Answer :
To find the inverse Laplace transform of F(s) = (4s + 5) / ((s^2 + 1)(8s^2 + 4s + 9)), we can use partial fraction decomposition and then refer to the Laplace transform table to find the inverse transforms of the individual terms.
First, let's factorize the denominator: (s^2 + 1)(8s^2 + 4s + 9) = (s^2 + 1)(2s + 1)^2 + 8.
Now, we can rewrite the F(s) using partial fraction decomposition:
F(s) = A/(s^2 + 1) + (Bs + C)/(2s + 1) + D/(2s + 1)^2 + E/(s^2 + 1)
Multiplying through by the denominator, we have:
4s + 5 = A(2s + 1)^2 + (Bs + C)(s^2 + 1) + D(s^2 + 1)(2s + 1) + E(s^2 + 1)
Expanding and equating coefficients, we get the following system of equations:
4 = 4A + D + E
5 = A + C + D
0 = 2A + C + 2D
0 = 2A + B
0 = A
Solving the system of equations, we find A = 0, B = -2, C = 4, D = -1, and E = 4.
Therefore, F(s) can be written as:
F(s) = (-2s + 4)/(2s + 1) - 1/(2s + 1)^2 + 4/(s^2 + 1)
Now, we can look up the inverse Laplace transforms of each term in the Laplace transform table:
Inverse Laplace transform of (-2s + 4)/(2s + 1) is -2e^(-t) + 4e^(-t/2)
Inverse Laplace transform of -1/(2s + 1)^2 is -te^(-t/2)
Inverse Laplace transform of 4/(s^2 + 1) is 4sin(t)
Combining these results, the inverse Laplace transform of F(s) is:
f(t) = -2e^(-t) + 4e^(-t/2) - te^(-t/2) + 4sin(t)
Therefore, the inverse Laplace transform of F(s) is given by f(t) = -2e^(-t) + 4e^(-t/2) - te^(-t/2) + 4sin(t).
Learn more about Laplace transform here -: brainly.com/question/29583725
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