Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)] $
System, then time-delay:
$ T_k[n]=(k+1)^2\delta[n-k] $
$ Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)] $
Both $ Y_k $ yield the same output; therefore, the system is time invariant.
