Chapter 5 - Laplace Transforms

Poularikas, A. D., Seely, S. “Laplace Transforms.”The Transforms and Applications Handbook: Second Edition.Ed. Alexander D. PoularikasBoca Raton: CRC Press LLC, 2000

5Laplace Transforms

5.1 Introduction 5.2 Laplace Transform of Some Typical Functions 5.3 Properties of the Laplace Transform 5.4 The Inverse Laplace Transform5.5 Solution of Ordinary Linear Equations

with Constant Coefficients 5.6 The Inversion Integral 5.7 Applications to Partial Differential Equations5.8 The Bilateral or Two-Sided Laplace Transform Appendix

5.1 Introduction*

The Laplace transform has been introduced into the mathematical literature by a variety of procedures.Among these are: (a) in its relation to the Heaviside operational calculus, (b) as an extension of theFourier integral, (c) by the selection of a particular form for the kernel in the general Integral transform,(d) by a direct definition of the Laplace transform, and (e) as a mathematical procedure that involvesmultiplying the function f(t) by e–s t dt and integrating over the limits 0 to ∞. We will adopt this latterprocedure.

Not all functions f(t), where t is any variable, are Laplace transformable. For a function f(t) to beLaplace transformable, it must satisfy the Dirichlet conditions — a set of sufficient but not necessaryconditions. These are

1. f(t) must be piecewise continuous; that is, it must be single valued but can have a finite numberof finite isolated discontinuities for t > 0.

2. f(t) must be of exponential order; that is, f(t) must remain less than Me –aot as t approaches ∞,where M is a positive constant and ao is a real positive number.

For example, such functions as: tan βt, cot βt, et2 are not Laplace transformable. Given a function f(t)that satisfies the Dirichlet conditions, then

(1.1)

is called the Laplace transformation of f(t). Here s can be either a real variable or a complex quantity.Observe the shorthand notation �{f(t)} to denote the Laplace transformation of f(t). Observe also thatonly ordinary integration is involved in this integral.

*All the contour integrations in the complex plane are counterclockwise.

F s f t e dt f ts t( ) = ( ) ( ){ }∞−∫0

written �

Alexander D. PoularikasUniversity of Alabama in Huntsville

Samuel Seely(Deceased)

© 2000 by CRC Press LLC

To amplify the meaning of condition (2), we consider piecewise continuous functions, defined for allpositive values of the variable t, for which

Functions of this type are known as functions of exponential order. Functions occurring in the solutionfor the time response of stable linear systems are of exponential order zero. Now we can recall that the

integral e–st dt converges if

If our function is of exponential order, we can write this integral as

This shows that for σ in the range σ > 0 (σ is the abscissa of convergence) the integral converges; that is

The restriction in this equation, namely, Re(s) = c , indicates that we must choose the path of integrationin the complex plane as shown in Figure 5.1.

FIGURE 5.1 Path of integration for exponential order function.

lim ,t

c tf t e c→∞

−( ) = =0 real constant .

f t( )∞

∫0

0

∞−∫ ( ) < ∞ = +f t e dt s js t , σ ω

0

∞− − −( )∫ ( )f t e e dtc t c tσ

.

0

∞−∫ ( ) < ∞ ( ) >f t e dt s cs t , Re .


5.2 Laplace Transform of Some Typical Functions

We illustrate the procedure in finding the Laplace transform of a given function f(t). In all cases it isassumed that the function f(t) satisfies the conditions of Laplace transformability.

Example 5.2.1

Find the Laplace transform of the unit step function f(t) = u(t), where u(t) = 1, t > 0, u(t) = 0, t < 0.

SolutionBy (1.1) we write

(2.1)

The region of convergence is found from the expression �e–s t � dt = e–σ t dt < ∞, which is the

entire right half-plane, σ > 0.

Example 5.2.2

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

(2.2)

To carry out the integration, define the quantity x = , then dx = dt, from which dt = dx =

2x dx. Then

But the integral

Thus, finally,

(2.3)

� u t u t e dt e dte

s ss t s t

s t

( ){ } = ( ) = = =∞

−∞

−− ∞

∫ ∫0 00

1.

0

∞

∫ 0

∞

∫

2t

π

F s t e dts t( ) =∞

−∫2

0

1

2

π.

t

1

21

2

1

2t−

2

1

2t

F s x e dxs x( ) =∞

−∫4

0

2 2

π.

0

2

3 2

2

4

∞−∫ =x e dx

ss x π

.

F ss

( ) = 13 2

.


Example 5.2.3

Find the Laplace transform of f(t) = erfc , where the error function, erf t , and the complementary

error function, erfc t , are defined by

SolutionConsider the integral

(2.4)

Change the order of integration, noting that u = , t =

The value of this integral is known

which leads to

(2.5)

Example 5.2.4

Find the Laplace transform of the function f(t) = sinh at.

SolutionExpress the function sinh at in its exponential form

The Laplace transform becomes

k

t2

erf erfct e du t e duo

tu

t

u= =∫ ∫−∞

−2 22 2

π π, ,

I e e du dtks t

t

u=

=∞

−∞

−∫ ∫2

20

2

π λλwhere .

λ

t

λ 2

2u

I e e dt dus

us

uduu

u

s t=

= − −

∞

−∞

−∞

∫ ∫ ∫2 2

0 0

22

2

2

2

2π πλ

λexp

= ⋅ −2

22

se s

ππ λ ,

� erfck

t sk s

2

1

= −{ }exp .

sinh .ate eat at

= − −

2


(2.6)

A moderate listing of functions f(t) and their Laplace transforms F(s) = �{f (t)} are given in Table5.1, in the Appendix.

5.3 Properties of the Laplace Transform

We now develop a number of useful properties of the Laplace transform; these follow directly from (1.1).Important in developing certain properties is the definition of f(t) at t = 0, a quantity written f(0+) todenote the limit of f(t) as t approaches zero, assumed from the positive direction. This designation isconsistent with the choice of function response for t > 0. This means that f(0+) denotes the initialcondition. Correspondingly, f (n) (0+) denotes the value of the nth derivative at time t = 0+, and f (–n)

(0+) denotes the nth time integral at time t = 0+. This means that the direct Laplace transform can bewritten

(3.1)

We proceed with a number of theorems.

Theorem 5.3.1 LinearityThe Laplace transform of the linear sum of two Laplace transformable functions f(t) + g(t) with respectiveabscissas of convergence σ f and σ g , with σg > σ f , is

�{ f (t ) + g (t )} = F (s ) + G (s ) . (3.2)

ProofFrom (3.1) we write

Thus,

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

As a direct extension of this result, for K1 and K2 constants,

�{ K 1 f (t ) + K 2 g (t )} = K 1 F (s ) + K 2 G (s ) . (3.3)

Theorem 5.3.2 DifferentiationLet the function f(t) be piecewise continuous with sectionally continuous derivatives df(t)/dt in everyinterval 0 ≤ t ≤ T. Also let f(t) be of exponential order ec t as t → ∞. Then when Re(s) > c , the transformof df(t)/dt exists and

� sinh

.

at e e dt

a

s a

s a t s a t{ } = −

=−

∞− −( ) − +( )∫1

2 0

2 2

F s f t e dt R aRa

a

Rs t( ) = ( ) > >

→∞→ +

−∫lim , , .0

0 0

� f t g t f t g t e dt f t e dt g t e dt

s

s t s t s t

g

( ) + ( ){ } = ( ) + ( )[ ] = ( ) + ( )( ) >

∞−

∞−

∞−∫ ∫ ∫0 0 0

,

Re .σ


(3.4)

ProofBegin with (3.1) and write

Write the integral as the sum of integrals in each interval in which the integrand is continuous. Thus,we write

Each of these integrals is integrated by parts by writing

with the result

But f(t) is continuous so that f(t1 – 0) = f(t1 + 0), and so forth, hence

However, with lim t→∞ f(t)e–st = 0 (otherwise the transform would not exist), then the theorem isestablished.

Theorem 5.3.3 DifferentiationLet the function f(t) be piecewise continuous, have a continuous derivative f (n–1)(t) of order n – 1 anda sectionally continuous derivative f (n)(t) in every finite interval 0 ≤ t ≤ T. Also, let f(t) and all its derivativesthrough f (n–1)(t) be of exponential order ect as t → ∞. Then the transform of f (n)(t) exists when Re(s) >c and it has the following form:

�{ f (n)(t )} = s nF (s ) – s n–1f (0+) – s n–2f (1)(0+) – L – s n–n f (n–1)(0+) . (3.5)

ProofThe proof follows as a direct extension of the proof of Theorem 5.3.2.

Example 5.3.1

Find �{tm} where m is any positive integer.

� �df t

dts f t f sF s f

( )

= ( ){ } − +( ) = ( ) − +( )0 0 .

�df t

dt

df t

dte dt

T

Ts t( )

=

( )→∞

−∫lim .0

0

1

0

1

1

2

1

Ts t

t

t

t

t

T

e f t dtn

∫ ∫ ∫ ∫− ( )( ) = [ ] + [ ] + + [ ]−

L .

u e du se dt

ddf

dtdt f

s t s t= = −

= =

− −

ν ν

e f t e f t e f t s e f t dts t t s tt

t s ttT

Ts t

n

− − − −( ) + ( ) + + ( ) + ( )− ∫0

0

1

1

2

1L .

0

1

0

0T

s t sTT

s te f t dt f e f T s e f t dt∫ ∫− ( ) − −( ) = − +( ) + ( ) + ( ) .


Solution The function f(t) = tm satisfies all the conditions of Theorem 5.3.3 for any positive c . Thus,

f (0+) = f (1)(0+) = L = f ( m–1)(0+) = 0

f (m )(t) = m!, f (m +1)(t) = 0 .

By (3.5) with n = m + 1 we have

�{ f (m +1)(t )} = 0 = s m +1�{t m} – m ! .

It follows, therefore, that

Theorem 5.3.4 Integration

If f(t) is sectionally continuous and has a Laplace transform, then the function dξ has the Laplace

transform given by

(3.6)

ProofBecause f(t) is Laplace transformable, its integral is written

This is integrated by parts by writing

Then

from which

� tm

sm

m{ } =+

!.

1

fo

t

ξ( )∫

�0

110

t

f dF s

s sf∫ ( )

=( )

+ +( )−( )ξ ξ .

�−∞

∞

−∞

−∫ ∫ ∫( )

= ( )

t ts tf d f d e dtξ ξ ξ ξ

0

.

u f d du f d f t dt

d e dts

e

t

s t s t

= ( ) = ( ) = ( )= = −

−∞

− −

∫ ξ ξ ξ ξ

ν ν 1.

�−∞

−

−∞

∞∞

−

∞−

−∞

∫ ∫ ∫

∫ ∫

( )

= − ( )

+ ( )

= ( ) + ( )

t s t ts t

s t

f de

sf d

sf t e dt

sf t e dt

sf d

ξ ξ ξ ξ

ξ ξ

00

0

0

1

1 1


where [f (–1)(0+)/s] is the initial value of the integral of f(t) at t = 0+. The negative number in the bracketedexponent indicates integration.

