Tài liệu Process Systems Analysis And Control P2 pptx

A rigorous definition of the class of functions possessing Laplace transforms is beyond the scope of this book, but readers will note that every function of interest to us does satisfy t

1 4 THE LAPLACE TRANSIVRh4

Example 2.1 Find the Laplace transform of the function

f(t) = 1

According to Eq (2 l),

f(s) = J-ow(l)e-“‘df = - e-S’ t=cu

S t=O = f

Thus,

L(1) = f There am several facts worth noting at this point:

1 The Laplace transform f(s) contains no information about the behavior of f(t)

for t < 0 This is not a limitation for control system study because t will represent the time variable and we shall be interested in the behavior of systems only for positive time In fact, the variables and systems are usually defined so that f (t) = 0 for t < 0 This will become clearer as we study specific examples

2 Since the Laplace transform is defined in Eq (2.1) by an improper integral,

it will not exist for every function f(t) A rigorous definition of the class of functions possessing Laplace transforms is beyond the scope of this book, but readers will note that every function of interest to us does satisfy the requirements

for possession of a transform.*

3 The Laplace transform is linear In mathematical notation, this means:

~bfl(~) + bf2Wl = 4fl(O) + mf2Wl

where a and b are constants, and f 1 and f2 am two functions of t

Proof Using the definition,

Uafl(t) + bfdt)) = lom[aflO) + bf2(~)le-s’d~

=a Iomfl(r)e-stdt + blom f2(t)e-S’dr

= &flW) + bUM)l

4 The Laplace transform operator transforms a function of the variable I to a func-tion of the variable s The I variable is eliminated by the integrafunc-tion

Tkansforms of Simple Fhnctions

We now proceed to derive the transforms of some simple and useful functions

*For details on this and related mathematical topics, see Churchill (1972)

1 The step function

.m = i y t<O t>O

This important function is known a$ the unit-step function and will henceforth

be denoted by u(t) From Example 2.1, it is clear that

L{u(t)} = f

As expected, the behavior of the function for t < 0 has no effect on its Laplace transform Note that as a consequence of linearity, the transform of any constant A, that is, f(t) = Au(t), is just f(s) = A/s

2 The exponential function

f(t) = _“,,

I

t<O t>O I = u(t)e-“’

where u(t) is the unit-step function Again proceeding according to definition,

I

0

e-(s+a)rdt = _ Ae-(s+a)t

provided that s + a > 0, that is, s > -a In this case, the convergence of the integral depends on a suitable choice of S In case s is a complex number,

it may be shown that this condition becomes

Re(s) > - a For problems of interest to us it will always be possible to choose s so that these conditions are satisfied, and the reader uninterested in mathematical niceties can ignore this point

3 The ramp function

Integration by parts yields

L{tu(t)} = -eesf f + f

cc

4 The sine function

= u(t)sin k t

L{u(t)sin k t } = sin kt ems’dt

16 THE LAPLACE TRANsmRM

TABLE 2.1

1 u(t)

W )

Pu(t)

e-=“u(t)

sin kt u(t)

-4

4 1

4

1 7

?I! sn+l

1

s+a

n! (s + a)“+l

k s2 + k2

THE LAPLACE TRANSFORM 17

TABLE 2.1 (Continued)

coskt u(t)

sinhkt u(t)

coshkr u(r)

e-=’ Sink? u(r)

e- cos kt u(t)

*

S(f), unit impulse

Area = 1

+-s

s2 + k2

k

s2 - k2

S

s2 - k2

k (s + a)? + k2

s+a

(s + a)2 + k2

1

I

Integrating by parts,

L{u(t)sin

k t } = -(s sin kt + k cos kt)

k

=-s* + k*

In a like manner, the transforms of other simple functions may be derived Table 2.1 is a summary of transforms that will be of use to us Those which have not been derived here can be easily established by direct integration, except for the transform of 6(t), which will be discussed in detail in Chap 4

Transforms of Derivatives

At this point, the reader may wonder what has been gained by introduction of the Laplace transform The transform merely changes a function of t into a function

of S The functions of s look no simpler than those of t and, as in the case

of A , A/s, may actually be more complex In the next few paragraphs, the motivation will become clear It will be shown that the Laplace transform has the remarkable property of transforming the operation of differentiation with respect

to t to that of multiplication by s Thus, we claim that

where

f(s) = u.f(t)1 and f(0) is f(t) evaluated at t = 0 [It is essential not to interpret f(0) as f(s) with s = 0 This will be clear from the following proof.]*

Proof.

To integrate this by parts, let

u = e-S’ dv = dfdt

d t

Then

du = sems’dt v = f(t)

* If f(t) is discontinuous at t = 0, f(0) should he evaluated at t = O’, i.e., just to the right of the origin Since we shall seldom want to differentiate functions that are discontinuous at the origin, this detail is not of great importance However, the reader is cautioned to watch carefully for situations

in which such discontinuities occur.

