The entries in our Laplace transform table also allow us to transform F(s) back into a time-domain function f(t).
Zeros and Poles
Consider a Laplace transform F(s) in the following form: $$ F(s) = \frac{N(s)}{D(s)} $$ By setting the numerator of F(s) to zero we can determine the zeros of the function. By setting the denominator to zero, we obtain the poles of the function. $$ let\;N(s) = 0 \;to\;find\;zeros $$ $$ let\;D(s) = 0 \;to\;find\;poles $$ By using partial fraction decomposition, we can break F(s) down into simpler terms that resemble known properties of the Lapace transform. Let's take a look at three possible forms of F(s)
1) Simple Poles
A simple pole is a first-order pole. If F(s) has only a simple pole, then D(s) becomes a product of factors: $$ F(s) = \frac{N(s)}{(s+p_1)(s+p_2)...(s+p_n)} $$ If the degree of N(s) is less than the degree of D(s), partial fraction can be used to break the function down into simpler terms. $$ F(s) = \frac{k_1}{s+p_1} + \frac{k_2}{s+p_2} + \frac{k_n}{s+p_n} $$ ...where: $$ k_1, k_2, k_n = expansion\;coefficients $$
Determining expansion coefficients
Given: $$ F(s) = \frac{k_1}{s+p_1} + \frac{k_2}{s+p_2} + \frac{k_n}{s+p_n} \qquad,(Eqn\;1) $$ We can use the Heaviside Theorem to obtain the expansion coefficients
The Heaviside Theorem states that: $$ k_i = (s+p_i) F(s) \Big|_{s=-p_i} $$
Once the expansion coefficients are determined, we can determine the inverse Laplace transform of each term in equation #1 using property #2 in our table as follows: $$ \mathcal{L}^{-1}\Big[ \frac{k}{s+a} \Big] = ke^{-at} $$ Therefore, for equation #1 we obtain:
$$ f(t) = (k_1e^{-p_1t} + k_2e^{-p_2t} + ... + k_ne^{-p_nt} $$
2) Repeated Poles
Consider a Laplace transform with the following form that has "n" repeated poles at s = -p:
$$ F(s) = \frac{k_n}{(s+p)^n} + \frac{k_{n-1}}{(s+p)^{n-1}} + ... + \frac{k_2}{(s+p)^2} + \frac{k_1}{s+p} + F_1(s) \qquad,(Eqn\;2) $$
...where F_1(s) is the remaining part of F(s) that does not have a pole at s = -p
Determining k_n
To determine k_n, we use the Heaviside theorem that was described above: $$ k_n = (s+p_n) \; F(s) \Big|_{s=-p} $$
Determining k_n-1, k_n-2...
To get the coefficient k_n-1, we begin by multiplying each term in equation #2 by (s+p)^n and then differentiate to get rid of k_n. Next, we evaluate the result at (s = -p) to get rid of the other coefficients (except k_n-1). The end result is the following: $$ k_{n-1} = \frac{d}{ds} [(s+p)^n \; F(s)] \Big|_{s=-p} $$ Proceeding onwards to k_n-2, we have: $$ k_{n-2} = \Big(\frac{1}{2!}\Big) \frac{d^2}{ds^2}[(s+p)^n \; F(s)]\Big|_{s=-p} $$ A more general expression for this technique (called the residue method) is shown below:
The Residue Method
$$ k_{n-m} = \Big(\frac{1}{m!}\Big) \frac{d^m}{ds^m}[(s+p)^n \; F(s)]\;\Big|_{s=-p} $$
As you might imagine, differentiation becomes more cumbersome as the value of "m" increases. Once the values of "k" are determined, we obtain the inverse Laplace transform for each term on the right-hand side of equation #2 (using property #23 in our table): $$ \mathcal{L}^{-1} \Big[ \frac{1}{(s+a)^n} \Big] = \frac{t^{n-1}e^{-at}}{(n-1)!} $$ ...and obtain the following result:
$$ f(t) = k_1e^{-pt} + k_2te^{-pt} + \frac{k_3}{2!}t^2e^{-pt} + ... + \frac{k_n}{(n-1)!}t^{n-1}e^{-pt} + f_1(t) $$
Complex Poles
A pair of complex poles is simple if it is not repeated. It is a double/multiple pole if it is repeated. Although simple complex poles can be handled the same way as simple real poles, the result is cumbersome due to complex algebra. For this reason, an approach that involves "completing the square" makes life easier. We will deal with this when we get to some example problems.
Objective when completing the square:
The goal is to express each complex pole pair in the denominator (which are quadratic expressions) as a complete square in the form of: $$ (s+a)^2 + B $$ This form allows us to use our table of Laplace transforms to find the inverse.
Now, let's take a look at our first example problem involving the inverse Laplace transform...