Problems with inverse laplace transform

You don't need to involve P(s) to find h(t) -- you just need to get the inverse laplace transform of H(s).

Whose idea is it to get it in terms of p(t) and its derivatives?

Well, as I understood my professor, to solve the PD controller, you have to get the value of some mius (u0, u1, u2, ...), which you get from solving the equations

h(t) * p(t) = p(t) u0 + (-T) p'(t) u1 + ((-T)^2/2!) p''(t)u2 + ... + ((-T)^k/k!) p(kth)(t)uk

where p(t) is a kth degree polinomial.

Tim Wescott wrote in news: snipped-for-privacy@corp.supernews.com:

My guess would be some differential equation prof who's trying to teach the relationship between laplace operators and differential equations. It's usually taught near the end of the semester.

Y/P=(s/[(s+1)(s+50)]). Multiply out the denominator, then cross multiply. Take it back to the time domain, knowing that X(s) transforms to x(t), sX(s) transforms to dx(t)/st, and s^2X(s) transforms to the second derivative. If you need to carry through initial conditions, the OP should go look it up in a diffeq text.

Scott

Scott

Scott Seidman wrote in news:Xns964D51D6A5D6Escottseidmanmindspri@130.133.1.4:

oops, typo.

sX(s) dx(t)/dt