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Prove that a function is periodic.

  1. Oct 1, 2012 #1
    Hey all, I want to prove that a function is periodic using the formula:

    x(t) = x(t+T)

    where T is the supposed period.

    An example equation would be:

    x(t) = 7sin(3t)

    I would set up the equation like so:

    7sin(3t) = 7sin(3*(t+T))

    assuming that they equal each other:

    3*t = 3*(t+T)

    solving for T gives:

    T = 0

    but T = 0 = 2∏

    am I right in assuming that it is periodic with a period of 2∏ or since T = 0 is it aperiodic?
    Last edited: Oct 1, 2012
  2. jcsd
  3. Oct 1, 2012 #2


    User Avatar

    Staff: Mentor

    Welcome to the PF. I have changed the bolded part slightly to state the solution better...
  4. Oct 1, 2012 #3
    Thank you for clearing that up for me, so from what you have corrected I believe that it's safe to assume that the equation is periodic with a period of:

    T = 2∏
  5. Oct 1, 2012 #4


    User Avatar

    Staff: Mentor

    Oops, no sorry. I skimmed what you wrote, and missed the multiplier of 3 inside: x(t) = 7sin(3t)

    the sin() function is periodic in 2∏. When you have sin(3t), what does t have to be to make 2∏ ?
  6. Oct 1, 2012 #5


    Staff: Mentor

    No, not quite. The function y = sin(t) is periodic with period ##2\pi##, but the function y = sin(3t) has a shorter period.
  7. Oct 1, 2012 #6
    Assuming that it has the same effect as time scaling, multiplying the period of the function by 3 should create a new period of:

    T = [itex]\frac{2}{3}[/itex]∏
  8. Oct 1, 2012 #7


    Staff: Mentor

  9. Oct 1, 2012 #8
    Awesome! Makes a lot more sense now! Thank you for all your help!
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