Prove that a function is periodic.

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Homework Help Overview

The discussion revolves around proving that a function, specifically a sine function, is periodic. The original poster presents the function x(t) = 7sin(3t) and explores the implications of its periodicity using the formula x(t) = x(t+T), where T is the period.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the periodicity condition and the implications of solving for T. There is a focus on the relationship between the function's period and the multiplier within the sine function.

Discussion Status

The discussion is ongoing, with participants providing clarifications and corrections regarding the periodicity of the function. Some participants suggest that the function is periodic with a period of 2π, while others note that the presence of the multiplier affects the period, leading to further exploration of the correct period.

Contextual Notes

There is some confusion regarding the implications of T = 0 and how it relates to periodicity. Participants are also considering the effect of the multiplier in the sine function on the overall period.

Mr.Tibbs
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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?
 
Last edited:
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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 = [itex]\frac{2}{3}[/itex]∏
 
Awesome! Makes a lot more sense now! Thank you for all your help!
 

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