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Even and odd signals

  1. Apr 18, 2007 #1
    Hi i am stuck with something really simple :(
    I know that we can express a signal with the even and odd signal
    (xe(t) means even signal and xo(t) means odd signal)

    x(-t)=x(t) for even signals and (1)
    x(-t)=-x(t) for odd signals (2)

    where even signal is
    xe(t)=1/2[x(t)+x(-t)] (3) and the odd one is
    xo(t)=1/2[x(t)-x(-t)] (4)
    i can validate that x(t)=xe(t)+xo(t) if i use equations 3 and 4
    x(t)=1/2[x(t)+x(-t)]+1/2[x(t)-x(-t)]= 1/2x(t)+1/2x(t)+1/2x(-t)-1/2x(-t)= x(t) done

    My problem arise when i try to use (1)+(2) to (3)+(4) to prove what i want
    using (1) to (3) we have xe(t)=1/2[x(t)+x(t)] =2/2x(t)
    using (2) to (4) we have xo(t)=1/2[x(t)-(-x(t))] =2/2x(t) and that means that i have proved that x(t)=4x(t)

    P.S Plz tell me where i am wrond and correct my bad english mathematical phrases
  2. jcsd
  3. Apr 18, 2007 #2
    Why would you consider equations 1 and 2 at all? A generic function x(t) will not have those properties always. The only function for which both of those equations can be true is x(t) = 0, which is consistent with your result x(t) = 4x(t).
  4. Apr 18, 2007 #3
    I cant understand what u are saying me . Plz try to clarify where i am wrong.
    Thx a lot
  5. Apr 18, 2007 #4
    Ok here is what makes your thinking wrong: You try to use x(-t) = x(t) and x(-t) = -x(t), two VERY SPECIFIC conditions, to show something for any x(t) which could have any sort of shape. What you meant to use for equations 1 and 2 was:

    xe(-t) = xe(t) (1)
    xo(-t) = -xo(t) (2)

    If you assume equations 1 and 2 are true then you are implying x(t) = 0 for all t

    I hope I made myself more understandable.
  6. Apr 19, 2007 #5
    Yes thx a lot i have clearly understand my wrong. But if someone ask me to prove that
    xe(t)=1/2[x(t)+x(-t)] how shouldi think to prove that?
  7. Apr 19, 2007 #6
    I would use the definition of an even function xe(t) = xe(-t), but recall that xe(t) = 1/2[x(t) + x(-t)], so what would xe(-t) be?
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