Visual walkthrough — Inverse Laplace transform — partial fractions, tables

The fraction we will chase the whole way down: $F(s)=\frac{3s+1}{(s-1)(s+2)}.$

Here $s$ is just a variable — think of it as the "code-world coordinate", the axis we work on after the Laplace transform has turned a differential equation into algebra. Our goal is the function $f(t)$ whose transform is exactly this $F(s)$. We recover it with partial fractions plus a lookup table.

---

Step 1 — Where does $F(s)$ even come from? (the round trip)

WHAT. Before touching the algebra, let us place $F(s)$ on the map. A problem starts as a differential equation in real time $t$. The Laplace transform $\mathcal{L}$ carries it rightward into the $s$-world where it becomes plain algebra. We solve there and get $F(s)$. The inverse ${\mathcal{L}}^{-1}$ is the return arrow that lands us back on $f(t)$.

WHY. You cannot invert a thing you cannot locate. Seeing the loop tells us why recognition (table lookup) is enough: the forward map is one-to-one (Lerch's theorem), so each $F(s)$ has exactly one honest partner $f(t)$. No ambiguity — a dictionary suffices. This is exactly the machinery behind solving ODEs with Laplace.

PICTURE. Two vertical lanes: the $t$-world (left, blue) and the $s$-world (right, orange). The top arrow $\mathcal{L}$ points right; the bottom arrow ${\mathcal{L}}^{-1}$ — the one this whole page is about — points left.

---

Step 2 — Why we cannot invert $F(s)$ as-is: the pole map

WHAT. Look at the denominator $(s-1)(s+2)$. The values $s=1$ and $s=-2$ make it zero — the fraction blows up there. Those two spots are called poles. Our fraction has two poles glued into one expression.

WHY. The lookup table only knows fractions with a single pole at a time (one $\frac{1}{s-a}$). A two-pole fraction is not in the table — it is a compound, not an atom. So the very first obstacle is: this object is too complicated to recognise. We must break it apart.

PICTURE. The $s$-axis (a horizontal number line) with two marked poles: a red dot at $s=1$ and a red dot at $s=-2$. Above each, a vertical spike showing $|F(s)|\to \infty$. The caption reads: "one fraction, two blow-ups — not a table atom."

$\underset{\text{numerator }P(s)}{\underbrace{3s+1}}/\underset{\text{denominator }Q(s)}{\underbrace{(s-1)(s+2)}}\mathrm{deg}P=1<\mathrm{deg}Q=2.$

• $P(s)=3s+1$ — the top polynomial. Its degree is $1$.
• $Q(s)=(s-1)(s+2)$ — the bottom. Its degree is $2$.
• Because $\mathrm{deg}P<\mathrm{deg}Q$, the fraction is proper — the precondition for partial fractions. (If it were not, we would long-divide first.)

---

Step 3 — Split into atoms (partial fractions)

WHAT. We propose that our compound is secretly a sum of two single-pole atoms, with unknown weights $A$ and $B$: $\frac{3s+1}{(s-1)(s+2)}=\frac{A}{s-1}+\frac{B}{s+2}.$

WHY. Each factor of $Q(s)$ owns exactly one pole. A distinct linear factor $(s-a)$ contributes one atom $\frac{A}{s-a}$ — a shape the table does recognise. We are guessing the recipe; the weights $A,B$ are what we still must find.

PICTURE. The single tall fraction on the left splits (a scissors icon) into two separate small fractions on the right, each drawn over its own single pole — the compound cracked into atoms.

• $\frac{A}{s-1}$ — the atom carrying the pole at $s=1$. Weight $A$ still unknown.
• $\frac{B}{s+2}$ — the atom carrying the pole at $s=-2$. Weight $B$ still unknown.

---

Step 4 — Weigh each atom by the cover-up trick

WHAT. To find $A$, multiply the whole identity by $(s-1)$ and then set $s=1$: $A=[(s-1)F(s){]}_{s=1}=\frac{3s+1}{s+2}{|}_{s=1}=\frac{3(1)+1}{1+2}=\frac{4}{3}.$ For $B$, multiply by $(s+2)$ and set $s=-2$: $B=\frac{3s+1}{s-1}{|}_{s=-2}=\frac{3(-2)+1}{-2-1}=\frac{-5}{-3}=\frac{5}{3}.$

WHY. Multiplying by $(s-1)$ cancels the $(s-1)$ under $A$'s own term, so $A$ stands alone. Every other term still carries a factor $(s-1)$, which becomes $0$ at $s=1$ — those neighbours vanish. So evaluating at the pole isolates precisely one weight. No linear system to solve.

PICTURE. The atom $\frac{A}{s-1}$ highlighted; a "cover" (gray patch) hides the $(s-1)$ factor in the original fraction; an arrow drops down to $s=1$ on the axis and reads off $4/3$. The other term is shown fading to zero.

• $(s-1)F(s)$ — removes the pole we are targeting.
• ${|}_{s=1}$ — the substitution that kills every rival term.
• $A=\frac{4}{3},B=\frac{5}{3}$ — the two residues, the "amount" of each atom.

