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The Laplace Transform

Module by: Steven J. Cox. E-mail the author

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As one goes through life one transforms each hard problem into an easy problem, solves the easy one, and then applies the inverse transform to arrive at the solution of the hard problem.

The Laplace transform is the tool used to take dynamical (hard) problems to easy (static) problems. It does so by multiplying a given function, say ht h t , by a complex exponential, ⅇ-st s t , and then integrating over all positive tt. That is,

ℒhs≡∫0∞htⅇ-stdt ℒ h s t 0 h t s t (1)
You should work a few examples by hand before trusting yourself to Matlab.


      

      >> syms t
      
      >> laplace (exp(t))
         ans = 1/(s-1)

      >> laplace (t*exp(-t))
         ans = 1/(s+1)^2

This transform enjoys many wondrous properties. The two most important of these are:

  • Linearity: if aa and bb are constants and ff and gg are functions then
    ℒaf+bg=aℒf+bℒg ℒ a f b g a ℒ f b ℒ g (2)
  • Differentiation is simply Multiplication:
    ℒ f ′=sℒf−f0 ℒ f s ℒ f f 0 (3)
The former property is a direct consequence of the linearity of integration while the latter property follow from integration by parts. Namely, ℒ f ′=∫0∞ f ′tⅇ-stdt=ftⅇ-st|t=0∞+s∫0∞ftⅇ-stdt ℒ f t 0 f t s t t 0 f t s t s t 0 f t s t Supposing that ff and ss are such that ftⅇ-st→0 f t s t 0 as t→∞ t we arrive at Equation 3.

These two properties are instrumental in turning linear differential equations into linear algebraic equations. Once the latter are solved we must then return from the ss world to the tt world. This is accomplished with the inverse Laplace transform, a decidedly more cumbersome object than Equation 1. For most applications it suffices to invert ss functions of the form hs=fsps h s f s p s where ff is smooth and bounded and pp is a polynomial. In this case the inverse Laplace transform of hh is

ℒ-1ht=12πⅈ∫Cfspsⅇstds ℒ h t 1 2 s C f s p s s t (4)
where ⅈ≡-1 -1 and CC is a circle in the complex plane that encloses all of the zeros of pp. Such complex integrals are easily computed via the residue theorem in module of Cauchy's Integral Formula or, even easier, via matlab. For example 


      
      
      
      >> syms s
      
      >> ilaplace(1/(s-1))
         ans = exp(t)

      >> ilaplace(1/(s+1)^2)
         ans = t*exp(-t)

For application of this transform to concrete dynamical systems please visit module of The Matrix Exponential via the Laplace Transform and module of Laplace Example in course ELEC 302.

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