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1 +-l+j 2 s+l+j
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This is the desired result. To invert n(s), we may now use the fact that l/(s + a) is the transform of e - r. The fact that Q is complex does not invalidate this result, as can be seen by returning to the derivation of the transform of e --ar. The result is
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x(t) = 1 + -1 - j -(l+j)t -e I -1 + j,-(l-j)r
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Using the identity
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da+jbjr = eaf(cos bt + j sin bt)
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x(t) = 1 - e- (cos t + sin t)
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The details of this conversion are recommended as an 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, it was found that the complex terms canceled, leaving a real result for x(t). Of course, it is necessary that x(t) be real, since the original differential equation and initial conditions are real. This information may be utilized as follows: the general case of complex conjugate roots arises in the form x(s) = F(s) (s + kl + jk2)(S + kl - jk2) 43.9)
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where F(s) is some real function of s.
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INVERSION
PAR T IA L
FRACTIONS
For instance, in Example 3.3 we had
F(s) = 5
Expanding (3.9) in partial fractions,
kr=l
k2=1
(s + kl + jkz)(s + kl - jk2) = Fl(s)
+ a1 + jbl s +kl+ jk2 + a2 + jb2 s + kl - jk2
(3.10)
where at, ~2, bl, b2 are the constants to be evaluated in the partial-fraction expansion and Fl(s) is a series of fractions arising from F(s). Again, in Example 3.3, 1
a1 = -
= - 2
bl = -;
b2 = ;
Fl(s)
Now, since the left side of Eq. (3.10) is real for all real s, the right side must also be real for all real s. Since two complex numbers will add to form a real number if they are complex conjugates, it is seen that the right side will be realfir all real s if and only if the two terms are complex conjugates. Since the denominators of the terms are conjugates, this means that the numerators must also be conjugates, or
a2 = al b2 = -bl
This is exactly the result obtained in the specific case of Example 3.3. With this information, Eq. (3.10) becomes F(s) = Fl(s) (s + kl + jk2)(s + kl - jk2) ~1 + jh + al - jh (3.11) s + kl + jk2 s + kl - jk2 i i Hence, it has been established that terms in the inverse transform arising from the complex conjugate roots may be written in the form +
(al + jbl)e(-kl-jk2) + (al _ jbl)e(-kl+jk2)t
Again, using the identity
e(Cl+jC2)r = eC~t
(cos C2t + j sin C2t)
this reduces to
2eeklr(alcos kzt + bl sin kg)
(3.12)
LAPLACE
TRANSFORM
Let us now rework Example 3.3 using Eq. (3.12). We return to the point at which we arrived, by our usual techniques, with the 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
= -;
bI = ;
k, = 1 k2 = 1 kl = 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. In either case, substitution of these constants into Eq. (3.12) yields --e- (cos t + sin t) which is, as we have discovered, the correct term in x(t). What this means is that one can proceed directly from the evaluation of one of the partial-fraction constants, in this case B, to the complete term in the inverse transform, in this case -e- (cos t + sin t). It is not necessary to perform all the algebra, since it has been done in the general case to arrive at Eq. (3.12). Another example will serve to emphasize the application of this technique.
Example 3.4. Solve d2x - +4x = 2e- dt2 x(0) = x (0) = 0
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