From: mailbox@dmitry-kazakov.de   
      
   On 20 Apr 2006 07:09:55 -0700, ann.tighe@gmail.com wrote:   
      
   > The original author's experiments use a transfer function   
   >   
   > X(s)= (0.00375/((s+0.15)*(s+0.25)))*U   
   >   
   > where U is the controller input to the system and X is the system   
   > output.   
   >   
   >>From my reading I believe that to graph this in a time domain I have to   
   > multiply it by 1/s and then get the inverse Laplace transform, am I on   
   > the right track?   
      
   Can't really say anything to the topic. Just regarding the Laplace   
   transformation, the original of   
      
   F(s) = 0.00375 / ((s+0.15)*(s+0.25))   
      
   should be   
      
   f(t) = exp(-0.25t) - exp(-0.15t)   
      
   In general:   
      
   L(exp(-t/a)-exp(-t/b) / (a-b)) = 1/(1+as)(1+bs)   
      
   --   
   Regards,   
   Dmitry A. Kazakov   
   http://www.dmitry-kazakov.de   
      
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