What is the Laplace inverse of F(s) = [3*e^(-4s)] / [s*(s^2 + s + 5/4)]?

[Yura]

Homework Statement

just a small part of a large problem ><a I am not even sure if this is in the right place. the first two questions i have to ask are really minor so i didnt want to make separate threads for each one. the third one is an actual problem itself tho.

(1) i need the laplace inverse of :
[s^2]/[s^2 - 3^2] -solved, there was an error at the start that threw the rest of my eqn off. there is no s^2

(2) laplace transformation of:
u(t-3) - solved

(3) find laplace inverse of:
F(s) = [3*e^(-4s)] / [s*(s^2 + s + 5/4)]

Homework Equations

for the second one, the u comes from working with time intervals using the second shift theorem
which is: (L(g(t - k)u(t - k)) = e^(-k*s)*G(s)
where G(s) is the laplace transform of of g(t)
the third question also uses

The Attempt at a Solution

for the first one, i have no idea of how to even look at this because I've been using a laplace conversion sheet to solve them all up until now. the only laplace transformation I've seen with s^2 in the numerator looks just like the problem i have here but the intire denominator is squared awell.. other than that i can't find a laplace transform to matchit. -solved


the second one seems simple enough but I am stuck. it was originally part of another function i had to laplace, but it was too complicated to laplace as it was and i singled that much out. the original term was:

(t - 2)u(t - 3) and so i broke this into smaller terms so thati could use the second shifting theorem on one of the termsand got:

(t - 3)u(t - 3) + u(t - 3)
[im leaving out the multiplication "*" signs so its easier to look at]

then that laplaced to:

e^(-3*s)/[s^2] + [the laplace of the second term]

im not quite sure whether this just counts as one and laplaces to (1/s) or something. -solved

for question three i took out the e and its power because the -4*s would be used for the second shifting theorem. leaving 3/[s*(s^2 + s + 5/4)]

from here i tried to use partial fractions on the equation to make it easier, but i keep ending up getting 2 different values for my first variable.

also with the (s^2 + s + 5/4) i perfected the square to:
[(s + 1/2)^2 + 1]

if i could work the partial fraction for this, i might beable to solve the laplace inverses for the single terms, using the laplace transformation chart. so the main problem is the partial fraction.
i had to "un-perfect the square" to make this part easier

what i have for that is:
3/[s*(s^2 + s + 5/4)] = A/s + B*(s^2 + s + 5/4)
3/[s*(s^2 + s + 5/4)] = [A*(s^2 + s + 5/4) + B*s] / [s*(s^2 + s + 5/4)]
3 = A*(s^2 + s + 5/4) + B*s
3 = A*s^2 + A*s + A*5/4 + B*s
3 = A*s^2 + (A+B)*s + A*5/4

this is what is giving me two separate values for A,
where due to the first term, A must equal zero.
but due to the second term A must equal to 12/5.

thanks if you can help

edit: i clicked off of the form without realising and accidentally submit. i think.

 

[Nino MMP]

The solution to (1) is that there is no s^2 in the denominator, so it is just 1/[s^2 - 3^2]. The Laplace inverse of this is e^(3t). For (2), the Laplace transform of u(t-3) is e^(-3s). For (3), using partial fractions, we can write F(s) = 3/[s*(s^2 + s + 5/4)] = A/s + B*(s^2 + s + 5/4)Multiplying both sides by s*(s^2 + s + 5/4) gives 3 = A*(s^2 + s + 5/4) + B*sSubstituting s = 0 yields B = -3/5. Substituting s = -1/2 yields A = 12/5. Therefore, F(s) = [12/5]/s + [-3/5]*(s^2 + s + 5/4) The Laplace inverse of F(s) is 12/5*e^(-t/2) + [-3/5]*(cos(sqrt(5)*t/2)+sin(sqrt(5)*t/2)/sqrt(5)).