The Laplace transform of et is Lſe = . If the Laplace transform of f(t) is F(s) and if U(t) is the unit step function, a shift theorem tells us that L[f(t – b)U(t – b)] = eF(s). From this we can deduce that the inverse s-a transform of F(s) = e4 5 U(t – 4)ešt 5 U(t – 4)e*-4 5 U(t – 4)e(e-4) 5 U(t – 5)e“ o o

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.6: Additional Trigonometric Graphs
Problem 77E
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If the Laplace transform of f(t) is F(s) and if U(t) is the unit step function, a shift theorem tells us that L f(t – b)U(t – b)| = e
The Laplace transform of et is L e|
8 — а
0$ F(s). From this we can deduce that the inverse
transform of F(s) = e-48
is:
S -
5 U(t – 4)ešt
5 U(t – 4)ešt-4
5 U(t – 4)e5(t-4)
5 U(t – 5)et
Transcribed Image Text:1 If the Laplace transform of f(t) is F(s) and if U(t) is the unit step function, a shift theorem tells us that L f(t – b)U(t – b)| = e The Laplace transform of et is L e| 8 — а 0$ F(s). From this we can deduce that the inverse transform of F(s) = e-48 is: S - 5 U(t – 4)ešt 5 U(t – 4)ešt-4 5 U(t – 4)e5(t-4) 5 U(t – 5)et
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