2 Inversion of the Laplace Transform

Author

Affiliation

Enrico Bertolazzi

University of Trento, Department of Industrial Engineering

2.1 Partial Fraction Expansion

The inversion of a Laplace transform is a crucial step that allows us to transport the algebraic solution found in the s domain back into a time-domain description. This process is represented by the operator \mathscr{L}^{-1}.

Rational Polynomial Representation

When dealing with (systems of) ordinary differential equations, the result of the Laplace transform (such as \hat{x}(s)) is typically expressed as a rational function in the form:

\frac{P(s)}{Q(s)},

where P(s) and Q(s) are polynomials with real coefficients in the variable s.

An important observation is that, in many practical applications, the degree of the numerator polynomial \partial P is less than that of the denominator polynomial \partial Q. If this condition does not hold naturally, we can apply polynomial long division. Specifically, if P(s) can be expressed as:

P(s) = T(s)Q(s) + R(s),

where T(s) is the quotient and R(s) is the remainder, we can rewrite the rational function as:

\frac{P(s)}{Q(s)} = T(s) + \frac{R(s)}{Q(s)}.

The term represented by the polynomial T(s), which has real coefficients, corresponds to the Laplace transform of Dirac delta functions. For example:

\begin{aligned} \mathscr{L}\{\delta(t)\}(s) & = \int_{0^-}^\infty \delta(t) e^{-st} \, \mathrm{d}t = e^{-st} \Big|_{t=0} = 1, \\ \mathscr{L}\{\delta'(t)\}(s) & = -\int_{0^-}^\infty \delta'(t) e^{-st} \, \mathrm{d}t = -\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{-st}\right) \Big|_{t=0} = s. \end{aligned}

2.2 Inverting the Transform

Assuming we have a transform solution expressed as a simplified rational polynomial

\frac{P(s)}{Q(s)} \quad\textrm{such that}\quad \partial P < \partial Q

we can compute its inverse to express it in the time domain. This is commonly done using partial fraction expansion, which allows us to rewrite the solution as a sum of simpler elements that can be easily inverted using a Laplace transform table.

To properly perform partial fraction expansion, it is essential to rewrite the denominator of the polynomial ratio in factored form:

\frac{P(s)}{Q(s)} = \frac{b_0 + b_1 s + b_2 s^2 + \dots + b_m s^m} {(s - p_1)^{m_1} (s - p_2)^{m_2} \cdots (s - p_n)^{m_n}},

where p_i are the roots of the denominator and each root has a multiplicity m_i.

Depending on the multiplicity and nature (real or complex conjugate) of these roots, the procedure for performing partial fraction expansion may vary.

In the following sections, we will present examples illustrating each case of partial fraction expansion based on different root types and multiplicities.

Real roots

Single real root

Let’s consider the following rational polynomial with no multiple roots:

G(s) = \frac{s}{(s-1)(s+3)(s-4)}

it’s factorization is based on determining the coefficients \alpha_i such that

\frac{\alpha_1}{s-1} + \frac{\alpha_2}{s+3} + \frac{\alpha_3}{s-4} = \frac{s}{(s-1)(s+3)(s-4)} \tag{2.1}

This can be accomplished in two ways. The first method involves computing the sum of the three basic elements in a parametric form that depends on \alpha_i, and then solving the resulting linear system to determine the parameters that satisfy the equality.

A second, more efficient approach is to recognize that each coefficient can be determined as follows:

\alpha_i = \lim_{s\rightarrow p_i} \big(s-p_i\big) G(s)

In the context of the example, to compute the coefficient \alpha_1 associated with the root p = 1, we can multiply Equation 2.1 by the factor (s - 1). This yields the following equation:

\frac{s\cancel{(s-1)}}{\cancel{(s-1)}(s+3)(s-4)} = \frac{\alpha_1}{\cancel{s-1}} \cancel{(s-1)} + {(s-1) \left( \frac{\alpha_2}{s+3} + \frac{\alpha_3}{s-4} \right)}

Evaluating this expression at the point s = 1 will yield an equation in which the only unknown coefficient is \alpha_1. This occurs because the coefficients \alpha_2 and \alpha_3 are effectively eliminated due to the multiplication by (s - 1), which equals zero at this point. Therefore, we have:

\xrightarrow{s=1} \quad \alpha_1 = \frac{1}{(1+3)(1-4)} = -\frac 1 {12}

By applying the same method to the other two roots, -3 and 4, we can derive the coefficients:

For the root -3, we obtain \alpha_2 = -\frac{3}{28}.
For the root 4, we find \alpha_3 = \frac{4}{21}.

