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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.
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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.
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Example 2.3. Solve $+3x =o
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x(0) = 2
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We number our steps according to the discussion in the preceding paragraphs:
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1. sx(s) - 2 + 3x(s) = 0 2. x(s) = -& = 2-& 3. x(t) = 2,-3r
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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 _ 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 partial-fraction 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)
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