Inverse Laplace transform — partial fractions, tables: Question bank

Every item points at a specific misconception about splitting $F (s)$ into atoms and reading the table. Prerequisites worth revisiting if a group hurts: Partial Fraction Decomposition (Algebra), First and Second Shift Theorems, Hyperbolic functions sinh/cosh vs sin/cos.

True or false — justify

True or false:

L^{- 1}

is linear, so

L^{- 1} {F (s) G (s)} = L^{- 1} {F} \cdot L^{- 1} {G}

False. Linearity only handles sums and scalar multiples; products in the

s

-world become a convolution in the

t

-world, not a product — see Convolution Theorem.

True or false:

\frac{1}{s ^{2} + 4}

inverts to a sine.

True, but you must fix the constant:

\frac{1}{s ^{2} + 4} = \frac{1}{2} \cdot \frac{2}{s ^{2} + 4} \to \frac{1}{2} sin 2 t

, because the table numerator must be

ω = 2

True or false:

\frac{1}{s ^{2} - 4}

inverts to a sine.

False. The sign is minus, so the poles are real (

s = \pm 2

), giving

\frac{1}{2} sinh 2 t

— a hyperbolic, growing function, not an oscillation.

True or false: every proper rational

F (s)

has a unique inverse

f (t)

True for piecewise-continuous functions of exponential order — Lerch's theorem makes the forward transform one-to-one, which is exactly what justifies a lookup table.

True or false: if

F (s) = \frac{2 s + 5}{s + 3}

you can go straight to partial fractions.

False. Degree of top

=

degree of bottom, so it is improper; do long division first (

= 2 - \frac{1}{s + 3}

), then invert the pieces.

True or false: a factor

(s + 1)^{3}

contributes exactly one partial-fraction term.

False. It contributes three terms, one for each power

(s + 1)^{1}, (s + 1)^{2}, (s + 1)^{3}

, because the numerator over it can be any polynomial of degree

< 3

True or false:

L^{- 1} {1}

is an ordinary function.

False.

F (s) = 1

is not a proper rational (constant top, constant bottom of degree 0), and its inverse is the Dirac delta

δ (t)

— see Heaviside Step & Dirac Delta.

True or false: completing the square is only cosmetic; you could invert

\frac{1}{s ^{2} + 2 s + 5}

without it.

False in practice. The table has no entry for a raw quadratic with a linear term; rewriting as

(s + 1)^{2} + 4

is what unlocks the shifted sine/cosine entries.

Spot the error

\frac{s}{s ^{2} + 9} \to sin 3 t

." Where's the slip?

s

on top means cosine:

\frac{s}{s ^{2} + 9} \to cos 3 t

. Only a lone

ω = 3

on top gives sine.

\frac{s}{( s + 1 ) ^{2}} = \frac{A}{( s + 1 ) ^{2}}

, one factor one term, so answer

= A t e^{- t}

The double pole needs both powers:

\frac{A}{s + 1} + \frac{B}{( s + 1 ) ^{2}}

. Dropping the first term loses the

e^{- t}

piece; the true answer is

(1 - t) e^{- t}

"For

(s - 1) (s + 2)

in the denominator, cover-up gives

A

by setting

s = 1

in the whole fraction including the

(s - 1)

You must remove the

(s - 1)

first (multiply through), then set

s = 1

; otherwise you divide by zero. Cover-up evaluates

(s - 1) F (s)

s = 1

(s + 1)^{2} + 4

, so

ω = 4

and I write

4 sin

The table uses

ω^{2}

, so

ω = 4 = 2

; the sine numerator and argument must both be

2

, giving

e^{- t} sin 2 t

\frac{2}{( s + 1 ) ^{2} + 4} \to e^{- t} cos 2 t

A constant numerator matches the sine atom, and it must equal

ω = 2

\frac{2}{( s + 1 ) ^{2} + 4} \to e^{- t} sin 2 t

. Cosine needs

(s + 1)

on top.

