Properties — linearity, first - second shift theorems, scaling: Question bank

True or false — justify

TRUE/FALSE:

L {f (t) + g (t)} = F (s) + G (s)

requires

f

and

g

to have the same order of growth.

False. Linearity comes purely from the integral splitting a sum; it needs only that both transforms exist separately, not that they match in any way.

TRUE/FALSE:

L {f (t) g (t)} = F (s) G (s)

False. Linearity handles sums, not products. The transform of a product of two time-functions is a convolution in

s

, not a product — see Convolution Theorem.

TRUE/FALSE: The first shift

L {e^{a t} f (t)} = F (s - a)

works for negative

a

too.

True. Nothing in the derivation assumed

a > 0

; e.g.

e^{- t} f (t)

gives

F (s + 1)

. The sign just decides which way the transform slides on the

s

-axis.

TRUE/FALSE: The second shift

L {f (t - a) u (t - a)} = e^{- a s} F (s)

is valid for

a < 0

False. The derivation splits the integral at

t = a

and needs

a \geq 0

so the delay lands inside

[0, \infty)

. A "negative delay" would ask for information before

t = 0

, outside the transform's domain.

TRUE/FALSE: Scaling

L {f (c t)} = \frac{1}{c} F (s / c)

holds for any real

c \neq = 0

False. It requires

c > 0

. If

c < 0

the substitution

τ = c t

flips the integration limits and

f (c t)

probes negative time, breaking the derivation.

TRUE/FALSE:

L {5} = \frac{5}{s}

is just linearity applied to

L {1} = \frac{1}{s}

True. A constant

5 = 5 \cdot 1

, and linearity pulls the

5

out:

L {5} = 5 \cdot \frac{1}{s}

TRUE/FALSE: Multiplying a function by

e^{a t}

changes the shape of its transform.

False. It slides the entire transform sideways (

s \to s - a

) without altering its form; every

s

simply becomes

s - a

TRUE/FALSE: A time delay changes the size (amplitude) of

F (s)

False. It only attaches the complex-magnitude-one-on-the-imaginary-axis tag

e^{- a s}

; the underlying

F (s)

is untouched, matching "delay doesn't change shape."

TRUE/FALSE: The first and second shift can be applied to the same function one after another.

True. They act on different domains (one on

s

, one via a time delay), so e.g.

L {e^{2 t} f (t - 3) u (t - 3)}

is a legitimate combination — just apply each rule carefully.

Spot the error

FIND THE ERROR: "

L {t^{2} u (t - 2)} = e^{- 2 s} \cdot \frac{2}{s ^{3}}

The second shift needs

f (t - a)

, but

t^{2} \neq = (t - 2)^{2}

. You must rewrite

t^{2} = (t - 2)^{2} + 4 (t - 2) + 4

first, then transform each piece. Only the delayed-argument form earns

e^{- a s} F (s)

FIND THE ERROR: "

L {e^{3 t} cos 2 t} = \frac{s}{( s - 3 ) ^{2} + 4}

The first shift replaces

s

s - 3

everywhere, including the numerator. The correct answer is

\frac{s - 3}{( s - 3 ) ^{2} + 4}

; leaving the top as

s

is the classic half-substitution slip.

FIND THE ERROR: "

L {sin 3 t} = \frac{1}{3} F (s /3) = \frac{1}{3 ( s ^{2} + 9 )}

where

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

The algebra dropped the

9

from clearing the fraction:

\frac{1}{3} \cdot \frac{1}{( s /3 ) ^{2} + 1} = \frac{1}{3} \cdot \frac{9}{s ^{2} + 9} = \frac{3}{s ^{2} + 9}

, not

\frac{1}{3 ( s ^{2} + 9 )}

FIND THE ERROR: "

L {f (c t)} = F (s / c)

— the

1/ c

is absorbed into

F

The

1/ c

comes from the Jacobian

d t = d τ / c

and sits outside the integral; it multiplies the whole result. Forgetting it gives an amplitude that is wrong by a factor of

c

FIND THE ERROR: "

L {3 f (t) + 5} = 3 F (s) + 5

The constant

5

is a function of time and must itself be transformed:

L {5} = \frac{5}{s}

. The correct answer is

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

; you cannot leave a raw constant in the

s

-world.

FIND THE ERROR: "Both shift theorems produce a factor

e^{- a s}

because both involve an exponential."

Only the second shift produces a factor

e^{- a s}

. The first shift produces a substitution

s \to s - a

with no external exponential factor at all.

