Chapter 9 · Section 9.1

9.1 Definition, Existence, and Basic Properties

The Laplace transform is a tool that turns a differential equation into an algebra problem. You take the transform of every term, the derivatives turn into multiplication by $s$ , the initial conditions slip in automatically, and what remains is an equation you can solve with ordinary algebra. At the end you transform back. Engineers use it constantly: it is the language of transfer functions, control systems and circuit analysis.

Before we can use it, we have to know what the transform is, when it exists and how it behaves. That is the job of this section. We will compute a handful of transforms directly from the definition, then collect a small set of rules that let us find almost every transform we need without integrating again.

One change of notation: in this chapter the independent variable is time $t$ , and functions are defined for $t \geq 0$ . A differential equation will look like $y^{''} + 4 y^{'} + 13 y = f (t)$ with $y = y (t)$ .

The definition: an improper integral

The transform takes a function $f (t)$ and produces a new function $F (s)$ of a new variable $s$ . It does this by multiplying $f$ by the decaying weight $e^{- s t}$ and adding up (integrating) over all $t \geq 0$ .

Definition 9.1.1 (Laplace transform)

Let $f (t)$ be defined for $t \geq 0$ . The Laplace transform of $f$ is the function $F$ given by

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

for every real $s$ for which this improper integral converges. We use capital letters for transforms: $L {f} = F$ , $L {g} = G$ , $L {y} = Y$ .

Recall what an improper integral means. We integrate up to a finite upper limit $T$ , and then let $T$ grow:

\int_{0}^{\infty} e^{- s t} f (t) d t = T \to \infty lim \int_{0}^{T} e^{- s t} f (t) d t .

If this limit exists (and is finite), the integral converges; otherwise it diverges and $F (s)$ is not defined for that $s$ . Typically the integral converges for all $s$ larger than some number, and diverges for smaller $s$ .

Figure 9.1.1 shows the idea for $f (t) = t$ . For each fixed $s$ , the product $e^{- s t} t$ is a curve, and $F (s)$ is the area under it. A larger $s$ makes the weight $e^{- s t}$ decay faster, so the area shrinks. The transform records how this area depends on $s$ .

Figure 9.1.1. The value

F (s)

is the area under

e^{- s t} f (t)

. For

f (t) = t

the areas are

4

1

and

1/4

s = 1/2, 1, 2

, which agrees with

F (s) = 1/ s^{2}

(Example 9.1.2).

First transforms from the definition

Start with the constant function $f (t) = 1$ . For $s > 0$ ,

\int_{0}^{T} e^{- s t} d t = [- \frac{1}{s} e^{- s t}]_{0}^{T} = - \frac{1}{s} e^{- s T} + \frac{1}{s} .

As $T \to \infty$ , the term $e^{- s T}$ goes to $0$ because $s > 0$ . Hence

L {1} = \frac{1}{s}, s > 0.

For $s \leq 0$ the integral diverges ( $e^{- s t} \geq 1$ for all $t$ , so the area is infinite). This is typical: a transform lives on a half-line $s > α$ .

Example 9.1.1 (The exponential)

Find $L {e^{a t}}$ , where $a$ is a real constant.

Solution

Combine the exponentials. $e^{- s t} e^{a t} = e^{- (s - a) t}$ . Assume $s > a$ , so that $s - a > 0$ .
Integrate up to $T$ . $\int_{0}^{T} e^{- (s - a) t} d t = [- \frac{e ^{- (s - a) t}}{s - a}]_{0}^{T} = \frac{1 - e ^{- (s - a) T}}{s - a} .$
Let $T \to \infty$ . Since $s - a > 0$ , $e^{- (s - a) T} \to 0$ . So $L {e^{a t}} = \frac{1}{s - a}, s > a .$
Check the other case. If $s \leq a$ , the integrand $e^{(a - s) t}$ does not decay, and the integral diverges. With $a = 0$ we recover $L {1} = 1/ s$ .

Example 9.1.2 (The ramp

f (t) = t

)

Find $L {t}$ .