With these coefficients in hand, we can now verify that:

-\frac{1}{12} \frac 1 {s-1} -\frac 3 {28} \frac 1 {s+3} + \frac 4 {21} \frac{1}{s-4} = \frac{s}{(s-1)(s+3)(s-4)}

Each basic element can be easily transform using the Laplace table, and in particular

\mathscr{L}^{-1}\{G(s)\}(t) \ : \quad -\frac 1 {12} \mathrm{e}^{-t} - \frac 3{28} \mathrm{e}^{3t} + \frac 4 {21} \mathrm{e}^{-4t}

Multiple real roots

Let’s now consider a case where the rational polynomial has multiple roots, specifically:

G(s) = \frac{1+s}{(s-4)^4}.

In this scenario, the partial fraction decomposition should not consist of four identical terms with the denominator s - 4. Instead, we need to consider four distinct exponential terms that increase in complexity, represented as follows:

\frac{\alpha_1}{s-4} + \frac{\alpha_2}{(s-4)^2} + \frac{\alpha_3}{(s-4)^3} + \frac{\alpha_4}{(s-4)^4} = \frac{1+s}{(s-4)^4} \tag{2.2}

This problem can be addressed by summing all the elements and solving the resulting linear system for the coefficients \alpha_i. However, a more efficient approach is to recognize that the element a_{n-k}—where n denotes the multiplicity of the root (in this case, n = 4)—can be expressed as follows:

\frac{1}{k!} \lim_{s\rightarrow p} \frac{\mathrm{d}^k}{\mathrm{d}s^k} \big[ (s-p)^n G(s) \big]

This process becomes clearer when we consider the example provided. The first step is to multiply Equation 2.2 by the polynomial (s - 4)^4, which leads us to the following relationship:

1+s = \alpha_1(s-4)^3 + \alpha_2(s-4)^2 + \alpha_3(s-4) + \alpha_4

Evaluating this expression at s = 4 allows us to determine the coefficient \alpha_4 = 5. To find the other coefficients, we can employ the proposed scheme, which involves differentiating the previous equation with respect to s and then evaluating it at the point s = 4.

\begin{aligned} & \downarrow \mathrm{d}/\mathrm{d}s \\[1em] 1 & = 3 \alpha_1 (s-4)^2 + 2 \alpha_2(s-4) + \alpha_3 \quad & \xrightarrow{s=4} \quad \alpha_3 = 1 \\[1em] & \downarrow \mathrm{d}/\mathrm{d}s \\[1em] 0 & = 6 \alpha_1 (s-4) + 2 \alpha_2 \qquad & \xrightarrow{s=4} \quad \alpha_2 = 0 \\[1em] & \downarrow \mathrm{d}/\mathrm{d}s \\[1em] 0 & = 6 \alpha_1 (s-4) \quad & \xrightarrow{s=4} \quad \alpha_1 = 0 \end{aligned}

Once the coefficients \alpha_i associated with the partial fraction expansion have been determined, we can compute the inverse transform of the original function G(s), which is given by:

\begin{aligned} \mathscr{L}^{-1}\{G(s)\}(t) &= \mathscr{L}^{-1}\left\{\frac{1}{(s-4)^3}\right\} +5 \mathscr{L}^{-1}\left\{\frac{1}{(s-4)^4} \right\} \\[1em] &= \frac{1}{2}\mathrm{e}^{4t}t^2 + 5 \frac{1}{6} \mathrm{e}^{4t}t^3 \end{aligned}

Example 2.1 (partial fraction expansion with different roots, one which is multiple) Let’s consider the following rational polynomial G(s) that’s already been split in the factor for the partial expansion:

\begin{aligned} G(s) & := \frac{1+s}{(s-1)(s+2)^3} \\ & = \frac{A}{s-1} + \frac{B}{s+2} + \frac{C}{(s+2)^2} + \frac{D}{(s+2)^3} \end{aligned}

To calculate the first coefficient A associated with the root s = 1, we can follow these straightforward steps:

Multiply by a Factor: Multiply the last equation by the factor (s - 1)(s + 2)^2. This step isolates the behavior of the function around the specified root.
Evaluate at s = 1: After performing the multiplication, evaluate the resulting expression at s = 1. This evaluation will yield the value of the coefficient A.

By following these steps, we can effectively determine the coefficient A related to the root s = 1.

\begin{aligned} 1+s & = A(s+2)^3 + (s+1) \Big[ B(s+2)^2 + C(s+2) + D \Big] \\ & \downarrow\quad s=1 \\ 2 & =A 3^3 \qquad \implies \quad A = \frac 2 {27} \end{aligned}

The next step is to calculate the remaining coefficients B, C, D associated with the multiple root s = -2. However, the current form of the equation does not allow us to apply the previously demonstrated method effectively.

