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1 7. H(s) 1 +es 2 +e-3 s2 3 | S S s2

1 7. H(s) 1 +es 2 +e-3 s2 3 | S S s2

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.5: The Binomial Theorem
Problem 15E
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In Exercise 27, use Theorem 8.4.2 to express the inverse transforms in terms of step functions, and then find distinct formulas the for inverse transforms on the appropriate intervals, as in Example 8.4.7. Where indicated by C/G , graph the inverse transform.
Example 8.4.7 Find the inverse Laplace transform h of H(s) = + -e and find distinct formulas for h on appropriate intervals. Solution Let 1 G1(s) = 1 2 G2(s) = 4 1 Go(s) = Then go(1) = 1, g1(t) = t + 2, g2(t) = 21² + 1. Hence, (8.4.12) and the linearity of L¬1 imply that h(t) L-' (Go(s)) – L¯' (e¬³G¡(s)) + L¬' (e¯*G2(s) 1 - u(t – 1) [(t – 1) + 2] + u(t – 4) [2(t – 4)² + 1] = t-u(t – 1)(t + 1) + u(t – 4)(2t² – 161 + 33), which can also be written as 0 <t < 1, 1s1<4, 212 – 16t + 32, t>4. h(t) = -1, Theorem 8.4.2 [Second Shifting Theorem] If t 2 0 and L(g) exists for s > So then L (u(t-t)g(t-t)) exists for s > So and L(u(/ - r)g(- r)) = e"L(g(). or, equivalently, if g(t) G(s). then u(t- -e"G(s). (8.4.12) REMARK: Recall that the First Shifting Theorem (Theorem 8.1.3 states that multiplying a function by eat corresponds to shifting the argument of its transform by a units. Theorem 8.4.2 states that multiplying aLaplace transform by the exponential e by z units. * corresponds to shifting the argument of the inverse transform
Transcribed Image Text:Example 8.4.7 Find the inverse Laplace transform h of H(s) = + -e and find distinct formulas for h on appropriate intervals. Solution Let 1 G1(s) = 1 2 G2(s) = 4 1 Go(s) = Then go(1) = 1, g1(t) = t + 2, g2(t) = 21² + 1. Hence, (8.4.12) and the linearity of L¬1 imply that h(t) L-' (Go(s)) – L¯' (e¬³G¡(s)) + L¬' (e¯*G2(s) 1 - u(t – 1) [(t – 1) + 2] + u(t – 4) [2(t – 4)² + 1] = t-u(t – 1)(t + 1) + u(t – 4)(2t² – 161 + 33), which can also be written as 0 <t < 1, 1s1<4, 212 – 16t + 32, t>4. h(t) = -1, Theorem 8.4.2 [Second Shifting Theorem] If t 2 0 and L(g) exists for s > So then L (u(t-t)g(t-t)) exists for s > So and L(u(/ - r)g(- r)) = e"L(g(). or, equivalently, if g(t) G(s). then u(t- -e"G(s). (8.4.12) REMARK: Recall that the First Shifting Theorem (Theorem 8.1.3 states that multiplying a function by eat corresponds to shifting the argument of its transform by a units. Theorem 8.4.2 states that multiplying aLaplace transform by the exponential e by z units. * corresponds to shifting the argument of the inverse transform
h(6) = ; ++e(+ 1 1 +e-s s2 27. 3 2 +e- s2 4 3s 3 S S s2
Transcribed Image Text:h(6) = ; ++e(+ 1 1 +e-s s2 27. 3 2 +e- s2 4 3s 3 S S s2
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