So: $F(s)=\frac{4/3}{s-1}+\frac{5/3}{s+2}.$

---

Step 5 — Look up each atom in the table

WHAT. The one table fact we need: ${\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}.$ Apply it to each atom, pulling the constant weight through by linearity: ${\mathcal{L}}^{-1}\left\{\frac{4/3}{s-1}\right\}=\frac{4}{3}{e}^{1\cdot t},{\mathcal{L}}^{-1}\left\{\frac{5/3}{s+2}\right\}=\frac{5}{3}{e}^{-2t}.$

WHY. A single pole at $s=a$ is the transform of a pure exponential $e^{at}$ — this comes straight from the forward integral ${\int }_0^{\infty }{e}^{-st}{e}^{at}dt=\frac{1}{s-a}$ (valid for $s>a$). So the location of the pole becomes the growth rate of the exponential. Pole at $s=1$ → growing $e^t$; pole at $s=-2$ → decaying $e^{-2t}$.

PICTURE. Each pole on the $s$-axis mapped by an arrow to its time-graph: the $s=1$ pole → a rising blue curve $\frac{4}{3}{e}^t$; the $s=-2$ pole → a falling orange curve $\frac{5}{3}{e}^{-2t}$. Pole position ↔ exponential rate made visual.

• $a=1$ (right of origin) → positive exponent → the term grows.
• $a=-2$ (left of origin) → negative exponent → the term decays.
• The weights $\frac{4}{3},\frac{5}{3}$ just scale each curve vertically.

---

Step 6 — Glue the pieces back (linearity)

WHAT. Add the two inverted atoms: $f(t)=\frac{4}{3}{e}^t+\frac{5}{3}{e}^{-2t}.$

WHY. Because ${\mathcal{L}}^{-1}$ is linear, the inverse of a sum is the sum of the inverses. We split to conquer, invert atom-by-atom, and now re-assemble. This sum is $f(t)$: the unique time-function whose Laplace transform was our original $F(s)$.

PICTURE. The blue growing curve and the orange decaying curve added pointwise into one green total curve $f(t)$ — early on the decaying term matters, but the growing $e^t$ dominates for large $t$.

---

Step 7 — The edge cases you must never trip on

WHAT / WHY / PICTURE. Three degenerate variations of the same machine, each drawn so you recognise the trap on sight.

Case A — repeated pole. If instead $Q(s)=(s+1{)}^2$, the two poles collapse onto one spot. One atom is not enough: a factor $(s-a{)}^k$ needs $k$ terms, $\frac{s}{(s+1{)}^2}=\frac{A}{s+1}+\frac{B}{(s+1{)}^2},$ because the numerator on top may be any polynomial of degree $<2$. Solving $s=A(s+1)+B$ gives $A=1,B=-1$, and with ${\mathcal{L}}^{-1}\{1/(s+1{)}^2\}=t{e}^{-t}$ we get $f(t)=(1-t)e^{-t}$. Picture: the two red poles slide together into one double pole; the answer grows a $t$-factor.

Case B — complex poles (irreducible quadratic). If $Q(s)=s^2+2s+5$ has no real roots, the poles leave the axis into the complex plane. Complete the square: $s^2+2s+5=(s+1{)}^2+4$, so $a=-1,\omega =\sqrt{4}=2$. Via the first shift theorem, ${\mathcal{L}}^{-1}\left\{\frac{s+3}{(s+1{)}^2+4}\right\}=e^{-t}\cos 2t+e^{-t}\sin 2t.$ Real poles gave exponentials; a complex pair gives an oscillation $\sin /\cos$ inside an $e^{at}$ envelope. Picture: poles lift off the axis; the time-graph becomes a decaying wiggle.

Case C — real symmetric poles: $\sinh /\cosh$, not $\sin$. For $\frac{1}{s^2-4}$ the poles are real at $s=\pm 2$ (denominator $s^2-4$, a minus). Split: $\frac{1}{s^2-4}=\frac{1}{4}\left(\frac{1}{s-2}-\frac{1}{s+2}\right)\Rightarrow f(t)=\frac{1}{4}(e^{2t}-e^{-2t})=\frac{1}{2}\sinh 2t.$ A minus sign in $s^2-{\omega }^2$ means real poles → hyperbolic $\sinh /\cosh$; a plus in $s^2+{\omega }^2$ means imaginary poles → ordinary $\sin /\cos$. Picture: the sign in the denominator flips the answer between oscillation and hyperbolic growth.

---

The one-picture summary

One diagram compresses the whole walkthrough: a compound fraction with two poles → scissors → two atoms → table arrows → two time-curves → a plus-sign gluing them into $f(t)$. The pole positions in the $s$-world set the exponential rates in the $t$-world.

Recall Feynman retelling (hidden)

A messy fraction in the $s$-world is a compound made of atoms we can't read whole. We find where it blows up — those are the poles. Each pole is one atom. We split the compound into its atoms (partial fractions) and weigh each atom by covering up its own factor and reading off the value at its pole. Then we open the dictionary: a pole at $s=a$ means the time-function has a piece $e^{at}$ — right of zero it grows, left of zero it decays. If two poles merge, we need an extra term and a $t$ sneaks in. If the poles fly off the real axis into complex space, the pieces become waves $\sin /\cos$ wrapped in an exponential. If the denominator carries a minus, real twins give $\sinh /\cosh$ instead. Finally we glue all the time-pieces back with a plus sign — because undoing a transform is linear — and that sum is the one true $f(t)$. Split, weigh, look up, glue. Home.

---

Prerequisites & neighbours: Laplace Transform — definition and existence · Solving ODEs with Laplace Transforms · Partial Fraction Decomposition (Algebra) · First and Second Shift Theorems · Convolution Theorem · Heaviside Step & Dirac Delta · Hyperbolic functions sinh/cosh vs sin/cos.