Properties — linearity, first - second shift theorems, scaling: Worked examples

Before anything, the one object every rule is built from — the definition from Laplace Transform — Definition and Existence:

L {f (t)} = F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t .

Here $t$ is time (always $\geq 0$ for us), $s$ is a helper number big enough to make the integral finite, and $F (s)$ is the "recipe" the machine spits out. The building-block recipes we reuse are from Laplace Transforms of Standard Functions (1, e^at, sin, cos, t^n):

L {1} = \frac{1}{s}, L {t^{n}} = \frac{n !}{s ^{n + 1}}, L {e^{a t}} = \frac{1}{s - a}, L {sin b t} = \frac{b}{s ^{2} + b ^{2}}, L {cos b t} = \frac{s}{s ^{2} + b ^{2}} .

The scenario matrix

Every problem this topic throws at you lands in exactly one of these cells. The worked examples below are labelled [C#] so you can see the whole grid gets covered.

Cell	Case class	What makes it tricky	Example
C1	Pure linearity, positive coefficients	just split & scale	Ex 1
C2	Linearity with a negative / sign-flipped term	keep the minus sign attached	Ex 1
C3	First shift, positive $a$ (growth $e^{a t}$ )	substitute $s \to s - a$ everywhere	Ex 2
C4	First shift, negative $a$ (decay $e^{- a t}$ )	$s \to s + a$ , sign flips	Ex 3
C5	Second shift, argument already $(t - a)$	plug straight in	Ex 4
C6	Second shift, argument NOT in $(t - a)$ form	rewrite by algebra first	Ex 5
C7	Scaling, general $c > 1$ or $0 < c < 1$	keep the $\frac{1}{c}$	Ex 6
C8	Degenerate input ( $c = 1$ , $a = 0$ , zero function)	rule must reduce to identity	Ex 7
C9	Combined rules (shift + linearity + step)	order of operations	Ex 8
C10	Real-world word problem (delayed switch-on)	translate words → $u (t - a)$	Ex 9
C11	Exam twist — reverse direction (inverse)	read the rule backwards	Ex 10

Ex 1 — Linearity, positive and negative terms [C1, C2]

Ex 2 — First shift, positive $a$ (growing exponential) [C3]

Ex 3 — First shift, negative $a$ (decaying exponential) [C4]

Ex 4 — Second shift, argument already $(t - a)$ [C5]

Ex 5 — Second shift, argument NOT in $(t - a)$ form [C6]

Worked example Find

$L {t u (t - 1)}$ .

Forecast: The step is $u (t - 1)$ so $a = 1$ , but the multiplier is $t$ , not $(t - 1)$ . What must we do before applying the rule? Guess.

Step 1. Rewrite $t$ in terms of $(t - 1)$ : $t = (t - 1) + 1$ . Why this step? The second shift theorem only works on $f (t - a)$ . We algebraically force the shifted variable to appear.

Step 2. Split by linearity: $t u (t - 1) = (t - 1) u (t - 1) + 1 \cdot u (t - 1) .$ Why this step? Now each piece has a clean $(t - 1)$ argument (the constant $1$ counts as $f = 1$ , and $L {u (t - a)} = \frac{e ^{- a s}}{s}$ ).

Step 3. Transform each: for $(t - 1) u (t - 1)$ use $f (t) = t$ , $F (s) = \frac{1}{s ^{2}}$ ; for $u (t - 1)$ use $F (s) = \frac{1}{s}$ . Both wear the tag $e^{- s}$ . $L {t u (t - 1)} = e^{- s} (\frac{1}{s ^{2}} + \frac{1}{s}) .$ Why this step? Same delay $a = 1$ ⇒ same $e^{- s}$ on both; only the base recipes differ.

Verify (sanity check the trap): The wrong answer $e^{- s} \frac{1}{s ^{2}}$ (grabbing the rule directly) misses the $\frac{1}{s}$ piece. Our version equals $e^{- s} \frac{s + 1}{s ^{2}}$ . Numerically at $s = 1$ : $e^{- 1} (1 + 1) = 2 e^{- 1} \approx 0.7358$ . ✔

Ex 6 — Scaling, both $c > 1$ and $0 < c < 1$ [C7]

Worked example Given

$L {cos t} = \frac{s}{s ^{2} + 1}$ , find $L {cos 5 t}$ and $L {cos \frac{1}{2} t}$ by scaling.

Forecast: Scaling says $L {f (c t)} = \frac{1}{c} F (s / c)$ . Predict which way the $s$ -axis stretches for $c = 5$ vs $c = \frac{1}{2}$ .

Step 1. For $c = 5$ : $F (s) = \frac{s}{s ^{2} + 1}$ , so $F (s /5) = \frac{s /5}{( s /5 ) ^{2} + 1} = \frac{s /5}{s ^{2} /25 + 1}$ . Why this step? Squeezing time by $5$ (faster wiggle) must stretch the $s$ -axis by $1/5$ — that's the substitution $τ = 5 t$ .

Step 2. Multiply by $\frac{1}{c} = \frac{1}{5}$ and simplify: $L {cos 5 t} = \frac{1}{5} \cdot \frac{s /5}{s ^{2} /25 + 1} = \frac{1}{5} \cdot \frac{5 s}{s ^{2} + 25} = \frac{s}{s ^{2} + 25} .$

Step 3. For $c = \frac{1}{2}$ (slower wiggle): $F (2 s) = \frac{2 s}{4 s ^{2} + 1}$ , times $\frac{1}{1/2} = 2$ : $L {cos \frac{1}{2} t} = 2 \cdot \frac{2 s}{4 s ^{2} + 1} = \frac{4 s}{4 s ^{2} + 1} = \frac{s}{s ^{2} + 1/4} .$ Why this step? $c < 1$ stretches time (slower), so the $s$ -axis compresses — pole moves inward to $s = \pm \frac{1}{2} i$ .

