Inverse Laplace transforms - System Analysis and Control

In the previous section, we learned how to decompose a rational function into a sum of simpler terms via partial fraction expansion. Now we develop the inverse Laplace transform pairs needed to convert each term back to the time domain.

The terms arising from PFE have specific forms depending on the pole type:

A simple real pole $s=r$ produces a term like $\displaystyle\frac{a}{s - r}$
Complex conjugate poles $s=p\pm qj$ produce a term like $\displaystyle\frac{as + b}{(s-p)^2 + q^2}$
Repeated poles produce terms like $\displaystyle\frac{a}{(s - r)^k}$ or $\displaystyle\frac{as + b}{((s-p)^2 + q^2)^k}$

We will derive the inverse Laplace transform for each of these cases.

Review of complex numbers¶

Before proceeding, we review some essential facts about complex numbers.

Complex number representations¶

We can write a number $z\in\C$ in two equivalent ways:

Rectangular form: $z = a + jb$ . We call $a$ is the real part and $b$ the imaginary part.
The notation used is $a = \text{Re}(z)$ and $b = \text{Im}(z)$ .
Polar form: $z = re^{j\theta}$ , where $r$ is the magnitude and $\theta$ is the angle (or argument).
The notation used is $r = |z|$ and $\theta = \angle z$ .

We can convert between rectangular and polar forms using the relationships:

\begin{aligned} a &= r \cos \theta, & b &= r \sin \theta, \\ r &= \sqrt{a^2 + b^2}, & \theta &= \arctan \bigl(\tfrac{b}{a}\bigr) \end{aligned}

(1)

We can view complex numbers as points in the complex plane, with the real part along the x-axis and the imaginary part along the y-axis. Here is a diagram showing the rectangular and polar representations of a complex number:

Complex number z = a + jb = re^{j\theta} shown in rectangular and polar forms. — Figure 1:Complex number $z = a + jb = re^{j\theta}$ shown in rectangular and polar forms.

Why different representations?

Different complex representations are useful in different contexts.

Rectangular is useful for addition/subtraction because complex numbers behave like vectors. If $z_1 = a_1 + jb_1$ and $z_2 = a_2 + jb_2$ , then
$z_1 \pm z_2 = (a_1 \pm a_2) + j(b_1 \pm b_2)$
(2)
In other words,
$\begin{aligned} \Re(z_1 \pm z_2) &= \Re(z_1) \pm \Re(z_2) \\ \Im(z_1 \pm z_2) &= \Im(z_1) \pm \Im(z_2) \end{aligned}$
(3)
Polar is useful for multiplication/division. If $z_1 = r_1 e^{j\theta_1}$ and $z_2 = r_2 e^{j\theta_2}$ , then
$z_1 z_2 = r_1 r_2 e^{j(\theta_1 + \theta_2)} \qquad\textsf{and}\qquad \frac{z_1}{z_2} = \frac{r_1}{r_2} e^{j(\theta_1 - \theta_2)}$
(4)
In other words,
$\begin{aligned} |z_1 z_2| &= |z_1| |z_2|\quad & \angle(z_1 z_2) &= \angle z_1 + \angle z_2 \\ \left|\frac{z_1}{z_2}\right| &= \frac{|z_1|}{|z_2|} \quad & \angle\left(\frac{z_1}{z_2}\right) &= \angle z_1 - \angle z_2 \end{aligned}$
(5)

Euler’s formula¶

The formulas (1) are derived from basic trigonometry and the diagram of Figure 1. However, they do not explain why $z = e^{j\theta}$ in polar form. This comes from Euler’s formula^[1]:

e^{j\theta} = \cos\theta + j\sin\theta

(6)

Proof of Euler’s formula

First we must address the question “what does it even mean to raise $e$ to an imaginary power?” The complex exponential function is defined via its Taylor series expansion, which converges for all complex numbers $z$ :

e^{z} = \sum_{n=0}^{\infty} \frac{z^n}{n!} = 1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots

(7)

Setting $z = j\theta$ gives:

\begin{aligned} e^{j\theta} &= 1 + \frac{j\theta}{1!} + \frac{(j\theta)^2}{2!} + \frac{(j\theta)^3}{3!} + \frac{(j\theta)^4}{4!} + \cdots \\ &= 1 + j\frac{\theta}{1!} - \frac{\theta^2}{2!} - j\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + \cdots \\ &= \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right) + j\left(\frac{\theta}{1!} - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right) \\ &= \cos\theta + j\sin\theta \end{aligned}

(8)

The last step follows from the Taylor series expansions of $\cos\theta$ and $\sin\theta$ .

