(New page: The system is not time invariant Let time delay = t then Xk[n]=δ[n-k] -> time delay -> system =X(k+t)[n]=δ[n-(k+t)]->system =X(k+t)[n]=δ[n-k-t]->system =Y(k+t)[n]=...) |
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| − | The system is | + | (a)The system is time invariant |
Let time delay = t | Let time delay = t | ||
| Line 7: | Line 7: | ||
Xk[n]=δ[n-k] -> time delay -> system | Xk[n]=δ[n-k] -> time delay -> system | ||
| − | + | equals to X(k+t)[n]=δ[n-(k+t)]->system | |
| − | + | equals to X(k+t)[n]=δ[n-k-t]->system | |
| − | =Y(k+t)[n]=<math>(k+t+1)^2</math> δ[n-(k+t+1)] | + | equals to Y(k+t)[n]=<math>(k+t+1)^2</math> δ[n-(k+t+1)] |
| + | |||
| + | while | ||
| + | |||
| + | Xk[n]=δ[n-k] -> system -> time delay | ||
| + | |||
| + | equals to Yk[n]=<math>(k+1)^2</math> δ[n-(k+1)]->time delay | ||
| + | |||
| + | equals to Y(k+t)[n]=<math>(k+t+1)^2</math> δ[n-(k+t+1)] | ||
| + | |||
| + | yields the same result | ||
| + | |||
| + | ---- | ||
| + | (b) | ||
| + | |||
| + | According to row 1 of the table, input x[n]=u[n] yields the output Y[n]=u[n-1]. | ||
Latest revision as of 15:42, 10 September 2008
(a)The system is time invariant
Let time delay = t
then
Xk[n]=δ[n-k] -> time delay -> system
equals to X(k+t)[n]=δ[n-(k+t)]->system
equals to X(k+t)[n]=δ[n-k-t]->system
equals to Y(k+t)[n]=$ (k+t+1)^2 $ δ[n-(k+t+1)]
while
Xk[n]=δ[n-k] -> system -> time delay
equals to Yk[n]=$ (k+1)^2 $ δ[n-(k+1)]->time delay
equals to Y(k+t)[n]=$ (k+t+1)^2 $ δ[n-(k+t+1)]
yields the same result
(b)
According to row 1 of the table, input x[n]=u[n] yields the output Y[n]=u[n-1].