Solution

Integration by parts. On $[0, T]$ take $u = t$ , $d v = e^{- s t} d t$ , so $d u = d t$ , $v = - e^{- s t} / s$ (with $s > 0$ ): $\int_{0}^{T} t e^{- s t} d t = [- \frac{t e ^{- s t}}{s}]_{0}^{T} + \frac{1}{s} \int_{0}^{T} e^{- s t} d t = - \frac{T e ^{- s T}}{s} + \frac{1}{s} \cdot \frac{1 - e ^{- s T}}{s} .$
Limits. $e^{- s T} \to 0$ , and also $T e^{- s T} \to 0$ (an exponential beats any power; use L'Hôpital's rule on $T / e^{s T}$ if in doubt).
Answer. $L {t} = \frac{1}{s ^{2}}, s > 0.$ Repeating the same integration by parts shows that each extra power of $t$ brings a factor $n / s$ : $L {t^{n}} = \frac{n}{s} L {t^{n - 1}}$ . Starting from $L {1} = 1/ s$ this gives $L {t^{n}} = n! / s^{n + 1}$ .

Example 9.1.3 (Cosine and sine in one stroke)

Find $L {cos b t}$ and $L {sin b t}$ for a real constant $b$ .

Solution

We could integrate by parts twice for each one. A shorter route uses Euler's formula $e^{ib t} = cos b t + i sin b t$ .

Transform of $e^{ib t}$ . Repeat Example 9.1.1 with $a = ib$ . The only step that needs care is the limit: $e^{- (s - ib) T} = e^{- s T} e^{ib T} = e^{- s T} \to 0$ for $s > 0$ . So $\int_{0}^{\infty} e^{- s t} e^{ib t} d t = \frac{1}{s - ib}, s > 0.$
Make the denominator real. Multiply top and bottom by $s + ib$ : $\frac{1}{s - ib} = \frac{s + ib}{( s - ib ) ( s + ib )} = \frac{s}{s ^{2} + b ^{2}} + i \frac{b}{s ^{2} + b ^{2}} .$
Take real and imaginary parts. The left side is $\int_{0}^{\infty} e^{- s t} cos b t d t + i \int_{0}^{\infty} e^{- s t} sin b t d t$ . Comparing parts: $L {cos b t} = \frac{s}{s ^{2} + b ^{2}}, L {sin b t} = \frac{b}{s ^{2} + b ^{2}}, s > 0.$

Piecewise continuous functions

In engineering, inputs are often switched on and off: a voltage applied for two seconds, a load removed at a certain moment. Such inputs have jumps. The Laplace transform handles them easily, because an integral does not care about a finite number of jumps.

Definition 9.1.2 (Piecewise continuous)

A function $f$ is piecewise continuous on a finite interval $[a, b]$ if $[a, b]$ can be split into finitely many subintervals such that $f$ is continuous inside each subinterval and has finite one-sided limits at the end points of each subinterval.

In words: finitely many jumps, every jump of finite size, nothing that blows up.

For example, a function equal to $2$ on $[0, 1)$ , to $- 1$ on $[1, 4)$ and to $0$ afterwards is piecewise continuous on every finite interval. The function $1/ (t - 1)$ is not piecewise continuous on $[0, 2]$ : it blows up at $t = 1$ . Neither is $sin (1/ t)$ on $[0, 1]$ : it has no limit as $t \to 0^{+}$ .

To transform a piecewise function from the definition, split the integral at the jump points.

Example 9.1.4 (A signal that is switched off)

Let $f (t) = e^{- t}$ for $0 \leq t < 2$ and $f (t) = 0$ for $t \geq 2$ . Find $L {f}$ .

Solution

Split the integral. The part over $[2, \infty)$ contributes nothing because $f = 0$ there: $F (s) = \int_{0}^{2} e^{- s t} e^{- t} d t + \int_{2}^{\infty} e^{- s t} \cdot 0 d t = \int_{0}^{2} e^{- (s + 1) t} d t .$
Integrate (for $s \neq = - 1$ ): $\int_{0}^{2} e^{- (s + 1) t} d t = [- \frac{e ^{- (s + 1) t}}{s + 1}]_{0}^{2} = \frac{1 - e ^{- 2 (s + 1)}}{s + 1} .$
Answer. $F (s) = \frac{1 - e ^{- 2 (s + 1)}}{s + 1}$ . Because the integral runs over a finite interval only, it converges for every $s \neq = - 1$ (and at $s = - 1$ the integral is simply $2$ ).

