Question

Just to be a bother, I'm throwing down a Laplace transform of a trig function and unit step function. I need the laplace of sin(pi*t)-sin(pi*t)[u(t-1)]. I believe that's the one... the function is sin(pi*t) from 0 to 1 and 0 after that. Go! Thanks.

Answer 1

You're not going to attempt it?

Answer 2

Well what I had was [pi/(s^2+pi^2)]-[(e^-s)pi/(s^2+pi^2)] but I don't know if that part is right. Ultimately I need the Laplace of a certain ODE but I didnt want to load all that on you guys, just check that part of it.

Answer 3

I think you have to offset the sine component first, don't you?

Answer 4

Like:\[
\sin(\pi t) = \sin(\pi t - \pi + \pi) = \sin(\pi(t-1)+\pi)
\]

Answer 5

I thought it was prob something along those lines. I will try it thanks.