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Laplace Transform with Double Roots

  • #1
I'm trying to prepare for an exam and we were warned that the Laplace transforms given to us may have Complex roots, double roots, or double complex roots.

I'm comfortable with complex roots, but I cant find a problem in the textbook that deals with double roots, and I haven't used them since Linear Algebra and Diff. Equations.

So hypothetically, If current I = (s+12)/[(s+3)2]
S1,2 = -3, -3

and after partial fractions A= 9 and B = 1

what is the inverse transform of I? how does the double root affect the expression? I vaguely remember something about multiplying e of the second double by t. Can someone show me the full expression?
 
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Answers and Replies

  • #2
vela
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I hope you realize we can only guess what A and B are supposed to stand for.
 
  • #3
Apologies, I've been staring at these equations for so long I've lost context.

A/(s+3)^2 + B/(s+3) = (s+12)/[(s+3)^2]
 
  • #4
Char. Limit
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Remember the identity:

[tex]\mathbb{L}\left(t f(t)\right) = -F'(s)[/tex]

Then let 9/(s+3)^2 = -F'(s).
 
  • #5
vela
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Also, chances are your table includes 1/s and 1/s2. Then use the delay property of the Laplace transform to account for the +3 in the denominator.
 

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