Deflating the Equation

To create a suitable equation for our analysis, we need to deflate it. This involves expanding the term (s + 2)^3, which corresponds to the coefficient A. By expanding this term and moving it to the left-hand side of the equation, we can simplify our calculations and isolate the coefficients B, C, D more effectively.

\begin{aligned} & 1+s - \frac{2}{27} \Big( s^2 + 6s^2 + 12s +8 \Big) \\ & \qquad = (s+1) \Big[ B(s+2)^2 + C(s+2) + D \Big] \\ & \qquad = \frac{1}{27} \Big( -2s^3 -12s^2+35s +11 \Big) \end{aligned}

Once we have deflated the equation properly, we can proceed with determining the coefficients associated with the multiple root. This will provide clarity on the polynomial’s behavior around s = -2.

At this stage, we can divide both sides of the equation by the factor s-1. If division on the left-hand side is not possible, it indicates that an error has occurred in the calculations leading up to this point. By performing polynomial division, we can simplify the equation and obtain the following result:

\frac{1}{27} \left( -2s^2 -14s -11 \right) = B(s+2)^2 + C(s+2) + D

From this point onward it’s possible to use the method shown to calculate the coefficient of the multiple roots; starting evaluating the previous expression at point s=-2 we get the value

D = \frac{-8+28-11}{27} = \frac{1}{3}

Continuing with differentiation we get

\begin{aligned} & \downarrow \mathrm{d}/\mathrm{d}s \\[1em] \frac{-4s -14}{27} & = 2B(s+2) + C \quad & \xrightarrow{s=-2} \quad C & = -\frac 2 9 \\[1em] & \downarrow \mathrm{d}/\mathrm{d}s \\[1em] \frac{-4}{27} & = 2B & \xrightarrow{s=-2} \quad B & = -\frac{2}{27} \end{aligned}

To end the inversion of the starting transform G(s) we can see that

\begin{aligned} \mathscr{L}^{-1}\{G(s)\}(t) & = \frac{2}{27} \mathscr{L}^{-1}\left\{\frac{1}{s-1}\right\}(t) - \frac{2}{27} \mathscr{L}^{-1}\left\{\frac{1}{s+2}\right\}(t) \\[1em] & - \frac{2}{9} \mathscr{L}^{-1}\left\{\frac{1}{(s+2)^2}\right\}(t) + \frac{1}{3} \mathscr{L}^{-1}\left\{\frac{1}{(s+2)^3}\right\}(t) \\[1em] & = \frac{2}{27} \mathrm{e}^t - \frac{2}{27}\mathrm{e}^{-2t} - \frac{2}{9} \mathrm{e}^{-2t}t + \frac{1}{3} \cdot \frac{1}{2}\mathrm{e}^{-2t} t^2 \end{aligned}

Complex roots

Handling complex roots in the denominator Q(s) of the solution to an ordinary differential equation presents greater computational challenges compared to dealing with real roots. One important observation is that if the complex value

\alpha = a +\mathrm{i}b \in\Bbb{C}

is a root of the denominator, so that Q(\alpha) = 0. It’s conjugate

\overline{\alpha} = a - \mathrm{i}b

is also a root, i.e. Q(\overline{\alpha}) = 0. Considering in fact that the polynomial denominator Q(s), evaluated in the point s = \alpha, can be expressed as

Q(\alpha) = q_0 + \alpha q_1 + \alpha^2 q_2 + \dots + \alpha^n q_n = 0

where q_i are real coefficients. At this point we have to consider the following property of the conjugate that are true for all couple of complex variables \alpha,\beta:

\begin{split} (i) & \qquad \overline{\big( \alpha + \beta \big)} = \overline{\alpha} + \overline{\beta} \\ (ii) & \qquad \overline{\big(\alpha \beta\big)} = \overline{\alpha}\overline{\beta} \\ (iii) & \qquad \alpha \overline{\alpha} = |\alpha|^2 \end{split}

Now known that Q(\alpha) = 0, than also the conjugate \overline{Q(\alpha)} should be equal to zero (because \overline{0} = 0) and so by applying the properties we can see that

\begin{aligned} 0 & = Q(\alpha) = \overline{Q(\alpha)} = \overline{q_0 + \alpha q_1 + \alpha^2 q_2 + \dots + \alpha^n q_n} \\ & = \overline{q_0} + \overline{\alpha} \overline{q_1} + \overline{\alpha^2q_2} + \dots + \overline{\alpha^nq_n} \\ & = q_0 + \overline{\alpha} q_1 + \overline{\alpha}^{2} q_2 + \dots + \overline{\alpha}^{n} q_n = Q\big(\overline{\alpha}\big) \end{aligned}

