Page 271 - Dynamics and Control of Nuclear Reactors

APPENDIX D Laplace transforms and transfer functions 273

This formidable-looking formula is quite easy to apply, especially for a first order
pole (n¼1).
For the first order case, the residue becomes

st
R s1 ¼ s s 1 ÞFsðÞe Š (D.11)
½
ð
s¼s 1
This can be applied to the general transform with all first order poles (a 1 ,a 2 , …)
ð s b 1 Þ s b 2 Þ… s b m Þ
ð
ð
FsðÞ ¼ (D.12)
ð s a 1 Þ s a 2 Þ… s a n Þ
ð
ð
Inversion by the residue theorem gives (by inspection)
ð
ð
ð a 1 b 1 Þ a 1 b 2 Þ… a 1 b m Þ a 1 t
ftðÞ ¼ e + ⋯
ð
ð
ð a 1 a 2 Þ a 1 a 3 Þ… a 1 a n Þ
(D.13)
ð a n b 1 Þ a n b 2 Þ… a n b m Þ a n t
ð
ð
+ e
ð
ð
ð a n a 1 Þ a n a 2 Þ… a n a n 1 Þ
Note that inversion for a specific pole involves removal of the term containing that
pole and substituting that pole’s value for s in the other terms. A little experience will
permit one to perform such an inversion very quickly.
Example D.4
Let us use the method of residues to invert the following Laplace transform
ð
ð s +1Þ s +4Þ
FsðÞ ¼ (D.14)
ð s +2Þ s +6Þ s +7Þ
ð
ð
The inversion gives:
ð
ð
ð 2+ 1Þ 2+ 4Þe 2t ð 6+ 1Þ 6+ 4Þe 6t ð 7+1Þ 7+4Þe 7t
ð
ftðÞ ¼ + +
ð 2+ 6Þ 2+ 7Þ ð 6+ 2Þ 6+ 7Þ ð 7+2Þ 7+6Þ (D.15)
ð
ð
ð
1 5 18
ft ðÞ ¼ e 2t e 6t + e 7t
10 2 5
Example D.5
Consider a second order pole
ð
ð s +1Þ s +4Þ
FsðÞ ¼ 2 (D.16)
ð s +2Þ s +6Þ
ð
The residue at the pole, s¼ 2, is given by
ð 2+1Þ 2+4Þ 2t
ð
R s¼ 2 ¼ 2 e
ð 2+6Þ
1 2t
R s¼ 2 ¼ e
8