Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Similar Threads for invariant Date
Time Invariance Sep 26, 2016
Maxwells Equations and Time Invariance Nov 26, 2011
Convolution in Contious Linear Time Invariant System Jan 21, 2010
Time invariancy Feb 10, 2008
Time Invariance Aug 25, 2007