Chapter 8

The Laplace Probe

Can one measured area tell us where a signal lives?

A probe for signals

Theo: Every time we solved something, we guessed $e^{st}$ , plugged it in, and prayed. Now add gravity, a starting height, a controller… my notebook is on fire.

June: When I build stuff, the trick is always to turn a hard job into an easy one. Hard: calculus with slopes of slopes. Easy: plain algebra with $+$ , $-$ , $\times$ , $\div$ .

Mika: So we need a machine that eats a curve and spits out something where slopes turn into multiplication? Does that exist?

It does, and we can build it from things we already own. Here's the idea. Take any signal, any curve that changes over time, $f(t)$ . Multiply it by a probe, the exponential $e^{-st}$ from Chapter 4 (for positive $s$ it shrinks). Then measure the total area under the product (Chapter 3's accumulated area), all the way out to $t = \infty$ .

F(s) = \int_0^\infty \out{f(t)}\; e^{-st}\,dt

That's it. A number $s$ goes in, an area comes out. Try different $s$ values and you get a whole new curve $F(s)$ . That curve is the Laplace transform of $f$ : a fingerprint of the signal. Two different (reasonable) signals never share the same $F(s)$ , which is why we'll be able to look answers up backwards later.

Predict first! Take the simplest signal, a step: $f(t) = 1$ forever. As we make $s$ bigger, what happens to the area under $1 \cdot e^{-st}$ ?

Chapter 8: The Laplace Probe ↗

Sweep $s$ in the widget and the dots trace out exactly that $1/s$ . Try the other signals too. A growing signal $e^{0.5t}$ only has a finite area when the probe shrinks faster than the signal grows, that is when $s > 0.5$ . Below that, the area runs off to infinity. The wiggle $\sin 2t$ fails differently: at $s = 0$ its area just sloshes back and forth between 0 and 1 (Chapter 3's negative area cancelling the positive), and never settles on one number.

The total area must settle to a finite value, so that each $s$ gets one number. If the probe fades too slowly, the area keeps growing and that $s$ has no answer.

The transform screams where the signal lives

Let's look closer at $f(t) = e^{at}$ . Probe it with $e^{-st}$ and the product is $e^{(a-s)t}$ , a plain exponential. As long as $s > a$ it fades, and its area is $\frac{1}{s-a}$ (we'll derive it properly when we build our table in the next chapter).

Chapter 8: The Laplace Probe ↗

Mika: As $s$ gets close to $a$ the number goes huge! At $s = a$ the probe exactly cancels the growth, the product is flat forever, and the area is infinite.

Theo: So the transform blows up exactly at the $s$ where the signal lives. If you gave me only $F(s)$ , I could find the signal's $a$ by looking for where it explodes.

Chapter 8: The Laplace Probe ↗

Mika: It's like a smoke alarm. Walk $s$ towards $a$ and it gets louder and louder. Step onto $a$ and it never stops.

Now make $s$ complex, $s = \sigma + i\omega$ , just like our spinning numbers from Chapter 5. Take the spring's spinner from Chapter 6, $e^{2it}$ (half of its twin pair), as the signal. The probe $e^{-st}$ shrinks it (that's the left-right part, $\sigma$ ) and spins it backwards (that's the up-down part, $\omega$ ). Below, the product's little pieces of area are added up head-to-tail as arrows, one arrow for every quarter second. Watch the running total.

Chapter 8: The Laplace Probe ↗

June: When the probe spins at exactly 2, the signal stops spinning, all the pieces point the same way and the total is as big as it gets: $1/\sigma$ , which runs off to infinity as σ shrinks to 0. When they don't match, the arrows curl around and cancel out.

Same lesson, one dimension up: the transform explodes at the complex $s$ where the signal lives. That's why from now on we'll draw every $s$ on a map with $\sigma$ across and $\omega$ up. Our "map of $s$ " from Chapter 6 finally gets its grown-up name: the s-plane.

New words

The s-plane: The map of complex numbers $s = \sigma + i\omega$ . Left means shrinking, right means growing, up and down means spinning. Every signal $e^{st}$ lives at one point on it, and its transform explodes there.

What we know now

What we figured out

The Laplace transform is a probe: multiply by $e^{-st}$ and measure the total area. $F(s) = \int_0^\infty f(t)e^{-st}dt$ .
It explodes exactly where the signal lives: $e^{at}$ gives $1/(s-a)$ , which blows up at $s = a$ . For complex $s$ the probe unspins spinning signals. We plot all of this on the s-plane.
An area over forever can still be finite, as long as the tail thins out fast enough. If the probe fades more slowly than the signal grows, that $s$ has no transform at all.

Theo: "So the probe tells us where a signal lives. But we wanted slopes to turn into multiplication. What does the probe do to a slope?"

Chapter 8: The Laplace Probe ↗