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7.1: Introduction to the Laplace Transform

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Definition of the Laplace Transform

To define the Laplace transform, we first recall the definition of an improper integral. If g is integrable over the interval [a,T] for every T>a, then the improper integral of g over [a,) is defined as

ag(t)dt=limTTag(t)dt.

We say that the improper integral converges if the limit in Equation ??? exists; otherwise, we say that the improper integral diverges or does not exist. Here’s the definition of the Laplace transform of a function f.

Defintion 7.1.1 : Laplace transform

Let f be defined for t0 and let s be a real number. Then the Laplace transform of f is the function F defined by

F(s)=0estf(t)dt,

for those values of s for which the improper integral converges.

It is important to keep in mind that the variable of integration in Equation ??? is t, while s is a parameter independent of t. We use t as the independent variable for f because in applications the Laplace transform is usually applied to functions of time.

The Laplace transform can be viewed as an operator L that transforms the function f=f(t) into the function F=F(s). Thus, Equation ??? can be expressed as

F=L(f).

The functions f and F form a transform pair, which we’ll sometimes denote by

f(t)F(s).

It can be shown that if F(s) is defined for s=s0 then it is defined for all s>s0 (Exercise 8.1.14b).

Computation of Some Simple Laplace Transforms

Example 7.1.1

Find the Laplace transform of f(t)=1.

Solution

From Equation ??? with f(t)=1,

F(s)=0estdt=limTT0estdt.

If s0 then

T0estdt=1sest|T0=1esTs.

Therefore

limTT0estdt={1s,s>0,,s<0.

If s=0 the integrand reduces to the constant 1, and

limTT01dt=limTT01dt=limTT=.

Therefore F(0) is undefined, and

F(s)=0estdt=1s,s>0.

This result can be written in operator notation as

L(1)=1s,s>0,

or as the transform pair

11s,s>0.

Note

It is convenient to combine the steps of integrating from 0 to T and letting T. Therefore, instead of writing Equation ??? and ??? as separate steps we write

0estdt=1sest|0={1s,s>0,s<0

We'll follow this practice throughout this chapter.

Example 7.1.2

Find the Laplace transform of f(t)=t.

From Equation ??? with f(t)=t,

F(s)=0esttdt.

If s0, integrating by parts yields

0esttdt=tests|0+1s0estdt=[ts+1s2]est|0={1s2,s>0,,s<0.

If s=0, the integral in Equation ??? becomes

0tdt=t22|0=.

Therefore F(0) is undefined and

F(s)=1s2,s>0.

This result can also be written as

L(t)=1s2,s>0,

or as the transform pair

t1s2,s>0.

Example 7.1.3 :

Find the Laplace transform of f(t)=eat, where a is a constant.

From Equation ??? with f(t)=eat,

F(s)=0esteatdt.

Combining the exponentials yields

F(s)=0e(sa)tdt.

However, we know from Example 7.1.1 that

0estdt=1s,s>0.

Replacing s by sa here shows that

F(s)=1sa,s>a.

This can also be written as

L(eat)=1sa,s>a, or eat1sa,s>a.

Example 7.1.4

[Find the Laplace transforms of f(t)=sinωt and g(t)=cosωt, where ω is a constant.

Define

F(s)=0estsinωtdt

and

G(s)=0estcosωtdt.

If s>0, integrating Equation ??? by parts yields

F(s)=estssinωt|0+ωs0estcosωtdt,

so

F(s)=ωsG(s).

If s>0, integrating Equation ??? by parts yields

G(s)=estcosωts|0ωs0estsinωtdt,

so

G(s)=1sωsF(s).

Now substitute from Equation ??? into this to obtain

G(s)=1sω2s2G(s).

Solving this for G(s) yields

G(s)=ss2+ω2,s>0.

This and Equation ??? imply that

F(s)=ωs2+ω2,s>0.

Tables of Laplace Transforms

Extensive tables of Laplace transforms have been compiled and are commonly used in applications. The brief table of Laplace transforms in the Appendix will be adequate for our purposes.

Example 7.1.5

Use the table of Laplace transforms to find L(t3e4t).

The table includes the transform pair

tneatn!(sa)n+1.

Setting n=3 and a=4 here yields

L(t3e4t)=3!(s4)4=6(s4)4.

We’ll sometimes write Laplace transforms of specific functions without explicitly stating how they are obtained. In such cases you should refer to the table of Laplace transforms.

Linearity of the Laplace Transform

The next theorem presents an important property of the Laplace transform.

Theorem 7.1.2 Linearity Property

Suppose L(fi) is defined for s>si, 1in). Let s0 be the largest of the numbers s1, s2, …,sn, and let c1, c2,…, cn be constants. Then

L(c1f1+c2f2++cnfn)=c1L(f1)+c2L(f2)++cnL(fn) for s>s0.