Exponential order and the existence theorem

Not every function has a Laplace transform. The weight $e^{- s t}$ can tame a function that grows, but only if the growth is not too fast. The next definition makes "not too fast" precise.

Definition 9.1.3 (Exponential order)

A function $f$ is of exponential order $α$ if there are constants $M > 0$ and $T \geq 0$ such that

∣ f (t) ∣ \leq M e^{α t} for all t \geq T .

We say simply "of exponential order" if such an $α$ exists.

Some examples:

Every bounded function ( $sin 5 t$ , $cos t$ , a constant, a step) is of exponential order $0$ : take $M$ as the bound.
$e^{2 t} cos 5 t$ is of exponential order $2$ , because $∣ e^{2 t} cos 5 t ∣ \leq e^{2 t}$ .
Every power $t^{n}$ is of exponential order $α$ for any $α > 0$ . For instance, the Taylor series $e^{t} = 1 + t + \frac{t ^{2}}{2} + \frac{t ^{3}}{6} + \dots$ has only positive terms when $t \geq 0$ , so $e^{t} \geq t^{3} /6$ , that is $t^{3} \leq 6 e^{t}$ for all $t \geq 0$ .
$e^{t^{2}}$ is not of exponential order: for any $α$ , $e^{t^{2}} / e^{α t} = e^{t^{2} - α t} \to \infty$ . No exponential bound can hold for large $t$ .

Figure 9.1.2. Left:

t^{3}

stays below the exponential bound

6 e^{t}

for all

t \geq 0

. Right:

e^{t^{2}}

eventually overtakes

10 e^{3 t}

(and any other bound

M e^{α t}

), so it has no Laplace transform.

Theorem 9.1.1 (Existence of the Laplace transform)

Suppose that

$f$ is piecewise continuous on $[0, b]$ for every $b > 0$ , and
$f$ is of exponential order $α$ .

Then $L {f} = F (s)$ exists for every $s > α$ . Moreover $F (s) \to 0$ as $s \to \infty$ .

Why this works

Split the integral at the point $T$ of Definition 9.1.3. On $[0, T]$ the function is piecewise continuous, so that finite piece is an ordinary integral. For $t \geq T$ ,

e^{- s t} f (t) \leq M e^{- (s - α) t},

and $\int_{T}^{\infty} M e^{- (s - α) t} d t$ converges when $s > α$ (Example 9.1.1). By the comparison test the tail converges too.

The conditions are sufficient, not necessary. For example, $t^{- 1/2}$ blows up at $t = 0$ (so it is not piecewise continuous on $[0, b]$ ), yet its transform exists: $L {t^{- 1/2}} = π / s$ . For the equations of this course the theorem covers every forcing we will meet.

The last statement of the theorem gives a quick test: if $F (s)$ does not tend to $0$ as $s \to \infty$ , it cannot be the transform of a piecewise continuous function of exponential order. For instance, neither $1$ nor $s / (s + 1)$ is such a transform. (In Section 9.4 we meet the delta "function", whose transform is a constant. It is not an ordinary function, so there is no contradiction.)

Linearity

The transform is an integral, and integrals respect sums and constant multiples. So the transform does too.

Theorem 9.1.2 (Linearity)

If $L {f}$ exists for $s > α_{1}$ and $L {g}$ exists for $s > α_{2}$ , then for any constants $c_{1}, c_{2}$ ,

L {c_{1} f (t) + c_{2} g (t)} = c_{1} L {f (t)} + c_{2} L {g (t)}

for $s > max (α_{1}, α_{2})$ .

