642 LAPLACE TRANSFORMSHenceL−^1{
5 s^2 + 8 s− 1
(s+3)(s^2 +1)}≡L−^1{
2
s+ 3+3 s− 1
s^2 + 1}≡L−^1{
2
s+ 3}
+L−^1{
3 s
s^2 + 1}−L−^1{
1
s^2 + 1}=2e−^3 t+3 cost−sint,
from (iii), (v) and (iv) of Table 66.1Problem 10. FindL−^1{
7 s+ 13
s(s^2 + 4 s+13)}7 s+ 13
s(s^2 + 4 s+13)≡A
s+Bs+C
s^2 + 4 s+ 13≡A(s^2 + 4 s+13)+(Bs+C)(s)
s(s^2 + 4 s+13)Hence 7s+ 13 ≡A(s^2 + 4 s+13)+(Bs+C)(s).Whens=0, 13= 13 A, from which,A=1.
Equatings^2 terms gives: 0=A+B, from which,
B=−1.
Equatingsterms gives: 7= 4 A+C, from which,
C=3.HenceL−^1{
7 s+ 13
s(s^2 + 4 s+13)}≡L−^1{
1
s+−s+ 3
s^2 + 4 s+ 13}≡L−^1{
1
s}
+L−^1{
−s+ 3
(s+2)^2 + 32}≡L−^1{
1
s}
+L−^1{
−(s+2)+ 5
(s+2)^2 + 32}≡L−^1{
1
s}
−L−^1{
s+ 2
(s+2)^2 + 32}+L−^1{
5
(s+2)^2 + 32}≡ 1 −e−^2 tcos 3t+5
3e−^2 tsin 3tfrom (i), (xiii) and (xii) of Table 66.1Now try the following exercise.Exercise 236 Further problems on inverse
Laplace transforms using partial fractionsUse partial fractions to find the inverse Laplace
transforms of the following functions:1.11 − 3 s
s^2 + 2 s− 3[2et−5e−^3 t]2.2 s^2 − 9 s− 35
(s+1)(s−2)(s+3)[4e−t−3e^2 t+e−^3 t]3.5 s^2 − 2 s− 19
(s+3)(s−1)^2[2e−^3 t+3et−4ett]4.3 s^2 + 16 s+ 15
(s+3)^3[e−^3 t(3− 2 t− 3 t^2 )]5.7 s^2 + 5 s+ 13
(s^2 +2)(s+1)
[
2 cos√
2 t+3
√
2sin√
2 t+5e−t]6.3 + 6 s+ 4 s^2 − 2 s^3
s^2 (s^2 +3)[2+t+√
3 sin√
3 t−4 cos√
3 t]7.26 −s^2
s(s^2 + 4 s+13)[2−3e−^2 tcos 3t−2
3e−^2 tsin 3t]66.4 Poles and zeros
It was seen in the previous section that Laplace trans-forms, in general, have the formf(s)=φ(s)
θ(s). This is
the same form as most transfer functions for engi-
neering systems, atransfer functionbeing one that
relates the response at a given pair of terminals to a
source or stimulus at another pair of terminals.Let a function in the s domain be given by:f(s)=φ(s)
(s−a)(s−b)(s−c)whereφ(s) is of lessdegree than the denominator.