THE LAPLACE TRANSFORM 1 9

Since

we have

I udv = uv- I vdu

= -f(O) + sf(s)

The salient feature of this transformation is that whereas the function of t was to be differentiated with respect to t, the corresponding function of s is merely multiplied by S We shall find this feature to be extremely useful in the solution

of differential equations

To find the transform of the second derivative we make use of the transform

of the first derivative twice, as follows:

L[$$ =L{-$(Z)} = sL{$f]-~I,=,

= s[sf(s) - ml - f’(O)

= s2f(s) - sf(0) - f’(0)

where we have abbreviated

df 0)

d t = f'(O) r=o

In a similar manner, the reader can easily establish by induction that repeated application of Eq (2.2) leads to

L d”f

I-1dt” = s*f(s) - s*-lf(~) _ p-*f(l)(o) _ _ sf(n-*)(o) - p-l)(o)

where f ’

t = 0 (‘)

(0) indicates the ith derivative of f(t) with respect to t, evaluated for Thus, the Laplace transform may be seen to change the operation of differen-tiation of the function to that of multiplication of the transform by S, the number

of multiplications corresponding to the number of differentiations In addition, some polynomial terms involving the initial values of f(t) and its first (n - 1) derivatives are involved In later applications we shall usually define our variables

so that these polynomial terms will vanish Hence, they are of secondary concern here

Example 2.2 Find the Laplace transform of the function n(t) that satisfies the

differential equation and initial conditions

dx (0) d*x(O) o x(O) = dt = - =

dt*

2 0 THE LAPLACE TRANSFORM

It is permissible mathematically to take the Laplace transforms of both sides of

a differential equation and equate them, since equality of functions implies equality

of their transforms Doing this, there is obtained

,3x(s) - 2x(O) - sir’(O) - x”(0) + 4[s%(s) - sx(0) - x’(O)]

+ S[sx(s) - x(O)] + 2x(s) = 5

where x(s) = L{x(r)} Use has been made of the linearity property and of the fact that only positive values of t are of interest Inserting the initial conditions and solving for x(s)

This is the required answer, the Laplace transform of x(t)

Solution of Differential Equations

There are two important points to note regarding this last example In the first place, application of the transformation resulted in an equation that was solved for the unknown function by purely algebraic means Second, and most important,

if the function x(t), which has the Laplace transform 2/s(s 3 + 4s2 + 5s + 2) were known, we would have the solution to the differential equation and bound-ary conditions This suggests a procedure for solving differential equations that is analogous to that of using logarithms to multiply or divide To use logarithms, one transforms the pertinent numbers to their logarithms and then adds or subtracts, which is much easier than multiplying or dividing The result of the addition or subtraction is the logarithm of the desired answer The answer is found by refer-ence to a table to find the number having this logarithm In the Laplace transform method for solution of differential equations, the functions are converted to their transforms and the resulting equations are solved for the unknown function

alge-braically This is much easier than solving a differential equation However, at the last step the analogy to logarithms is not complete We obviously cannot hope

to construct a table containing the Laplace transform of every function f(t) that possesses a transform Instead, we shall develop methods for reexpressing com-plicated transforms, such as x(s) in Example 2.2, in terms of simple transforms that can be found in Table 2.1 For example, it is easily verified that the solution

to the differential equation and boundary conditions of Example 2.2 is

x(t) = 1 - 2te-’ - e-2r

The Laplace transform of x, using Eq (2.4) and Table 2.1, is

(2.4)

Equation (2.3) is actually the result of placing Eq (2.5) over a common denomi-nator Although it is difficult to find x(t) from Eq (2.3), Eq (2.5) may be easily

THE LAPLACE TRANSFORM 21

inverted to Eq (2.4) by using Table 2.1 Therefore, what is required is a method for expanding the common-denominator form of Eq (2.3) to the separated form

of Eq (2.5) This method is provided by the technique of partial fractions, which

is developed in Chap 3

SUMMARY

To summarize, the basis for solving linear, ordinary differential equations with

constant coeficients with Laplace transforms has been established.

The procedure is:

1 Take the Laplace transform of both sides of the equation The initial conditions are incorporated at this step in the transforms of the derivatives

2 Solve the resulting equation for the Laplace transform of the unknown function algebraically

3 Find the function of t that has the Laplace transform.obtained in step 2 This function satisfies the differential equation and initial conditions and hence is the desired solution This third step is frequently the most difficult or tedious step and will be developed further in the next chapter It is called inversion of the transform Although there are other techniques available for inversion, the one that we shall develop and make consistent use of is that of partial-fraction expansion

A simple example will serve to illustrate steps 1 and 2, and a trivial case of step 3

Example 2.3 Solve

$+3x =o x(0) = 2

We number our steps according to the discussion in the preceding paragraphs:

1 sx(s) - 2 + 3x(s) = 0

2 x(s) = -& = 2-&

3 x(t) = 2,-3r

3

INVERSION

BY PARTIAL

FRACTIONS

Our study of the application of Laplace transforms to linear differential equations with constant coefficients has enabled us to rapidly establish the Laplace transform

of the solution We now wish to develop methods for inverting the transforms to obtain the solution in the time domain The first part of this chapter will be a series

of examples that illustrate the partial-fraction technique After a generalization of these techniques, we proceed to a discussion of the qualitative information that can be obtained from the transform of the solution without inverting it

The equations to be solved are all of the general form

d”x a’ dtn

+ a _ d”-lx

n ’ dt”-1 +

The unknown function of time is x(t), and an, an _ 1, , a 1, a 0, are constants.