This verifies that if \alpha is a root of Q(s), than also it’s conjugate \overline{\alpha} is a root of the same polynomial and so we can rewrite the denominator in the form

Q(s) = \dots \underbrace{\big(s-\alpha\big) \big(s-\overline{\alpha}\big)}_\textrm{conj. roots} \dots

Single conjugated complex roots

Let’s now assume that we have a rational polynomial that present two conjugated complex roots expressed in the form

\frac{P(s)}{Q(s)} = \frac{P(s)}{(s-\alpha)(s-\overline{\alpha})} = \frac A {s-\alpha} + \frac{B}{s-\overline{\alpha}}

By trying to apply the partial fraction expansion as for the real root previously described, we can demonstrate that the two coefficients A and B are conjugated, so such that B = \overline{A}. In fact by summing the two elements we can see that

\begin{aligned} \frac A {s-\alpha} + \frac{B}{s-\overline{\alpha}} & = \frac{A(s-\overline{\alpha}) + B(s-\alpha)}{(s-\alpha)(s-\overline{\alpha})} \\ &= \frac{(A+B)s - (A \overline{\alpha}+ B\alpha)}{(s-\alpha)(s-\overline{\alpha})} \end{aligned}

By expanding the definition of the complex variable (so expanding A = a+\mathrm{i}b, B = c+\mathrm{i}d…) and considering that the polynomial P(s) at the numerator has only real coefficients it’s possible to see that, in order to verify the equality, it must be a = c and b = - d, and so the two complex values A and B must be conjugated.

Proven that B = \overline{A} now the problem is to find such complex value A in order to satisfy the partial fraction expansion; in order to do so we can try to use the same approach shown for real roots by multiplying both members of the equation by Q(s) and then evaluating the result for s = \alpha:

\frac{P(s)}{(s-\alpha)(s-\overline{\alpha})} = \frac A {s-\alpha} + \frac{B}{s-\overline{\alpha}}

\xrightarrow[Q(s)]{\times} \qquad P(s) = A(s-\overline{\alpha}) + \overline{A}(s-\alpha)

\implies \qquad P(\alpha) = A (\alpha-\overline{\alpha}) + \cancel{\overline{A}(\alpha-\alpha)}

Considering now that \alpha-\overline{\alpha} conresponds to a purely imaginary number equals 2 times the imaginary part of the root \alpha, we can invert that relation in order to explicitly get A:

A =\frac{P(\alpha)}{2\mathrm{i}\Im(\alpha)} = -\mathrm{i}\frac{P(\alpha)}{2\Im(\alpha)}

Let’s now try to consider a real case scenario where we want to deal with the inversion of the following rational polynomial (that’s consequently expanded):

\frac{s}{ \big( s - (1+\mathrm{i}) \big)\big( s - (1-\mathrm{i}) \big) } = \frac A {s - (1+\mathrm{i})} + \frac{\overline{A}}{s-(1-\mathrm{i})}

We can now multiply both members of the equation by the denominator Q(s) and then evaluate the result for the point s = 1+\mathrm{i} (root of the denominator):

\begin{aligned} s &= A \big(s-(1-\mathrm{i})\big) +\overline{A}\big(s-(1+\mathrm{i})\big) \\[1em] & \downarrow\quad s = 1+\mathrm{i}\\[1em] 1+\mathrm{i} & = A \big(1+\mathrm{i} - 1 + \mathrm{i}\big) \\[2em] \implies \qquad A & = \frac{1+\mathrm{i}}{2\mathrm{i}} = \frac{1}{2}\big(1 - \mathrm{i} \big) \end{aligned}

With this value retrieved we can now state that

\frac{s}{ \big( s - (1+\mathrm{i}) \big)\big( s - (1-\mathrm{i}) \big) } = \frac{1}{2}\left(\frac {1-\mathrm{i}} {s - (1+\mathrm{i})} + \frac {1+\mathrm{i}}{s-(1-\mathrm{i})} \right)

