THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

C.3 Solving Differential Equations 639which also can be expressed asẍ + 2ξω n ẋ(t) + ω 2 n x(t) = f α(t)We now apply Equations (C.3) and (C.4)–(C.6) to obtains 2 X(s) − sx(0+) − ẋ(0+) + 2ξω n [sX(s) − x(0+)] + ω 2 n X(s) = F α(s)Solving for X(s):(C.6)X(s) = F α(x) + x(0+)(s + 2ξω n ) + ẋ(0+)(C.7)s 2 + 2ξω n s + ωn2Equation (C.7) can be written in the formX(s) = A(s)(C.8)B(s)Equation (C.7) and its variation (C.6) have been derived from a second-orderequation, and the techniques we describe below are valid for any order function.For equations with simple roots, we can write Equation (C.8) in the formX(s) = A(s)B(s) =A(s)(C.9)(s + a 1 )(s + a 2 ) ···(s + a n )It is assumed that B(s) is of higher order than A(s) throughout the discussion below.We can expand Equation (C.9) through the use of partial fraction expansions, whichwill result inwhereX(s) = C 1s + a 1+ C 2s + a 2+···+C k =lim (s + a k ) A(s)s→−a k B(s)C ns + a n(C.10)(C.11)Example Problem 1Find the inverse Laplace transform of the functionSolutionX(s) =13(s + 3)s(s + 1)(s + 2)The roots of the numerator (for example, s =−3) are called zeros of the function.The roots of the denominator (viz., 0, −1, −2) are called poles. A partial fractionexpansion of the poles is written as follows:(13(s + 3)s(s + 1)(s + 2) = 13 C1s + C 2s + 1 + C )3s + 2