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Chapter 5 Laplace Transform 5.1 IntroductionPowerPoint Presentation

Chapter 5 Laplace Transform 5.1 Introduction

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Chapter 5 Laplace Transform 5.1 Introduction

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The Laplace transform is an example of an integral

transform, namely, a relation of the form

Chapter 5 Laplace Transform

5.1 Introduction

(1)

which transforms a given function f(t) into another function

F(s); K(t,s) is called the kernel of the transform, and F(s)

is known as the transform of f(t). The most well known

Integral transform is the Laplace transform, where a = 0,

b =∞, and K(t,s) = e-st. In that case Eq. (1) takes the form

(2)

The Laplace transform of f(t) is also denoted as L{f(t)} or as .

The basic idea behind any transform is that the given problem

can be solved more readily in the “transform domain”. In this

chapter, we use t as the independent variable and consider the

interval 0 ≦ t <∞ because in most applications of the Laplace

transform the independent variable is the time t, with 0 ≦ t <∞.

5.2 Calculation of the Transform

5.3Properties of the transform

5.4 Application to the solution of differential equations

5.5 Discontinuous forcing functions; Heaviside step function

5.6 Impulsive forcing function; Diarac impulse function

5.7 Additional properties

5.2 Calculus of the transform

The first question to address is whether the transform F(s) of

a given function f(t) exist – that is, whether the integral

(1)

converge.

Exponential order: We say that f(t) is one of exponential order

as t → ∞ if there exist real constant K, c, and T such that

(2)

For all t≧T. That is, the set of functions of exponential order is

the set of functions that do not grow faster than exponentially,

which included most functions of engineering interest.

Example 1

Example 2

Example 3

Piecewise continuous: We say that f(t) is piecewise

continuous on a ≦ t ≦ b if there exist a finite number of

points t1, t2,…, tn such that f(t) is continuous on each open

subinterval a < t <t1, t1 < t < t2,…, tN < t < b, and has a finite

limit as t approaches end point from the interior of that

subinterval.

Theorem 5.2.1 Existence of the Laplace Transform

Let f(t) satisfy these conditions: (i) f(t) is

piecewise continuous on 0 ≦ t ≦ A, for

every A > 0, and (ii) f(t) is of exponential

order as t → ∞, so that there exist real

constant K, c, and T such that

for all t ≧ T. Then the Laplace transform of f(t),

namely, F(s) given by Eq. (1) exists for all s > c.

Proof:

Example 4 Find the Laplace transform of f(t) = 1

Example 5 Find the Laplace transform of f(t) = eat

Example 6 Find the Laplace transform of f(t) = sinat

Example 7 Find the Laplace transform of f(t) =

Example 8 Find the Laplace transform of f(t) =sinhat

For more information on the Laplace transform, please see

Laplace transform table in Appendix C. (p. 1271)

Define Laplace transform operator L and inverse transform

operator L-1 as:

(14)

and

(15)

Where γ is a sufficiently positive real number.

Theorem 5.3 Properties of the transform

Theorem 5.3.1 Linearity of the transform

Let u(t) and v(t) are any two functions such that the transforms

L{u(t)} and L{v(t)} both exist, then

(1)

for any constants a, b.

Example 1

Theorem 5.3.2 Linearity of the inverse transform

Let U(s) and V(s) such that the inverse transforms

L-1{U(s)} and L-1{V(s)} exist, then

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for any constants a, b.

Example 2

Theorem 5.3.3 Transform of the derivative

Let f(t) be continuous and f’(t) be piecewise continuous on

0 ≦ t ≦ t0 for every finite t0, and let f(t) be of exponential

order as t→∞ so that there are constants K, c, T such that

for all t > T. hen L{f’(t)} exists for all s > c, and

(8)

Proof:

Theorem 5.3.4 Laplace Convolution Theorem

(15)

or equivalently

(16)

Proof:

Example 3

Example 4

5.4 Application to solution of differential equations

Example 1

Example 2

Example 3

5.5 Discontinuous forcing functions: Heaviside step function

Definition: Heaviside step function

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Since H(t) is a unit step at t = 0, H(t-a) is a unit step shifted

to t = a, shown in Fig. (1b).

Fig. 1 Unit step function

The rectangular pulse shown in Fig. 2 is denoted as P(t;a,b),

we have

(2)

More generally, any piecewise continuous function

(3)

defined on 0 < t < ∞ can be given by the single expression

(4)

Example 1

Example 2

The function shown in Fig. 4 is called a ramp function which

can be expressed as

Fig. 4 The ramp function of Example 2

(8)

(9)

Example 3 LC Circuit

The differential equation governing the

charge Q(t) on the capacitor in the circuit is

Let R = 0 and E(T) be the rectangular pulse with the

magnitude of E0 and can be expressed as

(11a)

(11b)

5.6 Impulsive Forcing Functions; Dirac Impulse Function (Optional)

Besides forcing functions that are discontinuous, we are interested in ones that are impulsive – that is , sharply focused in time.

