(New page: ==Time Invariance== For a given input <math>X_i[n]=\delta[n-(i-1)]</math> it yields the output <math>Y_i[n]=(k+1)^2 \delta[n-(k+1)]</math> is NOT time invariant. Say we shifted the input...) |
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Revision as of 16:08, 11 September 2008
Time Invariance
For a given input $ X_i[n]=\delta[n-(i-1)] $ it yields the output $ Y_i[n]=(k+1)^2 \delta[n-(k+1)] $ is NOT time invariant.
Say we shifted the input at $ X_3[n] $ by +3, it would yield the output of $ Y[n]=\delta[n-1] $
However if we look at the output of $ X_3[n] $ and then shift it by 3, we get $ Y[n] = 16*\delta[n-4+3] = 16*\delta[n-1] $
Which is clearly 16 * the other input and thus they are not equal, so the system is not time invariant.
