Chapter 9

Calculus into Algebra

Can we turn calculus into algebra?

Slopes become multiplication

Mika: Remember the sticky note from Chapter 4? For $e^{at}$ , taking a slope was just $\times\, a$ . Now watch what the probe does with a slope.

New words

$f'$ and $f''$: Short for $\frac{df}{dt}$ , the slope of $f$ , and the slope of that slope. Chapter 7 wrote them with dots, $\dot x$ and $\ddot x$ .
$\mathcal{L}\{f\}$: "The transform of $f$ ": the same thing as $F(s)$ , handy when $f$ is a whole expression.
Capital letters: A capital letter is the transform of its small letter: $F$ for $f$ , $H$ for the height $h$ , $X$ for $x$ .

Here's the property that makes the whole thing worth it. What happens to the transform when we take the slope of the signal? Try it with $f = e^{at}$ . Its slope is $a e^{at}$ (Chapter 4), so

\begin{aligned} \mathcal{L}\{f'\} &= \frac{a}{s-a} = \frac{s}{s-a} - 1 \\ &= s\,F(s) - f(0) \end{aligned}

The slope turned into multiplying by $s$ , minus the starting value $f(0)$ . Calculus became algebra!

Chapter 9: Calculus into Algebra ↗

Theo: Hold on. That's a trick that works for $e^{at}$ because $e^{at}$ is special. Next you'll tell me it's a lookup rule we're supposed to memorise. I don't buy it for any curve.

June: Then let's test it on the ugliest curve you can draw.

Draw any signal you like in the first plot (or pick a preset). We compute its slope, then work out both sides of the rule with the area machine: the transform of the slope, and $s\,F(s) - f(0)$ . The second plot shows those two sides for every $s$ .

Chapter 9: Calculus into Algebra ↗

Theo: They match, even for my wobbly scribble. Not a lookup rule after all: it's what the area machine does to any slope.

Here is the missing algebra, as a picture. A product changes for two reasons. $f$ changes while the probe holds still, and the probe changes while $f$ holds still. Picture a rectangle with sides $f$ and $e^{-st}$ : it grows along one side, then along the other. This move has a name: integration by parts.

\big(f\,e^{-st}\big)' = f'\,e^{-st} - s\,f\,e^{-st}

Add up those changes from 0 to ∞ (Chapter 3: area undoes slope) and you get the value at the end minus the value at the start. That is what the bracket $\big[\,\cdot\,\big]_0^\infty$ means:

\big[f\,e^{-st}\big]_0^\infty = \int_0^\infty f'\,e^{-st}\,dt - s\,F(s)

At the far end the probe has squashed everything to zero (as long as $s$ beats the signal's growth), so the left side is just $0 - f(0)$ . Move $s\,F(s)$ across and there's the rule.

Theo: So $f(0)$ isn't a fudge. It's the rectangle we started with.

Apply the same rule once more, this time to $f'$ , whose starting value is $f'(0)$ , the starting speed:

\begin{aligned} \mathcal{L}\{f''\} &= s\,\mathcal{L}\{f'\} - f'(0) \\ &= s\big(s\,F(s) - f(0)\big) - f'(0) \\ &= s^2 F(s) - s\,f(0) - f'(0) \end{aligned}

Our own table

Textbooks print a "table of Laplace transforms" as if it came from the sky. We're going to earn ours, one row at a time. Each row is just the probe-and-area machine applied to one signal, with a check that the formula agrees with the actual area.

Chapter 9: Calculus into Algebra ↗

Mika: The sine row is my favourite. A sine is just two twin spinners, and each spinner's transform is the exponential row with $a = \pm i\omega$ . Chapter 6 added the twins to get a cosine. Subtract them instead and the along parts cancel, leaving $2i\sin\omega t$ . That's where the $2i$ comes from! And multiplying the twins' bottoms is Chapter 7's mirror product, $(s - i\omega)(s + i\omega) = s^2 + \omega^2$ .

(For a complex $a$ , the exponential row still works as long as the left-right part of $s$ beats the left-right part of $a$ . For the twins that's simply $\sigma > 0$ .)

And the last row needed no new work at all: multiplying by $e^{-\sigma t}$ simply shifts $s$ to $s+\sigma$ . Four rows, all derived, all checked.

One more rule we've been using quietly. The area machine is fair: make a signal 3 times taller and every slice of its area is 3 times taller, so the whole area is too. Add two signals and their areas add. Mathematicians call this linearity.

