Revision as of 20:06, 11 September 2008 by Jamorale (Talk)

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)] $

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010