Example 5.3.2

Deduce the value of �{sin at} from �{cos at} by employing Theorem 5.3.4.

SolutionBy ordinary integration

From Theorem 5.3.4 we can write, knowing that �{cos at} = .

so that

Theorem 5.3.5Division of the transform of a function by s corresponds to integration of the function between the limits0 and t

(3.7)

and so forth, for division by sn, provided that f(t) is Laplace transformable.

ProofThe proof of this theorem follows from Theorem 5.3.4.

Theorem 5.3.6 Multiplication by tIf f(t) is piecewise continuous and of exponential order, then each of the Laplace transforms: �{ f(t),�{tf(t), �{t2f(t),…is uniformly convergent with respect to s when s = c , where σ > c , and

(3.8)

�0

11 10

t

f ds

F ss

f∫ ( )

= ( ) + +( )−( )ξ ξ

0

t

ax dxat

a∫ =cossin

.

s

s a2 2+

�sinat

a s a

=+1

2 2

� sin .ata

s a{ } =

+2 2

�

�

−

−

( )

= ( )

( )

= ( )

∫

∫∫

1

0

1

200

F s

sf d

F s

sf d d

t

t

ξ ξ

ξξ

λ λ

� t f td F s

dsn n

n

n( ){ } = −( ) ( )1 .


Further

ProofIt follows from (3.1) when this integral is uniformly convergent and the integral converges, that

Further, it follows that

Similar procedures follow for derivatives of higher order.

Theorem 5.3.7 Differentiation of a TransformDifferentiation of the transform of a function f(t) corresponds to the multiplication of the function by–t ; thus

(3.9)

ProofThis is a restatement of Theorem 5.3.6. This theorem is often useful for evaluating some types of integrals,and can be used to extend the table of transforms.

Example 5.3.3

Employ Theorem 5.3.7 to evaluate ∂F(s)/∂s for the function f(t) = sinh at.

SolutionInitially we establish sinh at

By Theorem 5.3.7

from which

lim , , , , ,s

n

n sn

d F s

dst f t n

→∞ →∞

( )= ( ){ } = =0 0 1 2 3� K

∂ ( )∂

= −( ) ( ) = − ( ){ }∞−∫

F s

se t f t dt tf ts t

0

� .

∂ ( )∂

= −( ) ( ) = ( ){ }∞−∫

2

20

2 2F s

se t f t dt t f ts t � .

d F s

dsF s t f t n

n

n

n n( )= ( ) = −( ) ( )

= …( ) � , , , , .1 2 3

� sinh .at ee e

dta

s aF ss t

at at

{ } = −

=−

= ( )∞

−−

∫02 22

∂ ( )∂

= −( ) = ∂∂ −

= −

−( )∞

−∫F s

st at e dt

s

a

s a

as

s a

s t

02 2

2 22

2sinh

0 2 22

2∞−∫ = { } =

−( )e t at dt t at

as

s a

s t sinh sinh .�


We can, of course, differentiate F(s) with respect to a. In this case, Theorem 5.3.7 does not apply.However, the result is significant and is

Theorem 5.3.8 Complex Integration

If f(t) is Laplace transformable and provided that lim t→0+ exists, the integral of the function

F(s) ds corresponds to the Laplace transform of the division of the function f(t) by t,

(3.10)

ProofLet F(s) be piecewise continuous in each finite interval and of exponential order. Then

is uniformly convergent with respect to s . Consequently, we can write for Re(s) > c and any a > c

Express this in the form

Now if f(t)/t has a limit as t → 0, then the latter function is piecewise continuous and of exponentialorder. Therefore, the last integral is uniformly convergent with respect to a . Thus, as a tends to infinity

Theorem 5.3.9 Time Delay; Real TranslationThe substitution of t – λ for the variable t in the transform �{ f (t )} corresponds to the multiplicationof the function F(s) by e– λ s ; that is,

�{ f (t – λ )} = e – s λ F (s ) . (3.11)

ProofRefer to Figure 5.2, which shows a function f (t)u(t) and the same function delayed by the time t = λ ,where λ is a positive constant.

∂ ( )∂

= ( ) = { } = ∂∂ −

= +

−( )∞

−∫F s

ae t at dt t at

a

a

s a

s a

s a

s t

02 2

2 2

2 22

cosh cosh .�

f t

t

( )s

∞

∫

�f t

tF s ds

( )

= ( )

∞

∫0

.

F s e f t dts t( ) = ( )∞

−∫0

s

a

s

as tF s ds e f t dt ds∫ ∫∫( ) = ( )

∞−

0

.

= ( ) =( )

−( )∞−

∞− −∫ ∫ ∫0 0

f t e ds dtf t

te e dt

s

as t s t at .

s

F s dsf t

t

∞

∫ ( ) =( )

� .


We write directly

Now introduce a new variable τ = t – λ. This converts this equation to the form

because u(τ ) = 0 for –λ ≤ t ≤ 0.We would similarly find that

�{ f (t + λ ) u (t + λ )} = e s λ F (s ) . (3.12)

Example 5.3.4

Find the Laplace transform of the pulse function shown in Figure 5.3.

SolutionBecause the pulse function can be decomposed into step functions, as shown in Figure 5.3, its Laplacetransform is given by

FIGURE 5.2 A function f(t) at the time t = 0 and delayed time t = λ.

FIGURE 5.3 Pulse function and its equivalent representation.

� f t u t f t u t e dts t−( ) −( ){ } = −( ) −( )∞

−∫λ λ λ λ0

.

� f u e f u e d e f e d e F ss s s s sτ τ τ τ τ τ ττ τ( ) ( ){ } = ( ) ( ) = ( ) = ( )−

−

∞− −

∞− −∫ ∫λ

λ

λ λ

0


where the translation property has been used.

Theorem 5.3.10 Complex TranslationThe substitution of s + a for s, where a is a real or complex, in the function F(s + a), corresponds to theLaplace transform of the product e–atf(t).

ProofWe write

which is

F (s + a) = �{e – a t f (t )} . (3.13)

In a similar way we find

F (s – a) = �{ e a t f (t )} . (3.14)

Theorem 5.3.11 ConvolutionThe multiplication of the transforms of two sectionally continuous functions f1(t) (= F1(s)) and f2(t)(= F2(s)) corresponds to the Laplace transform of the convolution of f1(t) and f2(t).

F 1(s ) F 2(s) = �{ f 1(t ) ∗ f 2(t)} (3.15)

where the asterisk ∗ is the shorthand designation for convolution.

ProofBy definition, the convolution of two functions f1(t) and f2(t) is

(3.16)

Thus,

Now effect a change of variable, writing t – τ = ξ and therefore dt = dξ, then

� 2 1 5 21 1 2

11 5 1 5u t u ts s

es

es s( ) − −( )[ ]{ } = −

= −( )− −. . .

0 0

∞− −

∞− +( )∫ ∫( ) = ( ) ( ) > − ( )e f t e dt f t e dt s c aat s t s a t

for Re Re ,

f t f t f t f d f f t d1 20

1 20

1 2( )∗ ( ) = −( ) ( ) = ( ) −( )∞ ∞

∫ ∫τ τ τ τ τ τ .

� �f t f t f t f d

f t f d e dt

f d f t e dt

s

s t

s t

10

1 2

0 01 2

02

01

( )∗ ( ){ } = −( ) ( )

= −( ) ( )

= ( ) −( )

∞

∞ ∞−

∞ ∞−

∫

∫ ∫

∫ ∫

τ τ τ

τ τ τ

τ τ τ .

= ( ) ( )∞

−

∞− +( )∫ ∫0

2 1f d f e dsτ τ ξ ξ

τ

ξ τ.


But for positive time functions f 1(ξ ) = 0 for ξ < 0, which permits changing the lower limit of the secondintegral to zero, and so

which is

�{ f 1(t ) ∗ f 2 (t )} = F 1(s ) F 2(s ) .

Example 5.3.5

Given f1(t) = t and f2(t) = eat, deduce the Laplace transform of the convolution t ∗ eat by the use ofTheorem 5.3.11.

SolutionBegin with the convolution

Then

By Theorem 5.3.11 we have

and

Theorem 5.3.12The multiplication of the transforms of three sectionally continuous functions f1(t), f2(t), and f3(t)corresponds to the Laplace transform of the convolution of the three functions

�{ f 1(t ) ∗ f 2(t ) ∗ f 3(t )} = F 1(s ) F 2(s ) F 3(s ) . (3.17)

ProofThis is an extension of Theorem 5.3.11. The result is obvious if we write

F 1(s ) F 2(s ) F 3 (s ) = �{ f 1(t ) ∗ � –1{ F 2(s ) F 3(s )}} .

Example 5.3.6

Deduce the values of the convolution products: 1 ∗ f(t); 1 ∗ 1 ∗ f(t).

= ( ) ( )∞

−∞

−∫ ∫02

01f e d f e ds sτ τ ξ ξτ ξ ,

t e t e dt e

a

e

a

e

a ae atat

ta

at

a at

a∗ = −( ) = − −

= − −( )∫00

2

0

2

11τ τ ττ

τ τ ττ .

� t ea s a s s s s a

at∗{ } =−

− −

=−( )

1 1 1 1 1 12 2 2

.

F s f t ts

F s f t es a

at1 1 2 2 2

1 1( ) = ( ){ } = { } = ( ) = ( ){ } = { } =−

� � � �, .

� t es s a

at∗{ } =−( )

1 12

.


SolutionBy equations (3.14) and (3.16) we write directly

(a) For f1(t) = 1, f2(t) = f (t), �{1 ∗ f(t)} by equation (3.7)

(b) For f1(t) = 1, f2(t) = 1, f3(t) = f(t), �{1 ∗ 1 ∗ f(t)}

Theorem 5.3.13 Frequency Convolution — s-planeThe Laplace transform of the product of two piecewise and sectionally continuous functions f1(t) andf2(t) corresponds to the convolution of their transforms, with

(3.18)

ProofBegin by considering the following line integral in the z-plane:

This means that the contour intersects the x-axis at x1 > σ 2 (see Figure 5.4). Then we have

Assume that the integral of F2(z) is convergent over the path of integration. This equation is now writtenin the form

FIGURE 5.4 The contour C2 and the allowed range of s.

= =( ) ( )

∫

F s

sf d

t

�0

ξ ξ

= =( ) ( )

∫∫

F s

sf d d

t

200

� λ λ ξξ

� f t f tj

F s F s1 2 1 2

1

2( ) ( ){ } = ( )∗ ( )[ ]π

.

f tj

F z e dzC

z t2 2 2

1

2 2

( ) = ( ) =∫πσ, axis of convergence .

