Converting periodic function from radians to time?

Discussion in 'Homework Help' started by jean28, Dec 3, 2012.

  1. jean28

    Thread Starter Member

    Sep 5, 2012
    76
    0
    f(θ) = (80/∏2) θ, -∏/2 ≤ θ ≤ ∏/2
    (80/∏) - (80/∏2) θ, ∏/2 ≤ θ ≤ 3∏/2

    2. Relevant equations

    θ = ω0 t
    ω0 = 2∏/T

    3. The attempt at a solution

    After substituting the relevant equations into the original equation and using a bit of logic regarding the periods and time intervals, I got the following function:

    (160t)/(∏ * T) , -T/4 ≤ t ≤ T/4
    (80T -160)/(∏ * T) t, T/4 ≤ t ≤ 3T/4


    However, I don't know if this is the correct input. I would appreciate it if anyone could tell me if my calculations are correct.

    Thank you all very much.
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,795
    What have you done to check your own work? That is a critical skill that you must develop.

    Doing a dimensional analysis shows that you have a problem.

    Also, you have two definitions for f(θ) at θ=∏/2. Are the consistent with each other? In both domains?
     
    Last edited: Dec 5, 2012
  3. jean28

    Thread Starter Member

    Sep 5, 2012
    76
    0
    Yes. Both functions occur at θ=∏/2 since it is a piecewise function. Is there any way to graph this in MATLAB, for example?
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,795
    "Yes" as in, Yes, they are consistent in both domains?

    Are you sure?

    You have:
    (160t)/(∏ * T) , -T/4 ≤ t ≤ T/4
    (80T -160)/(∏ * T) t, T/4 ≤ t ≤ 3T/4

    Is (160t)/(∏ * T) evaluated at T/4 the same as (80T -160)/(∏ * T) t evaluated at T/4?
     
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