Laplace transform — definition, region of convergence: Question bank

Intuition How to use this page

Each line below is a Question ::: Answer reveal. Cover the right side, commit to a full answer with a reason, then check. These are conceptual traps — no heavy computation. They probe the four things people get wrong about the Laplace transform: what the kernel $e^{- s t}$ is doing, why the Region of Convergence (ROC) is a right half-plane, when the transform fails to exist, and the role of $Re (s)$ .

Reminders before you start:

$L {f (t)} = F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t$ — the defining integral.
$s = σ + iω$ , where $σ = Re (s)$ is the real part and $ω$ is the imaginary part.
Exponential order $a$ means $∣ f (t) ∣ \leq M e^{a t}$ for large $t$ — a ceiling on how fast $f$ can grow.
ROC = the set of $s$ where that integral gives a finite answer; it is a half-plane $Re (s) > a$ .

Prerequisite refreshers if a question stings: Improper Integrals, Exponential Order and Growth Rates. Parent: Laplace Transform — Definition & ROC.

True or false — justify

The transform of

f (t) = e^{5 t}

exists for

s = 3

(treating

s

as real).

False. The ROC is

Re (s) > 5

, and

3 < 5

, so the integral

\int_{0}^{\infty} e^{- (3 - 5) t} d t = \int_{0}^{\infty} e^{2 t} d t

blows up. You are inside the growth, not the decay.

f (t)

is bounded for all

t \geq 0

, its ROC is

Re (s) > 0

True. Bounded means

∣ f (t) ∣ \leq M = M e^{0 \cdot t}

, so

f

is of exponential order

a = 0

; the half-plane is

Re (s) > 0

Two different time functions can never share the same algebraic formula

F (s)

False. They can share the formula but with different ROCs — the ROC is part of the transform's identity, which is why it matters for inversion.

F (s) = \frac{1}{s - a}

is a complete answer on its own.

False. Without the condition

Re (s) > a

it is just an algebraic expression; the transform is only meaningful where the defining integral actually converged.

Every polynomial in

t

has a Laplace transform.

True. Any polynomial grows slower than

e^{t}

(indeed slower than any

e^{a t}

with

a > 0

), so it is of exponential order

0

and its ROC is

Re (s) > 0

f (t) = e^{t^{2}}

has a Laplace transform for very large

s

False. It grows faster than every

e^{a t}

, so

e^{- s t} e^{t^{2}} \to \infty

for every

s

; no ROC exists at all.

The ROC of a Laplace transform can be a vertical strip (bounded left and right).

False for the one-sided (from

0

\infty

) transform used here — it is always a right half-plane. (Two-sided transforms can give strips, but that is a different object.)

Convergence of the integral depends on the imaginary part

ω

s

False. Since

∣ e^{- s t} ∣ = e^{- σ t}

depends only on

σ = Re (s)

, the oscillating factor

e^{- iω t}

has magnitude

1

and never affects convergence.

f

is piecewise continuous and of exponential order, then

F (s)

exists and

F (s) \to 0

Re (s) \to \infty

True. This is exactly the existence theorem; the bound

∣ F (s) ∣ \leq \frac{M}{σ - a} \to 0

σ \to \infty

delivers both claims.

A discontinuous

f (t)

(e.g. a step) cannot have a Laplace transform.

False. The theorem only requires piecewise continuity (finitely many jumps), so step functions transform just fine.

Spot the error

L {1} = [\frac{- 1}{s} e^{- s t}]_{0}^{\infty} = 0 - 0 = 0

The

t = 0

end is wrong:

e^{- s \cdot 0} = 1

, not

0

. The correct value is

0 - (\frac{- 1}{s}) = \frac{1}{s}

. Never drop the lower limit — it usually carries the whole numerator.

"Since

e^{- s t} \to 0

\infty

for any

s

, the integral

\int_{0}^{\infty} e^{- s t} d t

always converges."

The claim

e^{- s t} \to 0

holds only when

Re (s) > 0

. If

σ \leq 0

the magnitude

e^{- σ t}

does not decay, so the integral diverges — convergence is a condition, not automatic.

L {e^{a t}} = \frac{1}{s - a}

so its ROC is

s > a

The condition is on the real part:

Re (s) > a

, not

s > a

. Writing

s > a

silently assumes

s

is real and ignores the half-plane geometry.

f (t) = e^{a t}

grows, so its integral diverges and it has no transform."

It has a transform on the half-plane

Re (s) > a

; the decaying kernel

e^{- s t}

beats the growth

e^{a t}

precisely when

σ > a

. Growth alone does not kill existence — growth faster than every exponential does.

