barcode reading in asp.net The Laplace transform method yields in Software

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The Laplace transform method yields
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2 x(s) = (s2 + 4)(s + 1)
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Factoring and expanding into partial fractions,
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2 A B c -+-----Y(s + l)(s + 2j)(s - 2j) = - + s + 2j s+l s - 2j (3.13)
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Multiplying Eq. (3.13) by (S + 1) and setting s = -1 yield
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2 2 (-1 +2j)(-1 -2j) = 5
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Multiplying Eq. (3.13) by (s + 2j) and setting s = -2 j yield B= Matching the term (-2 + j)/lO s + 2j with the first term of the conjugates of Eq. (3.11) requires that 2 -2 + =- j 10 (-2j + l)(-4j)
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bl = &, kl = 0 k2 = 2, Substituting in (3.12) results in -3cos2t+ Hence the complete answer is x(t) = 2e-r - 5 cos 2t + 1 sin 2t 5 Readers should verify that this answer satisfies the differential equation and boundary conditions. In addition, they should show that it can also be obtained by matching the term with the second term of the conjugates of Eq. (3.11) or by determining C instead of B.
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ALTERNATE METHOD USING QUADRATIC TERM. Another method for solv-
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isin2t
ing Example 3.3, which avoids some of the manipulation of complex numbers, is as follows. Expand the expression for x(s):
x(s) =
2 A Bs +C + s(s2 + 2s + 2) = s s2 + 2s + 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 over a common denominator; thus
x(s) =
2 = (s2 + 2s + 2)A + Bs2 + Cs s(s2 + 2s + 2) s(s2 + 2s + 2) 2 = (A + B)s2 + (2A + C)s + 2A
Equating the numerators on each side gives
T H E LAPLACE
TRANSFORM
We now equate the coefficients of like powers of s to obtain
A+B=O 2A+c=o
2A=2 Solving these equations gives A = 1, B = - 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 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 using the table of transforms, one can easily invert each term to give
x(t) = 1 - e- cos t - e- sin t kl[(s + cQ2 + k2] (s + a)/[(~ + cQ2 + k2]
(3.152) (3.15b)
which is the same result obtained before. A general discussion of this case follows. Consider the general expression involving a quadratic term
x(s) = F(s) s2 + as + p
(3.16)
where F(s) is some function of s (e.g. 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 can be found by solving for the roots of s 2 + (YS + p = 0 by the quadratic formula to give s 1 = -a + j k, sq = -a - j k . The quadratic term can now be written s2 + as + p = (s - sl)(s - s2) = (s + a - jk)(s + a + jk) = (s + u)~ + k2
INVERSION BY PARTIAL FRACTIONS
Equation (3.17) now becomes Bs + C
x(s) + Fib) + @ + a)* + p
The numerator of the quadratic term is now written to correspond to the transform pairs given by Eqs. (3.1% and b) Bs+C = B 1 (C/B) - a C-aBk k = B(s + a) + k s+a+ k k (s + ~2)~ + k*
Equation (3.18) becomes
X(S) = Fl(s) + B cs +
(s + a)
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