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Time invariancy

  1. Feb 10, 2008 #1
    For a past homework question that has been marked, I don't understand one thing.
    For the signal y[n] = 0.5*(x[n-1] + x[-n-1]), to prove its time invariancy, x1[n] = x[n-n0] is declared, with n0 not equal to 0. I understand that n-n0 should be substituted in place of n when I calculate y[n-n0] = 0.5*(x[n-n0-1] + x[-n+n0-1]). However, I don't understand why y2[n] = 0.5*(x1[n-n0]+x1[-n-n0]) = 0.5*(x[n-n0-1] + x[-n-n0-1]), i.e. substituting -n0 into the brackets directly, not like 0.5*(x[(n-n0)-1] + x[-(n-n0)-1]). Any quick help would be much appreciated as I have a midterm in a day.
     
  2. jcsd
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