Properties of the Laplace Transform | Shifting, Scaling, Derivatives

Three properties turn a short table into a large one. Exponential scaling says multiplying $f (t)$ by $e^{a t}$ shifts the transform to $F (s - a)$ . Time scaling says stretching time compresses frequency. And differentiating the transform corresponds to multiplying by $t$ . Together they generate almost every transform needed in practice from the handful computed in Section 7.1.

Quick Reference

Property	Statement	Valid for
Linearity	$ℒ {a f + b g} = a F (s) + b G (s)$	both transforms exist
Exponential scaling	$ℒ {e^{a t} f (t)} = F (s - a)$	$s > k + a$
Time scaling	$ℒ {f (a t)} = \frac{1}{a} F (\frac{s}{a})$ , $a > 0$	$s > k a$
Derivative of transform	\mathscr{L}\left\{t\,f(t)\right\} = -F'(s)	$s > k$
General form	$ℒ {t^{n} f (t)} = (- 1)^{n} \frac{d^{n}}{d s^{n}} F (s)$	$n = 1, 2, 3, \dots$

Transforms these generate:

$f (t)$	$F (s)$
$t^{n} e^{a t}$	$\frac{n!}{(s - a)^{n + 1}}$
$e^{a t} \sin k t$	$\frac{k}{(s - a)^{2} + k^{2}}$
$e^{a t} \cos k t$	$\frac{s - a}{(s - a)^{2} + k^{2}}$
$t \sin k t$	$\frac{2 k s}{{(s^{2} + k^{2})}^{2}}$
$t \cos k t$	$\frac{s^{2} - k^{2}}{{(s^{2} + k^{2})}^{2}}$

Exponential Scaling

In Section 7.1 we learned that $ℒ$ is a linear operator. We now study some further properties.

Exponential Scaling (also called the first shifting theorem). If $ℒ {f (t)} = F (s)$ exists for $s > k \geq 0$ and $a$ is a constant, then

ℒ {e^{a t} f (t)} = F (s - a)

for $s - a > k$ , that is, for $s > k + a$ .

Proof

\begin{aligned} \mathscr{L}\left\{e^{at}f(t)\right\} &= \int_0^{\infty}e^{-st}e^{at}f(t)\,dt \\ &= \int_0^{\infty}e^{-(s-a)t}f(t)\,dt \\ &= F(s-a) . \end{aligned}

The last step is just the definition of $F$ , read with $s - a$ in place of $s$ . Nothing is required beyond combining the two exponentials.

In words: multiplying by $e^{a t}$ in the $t$ domain translates the transform by $a$ in the $s$ domain. This is the single most useful property in the chapter. Read backwards, it says that whenever $s$ appears only in the combination $s - a$ , an exponential factor $e^{a t}$ is present in $f$ .

Find $ℒ {t^{2} e^{4 t}}$ .

Solution

We know $ℒ {t^{n}} = \frac{n!}{s^{n + 1}}$ for $s > 0$ . With $n = 2$ ,

ℒ {t^{2}} = \frac{2}{s^{3}} .

Applying the theorem with $a = 4$ , replace $s$ by $s - 4$ :

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

valid for $s - 4 > 0$ , that is, $s > 4$ .

Find $ℒ {e^{- t} \sin 3 t}$ .

Solution

Since $ℒ {\sin k t} = \frac{k}{s^{2} + k^{2}}$ for $s > 0$ , taking $k = 3$ gives

ℒ {\sin 3 t} = \frac{3}{s^{2} + 9} .

Applying the theorem with $a = - 1$ , replace $s$ by $s - (- 1) = s + 1$ :

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

valid for $s + 1 > 0$ , that is, $s > - 1$ .

Note that only the $s$ in the transform is replaced; the constant $k = 3$ in the numerator is untouched. Replacing $s$ everywhere, including inside $k^{2}$ , is a common slip.

Time Scaling

Time Scaling. If $ℒ {f (t)} = F (s)$ exists for $s > k \geq 0$ and $a$ is a positive constant, then

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

for $s / a > k$ , that is, $s > k a$ .

Proof

Substitute $τ = a t$ , so $t = τ / a$ and $d t = d τ / a$ :

ℒ {f (a t)} = \int_{0}^{\infty} e^{- s t} f (a t) d t = \int_{0}^{\infty} e^{- s τ / a} f (τ) \frac{d τ}{a} = \frac{1}{a} \int_{0}^{\infty} e^{- (s / a) τ} f (τ) d τ = \frac{1}{a} F (\frac{s}{a}) .

The requirement $a > 0$ keeps the limits of integration in order.

The theorem makes physical sense: if time is scaled by $a$ , frequency is scaled by $1 / a$ . Playing a recording twice as fast doubles every frequency, and the extra factor $1 / a$ out front conserves the total.

Given $ℒ {\sin t} = \frac{1}{s^{2} + 1}$ and $ℒ {\cos t} = \frac{s}{s^{2} + 1}$ , find $ℒ {\sin k t}$ and $ℒ {\cos k t}$ .

