Signals: Time Invariance

Discussion in 'Homework Help' started by jegues, Sep 30, 2012.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    1. The problem statement, all variables and given/known data

    Prove whether or not,

    y(t) = \frac{1}{2}\left( x(t) - x(-t) \right)

    Is time invariant or not

    2. Relevant equations



    3. The attempt at a solution

    Shifting the output by -T results in,

    y(t-T) =  \frac{1}{2}\left( x(t-T) - x(-(t-T)) \right)

    y(t-T) =  \frac{1}{2}\left( x(t-T) - x(-t+T) \right)

    Shifting the input by -T results in,

    \frac{1}{2}\left( x(t-T) - x(-t-T) \right)

    Since the last two lines are not the same they are not time invariant.

    I feel like this is wrong because,

    x(-t-T)

    shifts the input to the left while the other input (i.e. x(t)) is shifted to the right.

    What is the correct procedure to prove whether or not this is time invariant or not?
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Consider that a signal applied at a time close to t=0s produces an output that depends on two narrowly spaced points in the signal, which if the same signals are applied at a time well away from t=0s, the output signal depends on two widely spaced points in the signal.
     
  3. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    That's all fine and dandy, but this doesn't prove that whether or not it is time invariant.

    How would you prove whether or not it is time invariant.

    Just because the output depends on narrow and wider spaced points isn't to say that it isn't shifted as the input is shifted.

    Is there a short mathematical proof for when something is time invariant or not?
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    You already proved it!

    If

    y1(t) = f(t)

    then if

    y2(t) = f(t-T) is not equal to y1(t-T), the system is not time invariant.

    You sounded like you just didn't feel like the example that you had just proved to be time variant really was, so I was trying to give you an intuitive explanation because you had already done the mathematical one.
     
  5. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    Yes but is my proof in fact correct? I feel like its not because one input is shifted to the left (i.e. x(t-T)) and the other input is shifted to the right. (i.e. x(-t-T))

    Shouldn't both signals that comprise the input be shifted in the same direction in order to follow the definition of time invariance?

    (If I shift my input by -T is the result of said input also shifted by -T?)
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    Look carefully at what you are saying.

    You've applied the time-invariance test to a system and it failed, meaning that it is not timeinvariant. You are then stuck on trying to say that the results of the test must be wrong because it's not of the form that would be required for it to be time-invariant. Well.... it ISN"T time-invariant; that's what the test is all about! If you have a system that is not time-invariant, why would you expect it to be written in the form that would meant that it was?
     
  7. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    What I was asking was whether or not I applied the test correctly.

    I was worried I had not done the test correctly becuase one portion of the input is shifted to the left (i.e. x(t-T)) and the other portion of the input is shifted to the right. (i.e. x(-t-T))

    I thought in order to correctly apply the test both components of the input should be shifted in the same direction, no? Can you understand my confusion now?
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    The test is the test. You replace each occurance of 't' with 't-T'. From there you let the chips fall where they may. They system you are testing is what is causing one part to be shifted one way and another part to be shifted the other. This is what makes the system you are testing be time-variant.
     
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