## Heavyside function for Sine

1. The problem statement, all variables and given/known data

$$f(t) = \left\{ \begin{array}{rcl} 5sin(t) & \mbox{for} & 0 < t < 2\pi \\ 0 & \mbox{for} & t > 2\pi \end{array}\right.$$

Now, the problem is about rewriting f(t). My friend and I decided that it had to be

$$\dfrac{10 - 5e^{-2\pi s}}{s^2 + 1}$$

However, the answer turned out to be

$$\dfrac{5 - 5e^{-2\pi s}}{s^2 + 1}$$

Any help towards understanding this would be greatly appreciated! (We assumed they divided the first part by $$2\pi$$ when we extended $$5sin(t)$$ to $$5sin(t - 2\pi)$$, but we don't understand why!)
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 Recognitions: Gold Member Homework Help Did you begin by writing $$f(t) = 5\sin t(u(t) - u(t-2\pi) )$$?

 Quote by LCKurtz Did you begin by writing $$f(t) = 5\sin t(u(t) - u(t-2\pi) )$$?
Yeah, but we thought in this case that u(t) was $$2\pi$$, I guess that wasn't the case? :P

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## Heavyside function for Sine

 Quote by nicolayh Yeah, but we thought in this case that u(t) was $$2\pi$$, I guess that wasn't the case? :P
Nope, I guess not. u(t) is either 0 or 1.
 Thank you very much! :)