01 2

01 2

1

2 2

∞−

∞−( )∫ ∫ ∫( ) ( ) = ( ) ( )f t f t e dt

jf t dt F z e dzs t

C

z s t

π.


(3.19)

The Laplace transform of f1(t), the integral on the right, converges in the range Re(s – z) > σ 1, whereσ 1 is the abscissa of convergence of f1(t). In addition, Re(z) = σ 2 for the z-plane integration involved in(3.18). Thus, the abscissa of convergence of f1(t) f2(t) is specified by

Re(s) > σ 1 + σ 2 . (3.20)

This situation is portrayed graphically in Figure 5.4 for the case when both σ 1 and σ 2 are positive. Asfar as the integration in the complex plane is concerned, the semicircle can be closed either to the leftor to the right just so long as F1(s) and F2(s) go to zero as s → ∞.

Based on the foregoing, we observe the following:

• Poles of F1(s – z) are contained in the region Re(s – z) < σ 1

• Poles of F2(z) are contained in the region Re(z) < σ 2

• From (a) and (3.20) Re(z) > Re(s – σ 1) > σ 2

• Poles of F1(s – z) lie to the right of the path of integration• Poles of F2(z) are to the left of the path of integration• Poles of F1(s – z) are functions of s whereas poles of F2(z) are fixed in relation to s

Example 5.3.7

Find the Laplace transform of the function f(t) = f1(t) f2(t) = e–t e–2t u(t).

SolutionFrom Theorem 5.3.13 and the absolute convergence region for each function, we have

Further, f(t) = exp[–(2 + 1)t] u(t) implies that σ f = σ 1 + σ 2 = 3. We now write

To carry out the integration dictated by equation (3.19) we use the contour shown in Figure 5.5. If weselect contour C 1 and use the residue theorem, we obtain

The inverse of this transform is exp(–3t). If we had selected contour C2 , the residue theorem gives

01 2 2

01

2 1 1 2

1

2

1

2

2

2

2

2

∞−

− ∞

+ ∞ ∞− −( )

− ∞

+ ∞

∫ ∫ ∫

∫

( ) ( ) = ( ) ( )

= ( ) −( ) = ( ) ( ){ }

f t f t e dtj

F z dz f t e dt

jF z F s z dz f t f t

s t

j

js z t

j

j

π

π

σ

σ

σ

σ∆ � .

F ss

F ss

1 1

2 2

1

11

1

22

( ) =+

> −

( ) =+

> −

,

, .

σ

σ

F z F s zz s z s z s s z2 1

1

2

1

1

1

3

1

1

1

3

1

2( ) −( ) =

+ − +=

+ − +( ) −+ +

.

F sj

F z F s z dz j s F z F s zsC z

( ) = ( ) −( ) = ( ) −( )[ ] =+∫ = −

1

22

1

31

2 1 2 12π

π Re .


The inverse transform of this is also exp(–3t), as to be expected.

Theorem 5.3.14 Initial Value TheoremLet f(t) and f (1)(t) be Laplace transformable functions, then for case when lim sF(s) as s → ∞ exists,

(3.21)

ProofBegin with equation (3.6) and consider

Because f(0+) is independent of s, and because the integral vanishes for s → ∞, then

Furthermore, f(0+) = limt→0+ f(t) so that

If f(t) has a discontinuity at the origin, this expression specifies the value of the impulse f(0+). If f(t)contains an impulse term, then the left-hand side does not exist, and the initial value property does notexist.

FIGURE 5.5 The contour for Example 5.3.7.

F sj

F z F s z dz j s F z F s z

s s

C z s( ) = ( ) −( ) = − ( ) −( )[ ]

= − −+

=

+

∫ = +

1

22

1

3

1

3

2

2 1 2 11π

π Re

.

lim lim .s t

sF s f t→∞ → +

( ) = ( )0

lim lim .s

s t

s

df

dte dt sF s f

→∞

∞−

→∞∫ = ( ) − +( )[ ]0

0

lim .s

sF s f→∞

( ) − +( )[ ] =0 0

lim lim .s t

sF s f t→∞ → +

( ) = ( )0


Theorem 5.3.15 Final Value TheoremLet f(t) and f (1)(t) be Laplace transformable functions, then for t → ∞

(3.22)

ProofBegin with equation (3.6) and Let s → 0. Thus, the expression

Consider the quantity on the left. Because s and t are independent and because e–st → 1 as s → 0, thenthe integral on the left becomes, in the limit

Combine the latter two equations to get

It follows from this that the final value of f(t) is given by

This result applies F(s) possesses a simple pole at the origin, but it does not apply if F(s) has imaginaryaxis poles, poles in the right half plane, or higher order poles at the origin.

Example 5.3.8

Apply the final value theorem to the following two functions:

SolutionFor the first function from sF1(s ),

For the second function,

lim lim .t s

f t sF s→∞ →

( ) = ( )0

lim lim .s

s t

s

df

dte dt sF s f

→

∞−

→∫ = ( ) − +( )[ ]0 0 0

0

0

0∞

→∞∫ = ( ) − +( )df

dtdt f t f

tlim .

lim lim .t s

f t f sF s f→∞ →∞

( ) − +( ) = ( ) − +( )0 0

lim lim .t s

f t sF s→∞ →

( ) = ( )0

F ss a

s a bF s

s

s b1 2 2

2 2 2( ) = +

+( ) +( ) =

+, .

lim .s

s s a

s a b→

+( )+( ) +

=0 2 2

0

sF ss

s b( ) =

+

2

2 2.


However, this function has singularities on the imaginary axis at s = ±j b, and the Final Value Theoremdoes not apply.

The important properties of the Laplace transform are contained in Table 5.2 in the Appendix.

5.4 The Inverse Laplace Transform

We employ the symbol �–1{F(s)}, corresponding to the direct Laplace transform defined in (1.1), todenote a function f(t) whose Laplace transform is F(s). Thus, we have the Laplace pair

F (s ) = �{ f (t )}, f (t ) = �–1{ F (s )} . (4.1)

This correspondence between F(s) and f(t) is called the inverse Laplace transformation of f(t).Reference to Table 5.1 shows that F(s) is a rational function in s if f(t) is a polynomial or a sum of

exponentials. Further, it appears that the product of a polynomial and an exponential might also yielda rational F(s). If the square root of t appears on f(t), we do not get a rational function in s. Note alsothat a continuous function f(t) may not have a continuous inverse transform.

Observe that the F(s) functions have been uniquely determined for the given f(t) function by (1.1).A logical question is whether a given time function in Table 5.1 is the only t-function that will give thecorresponding F(s). Clearly, Table 5.1 is more useful if there is a unique f(t) for each F(s). This is animportant consideration because the solution of practical problems usually provides a known F(s) fromwhich f(t) must be found. This uniqueness condition can be established using the inversion integral.This means that there is a one-to-one correspondence between the direct and the inverse transform. Thismeans that if a given problem yields a function F(s), the corresponding f(t) from Table 5.1 is the uniqueresult. In the event that the available tables do not include a given F(s), we would seek to resolve thegiven F(s) into forms that are listed in Table 5.1. This resolution of F(s) is often accomplished in termsof a partial fraction expansion.

A few examples will show the use of the partial fraction form in deducing the f(t) for a given F(s).

Example 5.4.1

Find the inverse Laplace transform of the function

(4.2)

SolutionObserve that the denominator can be factored into the form (s + 2) (s + 3). Thus, F(s) can be writtenin partial fraction form as

(4.3)

where A and B are constants that must be determined.To evaluate A, multiply both sides of (4.3) by (s + 2) and then set s = –2. This gives

and B(s + 2)/(s + 3)� s = –2 is identically zero. In the same manner, to find the value of B we multiply bothsides of (4.3) by (s + 3) and get

F ss

s s( ) = −

+ +3

5 62.

F ss

s s

A

s

B

s( ) = −

+( ) +( ) =+

++

3

2 3 2 3.

A F s ss

sss

= ( ) +( ) = −+

= −=−

=−

23

35

22


The partial fraction form of (4.3) is

The inverse transform is given by

where entry 8 in Table 5.1, is used.

Example 5.4.2


SolutionThis function is written in the form

The value of A is deduced by multiplying both sides of this equation by (s + 3) and then setting s = –3.This gives

To evaluate B and C, combine the two fractions and equate the coefficients of the powers of s in thenumerators. This yields

from which it follows that

–(s 2 + 4s + 5) + B s 2 + (C + 3B )s + 2C = s + 1 .

B F s ss

sss

= ( ) +( ) = −+

==−

=−

33

36

33

.

F ss s

( ) = −+

++

5

2

6

3.

f t F ss s

e et t( ) = ( ){ } = −+

++

= − +− − − − −� � �1 1 1 2 351

26

1

35 6

F ss

s s( ) = +

+( ) +

+( )1

2 1 32

.

F sA

s

Bs C

s

s

s s( ) =

++ +

+( ) +

= +

+( ) +

+( )3 2 1

1

2 1 32 2

.

A s F ss

= +( ) ( ) = − +

− +( ) += −

=−3

3 1

3 2 11

3 2.

− +( ) +

+ +( ) +( )+( ) +

+( )= +

+( ) +

+( )1 2 1 3

2 1 3

1

2 1 3

2

2 2

s s Bs C

s s

s

s s


Combine like-powered terms to write

(–1 + B )s 2 + (–4 + C + 3B )s + (–5 + 3C ) = s + 1 .

Therefore,

–1 + B = 0, –4 + C + 3B = 1 , –5 + 3C = 1 .

From these equations we obtain

B = 1 , C = 2 .

The function F(s) is written in the equivalent form

Now using Table 5.1, the result is

f (t ) = –e – 3 t + e – 2 t cos t , t > 0 .

In many cases, F(s) is the quotient of two polynomials with real coefficients. If the numeratorpolynomials is of the same or higher degree than the denominator polynomial, first divide the numeratorpolynomial by the denominator polynomial; the division is carried forward until the numerator poly-nomial of the remainder is one degree less than the denominator. This results in a polynomial in s plusa proper fraction. The proper fraction can be expanded into a partial fraction expansion. The result ofsuch an expansion is an expression of the form

(4.4)

This expression has been written in a form to show three types of terms; polynomial, simple partialfraction including all terms with distinct roots, and partial fraction appropriate to multiple roots.

To find the constants A1, A2, … the polynomial terms are removed, leaving the proper fraction

F′(s) – (B0 + B1s + L) = F(s) (4.5)

where

To find the constants Ak that are the residues of the function F(s) at the simple poles sk, it is onlynecessary to note that as s → sk the term Ak(s – sk) will become large compared with all other terms. Inthe limit

(4.6)

F ss

s

s( ) = −

++ +

+( ) +

1

3

2

2 12

.

′ ( ) = + + +−

+−

+ +−

+−( )

+ +−( )

F s B B sA

s s

A

s s

A

s s

A

s s

A

s s

p

p

p

p

pr

p

r0 11

1

2

2

1 2

2L L L .