We can use Euler’s formula to write complex exponentials as a combination of a real exponential and trig functions. For a complex number $z = a + bj$ and time $t\in \R$ ,

\boxed{e^{zt} = e^{(a+jb)t} = e^{at}e^{jbt} = e^{at}(\cos bt + j\sin bt)}

(9)

This formula will be central to deriving the inverse Laplace transforms for complex poles.

Complex conjugate¶

Another useful concept is the complex conjugate of a complex number. The complex conjugate of $z = a + jb$ is denoted $\bar{z}$ and defined as:

\bar{z} = a - jb

(10)

In polar form, the complex conjugate is given by:

\bar{z} = re^{-j\theta}

(11)

Graphically, the complex conjugate is the reflection of $z$ across the real axis in Figure 1.

Simplifying fractions¶

When working with complex numbers, it is often necessary to simplify fractions involving complex numbers. The key technique is to multiply the numerator and denominator by the complex conjugate of the denominator. For example, let’s simplify the expression:

\begin{aligned} \frac{2+3j}{1 + 4j} &= \frac{(2+3j)(1 - 4j)}{(1 + 4j)(1 - 4j)} \\ &= \frac{2 + 3j - 8j - 12j^2}{1 - 16j^2} \\ &= \frac{2 - 5j + 12}{1 + 16} \\ &= \frac{14 - 5j}{17} \\ &= \frac{14}{17} - \frac{5}{17}j \end{aligned}

(13)

Simplifying powers¶

When raising complex numbers to powers, it is often easier to use the polar representation. For example, to compute $(1 + j)^{10}$ , we first convert to polar form:

1 + j = \sqrt{1^2 + 1^2}\, e^{j\arctan(1/1)} = \sqrt{2}\, e^{j\pi/4}

(14)

Next, we use the property of exponents:

(1 + j)^{10} = (\sqrt{2} e^{j\pi/4})^{10} = (\sqrt{2})^{10} e^{j10\pi/4} = 2^{5} e^{j5\pi/2} = 32 e^{j\pi/2} = 32j

(15)

The simplification $e^{j5\pi/2} = e^{j\pi/2} = j$ uses the fact that $e^{j2\pi} = 1$ , so we can subtract $2\pi$ from the angle without changing the value. Finally, Euler’s gives us $e^{j\pi/2} = \cos(\pi/2) + j\sin(\pi/2) = j$ .

Inverse Laplace transforms¶

Now that we have reviewed complex numbers, we can use their properties to derive the inverse Laplace transforms needed for PFE terms. Let’s start with exponential functions.

Exponentials¶

Let’s consider the Laplace transform of the exponential function $e^{zt}$ , where $z\in\C$ is a complex number. From the definition of the Laplace transform:

\Lap\{e^{zt}\} = \int_0^{\infty} e^{-st} e^{zt} \, dt = \int_0^{\infty} e^{(z-s)t} \, dt = \left[\frac{1}{z-s} e^{(z-s)t}\right]_{t=0}^{\infty} = \frac{1}{s-z}

(16)

If $z$ is a real number, say $z = -a$ with $a\in\R$ , then Eq. (16) becomes:

\boxed{\Lap\{e^{-at}\} = \frac{1}{s+a}}

(17)

No surprises here; this is what we previously derived in Table 1.

Sinusoids and damped sinusoids¶

Let’s now assume $z = -a+bj$ , with $a,b\in\R$ . Applying (16) gives:

\Lap\{e^{zt}\} = \frac{1}{s + a - bj}

(18)

The left-hand side of (18) is a complex exponential, which we can split into real and imaginary parts using Euler’s formula (9).

e^{zt} = e^{(-a+jb)t} = e^{-at}(\cos bt + j\sin bt)

(19)

Therefore: $\Lap\{e^{zt}\} = \Lap\{e^{-at}\cos bt\} + j\Lap\{e^{-at}\sin bt\}$ .

