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Time invariant

  1. Oct 9, 2006 #1
    i usually have such a hard time determining whether a signal is time invariant or not ...

    for example, why would x[-n] not be time-invariant?

    please dont just tell me why x[-n] would not be time invariant ...
    tell me techniques that I can apply to other signals too
  2. jcsd
  3. Oct 9, 2006 #2
    look at the parameters beside your function if they contain a t term then your signal is time varient while if the parameters are constants then the signal is time invarient.
  4. Oct 9, 2006 #3
    that makes sense ... but then tell me why x[-n] is not time invariant?
  5. Oct 9, 2006 #4
    do you see any t terms beside the function??
    it is time invarient. why r u sure it isn't time invarient?
    you can use this site to see the graph for check. http://www.jhu.edu/~signals/sys/resulta939.html [Broken]

    (i used unit step as an example)
    Last edited by a moderator: May 2, 2017
  6. Oct 9, 2006 #5
    i am so sure it isnt time invariant, because the solution manual said so ...
    Last edited by a moderator: May 2, 2017
  7. Oct 9, 2006 #6
    What is 'n'?
  8. Oct 9, 2006 #7
    n is just time, but it assumes discrete value only
  9. Oct 9, 2006 #8
    So, you've answered your own question.
  10. Oct 9, 2006 #9
    no i did not ... please enlighten me ...
  11. Oct 9, 2006 #10
    i am sure it is time invarient my mind says so.
  12. Sep 10, 2009 #11
    The key is in understanding the test for time invariance.

    To test: x[n] > DELAY > x[n-n0] > SYSTEM > w[n]
    >>> SYSTEM > y[n] > DELAY > y[n-n0]

    w[n] and y[n-n0] are equal if the system is time invariant

    in the case of y[n]=x[-n], for the top approach, delaying the system results in n-n0 then we apply the system's effect of reversing JUST n, so w[n]=x[-n-n0]. With the second path, we apply the system and get y[n]=x[-n] and then apply the delay to get y[n-n0]=x[-(n-n0)]=x[-n+n0].

    Since x[-n-n0] is not the same as x[-n+n0] the system is time VARIANT.
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