F sA

s s

A

s s

A

s s

A

s s

A

s s

A

s s

k

k

p

p

p

p

pr

p

r( ) =−

+−

+ +−

+−

+−( )

+ +−( )

1

1

2

2

1 2

2L L .

A s s F sk s s kk

= ( ) ( )→

lim – .


Upon taking the inverse transform for each simple pole, the result will be a simple exponential of theform

(4.7)

Note also that because F(s) contains only real coefficients, if sk is a complex pole with residue Ak, therewill also be a conjugate pole with residue . For such complex poles

These can be combined in the following way:

(4.8)

where θ k = tan–1 (bk/ak) and Ak = ak /cos θ k .When the proper fraction contains a multiple pole of order r , the coefficients in the partial-fraction

expansion Ap1, Ap2, …, Ap r that are involved in the terms

must be evaluated. A simple application of (4.6) is not adequate. Now the procedure is to multiply bothsides of (4.5) by (s – sp ) r , which gives

(4.9)

In the limit as s = sp all terms on the right vanish with the exception of Apr. Suppose now that thisequation is differentiated once with respect to s. The constant Apr will vanish in the differentiation butAp(r–1) will be determined by setting s = s p. This procedure will be continued to find each of the coefficientsApk. Specifically, the procedure is specified by

(4.10)

�−

−

=1 A

s sA ek

kk

s tk .

sk

∗ Ak

∗

�−∗

∗∗

−+

−

= +

∗1 A

s s

A

s sA e A ek

k

k

k

ks t

ks tk k .

response = +( ) + −( )= +( ) +( ) + −( ) +( )[ ]= −( )=

+( ) −( )a jb e a jb e

e a jb t j t a jb t j t

e a t b t

A e

k k

j t

k k

j t

tk k k k k k k k

tk k k k

kt

k k k k

k

k

k

σ ω σ ω

σ

σ

σ

ω ω ω ω

ω ω

cos sin cos sin

cos sin2

2 coscos ω θk kt +( )

A

s s

A

s s

A

s s

p

p

p

p

pr

p

r

1 2

2−( ) +−( )

+ +−( )

L

s s F s s sA

s s

A

s s

A

s sA s s

A s s A

p

r

p

r k

kp p

r

p r p pr

−( ) ( ) = −( ) −+

−+ +

−

+ −( ) +

+ −( ) +

−

−( )

1

1

2

21

1

1

L L

Ar k

d

dsF s s s k rpk

r k

r k p

r

s s p

=−( ) ( ) −( )

=

−

−=

11 2

!, , , , .K


Example 5.4.3

Find the inverse transform of the following function:

SolutionThis is not a proper fraction. The numerator polynomial is divided by the denominator polynomial bysimple long division. The result is

The proper fraction is expanded into partial fraction form

The value of A2 is deduced using (4.6)

To find A11 and A12 we proceed as specified in (4.10)

Therefore,

From Table 5.1 the inverse transform is

f (t ) = δ (t ) + 4 + t – e – t , for t ≥ 0 .

If the function F(s) exists in proper fractional form as the quotient of two polynomials, we can employthe Heaviside expansion theorem in the determination of f(t) from F(s). This theorem is an efficientmethod for finding the residues of F(s). Let

F ss s s

s s( ) = + + +

+( )3 2

2

2 3 1

1.

F ss s

s s( ) = + + +

+( )13 1

1

2

2.

F ss s

s s

A

s

A

s

A

sp ( ) = + ++( ) = + +

+

2

211 12

223 1

1 1.

A s F ss s

sp

ss

21

2

2

1

13 1

1= +( ) ( )[ ] = + + = −=−

=−

.

A s F ss s

s

Ad

dss F s

d

ds

s s

s

s s

s

s

s

ps

s

p

s ss

122

0

2

0

112

0

2

0

2

2

0

3 1

11

1

1

3 1

1

3 1

1

2 3

14

= ( )[ ] = + ++

=

= ( )

= + ++

= + +

+( )+ +

+=

==

= ==

!.

F ss s s

( ) = + + −+

14 1 1

12.

F sP s

Q s

A

s s

A

s s

A

s sk

k

( ) =( )( ) =

−+

−+ +

−1

1

2

2

L


where P(s) and Q(s) are polynomials with no common factors and with the degree of P(s) less than thedegree of Q(s).

Suppose that the factors of Q(s) are distinct constants. Then, as in (4.6) we find

Also, the limit P(s) is P(s k). Now, because

then

Thus,

(4.11)

From this, the inverse transformation becomes

This is the Heaviside expansion theorem. It can be written in formal form.

Theorem 5.4.1 Heaviside Expansion TheoremIf F(s) is the quotient P(s)/Q(s) of two polynomials in s such that Q(s) has the higher degree andcontains simple poles the factor s – s k, which are not repeated, then the term in f(t) corresponding to

this factor can be written .

Example 5.4.4

Repeat Example 4.1 employing the Heaviside expansion theorem.

SolutionWe write (4.2) in the form

As s

Q sP sk s s

k

k

=−

( ) ( )

→

lim .

lim lim ,s s

k

s sk

k k

s s

Q s Q s Q s→ → ( ) ( )−

( ) =( )

=( )

1 11 1

AP s

Q sk

k

k

=( )( )( )1

.

F sP s

Q s

P s

Q s s s

n

n nn

k

( ) =( )( ) =

( )( )

⋅−( )( )

=∑ 1

1

1.

f tP s

Q s

P s

Q se

n

nn

ks tn( ) =

( )( )

=

( )( )

−

( )=

∑� 1

11

.

P s

Q se

k

k

s tk( )( )( )1

F sP s

Q s

s

s s

s

s s( ) =

( )( ) = −

+ += −

+( ) +( )3

5 6

3

2 32.


The derivative of the denominator is

Q (1)(s ) = 2s + 5

from which, for the roots of this equation,

Q (1)(–2) = 1 , Q (1)(–3) = –1 .

Hence,

P (–2) = –5 , P (–3) = –6 .

The final value for f(t) is

f (t ) = –5e – 2 t + 6 e – 3 t .

Example 5.4.5

Find the inverse Laplace transform of the following function using the Heaviside expansion theorem:

SolutionThe roots of the denominator are

That is, the roots of the denominator are complex. The derivative of the denominator is

Q (1)(s ) = 2s + 4 .

We deduce the values P(s)/Q(1)(s) for each root

Then

�− ++ +

1

2

2 3

4 7

s

s s.

s s s j s j2 4 7 2 3 2 3+ + = + +( ) + −( ) .

For

For

s j Q s j P s j

s j Q s j P s j

1

1

1 1

2

1

2 2

2 3 2 3 1 2 3

2 3 2 3 1 2 3

= − − ( ) = − ( ) = − −

= − + ( ) = + ( ) = − +

( )

( ) .

f tj

je

j

je

ej

je

j

je

ee e

je

j t j t

t j t j t

t

j t j t

j

( ) = − −−

+ − +

= − −−

+ − +

=−( )

+

− −( ) − +( )

− −

−

−

−

1 2 3

2 3

1 2 3

2 3

1 2 3

2 3

1 2 3

2 3

2 3

2 2 3 2 2 3

2 2 3 2 3

2

2 3 2 3

2 3tt j t

t

e

e t t

+( )

= −

−

2 3

2 2 2 31

32 3cos sin


5.5 Solution of Ordinary Linear Equations with Constant Coefficients

The Laplace transform is used to solve homogeneous and nonhomogeneous ordinary different equationsor systems of such equations. To understand the procedure, we consider a number of examples.

Example 5.5.1

Find the solution to the following differential equation subject to prescribed initial conditions: y(0+);(dy/dt) + ay = x(t).

SolutionLaplace transform this differential equation. This is accomplished by multiplying each term by e–stdt andintegrating form 0 to ∞. The result of this operation is

sY (s ) – y (0+) + aY (s ) = X (s ) ,

from which

If the input x(t) is the unit step function u(t), then X(s) = 1/s and the final expression for Y(s) is

Upon taking the inverse transform of this expression

with the result

Example 5.5.2

Find the general solution to the differential equation

subject to zero initial conditions.

SolutionLaplace transform this differential equation. The result is

Y sX s

s a

y

s a( ) =

( )+

++( )

+0

.

Y ss s a

y

s a( ) =

+( ) ++( )

+1 0

.

y t Y sa s s a

y

s a( ) = ( ){ } = −

+

+

+( )+

− −� �1 1 1 1 1 0

y ta

e y eat at( ) = −( ) + +( )− −11 0 .

d y

dt

dy

dty

2

25 4 10+ + =

s Y s sY s Y ss

2 5 510( ) + ( ) + ( ) = .


Solving for Y(s), we get

Expand this into partial-fraction form, thus

Then

and

The inverse transform is

Example 5.5.3

Find the velocity of the system shown in Figure 5.6a when the applied force is f(t) = e–tu(t). Assumezero initial conditions. Solve the same problem using convolution techniques. The input is the force andthe output is the velocity.

SolutionThe controlling equation is, from Figure 5.6b,

Laplace transform this equation and then solve for F(s). We obtain

Y ss s s s s s

( ) =+ +( ) =

+( ) +( )10

5 4

10

1 42.

Y sA

s

B

s

C

s( ) =

++

++

1 4.

A Y s ss s

B Y s ss s

C sY ss s

s

s

s

s

s

s

= ( ) +( ) =+( ) = −

= ( ) +( ) =+( ) =

= ( ) =+( ) +( ) =

= −= −

= −= −

==

110

4

10

3

410

1

10

12

10

1 4

10

4

1

1

4

4

0

0

Y ss s s

( ) = −+( ) +

+( ) +

101

3 1

1

12 4

1

4.

x t e et t( ) = − + +

− −101

3

1

12

1

44 .

d

dtdt e u t

ttν ν ν+ + = ( )∫ −5 4

0

.


Write this expression in the form

where

The inverse transform of V(s) is given by

To find ν (t) by the use of the convolution integral, we first find h(t), the impulse response of thesystem. The quantity h(t) is specified by

where the system is assumed to be initially relaxed. The Laplace transform of this equation yields

FIGURE 5.6 The mechanical system and its network equivalent.

V ss

s s s

s

s s( ) =

+( ) + +( ) =+( ) +( )1 5 4 1 4

2 2.

V sA

s

B

s

C

s( ) =

++

++

+( )4 1 12

As

s

Bd

ds

s

s

Cs

s

s

s

s

=+( )

= −

=+

=

=+

= −

=−

=−

=−

1

4

9

1

1 4

4

9

4

1

3

2

4

1

1

!

.

ν t e e te tt t t( ) = − + − ≥− − −4

9

4

9

1

304 , .

dh

dth hdt t+ + = ( )∫5 4 δ


The inverse transform of this expression is easily found to be

The output of the system to the input e–tu(t) is written

This result is identical with that found using the Laplace transform technique.

