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If we use the same procedure as we did in HW2D then we will find out it is not time variant. The we get | If we use the same procedure as we did in HW2D then we will find out it is not time variant. The we get | ||
Revision as of 20:06, 11 September 2008
Part A
If we use the same procedure as we did in HW2D then we will find out it is not time variant. The we get
$ \,\ y(t)= (k+1)^2 \delta[(n-no)-(k + 1)] $
$ \,\ z(t) = (k+1+no)^2 \delta[(n-no)-(k + 1)] $
