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== Part B: Find input given output == | == Part B: Find input given output == | ||
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| + | The given output is: | ||
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| + | <math>\,Y[n]=u[n-1]\,</math> | ||
| + | |||
| + | |||
| + | This can be re-written as: | ||
| + | |||
| + | <math>\,Y[n]=\sum_{k=0}^{\infty}\delta [n-(k+1)]\,</math> | ||
| + | |||
| + | <math>\,Y[n]=\delta [n-1]+\delta [n-2]+\delta [n-3]+\ldots +\delta [n-(k+1)]\,</math> | ||
| + | |||
| + | <math>\,Y[n]=Y_0[n]+\frac{1}{4}Y_1[n]+\frac{1}{9}Y_2[n]+\ldots +\frac{1}{(k+1)^2}Y_k\,</math> | ||
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| + | Because the system is assumed to be linear, we can write the input as | ||
Revision as of 21:02, 11 September 2008
Part A: Can the system be time invariant?
The system cannot be time invariant.
For instance, the input
$ \,X_0[n]=\delta [n]\, $
yields the output
$ \,Y_0[n]=\delta [n-1]\, $
Thus,
$ \,Y_0[n-1]=\delta [n-2]\, $
However, the input
$ \,X_0[n-1]=\delta [n-1]=X_1[n]\, $
yields the output
$ \,Y_1[n]=4\delta[n-2]\, $
Since these two are not equal
$ \,\delta [n-2]\not= 4\delta[n-2]\, $
the system is time variant (by not fitting the definition of time invariance).
Part B: Find input given output
The given output is:
$ \,Y[n]=u[n-1]\, $
This can be re-written as:
$ \,Y[n]=\sum_{k=0}^{\infty}\delta [n-(k+1)]\, $
$ \,Y[n]=\delta [n-1]+\delta [n-2]+\delta [n-3]+\ldots +\delta [n-(k+1)]\, $
$ \,Y[n]=Y_0[n]+\frac{1}{4}Y_1[n]+\frac{1}{9}Y_2[n]+\ldots +\frac{1}{(k+1)^2}Y_k\, $
Because the system is assumed to be linear, we can write the input as