Example 5.5.4

Find an expression for the voltage ν 2(t) for t > 0 in the circuit of Figure 5.7. The source ν 1(t), the currentiL(0–) through L = 2H, and the voltage ν c(0–) across the capacitor C = 1 F at the switching instant areall assumed to be known.

SolutionAfter the switch is closed, the circuit is described by the loop equations

FIGURE 5.7 The circuit for Example 5.5.4.

H ss

s s

s

s s s s( ) =

+ +=

+( ) +( ) =+

−+2 5 4 4 1

4

3

1

4

1

3

1

1.

h t e e tt t( ) = − ≥− −4

3

1

304 , .

ν τ τ τ τ

τ τ

τ τ τ

τ τ

t h f t d e e e d

e e d d e e t

tt

tt t

t

t

( ) = ( ) −( ) = −

= −

=

−

−

= −

−∞

∞− −( ) − −

− − − −

∫ ∫

∫ ∫

0

4

0

3

0

3

0

4

3

1

3

4

3

1

3

4

3

1

3

1

3

4

9ee e te tt t t− − −+ − ≥4 4

9

1

30, .

32

12

12

32

0

2

11

22

2

11

22

2

2 2

idi

dti

di

dtt

idi

dti

di

dti dt

t i t

+

− +

= ( )

− +

+ + +

=

( ) = ( )∫

ν

ν .


All terms in these equations are Laplace transformed. The result is the set of equations

The current through the inductor is

i L(t ) = i 1(t ) – i 2(t ) .

At the instant t = 0+

i L(0+) = i 1 (0+) – i 2(0+) .

Also, because

then

The equation set is solved for I2(s), which is written by Cramer’s rule

3 2 1 2 2 0 0

1 2 3 21

2 0 00

2

1 2 1 1 2

1 2 1 2

2

2 2

+( ) ( ) − +( ) ( ) = ( ) + +( ) − +( )[ ]− +( ) ( ) + + +

( ) = − +( ) + +( )[ ] −+( )

( ) = ( )

s I s s I s V s i i

s I s ss

I s i iq

s

V s I s .

1 1

1 10 0

2 2

0 02

0

2

Cq t

Ci t dt

Ci t dt

Ci t dt

t

t

t

c

( ) = ( )

= ( ) + ( ) = + −( )−∞

→ + −∞

∫

∫ ∫lim ,ν

q

Ci

qc c

2

2

1 200 0 0

0

1

+( )= +( ) = ( ) = ( ) =

+( )−( )∆ ν ν – .

I s

s V s i

s is

s s

s ss

s iv

ss V s i

L

L

c

L

c

2

1

1

3 2 2 0

1 2 2 00

3 2 1 2

1 2 3 21

3 2 2 00

1 2 2

( ) =

+ ( ) + +( )− +( ) − +( ) −

+( )

+ − +( )− +( ) + +

=

+( ) − +( ) −+( )

+ +( ) ( ) +

ν

LL

c L

ss s

ss

s s si s s V s

s s

0

3 22 3 1

1 2

2 3 0 4 0 2

8 10 3

2 2

2 21

2

+( )[ ]+( ) + +

− +( )

=− +( ) +( ) − +( ) + +( ) ( )

+ +

ν.


Further

V 2(s ) = 2 I 2(s ) .

Then, upon taking the inverse transform

ν 1(t ) = 2� –1{ I 2(s )} .

If the circuit contains no stored energy at t = 0, then iL(0+) = ν c(0+) = 0 and now

For the particular case when ν1 = u(t) so that V1(s) = 1/s

The validity of this result is readily confirmed because at the instant t = 0+ the inductor behaves asan open circuit and the capacitor behaves as a short circuit. Thus, at this instant, the circuit appears astwo equal resistors in a simple series circuit and the voltage is shared equally.

Example 5.5.5

The input to the RL circuit shown in Figure 5.8a is the recurrent series of impulse functions shown inFigure 5.8b. Find the output current.

SolutionThe differential equation that characterizes the system is

FIGURE 5.8 (a) The circuit, (b) the input pulse train.

ν 21

21

22

2

8 10 3t

s s V s

s s( ) =

+( ) ( )+ +

−� .

ν 21

2

1

1 3 4

22 1

8 10 32

2 1

81

23 4

1

2

1

3

4

1

20

ts

s s

s

s s

s

e tt

( ) = ++ +

= +

+

+( )

=+

= ≥

− −

− −

� �

� , .


For zero initial current through the inductor, the Laplace transform of the equation is

(s + 1)I(s) = V(s) .

Now, from the fact that �{δ (t)} = 1 and the shifting property of Laplace transforms, we can write theexplicit form for V(s), which is

Thus, we must evaluate i(t) from

Expand these expressions into

The inverse transform of these expressions yields

i (t ) = 2e – tu (t ) + 2e – ( t–2)u (t – 2) + 2e –( t– 4 )u (t – 4) + L

+ e– (t–1) u (t – 1) + e –( t–3) u(t – 3) + e – (t–5) u(t – 5) + L

The result has been sketched in Figure 5.9.

5.6 The Inversion Integral

The discussion in Section 5.3 related the inverse Laplace transform to the direct Laplace transform bythe expressions

F (s ) = �{ f (t )} (6.1a)

f (t ) = �–1{ F(s )} . (6.1b)

The subsequent discussion indicated that the use of equation (6.1b) suggested that the f(t) so deducedwas unique; that there was no other f(t) that yielded the specified F(s). We found that although f(t)represents a real function of the positive real variable t, the transform F(s) can assume a complex variableform. What this means, of course, is that a mathematical form for the inverse Laplace transform was notessential for linear functions that satisfied the Dirichlet conditions. In some cases, Table 5.1 is not adequate

di t

dti t t

( )+ ( ) = ( )ν .

V s e e e e

e e e

e

e

s s s s

s s s

s

s

( ) = + + + + +

= +( ) + + +( )= +

−

− − − −

− − −

−

−

2 2 2

2 1

2

1

2 3 4

2 4

2

L

L

.

I se

e s e s

e

e s

s

s s

s

s( ) = +

− +=

−( ) +( ) +−( ) +( )

−

− −

−

−

2

1

1

1

2

1 1 1 12 2 2

.

I ss

e e es

e e e es s s s s s s( ) =+

+ + + +( ) ++

+ + + +( )− − − − − − −2

11

1

12 4 6 3 5 7L L .


for many functions when s is a complex variable and an analytic form for the inversion process of (6.1b)is required.

To deduce the complex inversion integral, we begin with the Cauchy second integral theorem, whichis written

where the contour encloses the singularity at s. The function F(s) is analytic in the half-plane Re(s) ≥ c .If we apply the inverse Laplace transformation to the function s on both sides of this equation, we can write

But F(s) is the Laplace transform of f(t); also, the inverse transform of 1/(s – z) is ez t. Then it follows that

(6.2)

This equation applies equally well to both the one-sided and the two-sided transforms.It was pointed out in Section 5.1 that the path of integration (6.2) is restricted to value of σ for which

the direct transform formula converges. In fact, for the two-sided Laplace transform, the region ofconvergence must be specified in order to determine uniquely the inverse transform. That is, for the two-sided transform, the regions of convergence for functions of time that are zero for t > 0, zero for t < 0,or in neither category, must be distinguished. For the one-sided transform, the region of convergence isgiven by σ , where σ is the abscissa of absolute convergence.

The path of integration in (6.2) is usually taken as shown in Figure 5.10 and consists of the straightline ABC displayed to the right of the origin by σ and extending in the limit from –j∞ to +j ∞ withconnecting semicircles. The evaluation of the integral usually proceeds by using the Cauchy integraltheorem, which specifies that

FIGURE 5.9 The response of the RL circuit to the pulse train.

F z

s zdz j F s

( )−

= ( )∫ 2π

j F s F zs z

dzj

j

211 1π

ω σ ω

σ ω� �−

→∞ −

+−( ){ } = ( ) −

∫lim .

f tj

e F z dzj

e F z dzz t

j

jz t

j

j

( ) = ( ) = ( )→∞ −

+

− ∞

+ ∞

∫ ∫1

2

1

2π πω σ ω

σ ω

σ

σlim .


(6.3)

But the contribution to the integral around the circular path with R → ∞ is zero, leaving the desiredintegral along the path ABC , and

(6.4)

We will present a number of examples involving these equations.

Example 5.6.1

Use the inversion integral to find f(t) for the function

Note that by entry 15 of Table 5.1, this is sin wt /w.

SolutionThe inversion integral is written in a form that shows the poles of the integrand.

The path chosen is Γ1 in Figure 5.10. Evaluate the residues

FIGURE 5.10 The path of integration in the s-plane.

f tj

F s e ds

F s e ABC t

R

s t

s t

( ) = ( )

= ( ) >

→∞∫∑

1

2

0

1πlim

; .

Γ

residues of at the singularities to the left of

f tj

F s e ds

F s e ABC t

R

s t

s t

( ) = ( )

= − ( ) <

→∞∫∑

1

2

0

2πlim

; .

Γ

residues of at the singularities to the right of

F ss w

( ) =+1

2 2.

f tj

e

s jw s jwds

s t

( ) =+( ) −( )∫1

2π.


Therefore,

Example 5.6.2

Evaluate �–1{1/ }.

Solution

The function F(s) = 1/ is a double-valued function because of the square root operation. That is, ifs is represented in polar form by re j θ, then re j (θ + 2 π ) is a second acceptable representation,

and , thus showing two different values for . But a double-valued functionis not analytic and requires a special procedure in its solution.

The procedure is to make the function analytic by restricting the angle of s to the range –π < θ < πand by excluding the point s = 0. This is done by constructing a branch cut along the negative real axis,as shown in Figure 5.11. The end of the branch cut, which is the origin in this case, is called a branchpoint. Because a branch cut can never be crossed, this essentially ensures that F(s) is single valued. Now,however, the inversion integral (6.3) becomes for t > 0

FIGURE 5.11 The integration contour for .

Res –

Res +

s jwe

s w

e

s jw

e

wj

s jwe

s w

e

s jw

e

wj

s t

s j w

s t

s j w

j w t

s t

s j w

s t

s j w

j w t

( )+

=+

=

( )+

=−

=−

= =

=− =−

−

2 2

2 2

2

2.

f te e

jw

wt

w

j w t j w t

( ) = = − =∑−

Res2

sin.

s

s

s re rej j= = −+( )θ π θ2

s

�− { }1

1 s


(6.5)

which does not include any singularity.First we will show that for t > 0 the integrals over the contours BC and CD vanish as R → ∞, from

which = 0. Note from Figure 5.11 that β = cos–1(σ/R) so that the integral over the

arc BC is, because �e jθ � = 1,

But for small arguments sin–1(σ /R) = σ /R, and in the limit as R → ∞, I → 0. By a similar approach, wefind that the integral over CD is zero. Thus, the integrals over the contours Γ2 and Γ3 are also zero asR → ∞.

