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inverseL - 1 Simple roots The simple roots are placed in a...

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Unformatted text preview: 1. Simple roots The simple roots are placed in a panial fractions expansion F(S)= Ko(s+zl)~~(s+:m) = K1 + K2 +...+ K" (2) (5+P1X5+P2)'”(5+Pn) 5+P1 S+P2 5+Pn The constants, K., can be found using the method of residues K,- =(s + p.-) Help“. (3) Finally, the tabulated Laplace n‘ansfonn pails are used to invert the expression. This is a nice form since the solution is f0) = Kl e""’ + K2 e‘l’l’ + < ,. +K,, 2‘?“ (4) 2. Complex conjugate roots Complex roots result in a Laplace transform of the form K1 KI =m+m+‘“ |K1|zs + |Kl[z—e =s+(a—m) Home) F(s) (5) +0. The K1 can then be found using the same method as for simple roots; however, the “positive” root of the form — a + j fl must be the one used to find K1, that is K1 = (s + a — j p) F(S)ls=—a+;p (6) The °°”°5P°Dding time domain function is then f(r) = 2le l 9-“: mm: + .9)+ . .. (7) 3. Repeated roots When F(s) has a root of multiplicity r, then F(s) is vwitten as P K K K F(5)= 1(3) r: u + 12 2+__.+ Ir r+___ (8) Q1(5)(5+P1) “"PI (“‘Pl) (“1%) where the time domain function is then r—l f(t)=Ku 2P“ +K131‘ep" +~--+K1r We?" +... (9) That is, we obtain the exponentials multiplied by t‘s, where the Ky- terms are evaluated fmm 1 d” r K . = . s + F s 10 ., (r_j)!ds,_l[( pl) 0]: ( ) =’Pl This actually simplifies nicely until you reach :3 terms, that is, for a double root (5 — p, )2 K12 = (5 + P1 )2 17(5)L=_n (11) K11 =%[(“‘ P1 )2 F(5)] ’:'Pl Thus K]: is found just like for simple mots. Note this backward order of solving for the K 's, that is, we solve for the KS of the higher order terms first. ...
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