Solution

Applying the theorem with $a = k > 0$ ,

\begin{aligned} \mathscr{L}\left\{\sin kt\right\} &= \frac{1}{k}\cdot\frac{1}{(s/k)^{2}+1} \\ &= \frac{1}{k}\cdot\frac{k^{2}}{s^{2}+k^{2}} \\ &= \frac{k}{s^{2}+k^{2}} \end{aligned}

where the middle step multiplied numerator and denominator by $k^{2}$ .

Similarly,

\begin{aligned} \mathscr{L}\left\{\cos kt\right\} &= \frac{1}{k}\cdot\frac{s/k}{(s/k)^{2}+1} \\ &= \frac{1}{k}\cdot\frac{sk}{s^{2}+k^{2}} \\ &= \frac{s}{s^{2}+k^{2}} \end{aligned}

Both agree with the direct computations of Section 7.1, obtained here with far less work.

Derivatives of a Transform

Derivatives of a Transform. If $ℒ {f (t)} = F (s)$ , then

\mathscr{L}\left\{t\,f(t)\right\} = -F'(s)

and in general

ℒ {t^{n} f (t)} = (- 1)^{n} \frac{d^{n}}{d s^{n}} F (s) (n = 1, 2, 3, \dots) .

Proof

Differentiate both sides of $F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t$ with respect to $s$ , moving the derivative inside the integral:

\begin{aligned} \frac{d}{ds}F(s) &= \frac{d}{ds}\int_0^{\infty}e^{-st}f(t)\,dt \\ &= \int_0^{\infty}\frac{\partial}{\partial s}\left[e^{-st}f(t)\right]dt \\ &= \int_0^{\infty}(-t)e^{-st}f(t)\,dt \\ &= -\int_0^{\infty}e^{-st}\,t\,f(t)\,dt \\ &= -\mathscr{L}\left\{t\,f(t)\right\} \end{aligned}

The interchange of differentiation and integration is justified when $f$ satisfies the conditions of the existence theorem.

Applying the result to $t f (t)$ in place of $f$ ,

\begin{aligned} \mathscr{L}\left\{t^{2}f(t)\right\} &= \mathscr{L}\left\{t\cdot t f(t)\right\} \\ &= -\frac{d}{ds}\mathscr{L}\left\{t f(t)\right\} \\ &= -\frac{d}{ds}\left(-\frac{d}{ds}\mathscr{L}\left\{f(t)\right\}\right) \\ &= \frac{d^{2}}{ds^{2}}F(s) . \end{aligned}

Repeating gives

ℒ {t^{n} f (t)} = (- 1)^{n} \frac{d^{n}}{d s^{n}} F (s),

with the sign alternating because each application contributes one factor of $- 1$ .

Note the pleasing symmetry with Section 7.5: differentiating in $t$ corresponds (roughly) to multiplying by $s$ , and differentiating in $s$ corresponds to multiplying by $- t$ . The two variables trade roles.

Find $ℒ {t \sin k t}$ .

Solution

Since $ℒ {\sin k t} = \frac{k}{s^{2} + k^{2}}$ , the theorem gives

\begin{aligned} \mathscr{L}\left\{t\sin kt\right\} &= -\frac{d}{ds}\left(\frac{k}{s^{2}+k^{2}}\right) \\ &= -\frac{\dfrac{d}{ds}(k)\cdot\left(s^{2}+k^{2}\right) - k\cdot\dfrac{d}{ds}\left(s^{2}+k^{2}\right)}{\left(s^{2}+k^{2}\right)^{2}} \\ &= -\frac{0\cdot\left(s^{2}+k^{2}\right) - k(2s)}{\left(s^{2}+k^{2}\right)^{2}} \\ &= \frac{2ks}{\left(s^{2}+k^{2}\right)^{2}} \end{aligned}

Remember that $k$ is a constant while $s$ is the variable, so $\frac{d k}{d s} = 0$ and $\frac{d}{d s} (s^{2} + k^{2}) = 2 s$ .

Find $ℒ {t^{2} e^{- 3 t}}$ by two different routes.

Solution

Method (a): differentiate the transform. Start from $ℒ {e^{a t}} = \frac{1}{s - a}$ with $a = - 3$ :

ℒ {e^{- 3 t}} = \frac{1}{s + 3} .

Then

\begin{aligned} \mathscr{L}\left\{t^{2}e^{-3t}\right\} &= \frac{d^{2}}{ds^{2}}\left[(s+3)^{-1}\right] \\ &= \frac{d}{ds}\left[-(s+3)^{-2}\right] \\ &= 2(s+3)^{-3} = \frac{2}{(s+3)^{3}} \end{aligned}

Method (b): use exponential scaling. Start from $ℒ {t^{n}} = \frac{n!}{s^{n + 1}}$ with $n = 2$ :

ℒ {t^{2}} = \frac{2}{s^{3}} .