Unfortunately watching the Laplace table we can see there’s no possible match between the known transform (where the numerator is expressed as a pure real number) and the complex coefficient retrieved. But an interesting fact that we can note is that by merging the two components we can get a rational polynomial with real coefficient: in a general way we can in fact state that

\begin{aligned} \frac A {s-\alpha} + \frac{\overline{A}}{s-\overline{\alpha}} & = \frac{A(s-\overline{\alpha}) + \overline{A}(s-\alpha)}{(s-\alpha)(s-\overline{\alpha})} \\[1em] & = \frac{ s\big(A+\overline{A}\big) - A\overline{\alpha} - \overline{A}\alpha }{s^2 - s \big(\alpha+\overline{\alpha}\big) + |\alpha|^2} \end{aligned}

This rational polynomial can be matched with the following transform:

\begin{aligned} \mathscr{L}\{C\mathrm{e}^{at}\cos(\omega t) + D \mathrm{e}^{at}\sin(\Omega t)\} & =C \frac{s-a}{(s-a)^2+\omega^2} + D \frac{\omega}{(s-a)^2 + \omega^2} \\ & = \frac{Cs - \big(D\omega - Ca\big)}{(s-a)^2 + \omega^2 } \end{aligned}

Given now the values of \alpha,A, the problem is to find the expression of the coefficients a,\omega,C and D; in order to to so we can perform two steps:

match the denominator in order to find a and \omega; in order to compare the denominator it’s useful to expand the denominator (s-a)^2 + \omega^2 = s^2 - 2sa + a^2 + \omega^2. At this point by relating the function we can see that \cancel {s^2} - s \big(\alpha+\overline{\alpha}\big) + |\alpha|^2 = \cancel{s^2} - 2sa + a^2 + \omega^2

\begin{aligned} & \implies \quad \left\{\begin{aligned} -2sa &= - s\big(\alpha + \overline{\alpha}\big) \\ \alpha^2+\omega^2 &= |\alpha|^2 \end{aligned}\right. \quad \implies \quad a = \frac{\alpha + \overline{\alpha}}{2} = \Re(\alpha) \\[1em] & \implies \quad \omega = \sqrt{|\alpha|^2-a^2} = \sqrt{|\alpha|^2-\Re(\alpha)^2} = \sqrt{\Im(\alpha)^2} = \pm \Im(\alpha) \end{aligned}
at this point, known a and \omega, it’s possible to match the numerator in order to get the coefficients C and D: s\big(A+\overline{A}\big) - A\overline{\alpha} - \overline{A}\alpha = Cs - \big(D\omega - Ca\big)

\begin{aligned} \implies & \quad \begin{cases} A + \overline{A} = C \\ -A\overline{\alpha} - \overline{A}\alpha = D\omega - Ca \end{cases} \\[2em] \implies & \quad C = A+\overline{A} = 2 \Re(\alpha) \\[1em] \implies & \quad D = \frac{Ca - A \overline{\alpha} - \overline{A} \alpha}{\omega} \end{aligned}

Example 2.2 (inversion of a transform with two conjugated complex roots) Let’s consider the following rational polynomial that has to be inverted:

\frac{s+1}{s^2+s+10}

At this point we can verify that the roots of the denominator are complex conjugated by simply applying the quadratic formula to retrieve them

s_{1,2} = \frac{-1\pm\sqrt{-9}}{2}.

At this point the problem of the inversion if finding the parameters a,\omega,C,D such that

\frac{s+1}{s^2+s+10} = C \frac{s-a}{(s-a)^2+\omega^2} + D \frac{\omega}{(s-a)^2 + \omega^2}

By comparing the denominators we can observe that -2sa = s and so a = -\frac{1}{2} and so

a^2+ \omega^2 = 10 \qquad \implies \quad \omega = \sqrt{10-a^2} = \frac{\sqrt{39}}{2}

At this point it’s possible to compare the denominator (considering that now a,\omega are known) that’s lead to

s + 1 = C\left( s - a\right) + D\omega \quad \implies \quad C = 1,\quad D = \frac{1+Ca}{\omega} = \frac{1} {\sqrt{39}}

Retrieved all the coefficient we can explicitly state the inversion of the original rational polynomial:

\mathscr{L}^{-1}\left\{\frac{s+1}{s^2+s+10} \right\}(t) = \mathrm{e}^{-\frac 1 2t} \left[ \cos\left( \frac{\sqrt{39}}{2} t \right) + \frac{1}{\sqrt{39}} \sin\left( \frac{\sqrt{39}}{2} t \right) \right]

Multiple complex roots

Dealing with multiple complex roots requires even more computational steps; let’s consider the generic case of a multiple complex root of multiplicity of 2 that’s already expanded:

\frac {P(s)}{(s-\alpha)^2(s-\overline{\alpha})^2} = \frac{A}{s-\alpha} + \frac{B}{(s-\alpha)^2} + \frac{\overline{A}}{s-\overline{\alpha}} + \frac{\overline{B}}{(s-\overline{\alpha})^2}

As in all the previously cases in order to compute the first parameter B it’s possible to multiply both sides of the equation by the denominator Q(s) = (s-\alpha)^2(s-\overline{\alpha})^2 and then evaluate the relation at the root s = \alpha:

\begin{aligned} P(s) & = A (s-\alpha)(s-\overline{\alpha})^2 + B (s-\overline{\alpha})^2 \\ & + \overline{A}(s-\overline{\alpha})(s-\alpha)^2 + \overline{B}(s-\alpha)^2 \end{aligned}

\implies \quad P(\alpha) = B \underbrace{\big(\alpha-\overline{\alpha}\big)}_{=2\Im(\alpha)}^2 \qquad \implies \quad B = \frac{P(\alpha)}{4 \Im(\alpha)^2}

Now in order to calculate the second parameter A of the relation we have to deflate te elements associated to B and then, as for the multiple real roots, compute the derivative in respect to the variable s:

\begin{aligned} & \,\downarrow \textrm{deflation} \\[1em] P(s) - B(s-\overline{\alpha})^2 - \overline{B}(s-\alpha)^2 & = A(s-\alpha)(s-\overline{\alpha})^2 + \overline{A}(s-\overline{\alpha})(s-\alpha)^2 \\[1em] & \, \downarrow \ \mathrm{d}/\mathrm{d}s \\[1em] P'(s) - 2B\big(s-\overline{\alpha}\big) - 2\overline{B}\big(s-\alpha\big) & = A \big(s-\overline{\alpha}\big) \Big[ s-\overline{\alpha} + s \big(s-\alpha\big) \Big] \\[1em] & \ + \overline{A} \big(s-\alpha\big)\Big[ s-\alpha + 2\big(s-\overline{\alpha}\big) \Big] \end{aligned}

By now evaluating the expression at the point s = \alpha it’s possible to have a clear definition of the complex variable A:

P'(\alpha) - 2B \big(s-\overline{\alpha}\big) = A \big(\alpha-\overline{\alpha}\big)^2 \quad \implies \quad A = \frac{P'(\alpha) - 2B\big(s-\overline{\alpha}\big)}{4\textrm{Im}^2(\alpha)}

Example 2.3 (inversion of a transform with multiple complex roots) Let’s consider the following rational transform that has to be inverted:

\frac{s^2}{\big(s^2 + s + 1\big)^2}

The first thing to notice that the multiple roots are complex because calculating their value by using the quadratic formula it happens that

s_{1,2} = \frac{-1 \pm \sqrt{1-4}}{2} = -\frac{1}{2} \pm\mathrm{i}\frac{\sqrt{3}}{2}

For sake of semplicity of representation of the equation, from now on these roots value will be reported as z = -\frac{1}{2} + \frac{\sqrt{3}}{2} and \overline{z} = -\frac{1}{2}- \frac{\sqrt{3}}{2}. At this point it’s possible to express the parametric partial fraction expansion of the transform as

\frac{s^2}{\big(s^2 + s + 1\big)^2} = \frac{A}{s-z} + \frac{\overline{A}}{s-\overline{z}} + \frac{B}{(s-z)^2} + \frac{\overline{B}}{(s-\overline{z})^2}

In order to retrieve the value of the coefficient B we multiply both the sides by the factor

Q(s) = (s^2+s+1)^2 = (s-z)^2(s-\overline{z})^2

that, evaluated at the point s=z, will give the expression

z^2 = B \underbrace{\big(z-\overline{z}\big)}_{i\sqrt 3}^2 \quad \implies \quad B = -\frac{z^2}{3} = - \frac{\left(-\frac{1}{2} + \mathrm{i}\frac{\sqrt{3}}{2}\right)^2}{3} = - \frac{1}{6} - \mathrm{i} \frac{\sqrt{3}}{6}

By deflating the term presenting the parameter B just calculate we can get the relation

\begin{aligned} s^2 - B(s-\overline{z})^2 - \overline{B}(s-z)^2 &= A(s-z)(s-\overline{z})^2 + \overline{A}(s-\overline{z})(s-z)^2 \\[1em] & \downarrow\quad \mathrm{d}/\mathrm{d}s \\[1em] 2s -2B(s-\overline{z}) - 2\overline{B}(s-z) &= A(s-\overline{z})^2 + \overline{A}(s-z)^2\\ & +2(A + \overline{A})(s-z)(s-\overline{z}) \end{aligned}

Having computed the derivative of the relation in respect to the variable s it’s possible to compute the parameter A just by evaluating the result in the point s = z:

2z - 2B\big(z-\overline{z}\big) = A \big(z-\overline{z}\big)^2 \quad \implies \quad A = \frac{4iB \Im(z)-2z}{4\Im(z)^2} = - \frac{2\mathrm{i}}{3\sqrt 3}

Parametric solution for multiple complex roots

To solve the partial fraction expansion of multiple complex roots it’s possible to use a parametric formula where the numerator are defined as polynomials A,B in a generic variable t; in particular for the double complex root the expansion can be expressed as

\frac{P(s)}{\big(s^2-as +t\big)^2 Q(s)} = \frac{A'(t)s + B'(t)}{s^2-as+t} - \frac{A(t)s + B(t)}{\big(s^2-as+t\big)^2} + \frac{q_t(s,t)}{Q(s)}

where the term q_t(s,t) represent the derivative \partial q/\partial t of the remaining term on the numerator after the partial fraction expansion. If the polynomial has 3 complex roots than the parametric decomposition becomes

\begin{aligned} 2\frac{P(s)}{\big(s^2-2as +t\big)^3 Q(s)} & = 2\frac{A(t)s + B(t)}{\big(s^2-2as+t\big)^3} - 2\frac{A'(t)s + B'(t)}{\big(s^2-2as+t\big)^2} \\ & + \frac{A''(t)s + B''(t)}{s^2-as+t} + \frac{q_{tt}(s,t)}{Q(s)} \end{aligned} \tag{2.3}

where q_{tt}=\partial^2q/\partial t^2.

This process can be better seen by considering a numerical example like the one that follows:

\begin{aligned} \frac{s^2+1}{(s-1)(s^2-2s+2)^3} & = \frac{A}{s-1} + \frac{B_1s + B_2}{(s^2-2s+2)^3} \\[1em] &+\frac {C_1s + C_2}{(s^2-2s+2)^2} + \frac{D_1s+D_2}{s^2-2s+2} \end{aligned}

The first coefficient A of the partial fraction expansion (associated to a single real root) can be determined easily with the methods previously described and so

A= \left. \frac{s^2+1}{(s^2-2s+2)^3} \right|_{s=1} = 2

At this point it’s important to acknowledge hot compute the parametric functions A,B,q; in particular this can be done by writing the following relation:

\frac{s^2+1}{(s-1)(s^2-2s+2)} = \frac{A(t)s+B(t)}{s^2-2s+t} + \frac{q(t)}{s-1}

We can note that the left hand side is equal to the original rational polynomial with the only difference that the term s^2-2s+2 is now considered once; the right hand side is instead the correct partial fraction expansion of the other side with the difference that the coefficient A,B,q are now function of the variable t and the last known coefficient of the complex root (in this case the +) has been replaced with the variable t.

By multiplying both member of the equation by the denominator of the left hand side we derive the following equation:

s^2+1= \Big(A(t)s + B(t)\Big) (s-1) + q(t) \big(s^2-2s+t\big)

This expression allows now us to compute the parametric form of the function A,B,q; in particular considering that s^2-2s+t= 0 and so s^2 = 2s-t we can rewrite the previous expression as

2s-t+1 = A(t)(2s-t) - A(t) s + B(t)s -B(t) + \cancel{q(t) \big(2s-t - 2s +t\big)}

\begin{aligned} & \implies \quad \left\{\begin{aligned} 2 &= A(t) + B(t) \\ -t+1 &= -tA(t) - B(t) \end{aligned}\right. \\ &\implies \quad A(t) = 1 - \frac{2}{t-1} \ , \quad B(t) = 1 + \frac{2}{t-1} \end{aligned}

In order to have the complete the partial fraction expansion we need to compute the derivative (up to the second order) and evaluate them for t=3 (noting that $A(2) = -1 $ and B(2) = 3):

\begin{aligned} A' & = \left.\frac{2}{(t-1)^2}\right|_{t=2} = 2 & \quad B' & = \left.-\frac{2}{(t-1)^2}\right|_{t=2} = 2 \\ A'' & = \left. -\frac 4{(t-1)^3} \right|_{t=2} = -4 &\quad B'' & = \left. \frac 4{(t-1)^3} \right|_{t=2} = 4 \end{aligned}

With all the coefficient defined is possible to express numerically the partial fraction expansion by using Equation 2.3:

\frac 2 {s-1} + \frac{-s + 3}{(s^2-2s+2)^3} -2 \frac{s-1}{(s^2-2s+2)^2} + \frac 12 \frac{-4s+4}{s^2-2s+2}

Example 2.4 (computation of a partial fraction expansion) Considering the rational polynomial in s

\begin{aligned} \xi: \quad \frac{P(s)}{Q(s)} & = \frac{s(s^2+7)}{(s-1)(s+1)^3} \\ & = \frac A {s-1} + \frac B {s+1} + \frac C{(s+1)^2} + \frac D{(s+1)^3} \end{aligned}

finding the partial fraction expansion means determining the coefficients A,B,C,D \in\Bbb{R} that satisfies the equation. In order to solve this problem we can define \zeta as the multiplication of \xi by Q(s) and so

\zeta: \quad s(s^2+7) = A (s+1)^3 + B (s-1)(s+1)^2 + C (s-1)(s+1) + D (s-1)

In order to find some coefficient we can note that such polynomial equation must be verified if we evaluate \zeta for some point, in particular for s = 1 (that’s a root of Q) we have 8 = 2^3 A that determines A=1, while if we evaluate for s = -1 we retrieve -8 = -2D and so D=4. We can so inflate \zeta by moving on the left hand side the associated terms on A and D determining the rewritten expression:

\begin{aligned} \zeta_r: \quad & s^3 + 7s - A (s^3 + 3s^2 + 3s + 1) - D(s-1) \\ & \qquad\qquad = B (s-1)(s+1)^2 + C (s-1)(s+1) \\[1em] & s^3 + 7s -s^3 -3s^2 - 3s - 1 -4s + 4 \\ & \qquad\qquad = B (s-1)(s^2+2s +1) + C (s^2-1) \\[2em] & -3s^2 + 3 = B (s^3 + s^2 -s - 1) + C(s^2-1) \end{aligned}

In order to determine the value of the other coefficient B,C we can start deriving \zeta in the variable s:

\zeta' = \frac{\mathrm{d}\zeta}{\mathrm{d}s}: \qquad -6s = B(3s^2 + 2s -1) + 2Cs

Evaluating this expression for s = -1 (associated to the root of both B and C) we have that 6 = -2 C s determining C=-3; after inflation we so have

\begin{aligned} \zeta'_r&: \qquad 0 = B (3s^2+2s-1) \\[1em] \zeta'' = \frac{\mathrm{d}\zeta'}{\mathrm{d}s}&: \qquad 0 = B(6s + 2) \end{aligned}

With the last evaluation on s=-1 we obtain 0 = -4B requiring B=0; with that said the overall partial fraction expansion is

\frac{s(s^2+7)}{(s-1)(s+1)^3} = \frac 1 {s-1} - \frac 3{(s+1)^2} + \frac 4{(s+1)^3}

2.3 Table for the Laplace transform

	f(t)	\mathscr{L}\{f\}(s)
1	a\,f(t)+b\,g(t)	a f(s)+b g(s)
2	f(at)	\dfrac{1}{a}\,f\left(\dfrac{s}{a}\right)\qquad [a > 0]
3	1	\dfrac{1}{s}
4	t	\dfrac{1}{s^2}
5	t^k	\dfrac{k!}{s^{k+1}}
6	t^{n}f(t)	(-1)^{n}\dfrac{\mathrm{d}^{n}}{\mathrm{d}s^{n}}f(s)
7	f'(t)	s f(s)-f(0^{+})
8	f''(t)	s^2 f(s)-f'(0^{+})-sf(0^{+})
9	\dfrac{\mathrm{d}^{n}}{\mathrm{d}t^{n}}f(t)	s^{n}{f}(s)-\displaystyle\sum\nolimits\limits_{j=0}^{n-1}s^{n-j-1}f^{(j)}(0^{+})
10	e^{\alpha t}-e^{\beta t}	\dfrac{\alpha-\beta}{(s-\alpha)(s-\beta)}
11	e^{at}f(t)	f(s-a)
12	f(t-a)	e^{-as}f(s)
13	(f\star g)(t)	f(s)\,g(s)
14	e^{at}\cos\omega t	\dfrac{s-a}{(s-a)^{2}+\omega^{2}}
15	e^{at}\sin\omega t	\dfrac{\omega}{(s-a)^{2}+\omega^{2}}
16	e^{at}\cosh\omega t	\dfrac{s-a}{(s-a)^{2}-\omega^{2}}
17	a^{bt}	\dfrac{1}{s-b\log a}
18	e^{at}\sinh\omega t	\dfrac{\omega}{(s-a)^{2}-\omega^{2}}
19	\displaystyle\int_{0}^{t}f(z)\,\mathrm{d}z	\dfrac{1}{s}f(s)
20	e^{at}t^{n}	\dfrac{n!}{(s-a)^{n+1}}