What is the Inverse Fourier Transform of (3jw+9)/((jw)^2+6jw+8)?

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SUMMARY

The inverse Fourier transform of the function F(w) = (3jw + 9) / ((jw)^2 + 6jw + 8) can be approached by factoring the denominator and utilizing the convolution property of Fourier transforms. The roots of the denominator are found to be -2 and -4, which do not simplify the numerator directly. The discussion emphasizes the application of the exponential Fourier transform pair, specifically exp(-at)u(t) <-> 1/(jw + a), to facilitate the transformation process.

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Homework Statement


(part of a problem)
Find the inverse Fourier of F(w) = (3jw+9)/((jw)^2+6jw+8)
where w is the angular frequency, w=2pi * f = 2*pi/T

Homework Equations


The fourier transfrom and its properties i guess.
Also the exponential FT common pair exp(-at)u(t) <-> 1/(jw+a)
where exp is the exponential function and u(t) the unit step function


The Attempt at a Solution


I factored out the denominator in a hope that the 3jw+9 would cancel out with a possible root of jw=-3 , but the roots are -2 and -4.
I've been trying to separate F into a product of easy transformable parts, to take advantage of the convolution property : x(t) [convolve] f(t) = X(w)F(w) , but i can't get rid of the nominator to apply the exponential Fourier pair.
Any hints?
 
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If the nominator was a constant, i could break it into simple fractions.
Can you do it if it's not a constant?
 
^
Ah,nevermind, you can.
case closed.
 

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