The right-hand side of (18) can also be separated into real and imaginary parts by multiplying the numerator and denominator by the complex conjugate of the denominator:

\begin{aligned} \frac{1}{s + a - bj} &= \frac{(s+a+bj)}{(s+a-bj)(s+a+bj)}\\ &= \frac{s + a}{(s + a)^2 + b^2} + j\frac{b}{(s + a)^2 + b^2} \end{aligned}

(20)

Equating real and imaginary parts on both sides gives:

\boxed{ \begin{aligned} \Lap\{e^{-at}\cos bt\} &= \frac{s+a}{(s+a)^2 + b^2}\\ \Lap\{e^{-at}\sin bt\} &= \frac{b}{(s+a)^2 + b^2} \end{aligned} }

(21)

Since $\sin$ and $\cos$ are bounded between -1 and 1, the entire function lies between $-e^{-at}$ and $+e^{-at}$ . This is the exponential envelope.

The parameter $a$ controls the decay rate: if $a > 0$ , the oscillations decay; if $a < 0$ , they grow; if $a=0$ , they have constant amplitude.
The parameter $b$ controls the frequency of oscillation: the larger $b$ is, the faster the oscillations.

The interactive app below allows you to visualize the effects of changing $a$ and $b$ on the damped sinusoid $e^{-at}\cos bt$ .

Repeated poles¶

Let’s return to our exponential Laplace formula (16):

\Lap\{e^{zt}\} = \frac{1}{s - z}

(22)

We can actually differentiate both sides with respect to $z$ and obtain another true statement. This follows because the Laplace transform is an integral with respect to $t$ , so we can interchange the order of differentiation and integration^[2]

Differentiating with respect to $z$ gives:

\Lap\{t e^{zt}\} = \frac{1}{(s - z)^2}

(23)

We can repeat this process $n-1$ more times to obtain:

\Lap\{t^n e^{zt}\} = \frac{n!}{(s - z)^{n+1}}

(24)

So, for example, when $z=-a$ is a real number, we obtain:

\boxed{\Lap\{t^n e^{-at}\} = \frac{n!}{(s + a)^{n+1}}}

(25)

We can also obtain a formula for the case of repeated complex poles by substituting $z= -a + bj$ into (24) and separating real and imaginary parts using Euler’s formula (9) as we did when we derived the damped sinusoid formulas (21). We will skip this derivation as the formulas are quite complicated and rarely come up in practice.

Summary of Laplace transform pairs

The following table summarizes the Laplace transform pairs we’ve seen so far.

Time domain $f(t)$	Frequency domain $F(s)$	Pole type
$e^{-at}$	$\displaystyle\frac{1}{s+a}$	Simple real
$\displaystyle\frac{t^{n-1}}{(n-1)!}e^{-at}$	$\displaystyle\frac{1}{(s+a)^n}$	Repeated real (order $n$ )
$e^{-at}\cos bt$	$\displaystyle\frac{s+a}{(s+a)^2+b^2}$	Complex conjugate pair
$e^{-at}\sin bt$	$\displaystyle\frac{b}{(s+a)^2+b^2}$	Complex conjugate pair

For the repeated real pole, we rearranged (25) so that setting $n=1$ recovers the simple real pole case.

We also included a more comprehensive table in our appendix on Laplace transforms.

System stability¶

Now that we have seen what sorts of functions of time we can expect out of LTI systems, we can start to think about the notion of stability, which is important for control systems. We say a LTI system is:

stable if the states go to zero. This is also called asymptotically stable. For example, a decaying exponential $e^{-t}$ is stable.
unstable if the states go to infinity. For example, a growing exponential $e^{t}$ is unstable.
marginally stable if the states do not go to zero or infinity, but remain constant or oscillate indefinitely. For example, $\sin t$ is marginally stable.

We can characterize stability for the functions of time we’ve seen so far:

Time domain $f(t)$	Stable	Marginally stable	Unstable
$e^{-at}$	$a > 0$	$a = 0$	$a < 0$
$\frac{t^{n-1}}{(n-1)!}e^{-at}$	$a > 0$	$a = 0$ and $n = 1$	$a < 0$
$e^{-at}\cos bt$	$a > 0$	$a = 0$	$a < 0$
$e^{-at}\sin bt$	$a > 0$	$a = 0$	$a < 0$

Stability is governed by $a$ , which is the real part of the pole. Marginal stability only occurs when $a=0$ and if the pole is not repeated (i.e., $n=1$ for the real pole case).