For evaluating the integral over γ , let s = r e jθ = r(cos θ + j sin θ ) and

The remaining integrals in (6.5) are written

(6.6)

Along path l–, let s = uejπ = –u ; = j , and ds = –du , where u and are real positive quantities.Then

Along path l +, s = –uej2π = –u, = –j (not + j ), and ds = –du. Then

Combine these results to find

f tj

F s e dsj

F s e ds

j

R G A B

s t

j

js t

BC F G

( ) = ( ) = ( )

= − + + + + + +

→∞ − ∞

+ ∞

− +

∫ ∫

∫ ∫ ∫ ∫ ∫ ∫ ∫

lim

,

1

2

1

2

1

2 2 3

π π

π

σ

σ

γΓ Γl l

Γ ΓΓ

2 3∫ ∫ ∫= = =

BCFG

Ie e

R e

j d e R d e RR

e RR

BC

t j t

j

j t t

t

≤ = = −

=

∫ ∫ −

−

σ ω

θ

θ σ

β

πσ

σ

θ θ π σ

σ

1

2 2

21

1

2Re cos

sin

γ π

π θ θ

θθ θ∫ ∫( ) = = →

−

+( )F s e ds

e

rejre d rs t

r j t

j

jcos sin

.2

0 0as

f tj

F s e ds F s e dss t s t( ) = − ( ) + ( )

− +∫ ∫1

2π l l

.

s u u

l− ∞

− ∞ −

∫ ∫ ∫( ) = − =F s e dse

j udu

j

e

j udus t

u t u t0

0

1.

s u u

l+

∞ − ∞ −

∫ ∫ ∫( ) = −−

=F s e dse

j udu

j

e

j udus t

u t u t

0 0

1.


which is a standard form integral with the value

Example 5.6.3


Solution

The integrand in the inversion integral possesses simple poles at: s = 0 and s = jnπ , n = ±1,

±3, ±L (odd values). These are illustrated in Figure 5.12. We see that the function est/s(1 + e–s) is analyticin the s-plane except at the simple poles at s = 0 and s = jnπ . Hence, the integral is specified in termsof the residues in the various poles. We have, specifically

(6.7)

FIGURE 5.12 The pole distribution of the given function.

f tj j

u e du u e duu t u t( ) = −

=∞ − −

∞ − −∫ ∫1

2

2 1

0

1

2

0

1

2

π π,

f tt t

t( ) = = >1 10

ππ

π, .

F ss e s

( ) =+( )−

1

1.

e

s e

s t

s1 + −( )

Res for

Res for

se

s es

s jn e

s es jn

s t

s

s

s t

s

s jn

1

1

20

1

0

0

0

+( )

= =

−( )+( )

= =

−

=

−

=

π .


The problem we now face in this evaluation is that

where the roots of d(s) are such that s = a cannot be factored. However, we know from complex functiontheory that

because d(a) = 0. Combine this result with the above equation to obtain

(6.8)

By combining (6.8) with (6.7), we obtain

We obtain, by adding all of the residues,

This can be rewritten as follows

This assumes the form

(6.9)

Res s an s

d ss a

−( ) ( )( )

=

=

0

0

d d s

ds

d s d a

s a

d s

s as a

s a s a

( )[ ]=

( ) − ( )−

=( )−

=

→ →lim lim

Res s an s

d s

n s

d

dsd ss a

s a

−( ) ( )( )

=

( )( )[ ]=

=

.

Res odd .e

sd

dse

e

jnn

s t

s

s jn

jn t

1+( )

=−

= π

π

π

f te

jn

jn t

n

( ) = +=−∞

∞

∑1

2

π

π.

f te

j

e

j

e

j

e

j

j n t

jn

j t j t j t j t

nn

( ) = + +−

+−

+ + +

= +

− −

=

∞

∑

1

2 3 3

1

2

2

3 3

1

L Lπ π π π

π π π π

ππ

sin.

odd

f tk t

kk

( ) = +−( )−

=

∞

∑1

2

2 2 1

2 11

ππsin

.


As a second approach to a solution to this problem, we will show the details in carrying out the contourintegration for this problem. We choose the path shown in Figure 5.12 that includes semicircular hooksaround each pole, the vertical connecting line from hook to hook, and the semicircular path at R → ∞.Thus, we examine

(6.10)

We consider the several integrals in this equation.Integral I1. By setting s = re j θ and taking into consideration that cos θ = –cos θ for θ > π /2, the

integral I1 → 0 as r → ∞.Integral I2. Along the Y-axis, s = jy and

Note that the integrand is an odd function, whence I2 = 0.Integral I3. Consider a typical hook at s = jnπ . The result is

This expression is evaluated (as for (6.7)) and yields e jnπ t/ jnπ . Thus, for all poles

Finally, the residues enclosed within the contour are

which is seen to be twice the value around the hooks. Then when all terms are included in (6.10), thefinal result is

f tj

e

s eds

j

s t

s

IBC A

I

( ) =+( )

= + + −

−∫

∫ ∫ ∑∑∫

1

2 1

1

22 31

π

π vertical connecting lines HooksI

Res .

I je

jy edy

j y t

j yr

20 1

=+( )−−∞

→

∞

∫ .

lim .r

s jn

s t

s

s jn e

s e→→

−

−( )+( )

=0 1

0

0π

Ij

e

s eds

j

j

e

jn

n t

n

s t

s

jn t

nn

nn

3

2

2

1

1

2 1 2

1

2

1

2

1

2

2=+( ) = +

= +

−− =−∞

∞

=

∞

∫ ∑ ∑πππ π π

ππ

π π

odd odd

sin.

Res

odd odd

e

s e

e

jn

n t

n

s t

s

jn t

nn

nn

1

1

2

1

2

2

1+( ) = + = +−

=−∞

∞

=

∞

∑ ∑π

π ππsin

,

f tn t

n

k t

kn

nk

( ) = + = +−( )−

=

∞

=

∞

∑ ∑1

2

2 1

2

2 2 1

2 11 1

ππ

ππsin sin

.

odd


We now shall show that the direct and inverse transforms specified by (4.1) and listed in Table 5.1constitute unique pairs. In this connection, we see that (6.2) can be considered as proof of the followingtheorem:

Theorem 5.6.1Let F(s) be a function of a complex variable s that is analytic and of order O(s–k) in the half-plane Re(s)≥ c , where c and k are real constants, with k > 1. The inversion integral (6.2) written �t

–1{F(s)} alongany line x = σ , with σ ≥ c converges to the function f(t) that is independent of σ ,

whose Laplace transform is F(s),

F (s ) = �{ f (t )}, Re(s ) ≥ c .

In addition, the function f(t) is continuous for t > 0 and f(0) = 0, and f(t) is of the order O(ec t) for all t > 0.Suppose that there are two transformable functions f1(t) and f2(t) that have the same transforms

�{ f 1(t )} = �{ f 2 (t )} = F (s ) .

The difference between the two functions is written φ (t)

φ (t ) = f 1(t ) – f 2(t )

where φ(t) is a transformable function. Thus,

�{ φ (t )} = F (s ) – F (s ) = 0 .

Additionally,

Therefore, this requires that f1(t) = f2(t). The result shows that it is not possible to find two differentfunctions by using two different values of σ in the inversion integral. This conclusion can be expressedas follows:

Theorem 5.6.2Only a single function f(t) that is sectionally continuous, of exponential order, and with a mean valueat each point of discontinuity, corresponds to a given transform F(s).

5.7 Applications to Partial Differential Equations

The Laplace transformations can be very useful in the solution of partial differential equations. A basicclass of partial differential equations is applicable to a wide range of problems. However, the form of thesolution in a given case is critically dependent on the boundary conditions that apply in any particularcase. In consequence, the steps in the solution often will call on many different mathematical techniques.Generally, in such problems the resulting inverse transforms of more complicated functions of s occurthan those for most linear systems problems. Often the inversion integral is useful in the solution of suchproblems. The following examples will demonstrate the approach to typical problems.

f t F st( ) = ( ){ }−� 1

φ t tt( ) = { } = >−� 1 0 0 0, .


Example 5.7.1

Solve the typical heat conduction equation

(7.1)

subject to the conditions

C-1. ϕ (x, 0) = f(x), t = 0

C-2. =0, ϕ (x, t) = 0 x = 0.

SolutionMultiply both sides of (7.1) by e–s x dx and integrate from 0 to ∞.

Also

Equation (7.1) thus transforms, subject to C-2 and zero boundary conditions, to

The solution to this equation is

Φ = Ae s 2 t .

By an application of condition C-1, in transformed form, we have

The solution, subject to C-1, is then

Now apply the inversion integral to write the function in terms of x from s,

∂∂

= ∂∂

< < ∞ ≥2

20 0

ϕ ϕx t

x t, ,

∂∂ϕx

Φ s t e x t dxs x, , .( ) = ( )∞

−∫0

ϕ

0

2

2

2 0 0∞

−∫ ∂∂

= ( ) − +( ) − ∂∂

+( )ϕ ϕ ϕx

e dx s s t sx

s x Φ , .

d

dts

Φ Φ− =2 0 .

Φ = = ( )∞

−∫A f e ds

0

λ λ .λ

Φ s t e f e ds t s,( ) = ( )+∞

−∫2

0

λ λ .λ

ϕπ

π

x tj

e f e d e ds

jf d e ds

s t s s x

s t s sx

,

.

( ) = ( )

= ( )

−∞

∞+

∞−

−∞

∞ ∞− +

∫ ∫

∫ ∫

1

2

1

2

2

2

0

0

λ λ

λ λ

λ

λ


Note that we can write

Also write

Then

But the integral

Thus, the final solution is

Example 5.7.2

A semi-infinite medium, initially at temperature ϕ = 0 throughout the medium, has the face x = 0maintained at temperature ϕ0. Determine the temperature at any point of the medium at any subsequenttime.

SolutionThe controlling equation for this problem is

(7.2)

with the boundary conditions:

a. ϕ = ϕ 0 at x = 0, t > 0b. ϕ = 0 at t = 0, x > 0.

To proceed, multiply both sides of equation (7.2) by e–s t dt and integrate from 0 to ∞. The transformedform of equation (7.2) is

s t s x s tx

t

x

t2

2 2

2 4− −( ) = −

−( )

−

−( )λ

λ λ.

s tx

tu−

−( )=

λ

2.

ϕπ

x tj

fx

td e

du

t

u, exp .( ) = ( ) −−( )

−∞

∞ ∞−∫ ∫1

2 4

2

2

λλ

λ0

0

∞−∫ =e duu2

π .

ϕπ

x tt

f e d

x

t,( ) = ( )−∞

∞ −−( )

∫1

2

2

4λ λ .λ

∂∂

= ∂∂

2

2

1ϕ ϕx K t

d

dx

s

KK

2

20 0

Φ Φ– , .

