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Applying the transform pairs of Eqs. (3.1% and b) to the quadratic terms on the right gives x(t) = Fl(t) + Bematcos kt + -atsin kt (3.19)
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where F1 (t) is the result of inverting Fl(s). We now apply this method to the following example. Example 3.5. Solve x(s) = 1 A Bs + C s(s2 - 2s + 5) = s +s*-2,+5
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Applying the quadratic equation to the quadratic term gives:
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Using the method just presented, we find that a = - 1, k = 2. Solving for A, B, and C gives A = l/S, B = -l/5, C = 215. Introducing these values into the expression for x(s) and applying Eq. (3.19) gives 1 1 1 x(t) = 5 - Jetcos 2t + Ge sin 2t The reader should solve Example 3.4 with this alternate method, which uses Eq. (3.19). In the next example, an exceptional case is considered; the denominator of x(s) has repeated roots. The procedure in this case will vary slightly from that of the previous cases. Example 3.6. Solve
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Application of the Laplace
transform yields x(s) = 1 s(s3 + 3s* + 3s + 1) fractions, (3.20)
Factoring
expanding x(s) =
partial
1 A B c D ~ ~ s(s + 1)X = s + (s + 1)3 + (s + 1>* + s+l
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
+ 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 these values into Eq. (3.20) and place the right side of the equation over a common denominator; the result is: 1 = (s + 1)3 - s + Cs(s + 1) + Ds(s + l)* s(s + 1)3 s(s + 1)3 Expanding the numerator of the right side gives 1 = (1 + D)s3 + (3 + C + 2D)s* + (2 + C + D)s + 1 s(s + 1)s s(s + 1)s We now equate the numerators on each side to get 1 = (1 + D)s3 + (3 + C + 2D)s* + (2 + C + D)s + 1 Equating the coefficients of like powers of s gives l+D=O 3+C+2D=O 2+C+D=O Solving these equations gives C = - 1 and D = - 1. The final result is then 1 - 1 -Gs) = ; - (s +
(3.22)
(3.23)
1 -1 (s + 1)2 s+l
(3.24)
Referring to Table 2.1, this can be inverted to x(t) = 1 - e- (3.25)
INVERSION BY PARTIAL FRACTIONS
The reader should verify that Eq. (3.24) placed over a common denominator results in the original form 1 x(s) = s(s + 1)3 and that Eq. (3.25) satisfies the differential equation and initial conditions.
The result of Example 3.6 may be generalized. The appearance of the factor (S + u)~ in the denominator of X(S) leads to n terms in the partial-fraction expansion:
Cl ( s c2 C* + up ( s + a ) - . . . z-i
The constant Cl can be determined as usual by multiplying the expansion by (S + u)~ and setting s = -a. The other constants are determined by the method shown in Example 3.6. These terms, according to Table 2.1, lead to the following expression as the inverse transform:
Cl (n J-l + c2 ------p-2
(n - 2)!
*.- + CnmIt + C, evar
It is interesting to recall that in the classical method for solving these equations, one treats repeated roots of the characteristic equation by postulating the form of Eq. (3.26) and selecting the constants to fit the initial conditions.
(3.26)
Qualitative Nature of Solutions
If we are interested only in the form of the solution x(t), which is often the case in our work, this information may be obtained directlyffom the roots of the denominator ofx(s). As an illustration of this qualitative approach to differential equations consider Example 3.3 in which B 2 A C x(s) = + + s(s2 + 2s + 2) = s s+1+j s+1-j is the transformed solution of d2x 2dx - +2x=2 dt* + dt It is evident by inspection of the partial-fraction expansion, without evaluation of the constants, that the s in the denominator of x(s) will give rise to a constant in x(t). Also, since the roots of the quadratic term are - 1 k j , it is known that x(t) must contain terms of the form e - (C ices t + C2sin t). This may be sufficient information for our purposes. Alternatively, we may be interested in the behavior of x(t) as t + CQ. It is clear that the terms involving sin and cos vanish because of the factor e- . Therefore, x(t) ultimately approaches the constant, which by inspection must be unity. The qualitative nature of the solution x(t) can be related to the location of the roots of the denominator of X(S) in the complex plane. These roots are the roots
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