Laplace transform of f(x)= xcos(ax)

  • Thread starter jhat21
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  • #1
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Homework Statement



Laplace transform from f(x) = x cos(ax)
to ^f(s) = (s^2 - a^2) / (s^2 + a^2)^2

how do you get the -a^2 term in the numerator,
all i come up with is s^2?

Homework Equations



f(x) = x e^(ax) ----> ^f(s) = 1/(s-a)^2

2 cos(ax) = e^(aix) + e^(-aix)


The Attempt at a Solution



f(s) = integral[ x/2 (e^-(s-ai)x + e^-(s+ai)x) ]dx
= 1/2[1/(s-ai)^2 + 1/(s+ai)^2]
=1/2[ (s+ai)^2 + (s-ai)^2] / (s^2+a^2)^2
=1/2(s^2+a^2)^2 [ s^2 + 2asi - a^2 + s^2 - 2asi +a^2]
=(s^2 )/ (s^2 + a^2)^2

but the given solution is:
(s^2 - a^2) / (s^2 + a^2)^2

so where am i missing the -a^2 term?
 

Answers and Replies

  • #2
Avodyne
Science Advisor
1,396
88
=1/2[ (s+ai)^2 + (s-ai)^2] / (s^2+a^2)^2
=1/2(s^2+a^2)^2 [ s^2 + 2asi - a^2 + s^2 - 2asi +a^2]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~|wrong sign!
 

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