Proof

We give the proof for the case where n=2. If s>s0 then

L(c1f1+c2f2)=0est(c1f1(t)+c2f2(t)))dt=c10estf1(t)dt+c20estf2(t)dt=c1L(f1)+c2L(f2).

Example 7.1.6

Use Theorem 7.1.2 and the known Laplace transform

L(eat)=1sa

to find L(coshbt)(b0).

Solution

By definition,

coshbt=ebt+ebt2.

Therefore

L(coshbt)=L(12ebt+12ebt)=12L(ebt)+12L(ebt)(linearity property)=121sb+121s+b,

where the first transform on the right is defined for s>b and the second for s>b; hence, both are defined for s>|b|. Simplifying the last expression in Equation ??? yields

L(coshbt)=ss2b2,s>|b|.

The next theorem enables us to start with known transform pairs and derive others. (For other results of this kind, see Exercises 8.1.6 and 8.1.13.)

Theorem 7.1.3 First Shifting Theorem

If

F(s)=0estf(t)dt

is the Laplace transform of f(t) for s>s0, then F(sa) is the Laplace transform of eatf(t) for s>s0+a.

Proof

Replacing s by sa in Equation ??? yields

F(sa)=0e(sa)tf(t)dt

if sa>s0; that is, if s>s0+a. However, Equation 7.1.41 can be rewritten as

F(sa)=0est(eatf(t))dt,

which implies the conclusion.

Example 7.1.7

Use Theorem 7.1.3 and the known Laplace transforms of 1, t, cosωt, and sinωt to find

L(eat),L(teat),L(eλtsinωt),and L(eλtcosωt).

Solution

In the following table the known transform pairs are listed on the left and the required transform pairs listed on the right are obtained by applying Theorem 7.1.3 .

Table 7.1.1
f(t)F(s) eatf(t)F(sa)
11s,s>0 eat1(sa),s>a
t1s2,s>0 teat1(sa)2,s>a
sinωtωs2+ω2,s>0 eλtsinωtω(sλ)2+ω2,s>λ
cosωtss2+ω2,s>0 eλtsinωtsλ(sλ)2+ω2,s>λ

Existence of Laplace Transforms

Not every function has a Laplace transform. For example, it can be shown (Exercise 8.1.3) that

0estet2dt=

for every real number s. Hence, the function f(t)=et2 does not have a Laplace transform.

Our next objective is to establish conditions that ensure the existence of the Laplace transform of a function. We first review some relevant definitions from calculus.

Recall that a limit

limtt0f(t)

exists if and only if the one-sided limits

limtt0f(t)andlimtt0+f(t)

both exist and are equal; in this case,

limtt0f(t)=limtt0f(t)=limtt0+f(t).

Recall also that f is continuous at a point t0 in an open interval (a,b) if and only if

limtt0f(t)=f(t0),

which is equivalent to

limtt0+f(t)=limtt0f(t)=f(t0).

For simplicity, we define

f(t0+)=limtt0+f(t)and f(t0)=limtt0f(t),

so Equation ??? can be expressed as

f(t0+)=f(t0)=f(t0).

If f(t0+) and f(t0) have finite but distinct values, we say that f has a jump discontinuity at t0, and

f(t0+)f(t0)

is called the jump in f at t0 (Figure 7.1.1 ).

clipboard_e78f1a38a6c5d8fae002e0ac2e97ee81b.png
Figure 7.1.1 : A jump discontinuity.

If f(t0+) and f(t0) are finite and equal, but either f isn’t defined at t0 or it is defined but

f(t0)f(t0+)=f(t0),

we say that f has a removable discontinuity at t0 (Figure 7.1.2 ). This terminolgy is appropriate since a function f with a removable discontinuity at t0 can be made continuous at t0 by defining (or redefining)

f(t_0)=f(t_0+)=f(t_0-).\nonumber

Note

We know from calculus that a definite integral isn’t affected by changing the values of its integrand at isolated points. Therefore, redefining a function f to make it continuous at removable discontinuities does not change \cal{L}(f).

Theorem 7.1.4 Piecewise Continuous
  • A function f is said to be piecewise continuous on a finite closed interval [0,T] if f(0+) and f(T-) are finite and f is continuous on the open interval (0,T) except possibly at finitely many points, where f may have jump discontinuities or removable discontinuities.
  • A function f is said to be piecewise continuous on the infinite interval [0,\infty) if it is piecewise continuous on [0,T] for every T>0.

Figure 7.1.3 shows the graph of a typical piecewise continuous function.