The given function f(t) is called theforcingfunction In addition, for all problems

of interest in control system analysis, the initial conditions are given In other

words, values of x, dxldt, , d”-‘xldP-* are specified at time zero The

problem is to determine x(t) for all t 2 0

Partial Fkactions

In the series of examples that follow, the technique of partial-fraction inversion for solution of this class of differential equations is presented

22

UWERSION B Y PARTlAL FRACTIONS 23

Example 3.1 Solve

$+x = 1 x(0) = 0 Application of the Laplace transform yields

sx(s) + x(s) = 5

o r

1 n(s) = -s(s + 1) The theory of partial fractions enables us to write this as

where A and B are constants Hence, using Table 2.1, it follows that

Therefore, if A and B were known, we would have the solution The conditions on

A and B are that they must be chosen to make Eq (3.1) an identity in s.

To determine A, multiply both sides of Eq (3.1) by s

Since this must hold for all s, it must hold for s = 0 Putting s = 0 in Eq (3.3) yields

A=1

To find B, multiply both sides of Eq (3.1) by (s + 1).

1

S

Since this must hold for all s, it must hold for s = - 1 This yields

B = - 1

Hence,

-= -s(s + 1) s s+l and therefore,

x(f) = 1 -e-’

(3.5)

(3.6)

2 4 THE LAPLACE TRANSFORM

Equation (3.5) may be checked by putting the right side over a common denomi-nator, and Eq (3.6) by substitution into the original differential equation and initial condition

Example 3.2 Solve

~+2~-~-2x=4+e2’

x(0) = 1 x’(0) = 0 x”(0) = -1 Taking the Laplace transform of both sides,

1 [s3x(s) s* + 11 + 2[s%(s) s] [$X(S) 11 2.X(s) = % +

-s - 2 Solving algebraically for x(s),

x(s) = s4 - 6s2 + 9s - 8 s(s - 2)(s3 + 2s2 - s - 2) The cubic in the denominator may be factored, and x(s) expanded in partial fractions

x(s) = s4 - 6s2 + 9s - 8 A B

s(s - 2)(s + I)(s + 2)(s - 1) = s + +C+D+E

s - 2 s+1, s+2 s-1

(3.7)

To find A, multiply both sides of Eq (3.7) by s and then set s = 0; the result is

(-2)(1)(2)(-l) = -2 The other constants are determined in the same way The procedure and results are

To determine multiply (3.7) by and set s to Result

Accordingly, the solution to the problem is

x(t) = -2 ++2t + l&-t _ ++-2t + jet

A comparison between this method and the classical method, as applied to Example 3.2, may be profitable In the classical method for solution of differential equations we first write down the characteristic function of the homogenepus equation:

s3 + 2s2 - s - 2 = 0

INVERSION BY PARTIAL FRACTIONS 2 5

This must be factored, as was also required in the Laplace transform method, to obtain the roots -1, -2, and + 1 Thus, the complementary solution is

xc(t) = Cle-’ + C*e-*’ + C3e’

Furthermore, by inspection of the forcing function, we know that the particular solution has the form

x,(t) = A + Be2f

The constants A and B are determined by substitution into the differential equation and, as expected, are found to be -2 and A, respectively Then

*

x ( t ) = - 2 + fie 2t + Cle-’ + C2em2’ + Cse’

and the constants Cl, C2, and Cs are determined by the three initial conditions The Laplace transform method has systematized the evaluation of these constants, avoiding the solution of three simultaneous equations Four points are worth not-ing:

1 In both methods, one must find the roots of the characteristic equation The roots give rise to terms in the solution whose form is independent of the forcing function These terms make up the complementary solution.

2 The forcing function gives rise to terms in the solution whose form depends

on the form of the forcing function and is independent of the left side of the equation These terms comprise the particular solution.

3 The only interaction between these sets of terms, i.e., between the right side and left side of the differential equation, occurs in the evaluation of the con-stants involved

4 The only effect of the initial conditions is in the evaluation of the constants This is because the initial conditions affect only the numerator of x(s), as may

be seen from the solution of this example

In the two examples we have discussed, the denominator of x(s) factored into real factors only In the next example, we consider the complications that arise when the denominator of x(s) has complex factors

Example 3.3 Solve

2

$+2$+2x=2

x(0) = x’(0) = 0

Application of the Laplace transform yields

s(s2 + 2s + 2)

The quadratic term in the denominator may be factored by use of the quadratic formula The roots are found to be (-1 - j) and (-1 + j) This gives the partial-fraction expansion

s(s + 1 + j)(s + 1 - j) = s + (s + 1 + j) + (s + 1 - j) (3.8)