Tài liệu Process Systems Analysis And Control P2 pptx

14 THE Example 2.1. Find the transform of the function = According to Eq. (2. e-S’ S t=O Thus, = There several facts worth noting at this point: 1. The transform f(s) contains no information about the behavior of f(t) for 0. This is not a limitation for control system study because 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 0. This will become clearer as we study specific examples. 2. Since the 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 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 transform is linear. In mathematical notation, means: + = + where a and b are constants, and f and two functions of Proof. Using the definition, + = + I + = + 4. The transform operator transforms a function of the variable to a func- tion of the variable s. The variable is eliminated by integration. of Simple We now proceed to derive the transforms of some simple and useful functions. *For details on this and related mathematical topics, see Churchill (1972). TRANSFORM 1. The step function = This important function is known the unit-step function and will henceforth be denoted by u(t). From Example 2.1, it is clear that As expected, the behavior of the function for 0 has no effect on its transform. Note that as a consequence of linearity, the transform of any constant A, that is, = Au(t), is just f(s) = 2. The exponential function = I = where is the unit-step function. Again proceeding according to definition, 1 0 provided that + a 0, that is, -a. In this case, the convergence of the integral depends on a suitable choice of In case is a complex number, it may be shown that this condition becomes -a For problems of interest to us it will always be possible to choose 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 = + 0 4. The sine function = kt kt} = sin kt 16 TABLE 2.1 Graph S sin 1 1 (s + k + THE TRANSFORM TABLE 2.1 (Continued) Graph u(t) coshkr Sink? u(r) 1 Area 1 + k S k (s + + (s + + THE TRANSFORM Integrating by parts, kt} = sin kt + k cos kt) + 0 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 which will be discussed in detail in Chap. 4. of Derivatives At this point, the reader may wonder what has been gained by introduction of the transform. The transform merely changes a function of into a function of The functions of look no simpler than those of and, as in the case of A may actually be more complex. In the next few paragraphs, the motivation will become clear. It will be shown that the transform has the remarkable property of transforming the operation of differentiation with respect to to that of multiplication by s. Thus, we claim that where = and f(0) is evaluated at = 0. [It is essential not to interpret f(0) as f(s) with = 0. This will be clear from the following proof.]* Proof. To integrate this by parts, let dv = dt Then du = = * If f(t) is discontinuous at = 0, should he evaluated at t = i.e., just to the right 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 TRANSFORM 19 Since we have I udv = I vdu = -f(O) + The salient feature of this transformation is that whereas the function of was to be differentiated with respect to the corresponding function of is merely multiplied by 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: = = f’(0) where we have abbreviated dt = In a similar manner, the reader can easily establish by induction that repeated application of Eq. (2.2) leads to . . . where (0) indicates the ith derivative of with respect to evaluated for Thus, the transform may be seen to change the operation of differen- tiation of the function to that of multiplication of the transform by 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 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 transform of the function that satisfies the differential equation and initial conditions 20 THE TRANSFORM It is permissible mathematically to take the 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 x”(0) + x’(O)] + = where = Use has been made of the linearity property and of the fact that only positive values of of interest. Inserting the initial conditions and solving for x(s) = This is the required answer, the 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 Second, and most important, if the function x(t), which has the transform + + + 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 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 transform of every function f(t) that possesses a transform. Instead, we shall develop methods for reexpressing com- plicated transforms, such as 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 The transform of x, using Eq. (2.4) and Table 2.1, is 1 1 1 --- = + Equation (2.3) is actually the result of placing Eq. (2.5) over a common denomi- nator. Although it is difficult to find from Eq. Eq. (2.5) may be easily 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 with transforms has been established. The procedure is: 1. Take the 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 transform of the unknown function algebraically. 3. Find the function of that has the 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 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 x(0) = 2 We number our steps according to the discussion in the preceding paragraphs: 1. 