(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...) |
(→Part A) |
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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. | + | ===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 | ||
| − | <math>\,\ y(t)= (k + 1)^2 \delta[n - | + | <math>\,\ y(t)= (k+1)^2 \delta[(n-no)-(k + 1)] </math> |
| − | <math>\,\ z(t) = (k + 1 + | + | <math>\,\ z(t) = (k+1+no)^2 \delta[(n-no)-(k + 1)]</math> |
| + | ===Part B=== | ||
| + | If we set k = 0 then the answer seems easier to get | ||
| + | |||
| + | i.e. <math>\,\ \delta(n - 1) </math> and now <math>\,\ X[n] = u(n)</math> | ||
Latest revision as of 20:17, 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)] $
Part B
If we set k = 0 then the answer seems easier to get
i.e. $ \,\ \delta(n - 1) $ and now $ \,\ X[n] = u(n) $