As a first use, the hyperbolic functions follow from the exponential. Since $cosh b t = \frac{1}{2} (e^{b t} + e^{- b t})$ , for $s > ∣ b ∣$

L {cosh b t} = \frac{1}{2} (\frac{1}{s - b} + \frac{1}{s + b}) = \frac{1}{2} \cdot \frac{( s + b ) + ( s - b )}{s ^{2} - b ^{2}} = \frac{s}{s ^{2} - b ^{2}},

and in the same way $L {sinh b t} = \frac{b}{s ^{2} - b ^{2}}$ . Notice how these resemble the cosine and sine, with $+ b^{2}$ replaced by $- b^{2}$ .

Example 9.1.5 (Linearity)

Find $L {3 t^{2} - 5 e^{- 4 t} + 2 sin 6 t}$ .

Solution

Split by linearity. $3 L {t^{2}} - 5 L {e^{- 4 t}} + 2 L {sin 6 t}$ .
Use the known transforms. $L {t^{2}} = 2! / s^{3} = 2/ s^{3}$ , $L {e^{- 4 t}} = 1/ (s + 4)$ , $L {sin 6 t} = 6/ (s^{2} + 36)$ .
Answer. $L {3 t^{2} - 5 e^{- 4 t} + 2 sin 6 t} = \frac{6}{s ^{3}} - \frac{5}{s + 4} + \frac{12}{s ^{2} + 36}, s > 0.$ (Each term needs its own condition: $s > 0$ , $s > - 4$ , $s > 0$ . All hold when $s > 0$ .)

Common mistake

The transform of a product is not the product of the transforms. For example $L {t \cdot t} = 2/ s^{3}$ , while $L {t} \cdot L {t} = 1/ s^{4}$ . Products are handled by special rules: shifting (multiplication by $e^{a t}$ ), multiplication by $t^{n}$ , and, in Section 9.2, convolution.

A table of transforms

Here are the transforms we use throughout the chapter. Rows 1–7 come from the computations above; rows 8–12 follow from the rules proved below; the last two rows are explained in Section 9.4. Keep this table at hand: solving an equation by Laplace transforms is mostly a matter of reading it forwards and backwards.

Table 9.1.1: Basic Laplace transforms (

n = 0, 1, 2, \dots

;

a

b

real;

a \geq 0

in the last two rows)

	$f (t)$	$F (s) = L {f (t)}$	valid for
1	$1$	$\frac{1}{s}$	$s > 0$
2	$t^{n}$	$\frac{n !}{s ^{n + 1}}$	$s > 0$
3	$e^{a t}$	$\frac{1}{s - a}$	$s > a$
4	$sin b t$	$\frac{b}{s ^{2} + b ^{2}}$	$s > 0$
5	$cos b t$	$\frac{s}{s ^{2} + b ^{2}}$	$s > 0$
6	$sinh b t$	$\frac{b}{s ^{2} - b ^{2}}$	$s > ∣ b ∣$
7	$cosh b t$	$\frac{s}{s ^{2} - b ^{2}}$	$s > ∣ b ∣$
8	$t^{n} e^{a t}$	$\frac{n !}{( s - a ) ^{n + 1}}$	$s > a$
9	$e^{a t} sin b t$	$\frac{b}{( s - a ) ^{2} + b ^{2}}$	$s > a$
10	$e^{a t} cos b t$	$\frac{s - a}{( s - a ) ^{2} + b ^{2}}$	$s > a$
11	$t sin b t$	$\frac{2 b s}{( s ^{2} + b ^{2} ) ^{2}}$	$s > 0$
12	$t cos b t$	$\frac{s ^{2} - b ^{2}}{( s ^{2} + b ^{2} ) ^{2}}$	$s > 0$
13	$sin b t - b t cos b t$	$\frac{2 b ^{3}}{( s ^{2} + b ^{2} ) ^{2}}$	$s > 0$
14	$u (t - a)$ (unit step)	$\frac{e ^{- a s}}{s}$	$s > 0$
15	$δ (t - a)$ (unit impulse)	$e^{- a s}$	all $s$

Row 13 combines rows 4 and 12 by linearity: $L {sin b t} - b L {t cos b t}$ equals

\frac{b}{s ^{2} + b ^{2}} - \frac{b ( s ^{2} - b ^{2} )}{( s ^{2} + b ^{2} ) ^{2}} = \frac{b ( s ^{2} + b ^{2} ) - b ( s ^{2} - b ^{2} )}{( s ^{2} + b ^{2} ) ^{2}} = \frac{2 b ^{3}}{( s ^{2} + b ^{2} ) ^{2}} .

