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Laplace Transforms


Date: 05/28/97 at 11:54:17
From: Grahame Duke
Subject: Laplace Transforms

I'm stuck on inverse Laplace transform problems. Like everything else, 
the simple ones I can do, but as soon as they get harder, I can`t find 
any textbooks that can help me. 

Can you tell me the solution and step-by-step workings to the 
following:

  L-1 {2(s+1)/(s^2+2s+10)} and 

  L-1 {(9s^2 + 4s - 10)/s(s-1)(s+2)} 

With the first one, I tried using partial fractions, but the bottom 
expression factorizes as a complex root, which I'm not sure how to 
deal with in this situation.

Many thanks for your generous assistance, 
Grahame 


Date: 05/29/97 at 20:02:21
From: Doctor Keith
Subject: Re: Laplace Transforms

Hi Grahame,

I love Laplace transforms; they really make life easier (even though 
they don't seem to be doing that for you right now!).  Sorry to hear 
none of the texts is helping.  

Let's go through this step by step. I will assume you want the causal 
transform pair. If you are not sure what this means, write back and I 
will clarify or you can read a book on signals and systems (most books 
on control systems will not cover it).

1) First complete the square on the bottom:
      
     2(s+1)         2(s+1)
    ---------    = -----------
    s^2+2s+10      (s+1)^2 + 9
   

Now this is actually a standard form in many books:
      
       s+a
   -------------    <-->   e^(-at)cos(wt)u(t)
   (s+a)^2 + w^2

What if you didn't know this? Use the properties of Laplace transform:

    e^(-at)f(t)     <--->   F(s+a)

This shows how the form was figured out from the Laplace pair of 
cos(wt).  I leave the actual answer to you.


2) This can be solved by partial fractions, but note that there is an 
unstable component:

   9s^2+4s-10    a    b     c
   ----------- = - + --- + ---
   s(s-1)(s+2)   s   s-1   s+2

Multiply by s(s-1)(s+2)

   9s^2+4s-10 = a(s-1)(s+2) + bs(s+2) + cs(s-1)
       s=0  =>   -10 = -2a   =>  a=5
       s=1  =>     3 =  3b   =>  b=1
       s=-2 =>    18 =  6c   =>  c=3

So the answer is:

     5u(t) + e^(t)u(t) + 3e^(-2t)u(t)

As I stated earlier, this is an unstable system.

Note that to clear up the causal, non-causal, anti-causal problem I 
mentioned earlier, you will often see an additional piece of 
information called region of convergence or ROC.  It just specifies 
where the Laplace transform is to be evaluated and thus the causality 
of the system.  Hope this helps.  

-Doctor Keith,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus

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