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I'm using Mathematica to calculate the inverse Laplace transform (ILT) of various functions--functions that I do not know ahead of time. I was stepping through the process and decided to write my own code to simplify the expressions and calculate the inverse transform, but I stumbled upon the general equation that it appears Mathematica uses to calculate the transforms. If we have a function

$$Y(s)=\frac{P(s)}{Q(s)}$$

where P(s) and Q(s) are both polynomials and P(s) is a lower order than Q(s), I have been able to calculate the ILT by evaluating

$$y(t)=\sum_{c=1}^n\frac{P(c)e^{ct}}{Q^{'}(c)}$$

where c is a root of Q(s) and n the total number of roots. According to wikipedia, the formula

$$\mathrm{Res}=\frac{P(c)}{Q^{'}(c)}$$

is an acceptable way to calculate the residue of these irreducible quadratics as long as (according to wikipedia) $P(s)$ and $Q(s)$ are holomorphic and $Q(c)=0$ and $Q^{'}(c)\neq0$. Is there anything else I should be aware of? My irreducible quadratics should always be holomorphic, correct? I have very little knowledge of complex calculus.

Second, I did find a site (Theorem 12.21) that uses this formula residue calculation for irreducible polynomials, however they use go back and plug the result of the residue equation into a formula that is in the s domain still, do partial fraction decomposition, then take the ILT. Why is that necessary if the above equation for $y(t)$ works? If someone could help me understand I would greatly appreciate it. I can provide an example of the types of functions I'm looking at if that would be useful (they are basic).

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  • Let $H(s) = \int_0^\infty e^{at}e^{-st}dt = \frac{1}{s-a}$ so that $H^{(k-1)}(s)=\frac{(k-1)! (-1)^{k-1}}{(s-a)^k}=\int_0^\infty e^{at}(-t)^{k-1}e^{-st}dt$ is the Laplace transform of $(-t)^{k-1} e^{at}1_{t > 0}$.

  • If $\deg(P) < \deg(Q)$ then $\frac{P(s)}{Q(s)}=\frac{P(s)}{\prod_{j=1}^l (s-a_j)^{k_j}} = \sum_{j=1}^l\sum_{k=1}^{k_j} \frac{c_{j,k}}{(s-a_j)^k}$ (partial fraction decomposition)

  • Therefore $F(s)=\frac{P(s)}{Q(s)}$ is the Laplace transform of $f(t)=\sum_{j=1}^l\sum_{k=1}^{k_j} \frac{c_{j,k}}{(k-1)!} t^{k-1} e^{a_j t}1_{t > 0}$ (the domain of convergence is $\Re(s) > \sup_j \Re(a_j)$)

  • The residue theorem is really another way to compute the coefficients $c_{j,k}$ ie. to do the same thing, applying the Cauchy integral theorem to the inverse Laplace transform integral

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