It is useful in Section 9.2, where a squared quadratic factor appears in a denominator.

Translation in $s$ : the first shifting theorem

What happens to the transform when we multiply $f (t)$ by an exponential $e^{a t}$ ? Look at the definition:

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

The last integral is exactly $F$ , evaluated at $s - a$ instead of $s$ .

Theorem 9.1.3 (First shifting theorem)

If $L {f (t)} = F (s)$ for $s > α$ , then

L {e^{a t} f (t)} = F (s - a), s > α + a .

Multiplying by $e^{a t}$ in the $t$ -world shifts the transform by $a$ in the $s$ -world: replace every $s$ by $s - a$ .

Rows 8–10 of the table follow at once. For example, from $L {sin b t} = b / (s^{2} + b^{2})$ we get $L {e^{a t} sin b t} = b / ((s - a)^{2} + b^{2})$ .

Example 9.1.6 (Shifting in

s

)

Find (a) $L {t^{3} e^{- 2 t}}$ , (b) $L {e^{3 t} cos 4 t}$ , (c) $L {e^{- t} (2 cos 3 t - 5 sin 3 t)}$ .

Solution

(a) Without the exponential, $L {t^{3}} = 3! / s^{4} = 6/ s^{4}$ . Here $a = - 2$ , so replace $s$ by $s - (- 2) = s + 2$ : $L {t^{3} e^{- 2 t}} = \frac{6}{( s + 2 ) ^{4}}, s > - 2.$
(b) $L {cos 4 t} = s / (s^{2} + 16)$ . With $a = 3$ replace $s$ by $s - 3$ everywhere, including the numerator: $L {e^{3 t} cos 4 t} = \frac{s - 3}{( s - 3 ) ^{2} + 16}, s > 3.$
(c) First the unshifted transform: $L {2 cos 3 t - 5 sin 3 t} = \frac{2 s}{s ^{2} + 9} - \frac{15}{s ^{2} + 9} = \frac{2 s - 15}{s ^{2} + 9} .$ Now replace $s$ by $s + 1$ : $L {e^{- t} (2 cos 3 t - 5 sin 3 t)} = \frac{2 ( s + 1 ) - 15}{( s + 1 ) ^{2} + 9} = \frac{2 s - 13}{s ^{2} + 2 s + 10}, s > - 1.$

Transforms of derivatives

This is the property that makes the transform useful for differential equations. Let us compute $L {f^{'}}$ by integrating by parts, with $u = e^{- s t}$ and $d v = f^{'} (t) d t$ :

\int_{0}^{T} e^{- s t} f^{'} (t) d t = [e^{- s t} f (t)]_{0}^{T} + s \int_{0}^{T} e^{- s t} f (t) d t = e^{- s T} f (T) - f (0) + s \int_{0}^{T} e^{- s t} f (t) d t .

If $f$ is of exponential order $α$ and $s > α$ , then $e^{- s T} f (T) \to 0$ . Letting $T \to \infty$ :

L {f^{'} (t)} = s F (s) - f (0) .

Differentiation in $t$ becomes multiplication by $s$ , minus the initial value. Applying the rule to $f^{'}$ instead of $f$ gives the second derivative:

L {f^{''}} = s L {f^{'}} - f^{'} (0) = s (s F (s) - f (0)) - f^{'} (0) = s^{2} F (s) - s f (0) - f^{'} (0) .

