Prove that a function is periodic.

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To prove that the function x(t) = 7sin(3t) is periodic, the equation x(t) = x(t+T) is used, leading to the relationship 3t = 3(t+T). Solving for T reveals that T can take values of 0, 2π, and 4π, but the correct period must account for the multiplier of 3 in the sine function. The periodicity of sin(t) is 2π, but for sin(3t), the period is scaled down to 2π/3. Therefore, the function x(t) is periodic with a period of 2π/3.
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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?
 
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Mr.Tibbs said:
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 = 0,2∏,4∏...

but [STRIKE]T = 0 = 2∏ [/STRIKE] T = 0,2∏

am I right in assuming that it is periodic with a period of 2∏?

Welcome to the PF. I have changed the bolded part slightly to state the solution better...
 
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∏
 
Mr.Tibbs said:
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∏

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∏ ?
 
Mr.Tibbs said:
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∏

No, not quite. The function y = sin(t) is periodic with period ##2\pi##, but the function y = sin(3t) has a shorter period.
 
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}∏
 
Awesome! Makes a lot more sense now! Thank you for all your help!
 

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