(New page: == Questions == <math>Y(t) = x(t - 1) - x(1 - t)</math> It is Time Invariant? Justify. == Answer == No. <math>S_1 = Y(t) = x(t - 1) - x(1 - t)</math> <math>S_2 = Y(t) = x(t - t_o)...) |
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== Questions == | == Questions == | ||
| − | + | The input x(t) and the output Y(t) of a system are related by the equation | |
<math>Y(t) = x(t - 1) - x(1 - t)</math> | <math>Y(t) = x(t - 1) - x(1 - t)</math> | ||
| − | + | Is the system Time Invariant? Justify your answer. | |
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== Answer == | == Answer == | ||
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<math> x(t - t_o - 1) - x(1 - t + t_o) =/= x(t - t_o - 1) - x(1 - t - t_o)</math> | <math> x(t - t_o - 1) - x(1 - t + t_o) =/= x(t - t_o - 1) - x(1 - t - t_o)</math> | ||
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Latest revision as of 15:40, 23 April 2013
Questions
The input x(t) and the output Y(t) of a system are related by the equation
$ Y(t) = x(t - 1) - x(1 - t) $
Is the system Time Invariant? Justify your answer.
Answer
No.
$ S_1 = Y(t) = x(t - 1) - x(1 - t) $
$ S_2 = Y(t) = x(t - t_o) $
$ x(t) -> S1 -> S2 -> x(t - t_o - 1) - x(1 - t + t_o) $
$ x(t) -> S2 -> S1 -> x(t - t_o - 1) - x(1 - t - t_o) $
$ x(t - t_o - 1) - x(1 - t + t_o) =/= x(t - t_o - 1) - x(1 - t - t_o) $
