# Quick Refresher on Laplace Transform

** Published:**

### Why talk about the Laplace Transform?

One thing that I think is sorely missing from a lot of texts and online resources that talk about the Laplace transform is a high-level, not-so-mathy discussion on the intuition behind the Laplace transform. This sort of discussion might not only be helpful to a beginner to the subject of control, but also to a long-time practitioner that needs a refresher after several months on another project (like myself). In this blog post, I talk about several high-level concepts regarding the Laplace transform so that it hopefully becomes more intuitive and less abstract for anyone that works with them in control theory.

# What does the Laplace Transform do?

Before we jump into formal definitions, let’s quickly talk about the general idea behind transforms. Take the following vector that I’ve drawn:

How I’ve drawn it, the vector points equally in the x-direction and y-direction, so it might be define by $<1,1>$ (ignoring magnitude). But this definition is entirely defined by the basis. If instead I used a different basis:

In this prime basis, the vector clear is along the x’ direction more that the y’, so it might be defined by $<0.99,0.01>$ (again, ignoring magnitude). The point is the same vector can be described in different ways, and if I were to go from one basis to another, I would merely be *transforming* the information that is there.

That, in a very small nutshell, is the big idea behind transforms: having different ways of defining the exact same information. This is exactly what the Laplace transform does. In control theory, one way to define a signal is how it looks in time, and is given by a function $f(t)$. The rest of this section talks about what kind of basis the Laplace transform transforms to.

The formal definition of the Laplace transform is given below:

\begin{equation} F(s) = \int_0^\infty f(t)e^{-st} dt \end{equation}

Notice that the time variable $t$ is integrated out, leaving a new variable $s$ and even a new function $F(s)$ to describe the information contained in the signal. Again, the signal itself does not change, *we are just looking that information in a different basis*.

This definition of the Laplace transform is often connected with the Fourier transform, which is given by:

\begin{equation} H(\omega) = \int_{-\infty}^\infty f(t)e^{-i\omega t} dt \end{equation}

This connection is made by setting $s=i\omega$ and changing the lower bound on the integral to $0$ in the Laplace transform. Clearly, the jump from Fourier to Laplace is a very small one, so I often like to think of the Laplace in terms of Fourier. The Fourier transform takes a function that is defined in time or space and transforms it to a frequency space, so that the function $H(\omega)$ describes the magnitude of the sinusoidal signal at frequency $\omega$ that comprises the time signal f(t). This is definition that has a strong physical connection, so I find this the most intuitive way of approaching the Laplace transform.

The big difference between the two transforms is that $\omega$ is strictly real in the Fourier transform, but $s$ is allowed to be complex in the Laplace transform. What does this difference do? That is a very interesting question, and probably the subject of a future blog post. The answer lies in the fact that $e^{-i\omega t}$ is an *orthogonal* basis with respect to $\omega$, while $e^{-st}$ is not orthogonal with respect to a generally complex $s$ variable. Orthogonality has its roots in linear algebra, and basically asks how ``mathematically’’ similar are two basis functions. This allows me to say that if $H(\omega = 1 \text{Hz}) = 10$, the signal in the time domain has a frequency of $1$ Hz and the amplitude of that signal has a magnitude of $10$. The Fourier transform has a strong physical interpretation. If, however, I find that $F(s = 1) = 10$ in the Laplace domain, I can’t really give that a physical interpretation; that feature simply isn’t there in the Laplace transform. But what we gain with the Laplace transform is practicality when manipulating signals and systems. The Fourier transform simply doesn’t allow the kind of quick analysis of systems that not only oscillate (i.e. Fourier) but also exponentially decay/grow, which is a big component in determining stability.

So, in the end: the Fourier transform is a nice way of making physical connections, and is helpful to keep in mind, but the Laplace transform is easier to work with when determining stability of the system, so control theory works with Laplace.

# Applying Laplace Transforms in Control Theory

With an understanding of what the Laplace Transform does, let’s turn to how its used in control theory. There are two ways that a Laplace analysis can show up in control:

- Transforming a signal
- Finding a transfer function

Below is a picture of the simplest block diagram I can think of:

The idea is this: we have a signal $u(t)$ that is defined in time, and the system takes that signal and outputs another signal $y(t)$, also a function of time. What if instead of working exclusively with time, we work with the Laplace $s$ variable instead? We can describe the signal with $s$ (via the Laplace transformation) and we can describe how that signal is changed in the $s$ basis as it passes through the system (via the transfer function). A priori, we have no reason to believe that this would help. But it does help a lot.

Again, almost any text book or online source talks about these ideas individually, so I will not really dive into the details, but instead briefly talk about both steps.

# Transforming a Signal

In order to transform a signal $f(t)$, the above definition of the Laplace transform is applied. One basic example is a step function, a.k.a. heaviside function $\Theta(t)$, shown below:

It’s Laplace transform $\Theta(s)$ is given by:

\begin{equation} \Theta(s) = 1/s \end{equation}

If we input a heaviside function into our system above, then we say $\Theta(t) \equiv u(t)$ and accordingly $\Theta(s) \equiv U(s)$.

Not all signals in the time domain have an analytical form in the Laplace domain. In practice, you either (1) work with signals that have well-known Laplace transforms (there are many tables that are easy to google), or (2) you use software like Matlab to perform a numerical analysis in the Laplace domain.

# Finding a transfer function

How does the system take $u(t)$ and output $y(t)$? This is described by a mathematical model (usually a differential equation) that you either derive from first principles or you find empirically from a process called system identification. For example, suppose we look at the mass-on-spring equation of motion derived from Newton’s 2nd law:

\begin{equation} \ddot{y}(t) = -a\dot{y}(t) -by(t) + u(t) \end{equation}

Here, $y(t)$ is the position of the mass on a spring and $u(t)$ is the force applied to the system, all in the time domain. Using the Laplace transform, we can transform this differential relationship between $u(t)$ and $y(t)$ in the time domain to an algebraic relationship in the Laplace domain:

\begin{equation} \frac{Y(s)}{U(s)} = \frac{1}{s^2 + as + b} \end{equation}

Here, the algebraic relationship of $Y(s)/U(s)$ is often defined as the transfer function $G(s)$. Furthermore, because of this algebraic relationship, finding the output $Y(s)$ given an input signal $U(s)$ is a simple matter of multiplication: $Y(s) = \frac{Y(s)}{U(s)}U(s) = G(s)U(s)$.

# Discussion and Concluding Remarks

Notice how the input signal $U(s)$ and the transfer function $G(s)$ almost always have the same form: a fraction that has some polynomial of $s$ both in the numerator and in the denominator. To me, this is what makes the Laplace transform so confusing from time to time; because both signal and transfer function *look* the same, it is easy to conflate their purpose. But, remember that they have very different interpretations! Signals are what we ultimately care about, and the Laplace transform is a simple change of basis for that signal. Transfer functions tell us the relationship between input and output in the Laplace domain.

A lot of information (such as stability) can be extracted from $G(s)$ alone, and a lot of time is spent in text books examine this transfer function $G(s)$ in various setups. Always remember, though, that if I want to know what $Y(s)$ looks like, I must choose an input function $U(s)$ and let $Y(s) = G(s)U(s)$. I can’t know the exact output $Y(s)$ without knowing $U(s)$!

This is where I’ll end the very brief overview of the Laplace transform in control. Again, the purpose of this post is to provide a theoretical refresher on what the Laplace transform does the grand scheme of things. After this refresher, you can hopefully dive into any control theory text book and have this context in the back of your mind.