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Since both cascades produce different outputs, this system is NON-time invariant. | Since both cascades produce different outputs, this system is NON-time invariant. | ||
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| + | == What Input X[n] Would Yield the Output Y[n]=u[n-1] == | ||
Revision as of 14:34, 11 September 2008
Part E. Linearity and Time Invariance
A discrete-time system is such that when the input is one of the signals in the left column, then the output is the corresponding signal in the right column:
| Input | Output | |
| X0[n]=δ[n] | Y0[n]=δ[n-1] | |
| X1[n]=δ[n-1] | Y1[n]=4δ[n-2] | |
| X2[n]=δ[n-2] | Y2[n]=9 δ[n-3] | |
| X3[n]=δ[n-3] | Y3[n]=16 δ[n-4] | |
| ... | ... | |
| Xk[n]=δ[n-k] | Yk[n]=(k+1)$ ^{2} $ δ[n-(k+1)] For any non-negative integer k |
Can This System Be Time Invariant?
Let the system be defined according to the first line, input: X0[n]=δ[n] and output: Y0[n]=δ[n-1] and time delay of 3. Using the same method as in Part D, we can determine whether this system is time invariant or not.
δ[n] -> time delay -> δ[n-3] -> system -> 16δ[n-4]
δ[n] -> system -> δ[n-1] -> time delay -> δ[n-4]
Since both cascades produce different outputs, this system is NON-time invariant.