It is shown in calculus that if a function is piecewise continuous on a finite closed interval then it is integrable on that interval. But if f is piecewise continuous on [0,\infty), then so is e^{-st}f (t), and therefore

\int_0^T e^{-st}f(t)\,dt \nonumber

clipboard_e8a13bcff17c498b55d624b6011453b79.png
Figure 7.1.2
clipboard_ef2ae35576ae06c2ad537e4572cc6fc43.png
Figure 7.1.3

exists for every T>0. However, piecewise continuity alone does not guarantee that the improper integral

\label{eq:8.1.13} \int_0^\infty e^{-st}f(t)\,dt=\lim_{T\to\infty}\int_0^T e^{-st}f(t)\, dt

converges for s in some interval (s_0,\infty). For example, we noted earlier that Equation \ref{eq:8.1.13} diverges for all s if f(t)=e^{t^2}. Stated informally, this occurs because e^{t^2} increases too rapidly as t\to\infty. The next definition provides a constraint on the growth of a function that guarantees convergence of its Laplace transform for s in some interval (s_0,\infty).

Theorem 7.1.5 : of exponential order

A function f is said to be of exponential order s_0 if there are constants M and t_0 such that

\label{eq:8.1.14} |f(t)|\le Me^{s_0t},\quad t\ge t_0.

In situations where the specific value of s_0 is irrelevant we say simply that f is of exponential order.

The next theorem gives useful sufficient conditions for a function f to have a Laplace transform. The proof is sketched in Exercise 8.1.10.

Theorem 7.1.6

If f is piecewise continuous on [0,\infty) and of exponential order s_0, then {\cal L}(f) is defined for s>s_0.

Note

We emphasize that the conditions of Theorem 7.1.6 are sufficient, but not necessary, for f to have a Laplace transform. For example, Exercise 8.1.14(c) shows that f may have a Laplace transform even though f isn’t of exponential order

Example 7.1.8

If f is bounded on some interval [t_0,\infty), say

|f(t)|\le M,\quad t\ge t_0,\nonumber

then Equation \ref{eq:8.1.14} holds with s_0=0, so f is of exponential order zero. Thus, for example, \sin\omega t and \cos \omega t are of exponential order zero, and Theorem 7.1.6 implies that {\cal L}(\sin\omega t) and {\cal L}(\cos \omega t) exist for s>0. This is consistent with the conclusion of Example 7.1.4 .

Example 7.1.9

It can be shown that if \lim_{t\to\infty}e^{-s_0t}f(t) exists and is finite then f is of exponential order s_0 (Exercise 8.1.9). If \alpha is any real number and s_0>0 then f(t)=t^\alpha is of exponential order s_0, since

\lim_{t\to\infty}e^{-s_0t}t^\alpha=0,\nonumber

by L’Hôpital’s rule. If \alpha\ge 0, f is also continuous on [0,\infty). Therefore Exercise 8.1.9 and Theorem 7.1.6 imply that {\cal L}(t^\alpha) exists for s\ge s_0. However, since s_0 is an arbitrary positive number, this really implies that {\cal L}(t^\alpha) exists for all s>0. This is consistent with the results of Example 7.1.2 and Exercises 8.1.6 and 8.1.8.

Example 7.1.10

Find the Laplace transform of the piecewise continuous function

f(t)=\left\{\begin{array}{cl} 1,&0\le t<1,\\ -3e^{-t},&t\ge 1.\end{array}\right.\nonumber

Solution

Since f is defined by different formulas on [0,1) and [1,\infty), we write

F(s)=\int_0^\infty e^{-st} f(t)\,dt =\int_0^1e^{-st}(1)\,dt+\int_1^\infty e^{-st}(-3e^{-t})\,dt.\nonumber

Since

\int_{0}^{1}e^{-st}dt = \left\{\begin{array}{cl} {\frac{1-e^{-s}}{s}}&{s\neq 0} \\ {1}&{s=0} \end{array} \right. \nonumber

and

\int_1^\infty e^{-st}(-3e^{-t})\,dt=-3\int_1^\infty e^{-(s+1)t}\,dt=-{3e^{-(s+1)}\over s+1},\quad s>-1,\nonumber

it follows that

F(s) = \left\{\begin{array}{rl}{\frac{1-e^{-s}}{s}-3\frac{e^{-(s+1)}}{s+1}}&{s>-1, s\neq 0} \\ {1-\frac{3}{e}}&{s=0} \end{array} \right. \nonumber

This is consistent with Theorem 7.1.6 , since

|f(t)|\le 3e^{-t},\quad t\ge 1,\nonumber

and therefore f is of exponential order s_0=-1.

Note

In Section 8.4 we’ll develop a more efficient method for finding Laplace transforms of piecewise continuous functions.

Example 7.1.11

We stated earlier that

\int_0^\infty e^{-st} e^{t^2} dt=\infty \nonumber

for all s, so Theorem 7.1.6 implies that f(t)=e^{t^2} is not of exponential order, since

\lim_{t\to\infty} {e^{t^2}\over Me^{s_0t}}=\lim_{t\to\infty} {1\over M} e^{t^2-s_0t}=\infty, \nonumber

so

e^{t^2}>Me^{s_0t} \nonumber

for sufficiently large values of t, for any choice of M and s_{0} (Exercise 8.1.3).


7.1: Introduction to the Laplace Transform is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by LibreTexts.

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