= >


The solution of this differential equation is

But Φ must be finite or zero for infinite x ; therefore, B = 0 and

Apply boundary condition (a) in transformed form, namely

Therefore,

and the solution in Laplace transformed form is

(7.3)

To find ϕ (x, t) requires that we find the inverse transform of this expression. This requires evaluatingthe inversion integral

(7.4)

This integral has a branch point at the origin (see Figure 5.13). To carry out the integration, we selecta path such as that shown (see also Figure 5.11). The integral in (7.4) is written

As in Example 5.6.2

For the segments

Φ = +−Ae Bex s K x s K .

Φ s x Aes

Kx

, .( ) =−

Φ 0 00

00, .s e dt

sxs t( ) = = =

∞−∫ ϕ

ϕfor

As

=ϕ 0

Φ s xs

es

Kx

, .( ) =−ϕ 0

ϕϕπ σ

σx t

j

e e

sds

xs

K s t

j

j

, .( ) =−

− ∞

+ ∞

∫0

2

ϕϕπ γ

x tj BC l l FG

, .( ) = + + + + + +

∫ ∫ ∫ ∫ ∫ ∫ ∫− +

0

2 2 3Γ Γ

Γ Γ2 3

0∫ ∫ ∫ ∫= = = =BC FG

.

l l− +∫ ∫= =, , .let and for let s e s ej jρ ρπ π


Then for �– and �+, writing this sum I� ,

Write

Then we have

This is a known integral that can be written

Finally, consider the integral over the hook,

Let us write

FIGURE 5.13 The path of integration.

Ij

e e eds

se x

s

K

ds

ss t jx s K jx s K s t

l = −

= −∞

− −∞

−∫ ∫1

2

1

0 0π πsin .

us

Ks ku ds kudu= = =2 2, .

I e uxdu

uKu t

l = −∞

−∫2

0

2

πsin .

I e dulu

x

Kt= − −∫2 2

0

2

π.

Ij

ee

sdsy

s tx s K

= ∫1

2π γ.

s re ds jre dds

sjj j= = =θ θ θ θ, , ,


then

For r → 0, I γ = , then I γ = 1. Hence, the sum of the integrals in (7.3) becomes

(7.5)

Example 5.7.3

A finite medium of length l is at initial temperature ϕ0. There is no heat flow across the boundary at x= 0, and the face at x = l is then kept at ϕ 1 (see Figure 5.14). Determine the temperature ϕ (t).

SolutionHere we have to solve

subject to the boundary conditions:

a. ϕ = ϕ0 t = 0 0 ≤ x ≤ lb. ϕ = ϕ1 t > 0 x = l

c. = 0 t > 0 x = 0.

Upon Laplace transforming the controlling differential equation, we obtain

FIGURE 5.14 Details for Example 5.7.3.

Ij

je e dt r e x r K ej j

γ πθ

θ θ= ∫2

2

.

j

j

j

j

2

2

2

2

ππ

ππ

=

ϕ ϕπ

ϕt e dux

Kt

u

x

K t( ) = =

−

−∫0

0

201

21

2

2

– .erf

∂∂

= ∂∂

2

2

1ϕ ϕx k t

∂∂ϕx

d

dx

s

k

2

20

Φ Φ− = .


The solution is

By condition c

This imposes the requirement that B = 0, so that

Now condition b is imposed. This requires that

Thus, by b and c

Now, to satisfy c we have

Thus, the final form of the Laplace transformed equation that satisfies all conditions of the problem is

To find the expression for ϕ (x, t), we must invert this expression. That is,

(7.6)

Φ = ′ + ′ = +−

A e B e A xs

kB x

s

k

xs

kx

s

k cosh sinh .

d

dxx t

Φ = = >0 0 0 .

Φ = A xs

kcosh .

ϕ1

sA l

s

k= cosh .

Φ =ϕ1

cosh

cosh

.x

s

k

s ls

k

Φ = −ϕ ϕ0 0

s s

xs

k

ls

k

cosh

cosh

.

Φ = +−ϕ ϕ ϕ0 1 0

s s

xs

k

ls

k

cosh

cosh

.

ϕ ϕϕ ϕ

π σ

σx t

je

xs

k

ls

k

ds

ss t

j

j

,cosh

cosh

.( ) = +−

− ∞

+ ∞

∫01 0

2


The integrand is a single valued function of s with poles at s = 0 and s = , n = 1, 2, … .

We select the path of integration that is shown in Figure 5.15. But the inversion integral over the pathBCA(=Γ ) = 0. Thus, the inversion integral becomes

By an application of the Cauchy integral theorem, we require the residues of the integrand at its poles.There results

Res� s = 0 = 1

Combine these with (7.5) to write finally

(7.7)

FIGURE 5.15 The path of integration for Example 5.7.3.

−−

kn

l

2 1

2

2 2

2

π

1

2π σ

σ

je

xs

k

ls

k

ds

sst

j

j

− ∞

+ ∞

∫cosh

cosh

.

Re

cosh

cosh

.s

e j nx

l

sd

dsl

s

k

s k nl

k nl

s k nl

=− −

− −

=− −

=−

1

2

1

2

1

2

2 2

2

2 2

2

2 2

2

1

2π

π

π

π

ϕ ϕϕ ϕ

ππ

πx t

nl n x l

n

n

k n l

, cos .( ) = +−( ) −( )

−−

=

∞ − −

∑0

1 0

1

1

24 1

2 1

1

2

22 2


Example 5.7.4

A circular cylinder of radius a is initially at temperature zero. The surface is then maintained at temper-ature ϕ 0. Determine the temperature of the cylinder at any subsequent time t.

SolutionThe heat conduction equation in radial form is

(7.8)

And for this problem the system is subject to the boundary conditions

C-1. ϕ = 0 t = 0 0 ≤ r < aC-2. ϕ = ϕ0 t > 0 r = a.

To proceed, we multiply each term in the partial differential equation by e–stdt and integrate. We write

Then (7.7) transforms to

which we write in the form

This is the Bessel equation of order 0 and has the solution

Φ = A I 0( µr ) + BN 0( µr ) .

However, the Laplace transformed form of C-1 when z = 0 imposes the condition B = 0 because N0(0)is not zero. Thus,

Φ = A I 0( µr ) .

The boundary condition C-2 requires Φ(r , a) = when r = a , hence,

so that

∂∂

+ ∂∂

= ∂∂

≤ < >2

2

1 10 0

ϕ ϕ ϕr r r k t

r a t, , .

0

∞−∫ = ( )ϕ e dt r ss t Φ ,

kd

dr r

d

drs

2

2

10

Φ Φ Φ+

− = ,

d

dr r

d

dr

s

k

2

2

10

Φ Φ Φ+ − = =µ µ, .

ϕ0

s

As I a

= ( )ϕ

µ0 1


To find the function ϕ(r, t) requires that we invert this function. By an application of the inversionintegral, we write

(7.9)

Note that I0(ξ r)/I0(ξ a) is a single-valued function of λ. To evaluate this integral, we choose as the pathfor this integration that shown in Figure 5.16. The poles of this function are at λ = 0 and at the roots ofthe Bessel function J0(ξ a) (= I0(jξ a)); these occur when J0(ξ a) = 0, with the roots for J0(ξ a) = 0, namelyλ = , … . The approximations for I0(ξ r) and I0(ξ a) show that when n → ∞ the integral overthe path BCA tends to zero. The resultant value of the integral is written in terms of the residues at zeroand when λ = . These are

Therefore,


Φ =( )( )

ϕ µ

µ0 0

0s

I r

I a.

ϕϕπ

ξ

ξξ

σ

σr t

je

I r

I a

d

kt

j

j

, , .( ) =( )( ) =

− ∞

+ ∞

∫0 0

02

λ λλ

λ

− −k kξ ξ1

2

2

2,

kn

ξ 2

Res

Res

==

=( )

0

0

1

2

2

k

k

n

n

dI a

dξ

ξ

ξλλ

.

ϕ ϕξ

ξ

ξ

λ ξ

r t eJ r

d

dI a

k t

n

n

k

n

n

, .( ) = +( )( )

−

=

∑0

0

0

12

2

λλ


Further, . Hence, finally,

(7.10)

Example 5.7.5

A semi-infinite stretched string is fixed at each end. It is given an initial transverse displacement andthen released. Determine the subsequent motion of the string.

SolutionThis requires solving the wave equation

(7.11)

subject to the conditions

C-1. ϕ (x, 0) = f(x) t = 0, ϕ (0, t) = 0 t > 0C-2. limx→∞ ϕ (x, t) = 0.

To proceed, multiply both sides of (7.11) by e–s tdt and integrate. The result is the Laplace-transformedequation

(7.12)

C-1. Φ(0, s) = 0C-2. limx→∞ Φ(x, s) = 0.

To solve (7.12) we will carry out a second Laplace transform, but this with respect to x, that is �{Φ(x, s)} = N(z, s). Thus,

Apply this transformation to both members of (7.12) subject to Φ(0, s) = 0. The result is

We denote (0, s) by C. Then the solution of this equation is

λλd

dI a aI a

n0 0

11

2ξ ξ ξ( ) ( )= ( )

ϕ ϕξ

ξ ξξt

ae

J r

J a

k t

n

n

n n

n( ) = +( )

( )

−

=

∞

( )∑0

1

0

0

11

2 2

.

ax t

22

2

2

2

∂∂

= ∂∂

ϕ ϕ

ad

dxs s x2

2

2

2 0 0Φ Φ= − +( ) >ϕ , .

N z s x s e dxz x, , .( ) = ( )∞

−∫0

Φ

s N z s s z a z N z sx

s z2 2 200, , , , .( ) − ( ) = ( ) − ∂

∂ ( )

( ) = { }Φ Φ Φ � ϕ

∂∂Φx

N z sC

zs

a

s

az

zs

a

, .( ) =−

− ( )−2

2

2

22

2

2

1Φ


The inverse transformation with respect to z is, employing convolution.

To satisfy the condition limx→∞ Φ(x, s) = 0 requires that the sinh terms be replaced by their exponentialforms. Thus, the factors

Then we have the expression

But for this function to be zero for x → ∞ requires that

Combine this result with Φ(x, s) to get

Each integral in this expression is integrated by parts. Here we write

The resulting integrations lead to

We note by entry 61, Table 5.1 that

This function of ξ vanishes except when ξ ≤ x + at. Thus,

Φ x saC

s

sx

a a

s

ax d

x

, sinh sinh .( ) = − ( ) −( )∫1

0

ϕ ξ ξ ξ

sinh , sinh , .sx

a

s

ax

ex

s a

→ −( ) → → ∞−1

2 2ξ

ξ

Φ x saC

s ae d

xs a, .( ) = − ( )∫ −

2

1

2 0

ϕ ξ ξξ

aC

s ae d xs a= ( ) → ∞

∞−∫1

0

ϕ ξ ξξ , .

