The Laplace Transform

Overview of Inverse Laplace Transform

Great! We have made it halfway to where we want to be; we know how to apply the Laplace transform, and now we need to learn how to apply the inverse Laplace Transform. Unlike the Laplace Transform, the Inverse Laplace Transform does not have a long set of rules to memorize (hooray!). That being said, having a working knowledge of partial fraction decomposition from calculus is necesarry to solve the Inverse Laplace Transform.

Essentially, now that we have a function in terms of lambda, we want to get it back to a function of t to obtain our solution.
$ F(\lambda) => Inverse Laplace => f(t)H(t) $
When we find our function of time, we will want to multiply by the Heaviside function to define it for t > 0.
The inverse Laplace Tranform is denoted with the following noation:
$\mathcal{L}^{-1}[F(\lambda)]$

Finding the Inverse Laplace Transform

We will want to make sure our function of lambda as a fraction of 2 functions:
$F(\lambda) = \frac{P(\lambda)}{Q(\lambda)}$ where degree of Q > degree of P and a shifted function has the following property:
$F(\lambda)e^{-a\lambda} = \frac{P(\lambda)}{Q(\lambda)}e^{-a\lambda}$
This property suggests the following:
$\mathcal{L}^{-1}[\frac{P(\lambda)}{Q(\lambda)}] = f(t)H(t)$ and a shifted function yields: $\mathcal{L}^{-1}[\frac{P(\lambda)}{Q(\lambda)}e^{-a\lambda}] = f(t)H(t-a)$, showing how important multiplying by the Heaviside function is.

Solving Using Partial Fraction Decomposition

1. First you will split your fraction into a sum of simple fractions by using Partial Fraction Decomposition.
2. Next you will look up the Inverse Laplace for a given function of lambda in a table provided to write the answer (if you are working by hand).
Table of Inverse Laplace Transforms.
Example 1:
Find the Inverse Laplace Transform of $F(\lambda) = \frac{1}{(\lambda - a)(\lambda - b)}$ where a and b are constants.
$F(\lambda) = \frac{1}{(\lambda - a)(\lambda - b)} = \frac{A}{\lambda - a} + \frac{B}{\lambda - b} = \frac{A(\lambda-b)+(\lambda-a)B}{(\lambda-a)(\lambda-b)}$
$ A = \frac{1}{a-b}$
$ B = \frac{1}{b-a}$
$F(\lambda) = (\frac{1}{b-a}) \mathcal{L}^{-1}[\frac{1}{\lambda-b}-\frac{1}{\lambda-a}]$
From the table, we ascertain that:
$\mathcal{L}^{-1}[F(\lambda)] = (\frac{1}{b-a})(e^{bt}-e^{at})H(t)$

Example 2:
$F(\lambda) = \frac{2\lambda^2+6\lambda+10}{(\lambda-1)(\lambda^2+4\lambda+13)}$
$\frac{2\lambda^2+6\lambda+10}{(\lambda-1)(\lambda^2+4\lambda+13)} = \frac{1}{\lambda-1}+\frac{\lambda+2}{(\lambda+2)^2 + 3^2}+\frac{1}{(\lambda+2)^2+3^2}$
From the table we ascertain that:
$\mathcal{L}^{-1}[F(\lambda)] = e^t + e^{2t}\cos(3t) + \frac{1}{3}e^{-2t}\sin(3t)$

Example 3:
$\mathcal{L}^{-1}[F(\lambda)] = \frac{\lambda+3}{\lambda^2(\lambda-1)}$
$F(\lambda)=\frac{-3}{\lambda^2}-\frac{4}{\lambda}+\frac{4}{\lambda-1}$
From the Table we ascertain that:
$\mathcal{L}^{-1}[F(\lambda)] = (-3t -4+4e^t)H(t)$

Example 4:
$\mathcal{L}^{-1}[F(\lambda)] = \frac{\lambda+3}{\lambda^2(\lambda-1)}e^{-2\lambda}$
By recognizing that this is just a shifted version of Example 3, we obtain the answer:
$\mathcal{L}^{-1}[F(\lambda)] = (-3(t-2) -4+4e^{t-2})H(t-2)$