Theorem 9.1.4 (Transform of derivatives)

Suppose $f, f^{'}, \dots, f^{(n - 1)}$ are continuous on $[0, \infty)$ and of exponential order $α$ , and $f^{(n)}$ is piecewise continuous on $[0, b]$ for every $b > 0$ . Then for $s > α$ ,

L {f^{(n)} (t)} = s^{n} F (s) - s^{n - 1} f (0) - s^{n - 2} f^{'} (0) - \dots - s f^{(n - 2)} (0) - f^{(n - 1)} (0) .

In particular

L {f^{'}} L {f^{''}} L {f^{'''}} = s F (s) - f (0), = s^{2} F (s) - s f (0) - f^{'} (0), = s^{3} F (s) - s^{2} f (0) - s f^{'} (0) - f^{''} (0) .

A memory aid for the pattern: the powers of $s$ go down while the order of the derivatives goes up, and the two always add up to $n - 1$ .

As a quick check, take $f (t) = sin b t$ , with $f (0) = 0$ and $f^{'} (t) = b cos b t$ . The theorem says $L {b cos b t} = s \cdot \frac{b}{s ^{2} + b ^{2}} - 0$ , so $L {cos b t} = \frac{s}{s ^{2} + b ^{2}}$ , in agreement with Example 9.1.3.

Example 9.1.7 (A preview: transforming an initial-value problem)

Suppose $y (t)$ solves $y^{''} + 5 y^{'} + 4 y = 0$ with $y (0) = 2$ , $y^{'} (0) = - 1$ . Find $Y (s) = L {y}$ without solving the differential equation.

Solution

Transform each term. By Theorem 9.1.4 with the given initial values, $L {y^{''}} L {y^{'}} = s^{2} Y - s y (0) - y^{'} (0) = s^{2} Y - 2 s + 1, = s Y - y (0) = s Y - 2.$
Transform the equation (by linearity, and $L {0} = 0$ ): $(s^{2} Y - 2 s + 1) + 5 (s Y - 2) + 4 Y = 0.$
Collect the $Y$ terms. $(s^{2} + 5 s + 4) Y = 2 s + 10 - 1 = 2 s + 9.$
Answer. $Y (s) = \frac{2 s + 9}{s ^{2} + 5 s + 4} .$ The differential equation has become a linear algebraic equation for $Y$ , and the initial conditions are already built in. Notice that the factor in front of $Y$ is the characteristic polynomial $s^{2} + 5 s + 4$ . Section 9.2 shows how to turn $Y (s)$ back into $y (t)$ ; Section 9.3 makes this a complete method.

Multiplication by $t^{n}$ : derivatives of $F$

Multiplying $f$ by $e^{a t}$ shifted the transform. Multiplying by $t$ differentiates it. To see why, differentiate the definition with respect to $s$ (this is allowed for $s > α$ under the hypotheses of Theorem 9.1.1):

F^{'} (s) = \frac{d}{d s} \int_{0}^{\infty} e^{- s t} f (t) d t = \int_{0}^{\infty} (- t) e^{- s t} f (t) d t = - L {t f (t)} .

Repeating $n$ times gives the general rule.

Theorem 9.1.5 (Multiplication by

t^{n}

)

If $f$ is piecewise continuous on every $[0, b]$ and of exponential order $α$ , and $L {f} = F (s)$ , then for $s > α$ and $n = 1, 2, \dots$

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

Example 9.1.8 (Multiplying by

t

)

Find (a) $L {t cos 2 t}$ , (b) $L {t^{2} e^{3 t}}$ in two ways, (c) $L {t e^{- t} sin 2 t}$ .

