# 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
18,085
4,917
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
18,085
4,917
"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?