(New page: == IS THIS SYSTEM TIME INVARIANT? == '''<math>X_k[n] = d[n-k] \Rightarrow Y_k[n]=(k+1)2 d[n-(k+1)]</math>''' '''TEST''' '''<math>d[n-k] \Rightarrow (k+1)2 d[n-(k+1)] \rightarrow [time ...) |
(→IS THIS SYSTEM TIME INVARIANT?) |
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'''TEST''' | '''TEST''' | ||
− | '''<math>d[n-k] \Rightarrow (k+1)2 d[n-(k+1)] \rightarrow [time delay] \rightarrow (k+1)^2 d[n -(k+1) - t_o]</math>''' | + | '''<math>d[n-k] \Rightarrow (k+1)2 d[n-(k+1)] \rightarrow [time delay] \rightarrow Z(t) = (k+1)^2 d[n -(k+1) - t_o]</math>''' |
− | '''<math>d[n-k] \rightarrow [time delay] \rightarrow d[(n-k) - t_o] \Rightarrow (k+1)^2 d[n -(k+1) - t_o]</math>''' | + | '''<math>d[n-k] \rightarrow [time delay] \rightarrow d[(n-k) - t_o] \Rightarrow W(t) = (k+1)^2 d[n -(k+1) - t_o]</math>''' |
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+ | since '''<math>Z(t) = W(t)</math>''' the system is time invariant. | ||
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+ | |||
+ | '''PART B''' | ||
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+ | Assuming that this system is linear, what input '''<math>X[n]</math>''' would yield the output '''<math>Y[n]=u[n-1]</math>'''? | ||
+ | |||
+ | The input would have to be '''<math>u[n]</math>''' |
Latest revision as of 17:45, 12 September 2008
IS THIS SYSTEM TIME INVARIANT?
$ X_k[n] = d[n-k] \Rightarrow Y_k[n]=(k+1)2 d[n-(k+1)] $
TEST
$ d[n-k] \Rightarrow (k+1)2 d[n-(k+1)] \rightarrow [time delay] \rightarrow Z(t) = (k+1)^2 d[n -(k+1) - t_o] $
$ d[n-k] \rightarrow [time delay] \rightarrow d[(n-k) - t_o] \Rightarrow W(t) = (k+1)^2 d[n -(k+1) - t_o] $
since $ Z(t) = W(t) $ the system is time invariant.
PART B
Assuming that this system is linear, what input $ X[n] $ would yield the output $ Y[n]=u[n-1] $?
The input would have to be $ u[n] $