"Since

\frac{1}{s - a} \to e^{a t}

, then

\frac{1}{s + 2} \to e^{2 t}

Here

a = - 2

, so it is

e^{- 2 t}

; read the sign of the pole carefully — the exponent is the root of the denominator.

\frac{5}{s ^{2} + 2 s + 5}

has

s

nowhere on top, so the answer is a pure sine with no exponential."

Completing the square gives

(s + 1)^{2} + 4

; the shift

s \to s + 1

forces an

e^{- t}

envelope: the answer is

\frac{5}{2} e^{- t} sin 2 t

, not a bare sine.

Why questions

Why does the First Shift Theorem turn a shifted denominator into a damped oscillation?

Replacing

s

s - a

inside

\int e^{- s t} f d t

inserts an extra

e^{a t}

factor, so a completed-square quadratic

(s - a)^{2} + ω^{2}

carries an

e^{a t}

envelope multiplying the sine/cosine.

Why do we bother with partial fractions instead of a direct inverse integral?

The table only knows one pole at a time; partial fractions splits a compound

F (s)

into those single-pole atoms so linearity can invert and reassemble — recognition beats integration.

Why does the cover-up method work only for simple (non-repeated) poles?

Multiplying by

(s - a)

and setting

s = a

isolates that residue only if no other term still has an

(s - a)

; a repeated factor leaves surviving

(s - a)

powers, so you need matching/differentiation for the lower coefficients.

Why must

F (s)

be proper before we invert with the table?

Every table entry vanishes as

s \to \infty

; an improper fraction does not, so its polynomial part corresponds to

δ (t)

and its derivatives, which live outside the ordinary-function table.

Why does

s^{2} + ω^{2}

give oscillation while

s^{2} - ω^{2}

gives hyperbolics?

The sign fixes whether the poles are imaginary (

\pm iω

, oscillating

e^{\pm iω t}

) or real (

\pm ω

, growing/decaying

e^{\pm ω t}

) — see Hyperbolic functions sinh/cosh vs sin/cos.

Why is uniqueness of

L^{- 1}

needed at all — couldn't we just guess?

Without uniqueness, two different

f (t)

could share one

F (s)

and the table's "the answer is..." would be meaningless; Lerch's theorem guarantees the return ticket lands on exactly one function.

Edge cases

What is

L^{- 1} {1/ s}

, the

a \to 0

limit of

1/ (s - a) \to e^{a t}

It is the constant

1

(

= e^{0 \cdot t}

) — the degenerate exponential is a flat line, consistent with the table's first row.

What happens to

\frac{ω}{s ^{2} + ω ^{2}}

ω \to 0

The numerator vanishes and the pole collapses to a double pole at

0

; the sine

sin ω t

flattens toward

0

, matching

ω t

becoming negligible — no oscillation survives.

What is

L^{- 1} {1/ (s^{2} + ω^{2})^{2}}

conceptually — still a table lookup?

No single atom fits a squared quadratic; you either use Convolution Theorem (self-convolution of

\frac{1}{ω} sin ω t

) or differentiation-in-

s

, since the basic table stops at first-power denominators.

L^{- 1} {e^{- 2 s} / s}

just the shift

1/ s \to 1

No — an

e^{- a s}

factor is a time shift (Second Shift), giving the Heaviside step

u (t - 2)

, not the

s

-shift envelope; see First and Second Shift Theorems.

Does a purely real repeated pole

(s - a)^{2}

ever produce oscillation?

Never — real poles give

e^{a t}

and

t e^{a t}

only; oscillation requires a complex pole (an irreducible quadratic with negative discriminant relative to its vertex).

Recall One-line self-test

If you can state, for any $F (s)$ , why each atom is oscillatory / exponential / polynomial / delta before touching the table, you own this topic. My reason for $\frac{s}{( s + 1 ) ^{2} + 9}$ ? ::: Shifted quadratic $\Rightarrow$ $e^{- t}$ envelope; numerator has the shifted $(s + 1)$ hidden inside, so cosine-plus-correction: $e^{- t} (cos 3 t - \frac{1}{3} sin 3 t)$ .