Problems with inverse laplace transform

jorchi@gmail.com wrote:


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?

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

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.

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

sX(s) <---> dx(t)/dt