20 0 0

a x s e d e d e ds x a s x a

xs x aΦ , .( ) = ( ) − ( ) + ( )

∞− −( ) ∞

− +( ) − −( )∫ ∫ ∫ϕ ξ ξ ϕ ξ ξ ϕ ξ ξξ ξ ξ

u du d d e da

se

s x

a s k a= ( ) = ( ) = =( ) −−( )

− −( )ϕ ξ ϕ ξ ξ ν ξ νξ

ξ, ; , .

1

Φ x ss

xs

e ds

e d

se d

x

s x

a

s x

a

s x

a

,

.

( ) = ( ) + ( ) − ( )

− ( )

∞ ( ) −−( ) ∞ ( ) −

+( )

∞ ( ) −−( )

∫ ∫

∫

1 1

2

1

2

1

2

1

0

1

0

1

ϕ ϕ ξ ξ ϕ ξ ξ

ϕ ξ ξ

ξ ξ

ξ

�−− −( )

= > −

= < −

1 11

0

se at x

at x

s x

a

ξ

ξ

ξ

when

when .


Proceed in the same way for the term

Thus, the second term becomes

The final term becomes

The final result is

(7.13)

Example 5.7.6

A stretched string of length l is fixed at each end as shown in Figure 5.17. It is plucked at the midpointand then released at t = 0. The displacement is b. Find the subsequent motion.

SolutionThis problem requires the solution of

(7.14)

subject to

FIGURE 5.17 A stretched string plucked at its midpoint.

�−∞ ( ) − +( ) + ( )∫ ∫( )

= ( ) = +( ) − ( )1 1 11

2

1

2

1

2

1

2x

s x a

x

x at

e d d x at xϕ ξ ξ ϕ ξ ξ ϕ ϕξ.

�−∞ ( ) − +( )∫ ( )

= > +

= < +

1

0

11

21

0

ϕ ξ ξ ξ

ξ

ξe d at x

at x

s x awhen

when .

�−∞ ( ) − +( ) + ( )∫ ∫( )

= ( ) = −( )1 1 11

2

1

2

1

2x

s x a

x

x at

e d d x atϕ ξ ξ ϕ ξ ξ ϕξ.

− ( ) >

− ( ) + −( ) <

1

21

2

1

2

ϕ

ϕ ϕ

x at x

x x at at x

when

when .

ϕ x t f at x f at x t x a

f x at f x at t x a

,

.

( ) = +( ) − −( )[ ] >

= +( ) + −( )[ ] <

1

2

1

2

when

when

∂∂

= ∂∂

< < >2

2

22

20 0

y

yc

y

ty l t, ,


To proceed, multiply (7.13) by e–s tdt and integrate in t . This yields

or

(7.15)

subject to Y(0, s) = Y(l , s) = 0. To solve this equation, we proceed as in Example 5.7.5; that is, we applya transformation on x, namely �{Y(x, s)} = N(z, s). Thus,

This equation yields, writing sY(0) as Φ(x, s),

The inverse transform is

where

Combine these integrals with the known form of f(x) in C-1 and C-2. Upon carrying out the integrations,

the resulting forms become, with k = ,

12

0 2

22 1

2

0

3 0 0 0

4 0 0 0

.

.

.

. ; .

ybx

lx l

yb

ll x x l

t

y

tx l t

y x x l t

= < <

= −( ) < <

=

∂∂

= < < =

= = = >

s Y sy cd Y

dx2 2

2

20− ( ) =

cd Y

dxs Y sy s f x2

2

2

2 0− = ( ) = − ( )

s N z s sY c z N z sy s

x2 2 20

0, ,

,.( ) − ( ) = ( ) −

( )

N z sz s

zs

c

,,( ) =

( )−

Φ

22

2

Y x s N x sx s

cs

c

, ,,

sinh( ) = ( ){ } =

( )−� 1Φ

Φ x sl x s

cy

s

cd

xs

cy

l s

cd

x l

, sinh sinh sinh sinh .( ) =−( ) ( ) + ( ) −( )∫ ∫0 0

ξ ξ ξ ξξ

ξ

s

c


To find y(t), we must invert these expressions. Note that symmetry exists and so we need consider only

the first term. We use the inversion integral on the term . Thus, we consider the integral

We choose the path in the λ-plane as shown in Figure 5.18. The value of the integral over path Γ iszero. Thus, the value of the integral is given in terms of the residues. These occur at λ = 0 and at the

values for which cosh = 0, which exist where

Thus, we have, by the theory of residues


lY

b

x

s

c

s

kx

klx l

lY

b

l x

s

c k l x

skl

lx l

2

2

0 2

2

2

2

2

2

= − ≤ ≤

=−( )

−−( )

≤ ≤

sinh

cosh

,

sinh

cosh

, .

1

2

2s

kx

kl

sinh

cosh

Ij

e

x

cx

c

db

j

j

=− ∞

+ ∞

∫1

2

2

π σ

σλ

2

λ

λλ

λ

sinh

cosh

.

λ2c

λ λ =2

2 1

2

2 1

cj

n

lj

n

lc= − ± −π πor .


These poles lead to

Thus, the poles at ±j(2n – 1) lead to

Then we have

so that finally

For the string for which ≤ x < l, the corresponding expression is the same except that (l – x) replaces x.

Note that this equation can be written, with η = (2n – 1) ,

which shows the traveling wave nature of the solution.

Res

Res

λ=0

λλ λ

=

=−( )

−−( )

−( )

x

c

ecx

l

j nx

ld

d

l

c

jn

lc

j n

jn

lc

2 12 1

2

2 1

2 1

2

π

π

ππ

sinh

cosh

.

= −( )−( )−( )

−( )1

2 2 1

2 12 2

2 1n j nc

ltl

c

nx

l

ne

π

ππsin

.

πc

l

= −( )−( ) −( )

−( )1

4 2 1 2 1

2 12 2

n l

c

nx

ln

c

lt

nπ

π πsin cos

.

ly

bx x

l

nn

x

ln

ct

l

n

n2

4 1

2 12 1 2 1

2 21

= − +−( )

−( )−( ) −( )

=

∞

∑ππ π

sin cos

yb

nn

x

ln

ct

lx

ln

n

=−( )

−( )−( ) −( ) ≤ ≤

−

=

∞

∑8 1

2 12 1 2 1 0

22

1

21

ππ π

sin cos , .

l

2

πl

sin cossin sin

,η ηη η

x ctx ct x ct

=−( ) + +( )

2


5.8 The Bilateral or Two-Sided Laplace Transform

In Section 5.1 we discussed the fact that the region of absolute convergence of the unilateral or one-sidedLaplace transform is the region to the left of the abscissa of convergence. The situation for the two-sidedLaplace transform is rather different; the region of convergence must be specified if we wish to invert afunction F(s) that was obtained using the bilateral Laplace transform. This requirement is necessary becausedifferent time signals might have the same Laplace transform but different regions of absolute convergence.

To establish the region of convergence, write the bilateral Laplace transform in the form

(8.1)

If the function f(t) is of exponential order (eσ1t), then the region of convergence for t > 0 is Re(s) > σ 1.If the function f(t) for t < 0 is of exponential order exp(σ2t), then the region of convergence is Re(s) <σ2. Hence, the function F2(s) exists and its analytic in the vertical strip defined by σ 1 < Re(s) < σ 2 ,provided, of course, that σ 1 < σ 2. If σ 2 > σ 1 , no region of convergence would exist and the inversionprocess could not be performed. This region of convergence is shown in Figure 5.19.

Example 5.8.1

Find the bilateral Laplace transform of the signals f(t) = e–atu(t) and f(t) = –e–atu(t) and specify theirregions of convergence.

SolutionUsing the basic definition of the transform (8.1), we obtain

and its region of convergence is Re(s) > –a.For the second signal,

FIGURE 5.19 Region of convergence for the bilateral transform.

F s e f t dt e f t dt e f t dts t s t s t2

0

0

( ) = ( ) = ( ) + ( )−∞

∞−

∞−

−∞

−∫ ∫ ∫ .

F s e u t e dt e dts a

at s t s a t

20

1( ) = ( ) = =+−∞

∞− −

∞− +( )∫ ∫

F s e u t e dt e dts a

at s t s a t

2

0 1( ) = − −( ) = − =+−∞

∞− −

−∞

− +( )∫ ∫


and its region of convergence is Re(s) < –a.Clearly, the knowledge of the region of convergence is necessary to find the time functions unambiguously.

Example 5.8.2

Find the function f(t), if its Laplace transform is given by

SolutionThe region of convergence and the paths of integration are shown in Figure 5.20. For t > 0 we close thecontour to the left and we obtain

For t < 0, the contour closes to the right and now

Appendix

FIGURE 5.20 Illustrating Example 5.8.2.

F ss s s

s2

3

4 1 22 1( ) =

−( ) +( ) +( ) − < ( ) < −, Re .

f te

s se t

st

s

t( ) =−( ) +( ) = >

=−

−3

4 1

1

20

2

2 , .

f te

s s

e

s se

et

st

s

st

s

tt

( ) =−( ) +( ) +

+( ) +( ) = − + <=− =

−3

4 2

3

1 2

3

5 100

1 4

4

, .


TABLE 5.1 Laplace Transform Pairs


















Several additional transforms, especially those involving other Bessel functions, can be found in thefollowing sources:“Fourier Integrals for Practical Applications,” G. A. Campbell and R. M. Foster, Van Nostrand, 1948. Inthese tables, only those entries containing the condition 0 < g or k < g, where g is our t, are Laplace transforms.“Formulaire pour le calcul symbolique,” N. W. McLachlan and P. Humbert, Gauthier–Villars, Paris, 1947.“Tables of Integral Transforms,” Bateman Manuscript Project, California Institute of Technology, A.Erdélyi and W. Magnus, Eds., McGraw-Hill, 1954; based on notes left by Harry Bateman.

TABLE 5.2 Properties of Laplace Transforms


References

1. R. V. Churchill, Modern Operational Mathematics in Engineering, McGraw-Hill, New York, 1944.2. J. Irving and N. Mullineux, Mathematics in Physics and Engineering, Academic Press, New York,

1959.3. H. S. Carslaw and J. C. Jaeger, Operational Methods in Applied Mathematics, Dover Publications,

Dover, NH, 1963.4. W. R. LePage, Complex Variables and the Laplace Transform for Engineers, McGraw-Hill, New York,

1961.5. R. E. Bolz and G. L. Turve, Eds., CRC Handbook of Tables for Applied Engineering Science, 2nd ed.,

CRC Press, Boca Raton, FL, 1973.6. A. D. Poularikas and S. Seeley, Signals and Systems, corrected 2nd ed. Krieger Publishing Co.,

Melbourne, FL, 1994.


Documents

Chapter 5 - Laplace Transforms