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-

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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