Inverse Laplace transform — partial fractions, tables

WHAT is the inverse Laplace transform?

WHY is it well-defined? For functions that are piecewise continuous and of exponential order, the forward transform is one-to-one (Lerch's theorem), so "undoing" it gives a unique $f(t)$. That uniqueness is what lets us use a lookup table with confidence.

The core table (derive, don't memorise blindly)

Every entry comes from the forward integral $F(s)=\int_0^\infty e^{-st}f(t)\,dt$. Let me derive the key ones so the table feels earned.

Master table (memorise these 7):

HOW: partial fractions — the recognition machine

A proper rational $F(s)=\dfrac{P(s)}{Q(s)}$ (degree of $P$ < degree of $Q$) is split based on the factors of $Q(s)$:

• Distinct linear $(s-a)$ → term $\dfrac{A}{s-a}$
• Repeated linear $(s-a)^k$ → terms $\dfrac{A_1}{s-a}+\dfrac{A_2}{(s-a)^2}+\dots+\dfrac{A_k}{(s-a)^k}$
• Irreducible quadratic $(s^2+bs+c)$ → term $\dfrac{Bs+C}{s^2+bs+c}$ (then complete the square to match the shift table)

Cover-up method (fast coefficient finder for simple poles): to get $A$ for the factor $(s-a)$, multiply both sides by $(s-a)$ and set $s=a$: $A=\left.\big[(s-a)F(s)\big]\right|_{s=a}.$ WHY it works: multiplying kills the $(s-a)$ in the denominator of its own term, and at $s=a$ every other term still has $(s-a)$ as a factor → they vanish, isolating $A$.

Worked Example 1 — distinct linear factors

Find $\mathcal{L}^{-1}\left\{\dfrac{3s+1}{(s-1)(s+2)}\right\}$.

Step 1. Write $\dfrac{3s+1}{(s-1)(s+2)}=\dfrac{A}{s-1}+\dfrac{B}{s+2}$. Why this step? Two distinct linear poles → two simple-pole atoms.

Step 2. Cover-up: $A=\dfrac{3(1)+1}{1+2}=\dfrac{4}{3}$, $\;B=\dfrac{3(-2)+1}{-2-1}=\dfrac{-5}{-3}=\dfrac{5}{3}$. Why this step? Cover-up isolates each residue without solving a system.

Step 3. Invert each atom with $\mathcal{L}^{-1}\{1/(s-a)\}=e^{at}$: $f(t)=\frac{4}{3}e^{t}+\frac{5}{3}e^{-2t}.$ Why this step? Linearity lets us invert term-by-term.

Worked Example 2 — irreducible quadratic (complete the square)

Find $\mathcal{L}^{-1}\left\{\dfrac{s+3}{s^2+2s+5}\right\}$.

Step 1. Complete the square: $s^2+2s+5=(s+1)^2+4$. Why this step? The denominator has no real roots; the shift table needs the form $(s-a)^2+\omega^2$ with $a=-1,\omega=2$.

Step 2. Rewrite numerator around $(s+1)$: $\frac{s+3}{(s+1)^2+4}=\frac{(s+1)+2}{(s+1)^2+4}=\frac{s+1}{(s+1)^2+4}+\frac{2}{(s+1)^2+4}.$ Why this step? Match $\frac{s-a}{(s-a)^2+\omega^2}$ (gives $e^{at}\cos$) and $\frac{\omega}{(s-a)^2+\omega^2}$ (gives $e^{at}\sin$). The second term has $2=\omega$, perfect.

Step 3. Read the table with $a=-1,\omega=2$: $f(t)=e^{-t}\cos 2t + e^{-t}\sin 2t.$ Why this step? First-shift theorem converts the shifted denominator into the $e^{-t}$ envelope.

Worked Example 3 — repeated factor

Find $\mathcal{L}^{-1}\left\{\dfrac{s}{(s+1)^2}\right\}$.

Step 1. $\dfrac{s}{(s+1)^2}=\dfrac{A}{s+1}+\dfrac{B}{(s+1)^2}$. Why this step? A double pole needs both powers; only one term would underfit.

Step 2. Multiply by $(s+1)^2$: $s = A(s+1)+B$. Set $s=-1$: $-1=B$. Compare $s$-coefficient: $1=A$. Why this step? Cover-up gives the highest-power coefficient directly; matching gives the rest.

Step 3. Invert: $\mathcal{L}^{-1}\{1/(s+1)\}=e^{-t}$, and $\mathcal{L}^{-1}\{1/(s+1)^2\}=t e^{-t}$ (shift of $1/s^2\to t$): $f(t)=e^{-t}-t e^{-t}=(1-t)e^{-t}.$

Forecast-then-Verify drill

Before checking the table, predict the form of $\mathcal{L}^{-1}\left\{\frac{1}{s^2-4}\right\}$. Real poles at $\pm2$? → expect $e^{\pm2t}$, i.e. $\sinh/\cosh$, not $\sin$. Then verify: $\frac{1}{s^2-4}=\frac14\left(\frac1{s-2}-\frac1{s+2}\right)\Rightarrow \frac14(e^{2t}-e^{-2t})=\tfrac12\sinh 2t$. ✓ Sign of the constant inside ($-4$ vs $+4$) flips trig↔hyperbolic.

Flashcards

Connections

• 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

Concept Map

Hinglish (regional understanding)