This leads us to the following important conclusion about system stability:

Inverting real quadratic terms¶

When we have an irreducible quadratic denominator and the numerator does not fit neatly into one of the forms in Eq. (21), we must do some additional algebra to rewrite the expression in a suitable form. To illustrate, let’s find the inverse Laplace transform

\Lap^{-1}\left\{\frac{2s + 3}{s^2 - 2s + 10}\right\}

(26)

First, complete the square in the denominator:

s^2 - 2s + 10 = (s - 1)^2 + 9 = (s-1)^2 + 3^2

(27)

This reveals that $a = -1$ and $b = 3$ . We know from (21) that:

\begin{aligned} \Lap^{-1}\left\{\frac{s - 1}{(s - 1)^2 + 3^2}\right\} &= e^{t}\cos 3t \\ \Lap^{-1}\left\{\frac{3}{(s - 1)^2 + 3^2}\right\} &= e^{t}\sin 3t \end{aligned}

(28)

Our goal is to write the original fraction as a linear combination of these two fractions. So we need to find constants $A$ and $B$ such that:

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

(29)

If we can do this, then we can take the inverse Laplace transform term-by-term using the formulas above and obtain:

\Lap^{-1}\left\{\frac{2s + 3}{s^2 - 2s + 10}\right\} = A e^{t}\cos 3t + B e^{t}\sin 3t

(30)

To find $A$ and $B$ , we put everything over a common denominator and compare numerators. Then we match coefficients since the expression must be true for all values of $s$ . This is similar to the systematic approach we used for PFE:

\begin{aligned} 2s+3 &= A(s-1) + B\cdot 3 \\ &= As + (3B-A) \end{aligned}

(31)

Therefore, we need to solve the system of equations:

\begin{aligned} A &= 2 \\ 3B - A &= 3 \end{aligned}

(32)

Solving gives $A = 2$ and $B = \frac{5}{3}$ . Substituting this into Eq. (30) gives the final answer:

\boxed{\Lap^{-1}\left\{\frac{2s + 3}{s^2 - 2s + 10}\right\} = 2e^{t}\cos 3t + \frac{5}{3}e^{t}\sin 3t}

(33)

Complete worked example¶

Let’s put everything together with a complete worked example. Let’s find:

\Lap^{-1}\left\{ \frac{s+10}{s^4 + 2s^3 + 10s^2} \right\}

(36)

Solution: The denominator factors as $s^2(s^2 + 2s + 10)$ . The quadratic $s^2 + 2s + 10$ is irreducible since its discriminant is negative: $b^2 - 4ac = 2^2 - 4\cdot 1 \cdot 10 = -36 < 0$ . Completing the square, we have:

s^2 + 2s + 10 = (s+1)^2 + 3^2

(37)

Using our PFE shortcut, we look for constants $A$ , $B$ , and $C$ such that:

\frac{s+10}{s^2(s^2 + 2s + 10)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C(s+1) + 3D}{(s+1)^2 + 3^2}

(38)

Once we find $A$ , $B$ , $C$ , and $D$ , the solution will be the sum of the inverse Laplace transforms of each term. In this case:

\Lap^{-1}\left\{\frac{s+10}{s^4 + 2s^3 + 10s^2}\right\} = A \cdot 1 + B \cdot t + C e^{-t}\cos 3t + D e^{-t}\sin 3t

(39)

We can find $B$ using cover-up. Multiply both sides of Eq. (38) by $s^2$ and set $s=0$ :

B = \frac{0 + 10}{0^2 + 0 + 10} = 1

(40)

To find $A$ , $C$ , and $D$ , we multiply both sides of Eq. (38) by the common denominator $s^2(s^2 + 2s + 10)$ and compare coefficients:

\begin{aligned} s + 10 &= A s (s^2 + 2s + 10) + (s^2 + 2s + 10) + [C(s+1) + 3D] s^2 \\ &= A s^3 + 2A s^2 + 10A s + s^2 + 2s + 10 + C s^3 + C s^2 + 3D s^2 \\ &= (A + C) s^3 + (2A + 1 + C + 3D) s^2 + (10A + 2) s + 10 \end{aligned}