FIND THE ERROR: "

L {u (t - 2)} = e^{- 2 s} \cdot \frac{1}{s ^{3}}

Here

f (t) = 1

(since

u (t - 2) = 1 \cdot u (t - 2)

), so

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

, giving

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

. The writer wrongly used

\frac{1}{s ^{3}}

, the transform of

t^{2}

Why questions

WHY does linearity hold at all for the Laplace transform?

Because

L

is built from an integral, and integration lets you pull out constants and split sums — those two rules are linearity.

WHY does the first shift turn a time-multiply by

e^{a t}

into an

s

-shift?

Combining

e^{- s t} e^{a t} = e^{- (s - a) t}

makes the integral look exactly like the definition of

F

but with

s

replaced by

s - a

, so the result is

F (s - a)

WHY does the second shift need the unit step

u (t - a)

Without it the integral starts at

t = 0

, but the substitution

τ = t - a

that factors out

e^{- a s}

only works cleanly if the integrand is zero below

t = a

— which is exactly what

u (t - a)

guarantees.

WHY does scaling time by

c

stretch the

s

-axis by

1/ c

The substitution

τ = c t

forces

s

to appear as

s / c

inside the integral, so squeezing time (large

c

) spreads the transform out along

s

— a reciprocal trade-off like Fourier.

WHY does an extra

1/ c

appear in the scaling formula but not in the shift formulas?

Only the scaling substitution changes the differential (

d t = d τ / c

); the shift derivations either combine exponentials or shift limits without rescaling

d t

, so no Jacobian factor arises.

WHY is

L {e^{a t} f (t)} = F (s - a)

preferred over re-integrating from scratch?

It reuses a transform you already know and only substitutes

s \to s - a

, avoiding a fresh integral — that reuse is the entire point of building a table of properties.

WHY can't we just write

L {δ (t - a)}

using the second shift with

f = 1

The Dirac delta is not an ordinary delayed function

f (t - a) u (t - a)

; it's a distribution and is transformed directly as

L {δ (t - a)} = e^{- a s}

— see Unit Step and Dirac Delta Functions.

Edge cases

EDGE CASE: What is

L {f (t - a) u (t - a)}

when

a = 0

The step becomes

u (t) = 1

for

t \geq 0

and the factor is

e^{0} = 1

, so it reduces to plain

L {f (t)} = F (s)

— the "no delay" case, consistent with the general rule.

EDGE CASE: What does the first shift give when

a = 0

e^{0 \cdot t} f (t) = f (t)

and

F (s - 0) = F (s)

, so nothing changes — the identity case, a sanity check that the rule degrades gracefully.

EDGE CASE: What does scaling give at

c = 1

\frac{1}{1} F (s /1) = F (s)

: no squeeze, no change — again the identity, confirming the formula is consistent at its boundary.

EDGE CASE: Applying scaling as

c \to 0^{+}

(very slow time).

\frac{1}{c} F (s / c)

blows up: the

1/ c

prefactor diverges and

F (s / c)

evaluates near

s = 0

, reflecting that infinitely-slowed signals lose their transform's convergence. The formula still holds for every fixed

c > 0

but degenerates in the limit.

EDGE CASE:

L {f (t) u (t - a)}

when

f

is NOT written as

f (t - a)

— can you still use the second shift?

Not directly. You must first re-express

f (t)

in powers of

(t - a)

(e.g. Taylor-expand or algebraically shift), transform each delayed piece, then apply

e^{- a s}

to each. The raw form has no shortcut.

EDGE CASE: Does linearity let you transform

∣ f (t) ∣

from

F (s)

No. Absolute value is a nonlinear operation, so

L {∣ f ∣}

has no relation to

F (s)

through linearity — you must transform

∣ f ∣

as a new function from scratch.

EDGE CASE: If

F (s)

exists only for

Re (s) > 0

, where does

L {e^{2 t} f (t)} = F (s - 2)

converge?

For

Re (s - 2) > 0

, i.e.

Re (s) > 2

: the region of convergence shifts right by

2

along with the transform. The shift moves the convergence strip, not just the formula.

Recall One-line summary of every trap

Linearity: sums split, products don't. First shift: substitute $s \to s - a$ everywhere, any sign of $a$ . Second shift: needs $u (t - a)$ , needs $a \geq 0$ , needs the argument written as $(t - a)$ , and stamps $e^{- a s}$ . Scaling: $c > 0$ only, and never drop the $1/ c$ .

Connections

4.6.27 Properties — linearity, first - second shift theorems, scaling (Hinglish)
Laplace Transform — Definition and Existence
Laplace Transforms of Standard Functions (1, e^at, sin, cos, t^n)
Inverse Laplace Transform and Partial Fractions
Unit Step and Dirac Delta Functions
Convolution Theorem