The Laplace Transform

Main idea

The Laplace transform weighs a function of $t$ by $e^{-st}$ and adds it all up, which gives a new function of $s$ . It's worth the trouble because it turns derivatives into multiplication by $s$ , so calculus problems become algebra.

{\htmlData{sym=1}{F(s)}} = \mathcal{L}\{f(t)\} = \int_0^\infty {\htmlData{sym=0}{e^{-st}}}f(t)\,dt

What the symbols mean

$e^{-st}$
Weight
Decays fast enough to tame most functions as long as s is large.
$F(s)$
Transform
A function of the new variable s.

Explanation

The Laplace transform is simply a way to turn a function of time $t$ into a function of a new variable $s$ , by weighting it and adding it all up. It's called a "transform" because it changes one function into another, and it's named after Pierre-Simon Laplace, who used integrals like this. What it's good for: it turns differential equations into algebra, so you translate the problem, solve it in the easier language, and translate back. The formula $F(s) = \int_0^\infty e^{-st}f(t)\,dt$ looks intimidating, but it's just three simple parts: $f(t)$ (the function you start with), $e^{-st}$ (a weight that shrinks the far-off future, more strongly for bigger $s$ ), and $\int_0^\infty$ (add it all up from time $0$ onward). It looks like a lot, but bear with it.

Computing from the definition. For $f(t) = e^{at}$ :

\int_0^\infty e^{-st}e^{at}\,dt = \int_0^\infty e^{-(s - a)t}\,dt = \frac{1}{s - a}\quad(s > a).

Notice the condition $s > a$ . The integral only converges when $s$ is large enough to beat the growth of $f$ . That's the whole job of the weight $e^{-st}$ : it drags the integrand down to zero fast enough for the area to be finite.

Linearity. $\mathcal L\{af + bg\} = aF + bG$ , because integrals split over sums and let constants out. That means a table of a few basic transforms goes a long way.

Why bother? In the next few lessons you'll see $\mathcal L\{y'\} = sY - y(0)$ . Taking a derivative in $t$ turns into multiplying by $s$ in the new world. So a differential equation becomes an algebra problem in $Y(s)$ , with the initial conditions already built in.

See it

The transform of f(t) = 1 is just the area under the weight e^(−st), and that area is 1/s. As you slide s up, the curve dies off faster and the area shrinks: about 2 at s = 0.5, 1 at s = 1, and 1/3 at s = 3. So F(s) = 1/s, one number for each s.

Worked examples

Example 1

Find $\mathcal L\{1\}$ .

1
$\displaystyle\int_0^\infty e^{-st}\,dt = \Big[-\tfrac1se^{-st}\Big]_0^\infty = \frac1s$ for $s > 0$ , since that's when $e^{-st}$ dies off at infinity.

$\frac1s$

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find $\mathcal L\{1\}$ at $s = 2$ .

Problem 2+5 XP

Find $\mathcal L\{e^{3t}\}$ at $s = 5$ .

Problem 3+5 XP

Find $\mathcal L\{t\}$ at $s = 1$ .

Problem 4+5 XP

Find $\mathcal L\{\sin t\}$ at $s = 1$ .

Problem 5+5 XP

For which $s$ does $\mathcal L\{e^{4t}\}$ exist?

Problem 6+5 XP

A machine earns $\$ 100 $per year forever, paid continuously. At a continuous interest rate of$ 5% $, its value today is$ \displaystyle\int_0^\infty 100e^{-0.05t},dt $, which is a Laplace transform evaluated at$ s = 0.05$. Find it.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find $\mathcal L\{\sin^2 t\}$ at $s = 1$ .

Where it's used

Electrical engineers analyze circuits almost entirely in the s-domain: impedances, transfer functions and stability all come from Laplace transforms.

RLC CircuitsDiff Eq · Electrical Engineering Transfer Functions for a MotorDiff Eq · Robotics

Mechanical Vibrations Computing Laplace Transforms