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SHOW: x(at) is a periodic signal with period T/a (a>0)

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data

    If x(t) is a periodic signal with period T, show that x(at), a > 0, is a periodic signal with period [itex]\frac{T}{a}[/itex], and [itex]x\left(\frac{t}{b}\right)[/itex], b > 0, is a periodic signal with period bT.



    2. Relevant equations
    HINT: Define [tex]x_a(t)\,=\,x(at)[/tex] and [tex]x_b(t)\,=\,x\left(\frac{t}{b}\right)[/tex]. Show that [tex]x_a\left(t\,+\,T_a\right)\,=\,x_a(t)\,\forall\,t\,\in\,\mathbb{R}[/tex] and [tex]x_b\left(t\,+\,T_b\right)\,=\,x_b(t)\,\forall\,t\,\in\,\mathbb{R}[/tex], where [tex]T_a\,=\,\frac{T}{a}[/tex] and [tex]T_b\,=\,bT[/tex].


    3. The attempt at a solution

    I take the hint, and define

    [tex]x_a(t)\,=\,x(at)[/tex]

    Now, I assume that [itex]x_a(t)[/itex] is periodic, with a period [itex]\frac{T}{a}[/itex]

    [tex]x_a(t)\,=\,x_a\left(t\,+\,\frac{T}{a}\right)[/tex]

    [tex]x_a\left(t\,+\,\frac{T}{a}\right)\,=\,x\left[a\left(t\,+\,\frac{T}{a}\right)\right]\,=\,x\left(at\,+\,T\right)[/tex]

    [tex]\therefore\,x_a\left(t\,+\,\frac{T}{a}\right)\,=\,x_a(t)\,\forall\,t\,\in\,\mathbb{R}[/tex]


    Does this seem right?
     
    Last edited: Sep 9, 2007
  2. jcsd
  3. Sep 11, 2007 #2
    Anyone verify this?
     
  4. Sep 11, 2007 #3

    Avodyne

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    Science Advisor

    Yep, it's right!
     
  5. Sep 11, 2007 #4

    HallsofIvy

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    Staff Emeritus
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    No, it's not right!

    You can't assume that- that's what you are trying to prove!

    But from here on you are okay. You aren't using "[itex]x_a(t)= x_a(t + T/a)[/itex]" you are using a(x+ T)= a(x) which is your hypothesis.

     
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