Revision as of 11:29, 11 September 2008 by Mjwhitta (Talk)

Part a

System: $ X_{k}[n]=\delta[n-k] \to Y_{k}[n] = (k+1)^2 \delta [n-(k+1)] $

Time-delay: $ X_{k}[n]=\delta[n-k] \to X_{k}[n-N]=\delta[n-N-k] $


$ X_{k}[n] \to timedelay \to sys \to Z_{k}[n]=(k+1)^2 \delta [n-N-(k+1)] $

$ X_{k}[n] \to sys \to timedelay \to Z_{k}[n]=(k+1)^2 \delta [n-N-(k+1)] $


Since $ (k+1)^2 \delta [n-N-(k+1)] $ is equal to $ (k+1)^2 \delta [n-N-(k+1)] $, the system is time-invariant.

Part b

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn