Prove that a function is periodic.

1. Oct 1, 2012

Mr.Tibbs

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. Oct 1, 2012

Staff: Mentor

Welcome to the PF. I have changed the bolded part slightly to state the solution better...

3. Oct 1, 2012

Mr.Tibbs

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∏

4. Oct 1, 2012

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∏ ?

5. Oct 1, 2012

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.

6. Oct 1, 2012

Mr.Tibbs

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 = $\frac{2}{3}$∏

7. Oct 1, 2012

Right.

8. Oct 1, 2012

Mr.Tibbs

Awesome! Makes a lot more sense now! Thank you for all your help!