2 + 3x(s) = 0 2. = = 3. x(t) = CHAPTER INVERSION BY PARTIAL FRACTIONS Our study of the application of transforms to linear differential equations with constant coefficients has enabled us to rapidly establish the 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 + + The unknown function of time is x(t), and . . . , a a are constants. The given function is called In addition, for all problems of interest in control system analysis, the initial conditions are given. In other words, values of dxldt,. . . , are specified at time zero. The problem is to determine for all 0. Partial In the series of examples that follow, the technique of partial-fraction inversion for solution of this class of differential equations is presented. 22 BY Example 3.1. Solve = 1 x(0) = 0 Application of the transform yields + x(s) = or 1 n(s) = + 1) The theory of partial fractions enables us to write this as x(s) = 1 A B + 1) (3.1) where A and B are constants. Hence, using Table 2.1, it follows that = A + Be-’ (3.2) 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 1 Since this must hold for all s, it must hold for = 0. Putting = 0 in Eq. (3.3) yields To find B, multiply both sides of Eq. (3.1) by (s + 1). 1 = + 1) + B (3.4) Since this must hold for all s, it must hold for = 1. This yields B -1 Hence, 1 1 1 + 1) and therefore, = 1 [...]... -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... The cubic in the denominator may be factored, and x(s) expanded in partial fractions x(s) = s4 - 6s2 + 9s - 8 A + -B +C+D+E s+1, s+2 s-1 s(s - 2)(s + I)(s + 2)(s - 1) = s s - 2 (3.7) To find A, multiply both sides of Eq (3.7) by s and then set s = 0; the result is A = -8 (-2)(1)(2)(-l) = -2 The other constants are determined in the same way The procedure and results are ~’ summarized in the following... conclusion that - 1 - j B=2 Comparison of Eqs (3.8) and (3.11) and the result for B show that we have two possible ways to assign a 1, bl, kl, and k2 so that we match the form of Eq (3.11) They are bl = -; bI = ; or k, = 1 kl = 1 k2 = 1 k2 = - 1 The first way corresponds to matching the term involving B with the first term of the conjugates of Eq (3.1 l), and the second to matching it with the second term... The Laplace transform method yields 2 x(s) = (s2 + 4)(s + 1) Factoring and expanding into partial fractions, 2 A B c -+ -Y(s + l)(s + 2j)(s - 2j) = - + s + 2j s+l s - 2j Multiplying Eq (3.13) by (S + 1) and setting s = -1 yield A= 2 2 (-1 +2j)(-1 -2j) = 5 (3.13) INVERSION BY PARTIAL FRACTIONS 29 Multiplying Eq (3.13) by (s + 2j) and setting s = -2 j yield B= 2 -2 + =- j 10 (-2j + l)(-4j) Matching the... 1, and C = -2 Equation (3.14) now becomes x(s) = f - s - s2 + 2s + 2 2 s2+2s +2 We now rearrange the second and third terms to match the following transform pairs from Table 2.1: e -a’sin kt e -ardor kt kl[(s + cQ2 + k2] (3.152) (s + a)/[(~ + cQ2 + k2] (3.15b) The result of the rearrangement gives 1 1 s+l +) = s - (s + 1)2 + 12 - (s + 1)2 + 12 We see from the quadratic terms that a = 1 and k = 1, and. .. I/s) Expanding the terms on the right side gives x(s) = Fl(s) + Bs + C s2 + as + p (3.17) where Fl(s) represents other terms in the partial-fraction expansion First solve for B and C algebraically by placing the right side over a common denominator and equating the coefficients of like powers of s The next step is to express the quadratic term in the form s2 + crs + p = (s + u)~ + k2 The terms a and k... Factoring and expanding x(s) = in partial fractions, 1 A B c D ~ ~ s(s + 1)X = s + (s + 1)3 + (s + 1>* + s+l (3.20) As in the previous cases, to determine A, multiply both sides by s and then set s to zero This yields A=1 Multiplication of both sides of Eq (3.20) by (s + 1)3 results in 1 _ = 4s + II3 S s + B + C(s + 1) + D(s + l)* (3.21) Setting s = -1 in Eq (3.15) gives B = -1 Having found A and B, introduce... the procedure at all However, the computations may be slightly more tedious To obtain A, multiply Eq (3.8) by s and set s = 0: 2 A = (1 + j)(l - j) = l To obtain B, multiply Eq (3.8) by (s + 1 + j) and set s = (-1 - j): 2 -1-j Ep 2 (-1 - j)(-2j) B= To obtain C, multiply Eq (3.8) by (s + 1 - j) and set s = (-1 + j): C = 2 C-1 + j>Gj) - l + =-j 2 Therefore, -1-j x(s) = ; + ~ 2 1 +-l+j 2 s+l+j 1 s + l -... exercise for the reader A more general discussion of this case will promote understanding It was seen in Example 3.3 that the complex conjugate roots of the denominator of x(s) gave rise to a pair of complex terms in the partial-fraction expansion The constants in these terms, B and C, proved to be complex conjugates (- 1 - j)/2 and (- 1 + j)/2 When these terms were combined through a trigonometric identity,... 2 (3.14) In this expression, the quadratic term is retained and the second term on the right side is the most general expression for the expansion The reader will find this form of expansion for a quadratic term in books on advanced algebra Solve for A by multiplying both sides of Eq (3.14) by s and let s = 0 The result is A = 1 Determine B and C algebraically by placing the two terms on the right side . A and B are determined by substitution into the differential equation and, as expected, are found to be -2 and respectively. Then x(t) = -2 + + + + and. transform is linear. In mathematical notation, means: + = + where a and b are constants, and f and two functions of Proof. Using the definition, + = + I + =