Solution

(a) $F (s) = L {cos 2 t} = \frac{s}{s ^{2} + 4}$ . By the quotient rule, $F^{'} (s) = \frac{( s ^{2} + 4 ) \cdot 1 - s \cdot 2 s}{( s ^{2} + 4 ) ^{2}} = \frac{4 - s ^{2}}{( s ^{2} + 4 ) ^{2}} .$ So $L {t cos 2 t} = - F^{'} (s) = \frac{s ^{2} - 4}{( s ^{2} + 4 ) ^{2}}$ . This is row 12 of the table with $b = 2$ .
(b) With Theorem 9.1.5. $F (s) = L {e^{3 t}} = (s - 3)^{- 1}$ . Then $F^{'} (s) = - (s - 3)^{- 2}$ and $F^{''} (s) = 2 (s - 3)^{- 3}$ . With $n = 2$ , $(- 1)^{2} F^{''} (s) = \frac{2}{( s - 3 ) ^{3}}$ .
With shifting. $L {t^{2}} = 2/ s^{3}$ ; replace $s$ by $s - 3$ : $\frac{2}{( s - 3 ) ^{3}}$ . The two methods agree.
(c) First, by Theorem 9.1.5 (or row 11), $L {t sin 2 t} = - \frac{d}{d s} \frac{2}{s ^{2} + 4} = \frac{4 s}{( s ^{2} + 4 ) ^{2}}$ . Then the factor $e^{- t}$ replaces $s$ by $s + 1$ : $L {t e^{- t} sin 2 t} = \frac{4 ( s + 1 )}{( ( s + 1 ) ^{2} + 4 ) ^{2}} .$

Use the calculator below to transform your own combinations of $t^{k}$ , $e^{a t}$ , $cos b t$ and $sin b t$ . It shows which table row and which rule (linearity, shifting, multiplication by $t$ ) it uses for each term, so you can compare with your own steps. The preset is a mix of Examples 9.1.6 and 9.1.8; try replacing it with $t e^{- t} sin 2 t$ .

Periodic functions

Many engineering signals repeat: square waves, sawtooth waves, rectified sine waves. If $f (t + P) = f (t)$ for all $t \geq 0$ , we only need to integrate over one period.

Theorem 9.1.6 (Transform of a periodic function)

Let $f$ be piecewise continuous on $[0, P]$ and periodic with period $P > 0$ . Then for $s > 0$

L {f (t)} = \frac{1}{1 - e ^{- P s}} \int_{0}^{P} e^{- s t} f (t) d t .

Why this works

Split $[0, \infty)$ into periods $[k P, (k + 1) P)$ , $k = 0, 1, 2, \dots$ . On the $k$ -th period substitute $t = τ + k P$ ; because $f (τ + k P) = f (τ)$ , the $k$ -th piece equals $e^{- k P s} \int_{0}^{P} e^{- s τ} f (τ) d τ$ . Adding all pieces gives the geometric series $1 + e^{- P s} + e^{- 2 P s} + \dots = \frac{1}{1 - e ^{- P s}}$ times the one-period integral.

Example 9.1.9 (Half-wave rectified sine)

A half-wave rectifier passes the positive half of a sine wave and blocks the negative half. The output is $f (t) = sin t$ for $0 \leq t < π$ , $f (t) = 0$ for $π \leq t < 2 π$ , repeated with period $2 π$ (Figure 9.1.3). Find $L {f}$ .

Solution

One-period integral. Only $[0, π)$ contributes. Use $\int e^{- s t} sin t d t = - \frac{e ^{- s t} ( s sin t + cos t )}{s ^{2} + 1}$ (check by differentiating): $\int_{0}^{π} e^{- s t} sin t d t = [- \frac{e ^{- s t} ( s sin t + cos t )}{s ^{2} + 1}]_{0}^{π} = - \frac{e ^{- π s} ( 0 - 1 )}{s ^{2} + 1} + \frac{0 + 1}{s ^{2} + 1} = \frac{1 + e ^{- π s}}{s ^{2} + 1} .$
Apply Theorem 9.1.6 with $P = 2 π$ : $F (s) = \frac{1 + e ^{- π s}}{( s ^{2} + 1 ) ( 1 - e ^{- 2 π s} )} .$
Simplify. Factor $1 - e^{- 2 π s} = (1 - e^{- π s}) (1 + e^{- π s})$ and cancel: $F (s) = \frac{1}{( s ^{2} + 1 ) ( 1 - e ^{- π s} )}, s > 0.$

Figure 9.1.3. The half-wave rectified sine of Example 9.1.9. Only the shaded hump in the first period enters the integral of Theorem 9.1.6.

