(New page: If we use the same procedure as we did in HW2D then we will find out it is not time variant. We get <math>\,\ y(t)= (k + 1)^2 \delta[n - n_0 -(k + 1)] </math> <math>\,\ z(t) = (k + 1 + n...)
 
Line 1: Line 1:
If we use the same procedure as we did in HW2D then we will find out it is not time variant. 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
  
<math>\,\ y(t)= (k + 1)^2 \delta[n - n_0 -(k + 1)] </math>
+
<math>\,\ y(t)= (k+1)^2 \delta[(n-no)-(k + 1)] </math>
  
<math>\,\ z(t) = (k + 1 + n_0)^2 \delta[n - n_0 -(k + 1)]</math>
+
<math>\,\ z(t) = (k+1+no)^2 \delta[(n-no)-(k + 1)]</math>

Revision as of 20:06, 11 September 2008

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

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett