Inverse Laplace Transforms and Partial Fractions, differential equations lesson

Solving an equation with the Laplace transform usually ends with a fraction in $s$ , and with the question of which signal has that Laplace transform. A table lists only simple shapes such as $1/(s - a)$ and $1/(s - a)^2$ , so a large fraction has to be split into pieces of those shapes first. This lesson explains why the signal is determined by its Laplace transform, how partial fractions handle repeated and quadratic factors, and how completing the square exposes a damped oscillation. Throughout, $t \ge 0$ is time and $s$ is a complex number.

Key terms

Inverse Laplace transform

Definition Let $f$ be piecewise continuous on $[0, \infty)$ and of exponential order, with $\mathcal{L}\{f\}(s) = F(s)$ on some half-plane $\operatorname{Re} s > \sigma$ . Then $f$ is an inverse Laplace transform of $F$ , written $f = \mathcal{L}^{-1}\{F\}$ . By the uniqueness theorem, two such functions with the same Laplace transform on a half-plane agree at every $t > 0$ where both are continuous. So the inverse is determined except at jumps, and a continuous inverse, when one exists, is unique on all of $[0, \infty)$ . The inverse is linear, and it describes the signal only for $t \ge 0$ . Every such $F$ tends to $0$ as $\operatorname{Re} s \to \infty$ , so a nonzero polynomial in $s$ is not the Laplace transform of any piecewise continuous function of exponential order.

In plain words The Laplace transform keeps everything about a signal except its values at isolated jumps, so a table of pairs can be read backward. In practice, split $F$ into entries of the table and invert each one.

Example Linearity and the pairs $\mathcal{L}\{1\} = 1/s$ , $\mathcal{L}\{t^2\} = 2/s^3$ and $\mathcal{L}\{e^{2t}\} = 1/(s - 2)$ give

\begin{aligned} &\mathcal{L}^{-1}\left\{\frac{3}{s} + \frac{4}{s^3} - \frac{1}{s - 2}\right\} \\[0.4em] &\quad = 3 + 2t^2 - e^{2t}, \end{aligned}

where the Laplace transform converges for $\operatorname{Re} s > 2$ . This signal is continuous, so it is the only continuous inverse. The function equal to $1$ for $t \ne 1$ and to $7$ at $t = 1$ has the same Laplace transform $1/s$ as the constant $1$ , which is why a value at a single point can never be recovered.

Partial fractions

Definition Let $F = P/Q$ be a ratio of polynomials with real coefficients, with all common factors of $P$ and $Q$ cancelled. The zeros of $Q$ are then the poles of $F$ . If $F$ is the Laplace transform of a piecewise continuous function of exponential order, it is strictly proper, $\deg P < \deg Q$ , because it must tend to $0$ . When the degrees fail this, long division splits off a polynomial in $s$ , which belongs to impulses and their derivatives rather than to an ordinary function. For strictly proper $F$ , factor $Q$ over the reals into powers $(s - a)^m$ of linear factors and powers $(s^2 + ps + q)^k$ of quadratic factors with $p^2 < 4q$ . Then $F$ is the sum, over all the factors, of the terms

\begin{gathered} \sum_{j = 1}^{m} \frac{A_j}{(s - a)^j} \quad \text{and} \\[0.4em] \sum_{j = 1}^{k} \frac{B_j s + C_j}{(s^2 + ps + q)^j}, \end{gathered}

and these real constants exist and are unique. At a simple pole $a$ , multiplying $F$ by $s - a$ and setting $s = a$ gives the coefficient $A_1$ directly. The Laplace integral of the inverse converges exactly on the half-plane $\operatorname{Re} s > \sigma_0$ , where $\sigma_0$ is the largest real part of a pole.

In plain words Split a complicated fraction into the simplest fractions its denominator allows. Each simple fraction is a table entry, so its inverse can be read off.

Example To write $(5s + 1)/((s - 1)(s + 2))$ as $A/(s - 1) + B/(s + 2)$ , multiply by $s - 1$ and set $s = 1$ , which gives $A = 6/3 = 2$ . Multiplying by $s + 2$ and setting $s = -2$ gives $B = (-9)/(-3) = 3$ . So

\begin{aligned} &\frac{5s + 1}{(s - 1)(s + 2)} \\[0.4em] &\quad = \frac{2}{s - 1} + \frac{3}{s + 2}, \end{aligned}

and the inverse is $2e^{t} + 3e^{-2t}$ , with region of convergence $\operatorname{Re} s > 1$ set by the pole at $1$ .

Repeated poles

Definition A pole $a$ of $F = P/Q$ in lowest terms has multiplicity $m$ when $a$ is a root of $Q$ of multiplicity $m$ , and it is a repeated pole when $m \ge 2$ . In the partial fraction form it needs all $m$ terms $A_j/(s - a)^j$ for $j = 1, \ldots, m$ . Their inverses come from the pair

\begin{gathered} \mathcal{L}\{t^n e^{at}\} = \frac{n!}{(s - a)^{n + 1}}, \\[0.4em] \operatorname{Re} s > a, \quad n = 0, 1, 2, \ldots, \end{gathered}

so a pole of multiplicity $m$ contributes $e^{at}$ times a polynomial in $t$ of degree $m - 1$ . With $G(s) = (s - a)^m F(s)$ , the coefficients are

\begin{gathered} A_{m - j} = \frac{1}{j!}\,G^{(j)}(a), \\[0.4em] j = 0, \ldots, m - 1. \end{gathered}

In plain words Each extra power of the same factor puts one more power of $t$ in front of the exponential. The highest power is read off by evaluating, and the lower ones by differentiating first.

Example Take the fraction $F$ with numerator $3s^3 + 11s^2 + 39s + 61$ and denominator $(s + 1)^2(s^2 + 4s + 13)$ . The pole $-1$ is double, so the form is

\begin{aligned} F(s) &= \frac{A}{s + 1} + \frac{B}{(s + 1)^2} \\[0.4em] &\quad + \frac{Cs + D}{s^2 + 4s + 13}. \end{aligned}

Here $G(s) = (s + 1)^2F(s) = N(s)/Q(s)$ with $N(s) = 3s^3 + 11s^2 + 39s + 61$ and $Q(s) = s^2 + 4s + 13$ . Since $N(-1) = 30$ and $Q(-1) = 10$ , the top coefficient is $B = G(-1) = 3$ . For $A$ , differentiate. The quadratic term of $G$ keeps a factor $(s + 1)^2$ , so its derivative vanishes at $-1$ , and $A = G'(-1)$ . With $N'(-1) = 26$ and $Q'(-1) = 2$ , the quotient rule gives

\begin{aligned} A &= G'(-1) \\[0.4em] &= \frac{26 \cdot 10 - 30 \cdot 2}{10^2} = 2. \end{aligned}

The two terms invert to $2e^{-t}$ and $3te^{-t}$ .

Irreducible quadratic factor

Definition A real quadratic $s^2 + ps + q$ is irreducible when $p^2 - 4q < 0$ . It has no real roots, and its roots are the complex conjugate pair $\alpha \pm i\beta$ with

\begin{gathered} \alpha = -\frac{p}{2}, \\[0.4em] \beta = \sqrt{q - \frac{p^2}{4}} > 0. \end{gathered}

In partial fractions each power of such a factor gets a numerator of degree one, $B_j s + C_j$ . A constant numerator is not enough in general. The full form has exactly $\deg Q$ unknowns, one for each coefficient that a numerator of lower degree can have, and dropping one of them leaves some numerators out of reach. The pair of conjugate poles produces the oscillating signals $e^{\alpha t}\cos\beta t$ and $e^{\alpha t}\sin\beta t$ .

In plain words A quadratic that cannot be factored over the reals hides a pair of complex poles. Keep it whole, give it a numerator of the form $Cs + D$ , and expect a damped or growing oscillation.

Example Continue with the fraction $F$ of the previous example. The factor $s^2 + 4s + 13$ has $p^2 - 4q = 16 - 52 = -36 < 0$ , so it is irreducible, with roots $-2 \pm 3i$ . Clearing denominators gives

\begin{aligned} N(s) &= A(s + 1)Q(s) \\[0.4em] &\quad + BQ(s) \\[0.4em] &\quad + (Cs + D)(s + 1)^2. \end{aligned}

The coefficient of $s^3$ gives $A + C = 3$ , so $C = 1$ . The constant term gives $13A + 13B + D = 61$ , so $D = 61 - 26 - 39 = -4$ . Hence

\begin{aligned} F(s) &= \frac{2}{s + 1} + \frac{3}{(s + 1)^2} \\[0.4em] &\quad + \frac{s - 4}{s^2 + 4s + 13}. \end{aligned}

A constant numerator would fail here. With $C = 0$ the $s^3$ coefficient would force $A = 3$ , while the derivative computation forces $A = 2$ .

Completing the square

Definition For an irreducible quadratic, completing the square gives

\begin{aligned} &s^2 + ps + q \\[0.4em] &\quad = \Bigl(s + \frac{p}{2}\Bigr)^2 + q - \frac{p^2}{4} \\[0.4em] &\quad = (s - \alpha)^2 + \beta^2. \end{aligned}

Regroup the numerator around the same shift, $Cs + D = C(s - \alpha) + (D + C\alpha)$ , and use the pairs

\begin{aligned} &\mathcal{L}\{e^{\alpha t}\cos\beta t\} \\[0.4em] &\quad = \frac{s - \alpha}{(s - \alpha)^2 + \beta^2}, \\[0.4em] &\mathcal{L}\{e^{\alpha t}\sin\beta t\} \\[0.4em] &\quad = \frac{\beta}{(s - \alpha)^2 + \beta^2}, \end{aligned}

both valid for $\operatorname{Re} s > \alpha$ . They follow from $\mathcal{L}\{e^{(\alpha \pm i\beta)t}\} = 1/(s - \alpha \mp i\beta)$ , Euler’s formula and linearity. The constant part of the numerator must be written as a multiple of $\beta$ before the sine pair applies.

