Inverse Laplace Transform

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i am having trouble with some aspects of Laplace transform, could anyone help me with this problem

Find the inverse Laplace transform.
4(s+1)/s²-16

1/(s²+9)

1/(s+1)(s+2)
Laplace is new to me so your help would be greatly appreciated.
 

4(s+1)/s²-16 here you have poles in s=+4 and s=-4 so you aplly fraction expansion of the form A/(s+4) + B/(s-4) find A anb and take the inverse of A/(s+4) and B/(s-4) which is A exp(-4t)u(t) + Bexp(4t)u(t).

1/(s²+9) the inverse Laplace transform of this function can be found directly from tables: It is sin(3t)u(t).


1/(s+1)(s+2) Here again you should use fraction expansion : A/(s+1) + B/(s+2) find A and B and take the inverse and you will find Aexp(-t)u(t) + Bexp(-2t)u(t).

You shoul take a look at fraction expansion method. I hope have helped.
 

Intuition or experience is ok,better way is to use the definition of this kind of transformations,in the textbook
 

any rational function can be expanded and after that simple inverses exist.
 

hi

ilaplace(4*(s+1)/(s*s-16)) ---------> (5/2 exp(4 t) + 3/2 exp(-4 t))*u(t)

ilaplace(1/(s*s+9)) ---------> 1/9*(9^(1/2))*sin(9^(1/2)*t) *u(t)

ilaplace(1/(s+1)(s+2)) ----------> exp(-t)-exp(-2*t)

bye
 

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