Understanding the Heaviside Function and Rewriting Sine Homework

[nicolayh]

Homework Statement

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

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!)

 

[LCKurtz]

Did you begin by writing

$$f(t) = 5\sin t(u(t) - u(t-2\pi) )$$?

 

[nicolayh]

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

 

[LCKurtz]

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.

 

[nicolayh]

Thank you very much! :)