In plain words The square is centered at $\alpha$ , the real part of the complex poles, which becomes the exponential rate, and the number $\beta^2$ added to the square is the frequency squared. The numerator has to be rewritten around the same shift, so that one piece matches the cosine and the rest matches the sine.

Example For the quadratic piece of $F$ , $s^2 + 4s + 13 = (s + 2)^2 + 9$ , so $\alpha = -2$ and $\beta = 3$ . The numerator becomes $s - 4 = (s + 2) - 6 = (s + 2) - 2 \cdot 3$ , which inverts to $e^{-2t}(\cos 3t - 2\sin 3t)$ . The full inverse is

\begin{aligned} f(t) &= 2e^{-t} + 3te^{-t} \\[0.4em] &\quad + e^{-2t}(\cos 3t - 2\sin 3t). \end{aligned}

Two checks support it. The value $f(0) = 2 + 0 + 1 = 3$ matches the limit of $sF(s)$ as $s \to \infty$ , which is $3$ , the ratio of the leading coefficients. The poles are $-1$ and $-2 \pm 3i$ , and the largest real part is $-1$ , so $F$ is the Laplace transform of $f$ for $\operatorname{Re} s > -1$ .

Check yourself

Find the inverse Laplace transform of $F(s) = (s + 3)/(s + 1)^2$ .

Show the answer

Write the numerator around the pole as $s + 3 = (s + 1) + 2$ . Then

F(s) = \frac{1}{s + 1} + \frac{2}{(s + 1)^2},

and the pairs for $e^{-t}$ and $te^{-t}$ give $f(t) = (1 + 2t)e^{-t}$ for $t \ge 0$ , with region $\operatorname{Re} s > -1$ . As a check, $f(0) = 1$ and $sF(s) \to 1$ as $s \to \infty$ .

A classmate tries to write $1/(s^2(s + 1))$ as $A/s^2 + B/(s + 1)$ . Show that no constants work, then find the correct decomposition and the inverse.

Show the answer

Clearing denominators would require $1 = A(s + 1) + Bs^2$ for all $s$ . The $s^2$ coefficient gives $B = 0$ , the $s$ coefficient gives $A = 0$ , and the constant term gives $A = 1$ , a contradiction. The double pole at $0$ needs both $1/s$ and $1/s^2$ , so write $A/s + B/s^2 + C/(s + 1)$ . With $G(s) = s^2F(s) = 1/(s + 1)$ , the coefficient $B = G(0) = 1$ and $A = G'(0) = -1$ . Multiplying by $s + 1$ and setting $s = -1$ gives $C = 1$ . So

\begin{aligned} \frac{1}{s^2(s + 1)} &= -\frac{1}{s} + \frac{1}{s^2} \\[0.4em] &\quad + \frac{1}{s + 1}, \end{aligned}

and the inverse is $-1 + t + e^{-t}$ for $t \ge 0$ , with region $\operatorname{Re} s > 0$ .

Find the inverse Laplace transform of $F(s) = (2s + 1)/(s^2 + 2s + 5)$ and state the region of convergence.

Show the answer

The quadratic has $p^2 - 4q = 4 - 20 = -16 < 0$ , so it is irreducible. Completing the square gives $s^2 + 2s + 5 = (s + 1)^2 + 4$ , so $\alpha = -1$ and $\beta = 2$ . Regroup the numerator as $2s + 1 = 2(s + 1) - 1$ and write the constant as a multiple of $\beta = 2$ ,

\begin{aligned} F(s) &= \frac{2(s + 1)}{(s + 1)^2 + 4} \\[0.4em] &\quad - \frac{1}{2} \cdot \frac{2}{(s + 1)^2 + 4}. \end{aligned}

So $f(t) = e^{-t}\bigl(2\cos 2t - \tfrac{1}{2}\sin 2t\bigr)$ for $t \ge 0$ . The poles are $-1 \pm 2i$ , so the region of convergence is $\operatorname{Re} s > -1$ . As a check, $f(0) = 2$ , which matches the limit of $sF(s)$ , the ratio $2/1$ of the leading coefficients.

Inverse Laplace Transforms and Partial Fractions

Key terms

Inverse Laplace transform

Partial fractions

Repeated poles

Irreducible quadratic factor

Completing the square

Check yourself

How this lesson connects

Builds on

Used later in