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Time Invariance of a basic system?

  1. Feb 5, 2013 #1

    ElijahRockers

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    Gold Member

    1. The problem statement, all variables and given/known data

    I am supposed to determine wether or not the discrete time system

    [itex] x[n] \rightarrow y[n] = x[-n] [/itex]

    is time invariant or not.

    3. The attempt at a solution

    Let [itex] x_d[n] = x[n-n_0][/itex]

    [itex]y_d[n] = x_d[-n] = x[-(n-n_0)] = x[-n+n_0][/itex]

    [itex]y[n-n_0] = x[-(n-n_0)] = x[-n+n_0][/itex]

    Since [itex]y_d[n] = y[n-n_0][/itex], shouldn't this prove time invariance?

    The book says the answer is that it is not time invariant...

    From the more qualitative definition, a time invariant system is one for which the behavior does not change depending on when it is evaluated...
    Now, I see that for -ve values of n, the system looks ahead, and for +ve values of n the system looks behind. Would this be considered time variant because of this? If so, how do I go about showing that mathematically?
     
    Last edited: Feb 5, 2013
  2. jcsd
  3. Feb 5, 2013 #2

    ElijahRockers

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    Gold Member

    I have just run into a similar problem, where y[n] = Even{x[n-1]}.

    When I try shifting the input, then shifting the output and comparing them, the expressions are equal, but the book is telling me the system is not time invariant.
     
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