(1)

where f(t) is the force applied to the

mass. If the force is due to a hammer

blow, for instance, initiated at time

t = 0, then we expected f(t) to be

somewhat as sketched in Fig. 1a.

However, we do not know the functional

form of f(t) corresponding to such an

event as a hammer blow. Eq. (1) can

be solved once the f is well defined.

In working with impulsive forces one

normally tries to avoid dealing with the

detailed shape of f and tried to limit one’s

concern to a global quantity known as the

impulse Ιof f, the area under its graph.

The idea is that if e really is small, then the

response x(t), while sensitive to I, should

be rather insensitive to the detailed shape

of f.

This idea suggests that we replace the

unknown f by a simple rectangular pulse

having the correct impulse as shown in

Fig. 1b: f(t) = I / εfor 0≦t ≦ε, and 0 for t ＞ε.With f thus simplified we can proceed to solve for the response x(t). We suppose that since ε is very small, we might as well take the limit of the solution as ε→ 0, to eliminate ε.

(2)

(3)

Prove that

(4)

for any function g that is continuous at the origin.

Let us denote such a rectangular pulse having a unit impulse

as D(t;e)

(5)

Suppose that g is continuous on for some positive b. We can assume that ε < b because we are letting ε → 0. Thus, g is continuous on the integration interval , so the mean value theorem of the integral calculus tells us that there is a point τ1 in [0, ε], such that . Thus Eq. (5) gives

(6)

(7)

Where is known as the Dirac delta function, or unit

impulse function. We can think of as being zero

everywhere except at the origin and infinite at the origin, in

such a way as to have unite area.

Example1.

(8)

(12)

It should be understood that the result is g(a) for any limits A,B.

(13)

Since δ(t) is focused at t＝ 0, it follows that δ(t－ a) is

focused at t＝a, and (7) generalizes to

Example 2 RC Circuit

Considering the charge Q(t) on the capacitor of the RC circuit

is governed by the differential equation RQ’+(1/C)Q = E(T).

Let E(T) be an impulse voltage, with impulse I acting at t = T,

and let Q(0)=Q0.

In the Laplace transform, the independent variable has been the t , so the delta function has represented actions that are focused in time. But the argument of the delta function need not to be time. For instance, if ω(x)is the load distribution on a beam; δ(x－a)represents a point unit load at x＝a.

Fig. 3 Load distribution in a beam

Consider a metal plate loaded by pressing a metal coin against

it at the origin (Fig. 4a). The stress distribution within the plate

can be determined once the load distribution can be prescribed.

The coin will flatten slightly at the point of contact, the load w(x)

will be distributed over a short interval from x = -e to x = e.

Whether one needs to determine the exact w(x) or if it suffices

to represent it simply as an idealized point force of magnitude

P, w(x) = Pd(x). (near field and far field, Saint Venant’s principle)

Fig. 4 Delta function idealization

5.7 Additional Properties

Theorem 5.7.1 s-Shift

If L{f(t)} ＝F(s)exists for s＞s0, then for any real constant a,

Example 1. Determine

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Example 2. We can invert (2s＋1)/(s2 ＋2s＋4)

Example 1. Determine

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Example 2. We can invert (2s＋1)/(s2 ＋2s＋4)

Theorem 5.7.2 t-Shift

If L{f(t)} ＝F(s)exists for s＞s0, then for any constant a＞0

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for s＞s0

(7)

Theorem 5.7.3 Multiplication by 1/s

If L{f(t)} ＝F(s)exists for s＞s0, then

(8)

for s＞max {0, s0} or, equivalently,

(9)

Example 3. To evaluate L－1{1/[s(s2＋1)]}.

Theorem 5.7.4 Differentiation with Respect to s

If L{f(t)} ＝F(s)exists for s＞s0, then

(16)

For s ＞max {0, s0} or, equivalently,

(17)

Example 4.

(18)

(21)

For s＞s0 or, equivalently,

(22)

Theorem 5.7.5 Integration with Respect to s

If there is a real number s0 such that L {f(t)} ＝F(s) exists

for s＞s0, and limt→0 f(t)/t exists, then

- Theorem 5.7.6 Large s Behavior o F(s)
- Let f(t) be piecewise continuous on for each finite t0 and of exponential order as t → ∞. Then
- F(s) → 0 as s→ ∞,
- sF(s) is bounded as s→ ∞.

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Example 7. The function f shown in Fig. 1 is seen to be periodic with period T = 4.

Theorem 5.7.7 Initial-Value Theorem

Let f be continuous and f’ be piecewise continuous on

for each finite t0, and let f an f’ be of exponential

order as t → ∞. Then

Theorem 5.7.8 Transform of Periodic Function

If f is periodic with period T on and piecewise continuous on one period, then

(37)

Example 8. If f is the sawtooth wave shown in Fig. 2, then

T = 2,

Use the Laplace transform to solve