"For

L {t}

, the boundary term

[- \frac{t}{s} e^{- s t}]_{0}^{\infty}

\infty

because

t \to \infty

The exponential decay dominates the linear

t

, so

t e^{- s t} \to 0

for

Re (s) > 0

. The boundary term vanishes at both ends, leaving

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

sin (ω t)

grows without bound as

t

increases, so ROC is

Re (s) > ω

Sine is bounded (

∣ sin ω t ∣ \leq 1

), exponential order

a = 0

, so ROC is

Re (s) > 0

. The frequency

ω

sets the poles at

\pm iω

, not the growth rate.

Why questions

Why is the kernel specifically

e^{- s t}

and not, say,

e^{- t} f (t)

with a fixed rate?

A variable rate

s

lets us sweep through fade-speeds; and

\frac{d}{d t} e^{- s t} = - s e^{- s t}

is the exact property that turns differentiation into multiplication by

s

— see Laplace Transform of Derivatives.

Why does the ROC always sit to the right (larger

Re (s)

The integrand behaves like

e^{(a - σ) t}

; it decays only when

σ > a

. Bigger

σ

= faster fade = safer convergence, so the "good" side is to the right of

a

Why do we need exponential order at all — isn't piecewise continuity enough?

Continuity controls behaviour on any finite piece, but the integral runs to

\infty

. Exponential order caps the growth at large

t

so the tail

\int^{\infty} e^{- σ t} M e^{a t} d t

actually converges.

Why must we state the ROC when inverting a transform?

Because the same

F (s)

formula can arise from different signals with different ROCs; the ROC selects which signal is meant, guaranteeing a unique inverse — see Inverse Laplace Transform.

Why does

lim_{Re (s) \to \infty} F (s) = 0

for every valid transform?

From

∣ F (s) ∣ \leq \frac{M}{σ - a}

, sending

σ \to \infty

drives the bound to

0

. Physically: fade infinitely fast and only the instant

t = 0

survives, contributing nothing in the limit.

Why is the Fourier transform describable as "Laplace along a vertical line"?

Setting

σ = 0

(so

s = iω

) turns

e^{- s t}

into a pure oscillation

e^{- iω t}

; this is the boundary

Re (s) = 0

of the half-plane — see Fourier Transform.

Edge cases

What is the ROC of the zero function

f (t) = 0

The integral is

0

for every

s

, so the ROC is the entire complex plane — the only function with no convergence restriction.

Does

f (t) = e^{a t}

with

a < 0

(a decaying exponential) have a special ROC?

Still

Re (s) > a

, but now

a

is negative, so the half-plane includes part of the left side, e.g.

Re (s) > - 3

for

e^{- 3 t}

. Decay pushes the boundary leftward.

What happens exactly on the boundary line

Re (s) = a

The integral typically diverges there (e.g.

\int_{0}^{\infty} e^{0 \cdot t} d t = \infty

for

L {1}

σ = 0

), which is why the ROC is the open half-plane

Re (s) > a

, excluding the boundary.

f (t)

is only defined for

t \geq 0

, why do we not worry about

t < 0

The one-sided transform integrates from

0

\infty

by definition, so negative-time values simply never enter — the transform is built for signals that "switch on" at

t = 0

Can a function be of exponential order

a = 0

yet unbounded?

Yes — e.g.

f (t) = t

is unbounded but satisfies

∣ t ∣ \leq M e^{0 \cdot t}

? No;

t

needs

a > 0

strictly. Better example: any bounded function is order

0

; unbounded-but-polynomial functions need a tiny positive

a

. The subtlety: order

0

means "no faster than a constant," which forces boundedness for large

t

What if

f

has an infinite jump (a spike) at a single point but is otherwise nice?

A genuine infinite spike breaks piecewise continuity, so the existence theorem no longer applies directly; such objects need distribution theory (e.g. the Dirac delta) rather than the ordinary integral.

L {f}

affected by changing

f

at one isolated point?

No. An integral is blind to values on a set of zero length, so redefining

f

at finitely many points leaves

F (s)

and its ROC unchanged.

Recall One-line summary of every trap

Convergence depends on $Re (s)$ only, the ROC is an open right half-plane $Re (s) > a$ , existence needs piecewise continuity plus exponential order, and functions like $e^{t^{2}}$ that outgrow all exponentials have no transform at all.

Connections

Inverse Laplace Transform — why the ROC is needed for a unique return trip.
Laplace Transform of Derivatives — the $\times s$ property the kernel was chosen for.
Solving ODEs with Laplace Transforms — the payoff.
Improper Integrals — the convergence machinery under every "does it exist?" question.
Fourier Transform — the $Re (s) = 0$ boundary case.
Exponential Order and Growth Rates — the growth ceiling behind existence.

#flashcard