Method in a nutshell: finding a transform

Use linearity to split the function into simple terms.
Look up each basic term in Table 9.1.1.
A factor $e^{a t}$ : find the transform without it, then replace $s$ by $s - a$ (first shifting theorem).
A factor $t^{n}$ : find the transform without it, then differentiate $n$ times and multiply by $(- 1)^{n}$ .
Piecewise functions: integrate piece by piece from the definition (or, faster, use the step functions of Section 9.4). Periodic functions: integrate over one period and divide by $1 - e^{- P s}$ .

Exercises

Use the definition to find $L {f}$ when $f (t) = 0$ for $0 \leq t < 1$ , $f (t) = 3$ for $1 \leq t < 4$ , and $f (t) = 0$ for $t \geq 4$ .
Answer
$F (s) = \frac{3 ( e ^{- s} - e ^{- 4 s} )}{s}$
Use the definition to find $L {f}$ when $f (t) = t$ for $0 \leq t < 1$ and $f (t) = 1$ for $t \geq 1$ .
Answer
$F (s) = \frac{1 - e ^{- s}}{s ^{2}}$
Find $L {4 t^{3} - 2 e^{5 t} + 3 cos 4 t}$ .
Answer
$\frac{24}{s ^{4}} - \frac{2}{s - 5} + \frac{3 s}{s ^{2} + 16}$ , $s > 5$
Find $L {t^{4} e^{- 3 t}}$ .
Answer
$\frac{24}{( s + 3 ) ^{5}}$
Find $L {e^{- t} (3 cos 2 t - 4 sin 2 t)}$ .
Answer
$\frac{3 ( s + 1 ) - 8}{( s + 1 ) ^{2} + 4} = \frac{3 s - 5}{s ^{2} + 2 s + 5}$
Use Theorem 9.1.5 to find $L {t sinh 3 t}$ .
Answer
$\frac{6 s}{( s ^{2} - 9 ) ^{2}}$ , $s > 3$
Find $L {t e^{2 t} cos t}$ .
Answer
$\frac{( s - 2 ) ^{2} - 1}{( ( s - 2 ) ^{2} + 1 ) ^{2}}$
Decide whether each function is of exponential order: (a) $t^{5} e^{2 t}$ , (b) $e^{t^{3/2}}$ , (c) $cos (e^{t^{2}})$ .
Answer
(a) yes, of exponential order $α$ for any $α > 2$ ; (b) no, since $t^{3/2} - α t \to \infty$ for every $α$ ; (c) yes, it is bounded by $1$ (order $0$ ), although it oscillates wildly.
Let $f (t) = t e^{2 t}$ . Compute $f^{'} (t)$ , then use $L {f^{'}} = s F (s) - f (0)$ to find $F (s)$ without integrating.
Answer
$f^{'} = e^{2 t} + 2 t e^{2 t}$ , so $\frac{1}{s - 2} + 2 F = s F$ , giving $F (s) = \frac{1}{( s - 2 ) ^{2}}$ .
If $y^{''} - 2 y^{'} + 5 y = 0$ , $y (0) = 1$ , $y^{'} (0) = 3$ , find $Y (s) = L {y}$ .
Answer
$Y (s) = \frac{s + 1}{s ^{2} - 2 s + 5}$
Find the transform of the sawtooth wave $f (t) = t$ for $0 \leq t < 2$ , $f (t + 2) = f (t)$ .
Answer
$\frac{1}{s ^{2}} - \frac{2 e ^{- 2 s}}{s ( 1 - e ^{- 2 s} )}$
Explain why $F (s) = \frac{s}{s + 1}$ is not the Laplace transform of any piecewise continuous function of exponential order.
Answer
By Theorem 9.1.1 such a transform must tend to $0$ as $s \to \infty$ , but $s / (s + 1) \to 1$ .

The definition: an improper integral

First transforms from the definition

Piecewise continuous functions

Exponential order and the existence theorem

Linearity

A table of transforms

Translation in s: the first shifting theorem

Transforms of derivatives

Multiplication by tn: derivatives of F

Periodic functions

Exercises

Translation in $s$ : the first shifting theorem

Multiplication by $t^{n}$ : derivatives of $F$