Heavyside function for Sine

nicolayh

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

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

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

[tex] \dfrac{10 - 5e^{-2\pi s}}{s^2 + 1} [/tex]

However, the answer turned out to be


[tex] \dfrac{5 - 5e^{-2\pi s}}{s^2 + 1} [/tex]

Any help towards understanding this would be greatly appreciated! (We assumed they divided the first part by [tex]2\pi[/tex] when we extended [tex]5sin(t)[/tex] to [tex]5sin(t - 2\pi)[/tex], but we don't understand why!)

LCKurtz

Did you begin by writing

[tex]f(t) = 5\sin t(u(t) - u(t-2\pi) )[/tex]?

nicolayh

Yeah, but we thought in this case that u(t) was [tex]2\pi[/tex], I guess that wasn't the case? :P

LCKurtz

Nope, I guess not. u(t) is either 0 or 1.

nicolayh

Thank you very much! :)