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+‘“
K1zs + 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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 Spring '07
 Fuchs
 Physics

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