Suppose \(F(s)\) has the general form of
\(\displaystyle F(s) = \frac{N(s)}{D(s)}\)
where \(N(s)\) is the numerator polynomial and \(D(s)\) is the denominator polynomial. The roots of \(N(s) = 0\) are called the zeros of \(F(s)\), while the roots of \(D(s) = 0\) are the poles of \(F(s)\).
Steps to find the Inverse Laplace Transform:
- Decompose \(F(s)\) into simple terms using partial fraction expansion.
- Find the inverse of each term by matching entries in Table of Laplace Transform Pairs.
Three possible forms of \(F(s)\)
Simple Poles
If \(F(s)\) has only simple poles, then \(D(s)\) becomes a product of factors.
\(\displaystyle F(s) = \frac{N(s)}{(s + p_1)(s + p_2)\cdots(s + p_n)}\)
where \(s = -p_1,\,-p_2,\,\dots,\,-p_n\) are the simple poles, and \(p_i \ne p_j\) for all \(i \ne j\).
Use partial fraction to decompose \(F(s)\) as
\(\displaystyle F(s) = \frac{k_1}{(s + p_1)} + \frac{k_2}{(s + p_2)} + \cdots + \frac{k_n}{(s + p_n)}\)
The expansion coefficients \(k_1,\,k_2,\,\dots,\,k_n\) are known as the residues of \(F(s)\). There are many ways of finding the expansion coefficients. One way is using the residue method.
\(\boxed{k_i = \left.(s + p_i)F(s)\right|_{\displaystyle s = -p_i}}\)
This is known as Heaviside’s theorem. Once the values of \(k_i\) are known, find the inverse of \(F(s)\).
\(\displaystyle \mathcal{L}^{-1}\left[\frac{1}{s + a}\right] = e^{-at}\,u(t)\)
\(\boxed{\mathcal{L}^{-1}\left[F(s)\right] = f(t) = \left(k_1\,e^{-p_1 t} + k_2\,e^{-p_2 t} + \cdots + k_n\,e^{-p_n t}\right)\,u(t)}\)
Repeated Poles
Suppose \(F(s)\) has \(n\) repeated poles at \(s = -p\).
\(\displaystyle F(s) = \sum_{m = 0}^{n - 1}\left[\frac{k_{n - m}}{(s + p)^{n - m}}\right] + F_1(s)\)
where \(F_1(s)\) is the remaining part of \(F(s)\) that does not have a pole at \(s = -p\).
Determine the expansion coefficient \(k_n\) as
\(\boxed{k_n = \left.(s + p)^{n}F(s)\right|_{\displaystyle s = -p}}\)
Determine the expansion coefficient \(k_{n - m}\) as
\(\boxed{k_{n - m} = \frac{1}{m!}\left.\left[\frac{d^m}{ds^m}(s + p)^{n}F(s)\right]\right|_{\displaystyle s = -p}}\)
where \(m = 1,\,2,\,\dots,\,(n - 1)\)
Once the values of \(k_1,\,k_2,\,\dots,\,k_n\) are known, find the inverse of \(F(s)\).
\(\displaystyle \mathcal{L}^{-1}\left[\frac{1}{(s + a)^n}\right] = \frac{t^{n - 1}\,e^{-at}}{(n - 1)!}\,u(t)\)
\(\boxed{\mathcal{L}^{-1}\left[F(s)\right] = f(t) = \sum_{m = 0}^{n - 1}\left[\frac{k_{n - m}}{\left(n - m - 1\right)!}\,t^{n - m - 1}\,e^{-pt}\right]\,u(t) + f_1(t)}\)
Complex Poles
A pair of complex poles is simple if it is not repeated; it is a double or multiple pole if repeated. Simple complex poles may be handled the same way as simple real poles, but because complex algebra is involved the result is always cumbersome. An easier approach is a method known as completing the square.
\(F(s)\) may have the general form
\(\displaystyle F(s) = \frac{A_1 s + A_2}{s^2 + as + b} + F_1(s)\)
where \(F_1(s)\) is the remaining part of \(F(s)\) that does not have this pair of complex poles.
By using the method of completing the square.
\(\displaystyle \frac{A_1 s + A_2}{s^2 + as + b} = \frac{A_1 s + A_2}{\displaystyle s^2 + as + \left(\frac{a}{2}\right)^2 - \left(\frac{a}{2}\right)^2 + b}\)
Let \(\displaystyle \alpha = \frac{a}{2}\) and \(\displaystyle \beta^2 = b - \left(\frac{a}{2}\right)^2\)
\(\displaystyle \frac{A_1 s + A_2}{s^2 + 2\alpha s + \alpha^2 + \beta^2} = \frac{A_1 s + A_2}{(s + \alpha)^2 + \beta^2}\)
\(\displaystyle \frac{A_1 s + A_2}{(s + \alpha)^2 + \beta^2} = \frac{A_1 s + A_{1}\alpha - A_{1}\alpha + A_2}{(s + \alpha)^2 + \beta^2} = \frac{A_1 (s + \alpha) + A_2 - A_{1}\alpha}{(s + \alpha)^2 + \beta^2}\)
Then \(F(s)\) becomes
\(\boxed{F(s) = \frac{A_1 s + A_2}{s^2 + as + b} + F_1(s) = \frac{A_1(s + \alpha)}{(s + \alpha)^2 + \beta^2} + \frac{B_1\beta}{(s + \alpha)^2 + \beta^2} + F_1(s)}\)
where
\(\displaystyle \alpha = \frac{a}{2}\)
\(\displaystyle \beta = \sqrt{b - \alpha^2}\)
\(\displaystyle B_1 = \frac{A_2 - A_{1}\alpha}{\beta}\)
The inverse transform is
\(\boxed{\mathcal{L}^{-1}\left[F(s)\right] = f(t) = \left[A_1\,e^{-\alpha t}\cos(\beta t) + B_1\,e^{-\alpha t}\sin(\beta t)\right]\,u(t) + f_1(t)}\)
or
\(\boxed{\mathcal{L}^{-1}\left[F(s)\right] = f(t) = C_1\,e^{-\alpha t}\cos(\beta t - \theta)\,u(t) + f_1(t)}\)
where
\(\displaystyle C_1 = \sqrt{\left.A_{1}\right.^2 + \left.B_{1}\right.^2}\)
\(\displaystyle \theta = \tan^{-1}\left(\frac{B_1}{A_1}\right)\)