Laplace Transforms

Introduction

Recall in Proof for Euler's Identity using Maclaurin Series, and Proof for Euler's Identity using ODEs. We derived that $g (t) = e^{i t} = \cos t + i \sin t$ . where $g$ maps $t$ (time) to a position on the complex plane. $g : R \mapsto C$

Suppose a particle follows the path of $g (t)$ . It's velocity is given by $\frac{d}{d t} g (t) = i g (t)$ . Thus its velocity vector is always perpendicular to its position (Recall that $i$ is a $90$ degree rotation). As such, it will move in a circular motion. $x = \cos t$ and $y = \sin t$ .

Generalisation

Suppose a particle follows the path of $f (t) = e^{s t}$ where $s \in C$ and $f : R \mapsto C$ . Let $s = a + b i$ .

f (t) = e^{s t}

Intuition

Consider initial position at $f (0) = 1$ . The velocity of the particle is $\frac{d}{d t} f (t) = s f (t)$ . Thus, the particle at $1 + 0 i$ will have a velocity $s = a + i b$ (in yellow). If $a < 0$ , $s$ is pulling the particle inwards, causing it to spiral inwards. If $a > 0$ , $s$ is pushing the particle outwards, causing it to spiral outwards. As $| b |$ increases, it's angular frequency $ω = \frac{d θ}{d t} = \frac{d b t}{d t} = b$ increases, causing it to move faster in a circular motion. Thus $x = e^{a t} \cos (b t)$ , $y = e^{a t} \sin (b t)$ .

\begin{aligned} e^{(a + i b) t} & = \overset{Magnitude}{\overset{⏞}{e^{a t}}} \overset{ω}{\overset{⏞}{e^{i b t}}} \\ = e^{a t} [\cos (b t) + i \sin (b t)] \end{aligned}

$R e (s)$ determines the rate of change of magnitude.
$I m (s)$ determines the angular frequency of the particle ( $ω$ ).

Why do we care?

Consider a scenario of a mass on spring placed horizontally.
The force acting on the mass can be modelled as such.

F = - k x - μ v

$k$ : spring constant
$μ$ : damping term (friction, air resistance)
$x$ : displacement from equilibrium.
$v$ : velocity

It describes how the spring and damping will exert a force $F$ on the mass to resist its direction of motion. We can rewrite it as

m x^{″} (t) + k x (t) + μ x^{'} (t) = 0

Let's guess that the solution of $x (t)$ is $e^{s t}$ . Thus

\begin{aligned} m s^{2} e^{s t} + k e^{s t} + μ s e^{s t} & = 0 \\ e^{s t} (m s^{2} + μ s + k) & = 0 \\ m s^{2} + μ s + k & = 0 \end{aligned}

Solving for $s$ gives

s = \frac{- μ \pm \sqrt{μ^{2} - 4 (m k)}}{2 m}

For simplicity, let's assume that $μ$ is $0$ .

\begin{aligned} m s^{2} & = - k \\ s & = \pm \sqrt{- \frac{k}{m}} = \pm i \sqrt{\frac{k}{m}} = ω \end{aligned}

What does this mean? It basically means that the natural frequency of the system is $I m (s) = ω = \pm \sqrt{\frac{k}{m}}$ . Since $R e (s) = 0$ , its amplitude does not change.

BUT WAIT!!! There are two solutions for $s$ , but in fact, there are a whole family of solutions for $x (t)$ !

Linearity

If $x_{1}$ solves

m x_{1}^{″} (t) + k x_{1} (t) = 0

and $x_{2}$ solves

m x_{2}^{″} (t) + k x_{2} (t) = 0

Then $(c_{1} x_{1} + c_{2} x_{2})$ solves it as well.

m [c_{1} x_{1}^{″} (t) + c_{2} x_{2}^{″} (t)] + k [c_{1} x_{1} (t) + c_{2} x_{2} (t)] = 0

Thus, there are a whole family of equations $x (t)$ that satisfy the conditions $m x^{″} (t) + k x (t) = 0$

x (t) = c_{1} e^{s_{1} t} + c_{2} e^{s_{2} t}

Thus, the equation is just the linear combination of the complex exponential of $s$ .

In physics, a lot of functions can be expressed as

f (t) = \sum_{n = 1}^{N} c_{n} e^{s_{n} t}

Introduction

A Laplace transform is a "function" that maps one function in time-domain to another function in the s domain. where $t \in R, s \in C$

F (s) = \int_{0}^{\infty} f (t) e^{- s t} d t

And

L {f (t)} = F (s)

Lets consider the simplest case where $f (t) = 1$

\begin{aligned} \int_{0}^{\infty} e^{- s t} d t & = {[- \frac{1}{s} e^{- s t}]}_{0}^{\infty} \\ = \frac{1}{s} \end{aligned}

Consider now, $f (t) = e^{a t}$

e^{a t} \overset{L}{\to} \int_{0}^{\infty} e^{a t} e^{- s t} d t = \frac{1}{s - a}

The Laplace transform of $e^{a t}$ has a pole at $a$ where $F (a) = \infty$ . Thus we have figured out what values of $a$ are embedded in the function $f (t)$ .

Consider now $f (t) = \cos (ω t)$ . $\cos (ω t) = \frac{1}{2} e^{i ω t} + \frac{1}{2} e^{- i ω t}$ . (Recall Complex Numbers)
Its Laplace transform is

\begin{aligned} \cos (ω t) & \overset{L}{\to} \int_{0}^{\infty} [\frac{1}{2} e^{i ω t} + \frac{1}{2} e^{- i ω t}] e^{- s t} d t \\ L {\cos ω t} & = \int_{0}^{\infty} \frac{1}{2} e^{i ω t} e^{- s t} + \int_{0}^{\infty} \frac{1}{2} e^{- i ω t} e^{- s t} \\ = \frac{1}{2 (s - ω i)} + \frac{1}{2 (s + ω i)} \\ = \frac{s}{s^{2} + ω^{2}} \end{aligned}

The poles in its Laplace transform at $s = \pm i ω$ helps us identify the family of equations that describes $\cos (ω t)$ . Typically, the expression $\cos (ω t) = \frac{1}{2} e^{i ω t} + \frac{1}{2} e^{- i ω t}$ is unknown, thus integration by parts is usually used to get the poles of the equation.