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6A - Is the system time invariant?

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

Time-delay, then system: $ T_k[n]=\delta[n-(k+1)] $
$ Y_k[n]=(k+1)^2\delta[n-(k+1)] $

System, then time-delay: $ T_k[n]=(k+1)^2\delta[n-k] $
$ Y_k[n]=(k+1)^2\delta[n-(k+1)] $

Both $ Y_k $ yield the same output; therefore, the system is time invariant.

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett