Understanding the Heaviside Function and Rewriting Sine Homework

[nicolayh]

Homework Statement

f(t) = \left\{ \begin{array}{rcl}<br /> 5sin(t) &amp; \mbox{for}<br /> &amp; 0 &lt; t &lt; 2\pi \\<br /> 0 &amp; \mbox{for} &amp; t &gt; 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! :)