By the exponential scaling theorem, replace $s$ by $s + 3$ :

ℒ {t^{2} e^{- 3 t}} = {\frac{2}{s^{3}} |}_{s \to s + 3} = \frac{2}{(s + 3)^{3}} .

Method (b) is much shorter, and it generalizes at once to $ℒ {t^{n} e^{a t}} = \frac{n!}{(s - a)^{n + 1}}$ , a formula worth memorizing.

Find $ℒ {t \cos k t}$ .

Solution

Starting from $ℒ {\cos k t} = \frac{s}{s^{2} + k^{2}}$ and applying the quotient rule,

\begin{aligned} \mathscr{L}\left\{t\cos kt\right\} &= -\frac{d}{ds}\left(\frac{s}{s^{2}+k^{2}}\right) \\ &= -\frac{1\cdot\left(s^{2}+k^{2}\right) - s(2s)}{\left(s^{2}+k^{2}\right)^{2}} \\ &= -\frac{k^{2}-s^{2}}{\left(s^{2}+k^{2}\right)^{2}} = \frac{s^{2}-k^{2}}{\left(s^{2}+k^{2}\right)^{2}} \end{aligned}

This transform appears constantly in resonance problems, where the factor $t$ is what the Modification Rule of Section 4.3 predicted.

Exercises

Find $ℒ {e^{2 t} \cos 5 t}$ .

Solution

Since $ℒ {\cos 5 t} = \frac{s}{s^{2} + 25}$ , exponential scaling with $a = 2$ gives

ℒ {e^{2 t} \cos 5 t} = \frac{s - 2}{(s - 2)^{2} + 25},

valid for $s > 2$ . Note that the numerator becomes $s - 2$ , not $s$ : every $s$ in the transform is replaced.

Find $ℒ {t^{3} e^{- 2 t}}$ .

Solution

Since $ℒ {t^{3}} = \frac{6}{s^{4}}$ , replacing $s$ by $s + 2$ gives

ℒ {t^{3} e^{- 2 t}} = \frac{6}{(s + 2)^{4}} .

Find $ℒ {t^{2} \sin t}$ .

Solution

Start from $F (s) = \frac{1}{s^{2} + 1}$ and differentiate twice, using $n = 2$ so the sign is $(- 1)^{2} = + 1$ .

First derivative: F'(s) = -\dfrac{2s}{\left(s^{2}+1\right)^{2}}.

Second derivative, by the quotient rule:

F''(s) = -\frac{2\left(s^{2}+1\right)^{2} - 2s\cdot 2\left(s^{2}+1\right)(2s)}{\left(s^{2}+1\right)^{4}} = -\frac{2\left(s^{2}+1\right) - 8s^{2}}{\left(s^{2}+1\right)^{3}} = \frac{6s^{2}-2}{\left(s^{2}+1\right)^{3}} .

Therefore

ℒ {t^{2} \sin t} = \frac{6 s^{2} - 2}{{(s^{2} + 1)}^{3}} .

Given that $ℒ {f (t)} = \frac{1}{s^{2} + 1}$ , find $ℒ {f (3 t)}$ without identifying $f$ .

Solution

By time scaling with $a = 3$ ,

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

As a check, $f (t) = \sin t$ so $f (3 t) = \sin 3 t$ , whose transform is indeed $\frac{3}{s^{2} + 9}$ .

Frequently Asked Questions

What is the exponential scaling theorem?

It states that $ℒ {e^{a t} f (t)} = F (s - a)$ : multiplying by an exponential in $t$ shifts the transform in $s$ . It is often called the first shifting theorem, to distinguish it from the time-shifting theorem of Section 7.6.

How do I apply the shift correctly?

Compute $F (s)$ for the function without the exponential, then substitute $s - a$ for every occurrence of $s$ . For $e^{- t} \sin 3 t$ : first get $\frac{3}{s^{2} + 9}$ , then replace $s$ by $s + 1$ to get $\frac{3}{(s + 1)^{2} + 9}$ . The constant $3$ in the numerator is not an $s$ , so it stays.

Why does

ℒ {t f (t)}

carry a minus sign?

Because differentiating $e^{- s t}$ with respect to $s$ brings down a factor of $- t$ . Each further differentiation brings down another, which is why the general formula carries $(- 1)^{n}$ .

Which property should I use for

t^{n} e^{a t}

Exponential scaling, applied to $ℒ {t^{n}} = \frac{n!}{s^{n + 1}}$ , gives $\frac{n!}{(s - a)^{n + 1}}$ in one step. Differentiating the transform $n$ times reaches the same answer with far more algebra. When two routes exist, the shifting theorem is almost always the shorter one.

Why must

a

be positive in the time-scaling theorem?

Because the substitution $τ = a t$ must preserve the orientation of the interval $[0, \infty)$ . A negative $a$ would map it to $(- \infty, 0]$ , which lies outside the range the unilateral transform can see.

Further Properties of Laplace Transforms

Quick Reference

Exponential Scaling

Time Scaling

Derivatives of a Transform

Exercises

Frequently Asked Questions

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