(41)

Comparing coefficients gives the system of equations:

\begin{aligned} A + C &= 0 \\ 2A + 1 + C + 3D &= 0 \\ 10A + 2 &= 1 \\ 10 &= 10 \end{aligned}

(42)

Solving gives $A = -\frac{1}{10}$ , $C = \frac{1}{10}$ , and $D = -\frac{3}{10}$ . Substituting these into Eq. (39) gives the final answer:

\boxed{\Lap^{-1}\left\{\frac{s+10}{s^4 + 2s^3 + 10s^2}\right\} = -\frac{1}{10} + t + \frac{1}{10} e^{-t}\cos 3t - \frac{3}{10} e^{-t}\sin 3t}

(43)

Test your knowledge¶

Solution to Exercise 1 #

First, complete the square in the denominator:

s^2 + 4s + 13 = (s + 2)^2 + 3^2

(45)

This reveals that $a = -2$ and $b = 3$ . We know from (21) that:

\begin{aligned} \Lap^{-1}\left\{\frac{s + 2}{(s + 2)^2 + 3^2}\right\} &= e^{-2t}\cos 3t \\ \Lap^{-1}\left\{\frac{3}{(s + 2)^2 + 3^2}\right\} &= e^{-2t}\sin 3t \end{aligned}

(46)

Our goal is to write the original fraction as a linear combination of these two fractions. So we need to find constants $A$ and $B$ such that:

\frac{3s + 5}{(s + 2)^2 + 3^2} = A\left(\frac{s + 2}{(s + 2)^2 + 3^2}\right) + B\left(\frac{3}{(s + 2)^2 + 3^2}\right)

(47)

If we can do this, then we can take the inverse Laplace transform term-by-term using the formulas above and obtain:

\Lap^{-1}\left\{\frac{3s + 5}{s^2 + 4s + 13}\right\} = A e^{-2t}\cos 3t + B e^{-2t}\sin 3t

(48)

\begin{aligned} 3s + 5 &= A(s + 2) + B\cdot 3 \\ &= As + (2A + 3B) \end{aligned}

(49)

Therefore, we need to solve the system of equations:

\begin{aligned} A &= 3 \\ 2A + 3B &= 5 \end{aligned}

(50)

Solving gives $A = 3$ and $B = -\frac{1}{3}$ . Substituting this into the inverse Laplace expression gives the final answer:

\boxed{\Lap^{-1}\left\{\frac{3s + 5}{s^2 + 4s + 13}\right\} = 3e^{-2t}\cos 3t - \frac{1}{3}e^{-2t}\sin 3t}

(51)

Solution to Exercise 2 #

We start with the partial fraction expansion:

\frac{4}{(s+1)(s+2)(s+3)^2 } = \frac{A}{s+1} + \frac{B}{s+2} + \frac{C}{s+3} + \frac{D}{(s+3)^2}

(53)

We can find $A$ , $B$ , $D$ using cover-up:

\begin{aligned} A &= \frac{4}{( -1 + 2)( -1 + 3)^2} = 1 \\ B &= \frac{4}{( -2 + 1)( -2 + 3)^2} = -4 \\ D &= \frac{4}{( -3 + 1)( -3 + 2)} = 2 \end{aligned}

(54)

So far, we have:

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

(55)

To find $C$ , let’s set $s=0$ on both sides:

\frac{4}{18} = \frac{1}{1} - \frac{4}{2} + \frac{C}{3} + \frac{2}{9}

(56)

Solving gives $C = 3$ . Therefore, the complete PFE is:

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

(57)

Taking the inverse Laplace transform term-by-term gives:

\boxed{\Lap^{-1}\left\{\frac{4}{(s+1)(s+2)(s+3)^2}\right\} = e^{-t} - 4e^{-2t} + 3e^{-3t} + 2te^{-3t}}

(58)

Footnotes¶

Leonhard Euler was a prolific 18th-century Swiss mathematician who made significant contributions to many areas of mathematics and there are countless formulas named after him. Euler’s name is pronounced “oiler” (like “boiler” without the “b”).
↩
This is called the Leibniz integral rule.
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