solve for the roots of a_{n}s^{n}+…+a_{1}s+a_{0}=0 say: -r_{1}, -r_{2}, … , -r_{n }then
where
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The roots of
are: -5, -4+3i, -4-3i, -1, -1, -1
Converting to partial fractions,
where
Thus the inverse Laplace transform of f(s) is
+
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A Sketch
of the Derivation of the Short Cut Formulas for Obtaining Coefficients
in the Partial Fractions
Case 1. (For each real and distinct root).
Suppose (-r) is an unrepeated root in the denominator, and g(s) is obtained by lumping all terms except for 1/(s+r), i.e.
=
+ [other terms]
After multiplying both sides by (s+r),
and letting s be equal to -r, only A remains on the right hand side of the equation, so
Case 2. (For each root that is repeated k times).
Suppose (-r) is a root in the denominator that is repeated k times, and g(s) is obtained by lumping all terms except for 1/(s+r)k, i.e.
+ [other terms]
After multiplying both sides by (s+r)k,
+ [other terms](s+r)^{k}
Take the derivative of both sides with respect to s,
+ [other terms]k(s+r)^{k-1} + (s+r)^{k} (d/ds)[other terms]
and for the second derivative,
+ [other terms]k(k-1)(s+r)^{k-2 }+ (s+r)k(d^{2}/ds^{2}) [other terms]
+ 2((d/ds) [other terms] )k(s+r)^{k-1}
In general, we get for the ith derivative with respect to s,
so by choosing s = - r at the appropriate ith derivative, we have
Case 3. (For each two roots that are complex conjugate pairs).
Suppose (-a+ib) and (-a-ib) are complex conjugate roots in the denominator, and g(s) is obtained by lumping all terms except for 1/[ (s+a)2 + b2 ], i.e.
+ [other terms]
Multiplying both sides by ,
+ [other terms][ (s+a)2+b2 ]
By letting Q be g(s= -a+ib), the other terms disappear, leaving only A and B on the right hand side,
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This page is maintained by Tomas B. Co (tbco@mtu.edu). Last revised 1/12/2000.
Tomas B. Co
Associate Professor
Department of Chemical Engineering
Michigan Technological University
1400 Townsend Avenue
Houghton, MI 49931-1295