# Simple examples to illustrate the utility of the Laplace Transform

I am a math professor teaching an Intro ODEs course, and most of my students are going into engineering. As such, I like to give them lots of examples of how our material is applied.

We are coming up on the Laplace Transform and I'm looking for a relatively simple example from Control Theory. I was thinking of perhaps examining a PID control for the cruise control in a car or for a thermostat (these appear to be the classic examples) but I'm having trouble seeing why we really need the Laplace Transform, or transfer functions, at all: we already know how to solve linear ODEs with constant coefficients without Laplace.

I am hoping someone can show me why we really need (or are at least greatly helped by) the Laplace Transform and transfer functions in these examples, or point me toward a simple example where we do.

• A search gave : quora.com/… – Solar Mike Feb 9 '18 at 20:36
• In control engineering and control theory the transfer function is derived using the Laplace transform; for continuous time signals, transfer functions map the Laplace transform of the input to the Laplace transform of the output. More information found here. – AsymLabs Feb 11 '18 at 20:55
• Following from the above, there are several exercises in the pdf document here that are also practical applications and this github.com search provides 3 pages of software repositories. – AsymLabs Feb 11 '18 at 21:19

Consider $s$ as a derivative operator. Therefore, $$\frac{Y}{X}=\frac{s+2}{s^2+0.5s+3}$$

looks like $$\frac{Y}{X}=\frac{\frac{d}{dt}.+2}{\frac{d^2}{dt}.+0.5\frac{d}{dt}.+3}$$

$$(\frac{d^2}{dt}.+0.5\frac{d}{dt}.+3) y=(\frac{d}{dt}.+2)x$$

$$(\frac{d^2}{dt}y+0.5\frac{d}{dt}y+3y) =(\frac{d}{dt}x+2 x)$$

For linear systems, we more need derivatives and convolutions than time-variance and multiplication (except for scaling). Therefore, it worths moving to Laplace domain.

$s$ represents frequency as well. The beautiful examples are the initial and final value theorems which show how time and frequency are, in the limit, the inverse of each other

$$x(0)=\lim_{s\to \infty} sX(s)$$ $$x(\infty)=\lim_{s\to 0} sX(s)$$

Another useful application of Laplace transform is for the design of high-pass and low-pass filters. They are much more comfortable to work in $s$ domain than messy derivatives.

What's the response (to an arbitrary load) of a viscoelastic assembly consisting of a damper (with damping constant $\eta_1$) connected in series with a spring (with spring constant $k_2$) and another damper ($\eta_2$) that are connected in parallel?

In other words, the assembly looks like this: and we're pulling the right side; the left side is fixed. We wish to know the displacement of the right side.

(The individual displacement $u_i(t)$ of a single spring exposed to a load $F_i(t)$ is $u_i(t)=\frac{F_i(t)}{k_i}$; the individual displacement rate $\dot u_i(t)$ of a single damper exposed to the same load is $\dot u_i(t)=\frac{F_i(t)}{\eta_i}$.)

This type of problem arises all the time in the context of automotive engineering, metallurgy, polymer synthesis, and biomechanics, among other fields. But you'll have a tough time writing the response of this and much more complex assemblies if you keep the time derivatives around.

Instead, let's take the Laplace transforms for each component and assume zero displacement at $t=0$: $F_i(s)=k_iu_i(s)$ for a spring and $F_i(s)=s \eta_i u_i(s)$ for a damper. Now, realizing that displacements add when lumped components are connected in series, whereas forces add when the components are connected in parallel, we find that the total displacement is

$$u(s)=\frac{F(s)}{s\eta_1}+\frac{F(s)}{k_2+s\eta_2}$$

Thus, the transfer function of the assembly is

$$\frac{u(s)}{F(s)}=\frac{1}{s\eta_1}+\frac{1}{k_2+s\eta_2}$$

What happens if I apply a unit step load for 1 second and then let go? We'd write this load as $F(t)=u(t)-u(t-1)$, corresponding to $F(s)=\frac{1}{s}-\frac{\exp(-s)}{s}$ (we can look up these Laplace transforms in a table, for example, or use a symbolic tool such as Wolfram Alpha). The resulting response is

$$u(s)=\frac{1-e^{-s}}{s^2\eta_1}+\frac{1-e^{-s}}{s(k_2+s\eta_2)}$$

The corresponding time response (letting all system variables equal 1 for simplicity) is $$u(t, \eta_1=\eta_2=k_2=1)=e^{-t}\left[(e-te^t)u(t-1)+e^t(t+1)-1\right]$$

or, graphically (with time in seconds on the x-axis and displacement on the y-axis), For your approach you need to know the exact ODE of the system. However when designing a controller for a system that can be approximated as LTI you do not necessary need a fully parameterized and identified ODE model. You can either use some PID tuning algorithm or, maybe more related to your question, measure the frequency response function (FRF) of the system. You could fit a transfer function onto this FRF, but the FRF itself is already enough information to design a controller using loopshaping.