Verify: Both match the direct table form $\frac{s}{s ^{2} + b ^{2}}$ with $b = 5$ and $b = \frac{1}{2}$ respectively. ✔

Ex 7 — Degenerate inputs: $c = 1$ , $a = 0$ , and the zero function [C8]

Worked example Show each rule collapses to something obvious when its parameter is "off".

Forecast: A good rule must do nothing when its knob is at the neutral setting. Predict the three neutral settings.

Step 1 — Scaling at $c = 1$ : $L {f (1 \cdot t)} = \frac{1}{1} F (s /1) = F (s)$ . Why this step? No time-squeeze means no change — the rule must reduce to identity, and it does.

Step 2 — First shift at $a = 0$ : $L {e^{0} f (t)} = F (s - 0) = F (s)$ . Why this step? $e^{0} = 1$ ; multiplying by $1$ changes nothing, and the shift by $0$ agrees.

Step 3 — Second shift at $a = 0$ : $L {f (t) u (t)} = e^{0} F (s) = F (s)$ . Why this step? $u (t) = 1$ for $t \geq 0$ (our whole domain), so no delay, tag $e^{0} = 1$ .

Step 4 — Zero function: $L {0} = \int_{0}^{\infty} e^{- s t} \cdot 0 d t = 0$ . Why this step? Linearity with $a = b = 0$ forces the transform of nothing to be nothing.

Verify: All four reduce to the identity/zero as required — no rule "invents" a change from a neutral input. ✔

Ex 8 — Combined: linearity + first shift + step [C9]

Worked example Find

$L {e^{- t} (t - 3) u (t - 3)}$ .

Forecast: Two exponentials will appear — one a shift ( $e^{- t}$ ) and one a delay tag ( $e^{- 3 s}$ ). Guess which becomes $s \to s + 1$ and which becomes a factor.

Step 1. Handle the delay first: $g (t) = (t - 3) u (t - 3)$ has $f (t) = t$ , $F (s) = \frac{1}{s ^{2}}$ , so $L {(t - 3) u (t - 3)} = \frac{e ^{- 3 s}}{s ^{2}} \equiv G (s) .$ Why this step? Second shift acts on the time-delayed part; do it before touching the $e^{- t}$ .

Step 2. Now the extra $e^{- t}$ triggers the first shift with $a = - 1$ : replace $s \to s + 1$ in all of $G (s)$ , including the exponential. Why this step? First shift substitutes $s \to s - a = s + 1$ everywhere — the $e^{- 3 s}$ is part of $G$ , so it too gets shifted to $e^{- 3 (s + 1)}$ . $L {e^{- t} (t - 3) u (t - 3)} = \frac{e ^{- 3 (s + 1)}}{( s + 1 ) ^{2}} = \frac{e ^{- 3 s - 3}}{( s + 1 ) ^{2}} .$

Verify: At $s = 1$ : $\frac{e ^{- 6}}{4} = \frac{e ^{- 6}}{4} \approx 6.196 \times 1 0^{- 4}$ , positive and tiny (integrand decays fast). ✔

Ex 9 — Real-world word problem: a delayed voltage pulse [C10]

Worked example A circuit's input voltage is

off until $t = 2$ seconds, then equals $v (t) = 3$ volts constant from then on. Find $L {v (t)}$ .

Forecast: "Off then constant" screams a step $u (t - 2)$ . Predict the $e^{- 2 s}$ tag.

Step 1. Write the input in symbols: $v (t) = 3 u (t - 2)$ . Why this step? $u (t - 2)$ is $0$ before $t = 2$ and $1$ after — exactly "off until $2$ , then on."

Step 2. Base $f (t) = 3$ (a constant, so $f (t - 2) = 3$ too, since constants don't feel a shift), $F (s) = \frac{3}{s}$ . Attach the delay tag: $L {3 u (t - 2)} = e^{- 2 s} \cdot \frac{3}{s} = \frac{3 e ^{- 2 s}}{s} .$ Why this step? Second shift with $a = 2$ ; a constant's shape is untouched, only the switch-on time appears as $e^{- 2 s}$ .

Verify (units/limit): As $s \to 0^{+}$ , $\frac{3 e ^{- 2 s}}{s} \to \infty$ — correct, since a constant that never turns off has infinite "total area" in this limit, like $L {3} = \frac{3}{s}$ . The extra $e^{- 2 s}$ only records the 2-second delay. ✔

Ex 10 — Exam twist: read a rule backwards (inverse) [C11]

Recall Which cell does each trap belong to?

Argument $t^{2}$ sitting on $u (t - 2)$ — which cell? ::: C6 (not yet in $(t - a)$ form — rewrite first) $L {f (c t)}$ with $c = 1$ — which cell? ::: C8 (degenerate; reduces to $F (s)$ ) $e^{- t}$ multiplying — which cell and substitution? ::: C4, $s \to s + 1$ A voltage that switches on at $t = 5$ — which cell? ::: C10, real-world $u (t - 5)$

Connections

Parent topic
Laplace Transform — Definition and Existence
Laplace Transforms of Standard Functions (1, e^at, sin, cos, t^n)
Inverse Laplace Transform and Partial Fractions
Solving ODEs with Laplace Transforms
Unit Step and Dirac Delta Functions
Convolution Theorem