Laplace transform of (e^-s) / [ (s)(s-3) ]

foo9008 · Jun 11, 2016

Homework Statement

find the laplace transform of (e^-s) / [ (s)(s-3) ]

since there's (e^-s) which can be found in L { f(t-a) H(t-a) } = (e^-(as)) F(s) , so , i found a = 1 , then i found F(s) = 1/ [ (s)(s-3) ] , formula :

i have attached the working below , is it correct ? btw , the author's ans doesn't have H(t-a) , is the author's ans wrong ?
author's ans :

$Handwritten partial fraction result: -1/3 + (1/(3e^3)) e^(3t).$

Homework Equations

The Attempt at a Solution

my working :

$Handwritten partial fraction working: e^-s/(s(s-3)) rewritten as e^-s times 1/(s(s-3)), then 1/(s(s-3)) = A/s + B/(s-3), giving A = -1/3 and B = 1/3, and the inverse transform -1/3 + (1/3)e^3t, with a step showing the H(t-1) shifted form.$

which is correct ? the author ? or me ?

Ray Vickson · Jun 12, 2016

foo9008 said:

Homework Statement

find the laplace transform of (e^-s) / [ (s)(s-3) ]

since there's (e^-s) which can be found in L { f(t-a) H(t-a) } = (e^-(as)) F(s) , so , i found a = 1 , then i found F(s) = 1/ [ (s)(s-3) ] , formula :

$Photo of a printed textbook box reading Theorem 4.8 (Second Shift Property): if a > 0 and L{f(t)} = F(s), then L{f(t - a)H(t - a)} = e^(-as) L{f(t)} = e^(-as) F(s).$

i have attached the working below , is it correct ? btw , the author's ans doesn't have H(t-a) , is the author's ans wrong ?
author's ans :

Homework Equations
The Attempt at a Solution

my working :

which is correct ? the author ? or me ?

There is no way that an author would have asked you to find the Laplace transform of ##e^{-s}/(s(s-3))##. Are you sure it does not ask for the INVERSE Laplace transform? There is a huge difference between these two concepts.

Laplace transform of (e^-s) / [ (s)(s-3) ]

Homework Statement

Homework Equations

The Attempt at a Solution

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Homework Statement

Homework Equations

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