EXAMPLE 2-4 Obtain the inverse Laplace transform of G(s) s³ + 5s² + 9s + 7 = (s + 1)(s + 2) Here, since the degree of the numerator polynomial is higher than that of the denominator polynomial, we must divide the numerator by the denominator. $ + 3 G(s) = s +2+ (s + 1)(s + 2) Note that the Laplace transform of the unit-impulse function ε(t) is 1 and that the Laplace trans- form of dd(t)/dt is s. The third term on the right-hand side of this last equation is F(s) in Example 2-3. So the inverse Laplace transform of G(s) is given as g(t) = d dt 8(t) + 28(t) + 2e − e−21, for t≥ 0-
EXAMPLE 2-4 Obtain the inverse Laplace transform of G(s) s³ + 5s² + 9s + 7 = (s + 1)(s + 2) Here, since the degree of the numerator polynomial is higher than that of the denominator polynomial, we must divide the numerator by the denominator. $ + 3 G(s) = s +2+ (s + 1)(s + 2) Note that the Laplace transform of the unit-impulse function ε(t) is 1 and that the Laplace trans- form of dd(t)/dt is s. The third term on the right-hand side of this last equation is F(s) in Example 2-3. So the inverse Laplace transform of G(s) is given as g(t) = d dt 8(t) + 28(t) + 2e − e−21, for t≥ 0-
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Inverse laplace transform. I would like to understand the details of the method for doing this. Thank you in advance for your answer.
Transcribed Image Text:EXAMPLE 2-4
Obtain the inverse Laplace transform of
G(s)
s³ + 5s² + 9s + 7
= (s + 1)(s + 2)
Here, since the degree of the numerator polynomial is higher than that of the denominator
polynomial, we must divide the numerator by the denominator.
$ + 3
G(s) = s +2+
(s + 1)(s + 2)
Note that the Laplace transform of the unit-impulse function ε(t) is 1 and that the Laplace trans-
form of dd(t)/dt is s. The third term on the right-hand side of this last equation is F(s) in Example
2-3. So the inverse Laplace transform of G(s) is given as
g(t)
=
d
dt
8(t) + 28(t) + 2e − e−21,
for t≥ 0-
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