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Code 128 Code Set C Scanner In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. Code 128 Drawer In Visual C#.NET Using Barcode drawer for .NET framework Control to generate, create Code 128B image in VS .NET applications. inverted to Eq. (2.4) by using Table 2.1. Therefore, what is required is a method for expanding the commondenominator 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. Painting Code128 In .NET Using Barcode creator for ASP.NET Control to generate, create Code 128B image in ASP.NET applications. Printing Code 128 Code Set B In .NET Using Barcode drawer for .NET framework Control to generate, create Code 128A image in .NET framework applications. SUMMARY
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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 partialfraction 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 _ d lx n dt 1 a dtn 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 partialfraction inversion for solution of this class of differential equations is presented. UWERSION
PARTlAL
FRACTIONS
Example 3.1. Solve $+x = 1 x(0) = 0 Application of the Laplace transform yields sx(s) + x(s) = 5 or 1 n(s) = s(s + 1) The theory of partial fractions enables us to write this as 1 A B x(s) = =+s s+1 s(s + 1) where A and B are constants. Hence, using Table 2.1, it follows that n(t) = A + Be (3.2) (3.1) 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. =A+ 1 s+l
BZ s+1
(3.3) 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 + 1) + B (3.4) Since this must hold for all s, it must hold for s =  1. This yields B = 1 Hence, 1 1 1 =s(s + 1) s s+l and therefore, x(f) = 1 e (3.6) (3.5)