Chapter 9: Calculus into Algebra ↗

June: That's what lets us transform an equation one term at a time, numbers in front and all. And the drone's equation is exactly that kind of sum. Let's use the table!

Undoing a common denominator

Theo: Every answer we'll get comes out as one big fraction. But our table only knows small ones: $\frac1s$ , $\frac{1}{s-a}$ , the sines. How do we break a big fraction into table pieces?

At school you added fractions by finding a common bottom. Here we go backwards: start from one fraction and find the small pieces that add up to it. That is called splitting into partial fractions:

\frac{1}{s} + \frac{1}{s+2} = \frac{2s + 2}{s\,(s+2)} \qquad\text{so}\qquad \frac{2s+2}{s\,(s+2)} = \frac{A}{s} + \frac{B}{s+2}

What goes on top of each piece? **One unknown for every power of $s$ in its bottom.** A bottom like $s$ or $s + 2$ gets a plain number. A quadratic bottom like $s^2 + 2s + 5$ gets $B\,s + C$ : two unknowns, because a quadratic has room for two.

Chapter 9: Calculus into Algebra ↗

June: Each slider moves one piece, and the pieces just add up in time too. The $A$ piece is a step; the quadratic piece is a wiggle that dies away. And the big fraction on the right is what they add up to.

Finding the numbers has a shortcut for the first-power pieces, the cover-up trick. Multiply by the piece's bottom, then set $s$ to the value that makes it zero: every other piece vanishes and only its top is left.

A = \Big[s \cdot \frac{2s+2}{s\,(s+2)}\Big]_{s=0} = \frac{2}{2} = 1, \qquad B = \Big[(s+2) \cdot \frac{2s+2}{s\,(s+2)}\Big]_{s=-2} = \frac{-2}{-2} = 1

A quadratic bottom needs one more move before the table can read it: push it into the shape $(s + \sigma)^2 + \omega^2$ by completing the square, halving the middle number and squaring it. So $s^2 + 2s + 5 = (s + 1)^2 + 4$ : a decaying cosine and sine with $\sigma = 1$ and $\omega = 2$ .

Solving the drone with algebra

Here is the Chapter 2 drone under P control, with no free hover thrust this time, starting on a 1 m ledge. At $t=0$ we release it and ask for 2 m. In the calculation from that instant onward, the new setpoint and constant gravity force each transform like a step:

m\,\out{h}'' + c\,\out{h}' = \eff{K_p}\,(\sp{2} - \out{h}) - \dis{m g}

Step 1: transform every term. Because areas scale and add (the table section), we can transform each term on its own. A step of size $r$ becomes $r/s$ , and slopes become multiplication by $s$ . Here's what we wrote:

m\,s^2\out{H} + c\,s\,\out{H} = \eff{K_p}\Big(\frac{\sp{2}}{s} - \out{H}\Big) - \frac{\dis{m g}}{s}

Step 2: it's algebra now. Collect $\out{H}$ on one side, then divide:

\begin{gathered} \begin{aligned} &(m s^2 + c s + \eff{K_p})\,\out{H} \\ &\quad = \frac{\eff{K_p}\,\sp{2} - \dis{m g}}{s} \end{aligned} \\[8pt] \out{H}(s) = \frac{N(s)}{s\,\big(m s^2 + c s + \eff{K_p}\big)} \\[8pt] N(s) = \eff{K_p}\,\sp{2} - \dis{m g} \end{gathered}

Mika: Look at that. No slopes left, just letters being shuffled around. Theo, your notebook has stopped smoking.

Step 3: split it into table pieces, as in the section above: one number on top of the $s$ piece, and $B s + C$ on top of the quadratic one:

\out{H}(s) = \frac{A}{s} + \frac{B s + C}{m s^2 + c s + \eff{K_p}}

How do we find $A$ , $B$ , and $C$ ? Multiply both sides by the shared bottom $s\,(m s^2 + c s + \eff{K_p})$ :

\begin{aligned} N(s) &= A\,(ms^2+cs+\eff{K_p}) \\ &\quad + s\,(Bs+C) \end{aligned}

The cover-up trick

There's a shortcut for $A$ . Multiply $\out{H}$ by $s$ and every piece except $A/s$ still has an $s$ on top. Set $s = 0$ and they all vanish. Only $A$ survives:

A = \Big[s\,\out{H}(s)\Big]_{s=0} = \frac{N(0)}{\eff{K_p}} = \sp{2} - \frac{\dis{mg}}{\eff{K_p}}

And $s\,\out{H}$ near $s = 0$ is where the drone ends up in the long run, if it settles at all. The final height falls straight out of the algebra, no simulation needed.

