Impedances, Efforts and Flow in Passive Systems

While trying to design a controller for a passive system, I am a bit confused regarding the problem stated below:
All such functions are written like:
e(s) = Z(s).f(s) where e is effort, and f is flow, for example, a mechanical system would be described as
F(s) = Z(s).V(s) where F is force, and V is velocity.
I am getting confused that all this is in Laplace domain, what if I were to use this concept in time-domain? How would I translate a hybrid impedance transfer function from Laplace to time? Ofcourse I can take I.Laplace, but I feel I am missing something very basic.. Any help???
Thanks.
Tom
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
In addition to it.....
Lets say if I use Z(s) as a Simulink Block, could the inputs and outputs be treated as f(t) and v(t) instead of F(s) and V(s) or would I have to change something? I am really getting confused. What is the laplace transform of an input signal v(t) which is generated by a singnal generator in Simulink? ..... Help me....
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Tom wrote:

In Simulink a transfer function block Z(s) is turned into a time-domain differential equation and executed. Similarly a discrete-time H(z) is turned into a difference equation.
In Simulink the Laplace transform of an input signal is never taken -- everything stays in the time domain.
--

Tim Wescott
Wescott Design Services
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Ohh okay. Tim Thank you very much for your explanation. Now it makes sense to me, I think. But just as a matter of discussion, so it means laplace domain is for design and analysis only, it is never used for simulation!! Am I right here?
-Tom
Tim Wescott wrote:

would I

time-domain
is
--
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Yes. The main reason for this is that there's no reason to simulate an purely linear, time-invariant system, because you can get perfect, more general results much easier using Laplace. On the other hand you can't use Laplace at all to get results from a nonlinear system, so you have to solve the differential equations numerically -- and that's what you do when you simulate.
Tom wrote:

--

Tim Wescott
Wescott Design Services
  Click to see the full signature.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Thanks Tim. Now I understand it. I visited your site too. It is very nice, esp. enjoyed your articles on z-transforms. Good work.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.