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Even odd functions

  1. Sep 6, 2011 #1
    1)-why is x(t)+x(-t) always even??..no matter if x(t) even or odd???

    2)-when we talk about unit step function...u(t)..and we add..u(t)+u(-t)..the value of both is 1 at t=0..so does'n't that gets added twice??..and it becomes 2 at t=0...

    3)when we have x(-t) and we time shift it say x(-t-3) it shifts toward the -ve t axis.. where as x(t-3) the function is shifted on the + axis..why is it so??

    i would be really greatful if you can help me out with the above 3 doubts..
    Last edited by a moderator: Sep 28, 2011
  2. jcsd
  3. Sep 6, 2011 #2
    What did you try to answer this??

    For the first, fill in -a in x(t)+x(-t) and see if


    For the second one. It actually never matters what the unit step function is in 0. So saying that u(t)+u(-t)=2 in 0, is correct, but it doesn't matter.
    Note that a lot of people choose that the unit step function is 0, or 1/2 in 0.

    The third one. We actually have a function y(t)=x(-t). Then x(-t-3)=y(t+3). So it makes sense that it gets shifted to the other side.
  4. Sep 8, 2011 #3
    okh..thanks for answering..i got the first 2 parts...
    buh i m still a little confuse about the third...see when we say there are 2 functions x(t) and x(t-2) it means x(t-2) is delayed by 2 sec with respect to x(t) therefore it shifts in the positive x direction.. does'n't the same apply for x(-t) and x(-t-2) the second function is time delayed by 2 secs with respect to the first so even it should shift to the positive x axis..isn't it????
    Last edited by a moderator: Sep 28, 2011
  5. Sep 8, 2011 #4
    The - in front of the t reverses the direction. So shifting towards the positive axis becomes shifting towards the negative axis and vice versa.
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