Mika: Hang on. We can read where it settles without solving anything? That feels like cheating.

Theo: It is cheating, unless everything else fades away. Here the other pieces are the drone's own wiggle, and they die out, so $A$ is what's left. If the bottom had a root in the right half, $A/s$ would still sit in the fraction, but a growing piece would swamp it and the drone would never settle anywhere. Chapter 10 shows how to check.

Then compare both sides, one power of $s$ at a time. The plain numbers give $A$ , the $s^2$ terms give $B$ , and the $s$ terms give $C$ :

\begin{gathered} A = \sp{2} - \frac{\dis{mg}}{\eff{K_p}} \\ B = -m\,A, \qquad C = -c\,A \end{gathered}

No new physics appears here: we have simply undone a common denominator to reach pieces our table can take back to time. Notice that $A$ doesn't care where the drone started.

Theo: And we never had to guess $e^{st}$ once. The table did the guessing for us.

Step 4: look each piece up in the table we just derived: $A/s$ is a step, and the second piece is a decaying cosine plus a decaying sine (the side trip below shows how it fits those rows). Read off $h(t)$ .

Completing the square: Chapter 7 was hiding in the bottom

The second piece isn't in our table yet: its bottom is a plain quadratic. Push it into the shape $(s+\sigma)^2 + \omega^2$ (completing the square, as in the section above) and it matches the decaying-sine rows exactly. The top splits the same way, one part for the cosine row and one for the sine row:

\begin{gathered} 0.5\,s^2 + s + \eff{K_p} = 0.5\,\big((s+1)^2 + \omega_d^2\big) \\ \omega_d^2 = 2\eff{K_p} - 1 \\[6pt] B\,s + C = B\,(s+1) + (C - B) \end{gathered}

So $\sigma = 1$ , which is Chapter 7's $\zeta\omega_n$ for our drone, and $\omega$ is the damped wiggle $\omega_d$ .

Chapter 9: Calculus into Algebra ↗

Theo: Wait. The bottom is zero at exactly the two dots we put on the map in Chapter 7! Keep that thought.

June: Let's check it against the simulation!

Predict first! We start the drone on a 1 m ledge and run our formula next to the simulation. What will we see?

Chapter 9: Calculus into Algebra ↗

Mika: Uh-oh. The formula starts at zero, but the drone starts on the ledge at 1 m. The algebra is lying?

Theo: The algebra's fine. We lied. Look at Step 1: we wrote $\mathcal{L}\{h''\} = s^2 H$ and $\mathcal{L}\{h'\} = s\,H$ , and dropped every $h(0)$ . That's exactly the piece that remembers the ledge.

Flip the switch to put the starting-height terms back, and watch Steps 1 and 2 and the coefficients above change with it. The dashed formula and the blue simulation now sit exactly on top of each other: the biggest gap is less than a millionth of a metre, just rounding. Change $\eff{K_p}$ or the ledge height and they still agree.

Chapter 9: Calculus into Algebra ↗

Theo: In Chapter 2 we had to fly the drone to find its droop. Here it just fell out of a fraction.

What we know now

What we figured out

Slopes become multiplication: $\mathcal{L}\{f'\} = sF(s) - f(0)$ . The $f(0)$ remembers where the signal started. Forget it and your answer starts in the wrong place.
We derived our own table: $1/s$ , $1/(s-a)$ , $\omega/(s^2+\omega^2)$ , and the shifted version for decaying sines. Areas scale and add, so we can transform an equation term by term.
Big fractions split into partial fractions: one unknown per power of $s$ in each bottom, found with the cover-up trick. A quadratic bottom becomes a shrinking cosine and sine.
Transform, do algebra, split into table pieces, read off the answer. The $A/s$ piece is where the drone settles, droop and all. It matches the simulation exactly, as long as you keep the starting values.

June: "Every time we solved something, the answer was (something) ÷ (a polynomial in s), and the drone's behaviour came from where the bottom is zero. What if that fraction is the drone?"

Chapter 9: Calculus into Algebra ↗