e^(-5s)/(s^2+3s+2)
Answer: u_{5}(t)[-e^(-2(t-5))+e^(-(t-5))]
How do I get the answer?
e^(-5s)/(s^2+3s+2)
Answer: u_{5}(t)[-e^(-2(t-5))+e^(-(t-5))]
How do I get the answer?
Typically, it is extremely complicated to directly solve for the inverse laplace transform of a function. This is why we use tables. With that being said, we want to manipulate the function to match an equation in the table of laplace transforms.
e^(-5s)/(s^2+3s+2) = e^(-5s)*(1/(s^2+3s+2) = e^(-5s)*(1/s(s+3+2/s)) = (e^(-5s)/s)*(1/(s+3+2/s)) = (e^(-5s)/s)*(s/(s+1)(s+2))
Now, doing partial fractions:
H(s) = A/(s+1) + B/(s+2)
s/(s+1)(s+2) = A(s+1)+B(s+2)/(s+1)(s+2)
s = A(s+1)+B(s+2)
B = -1
A = 1
H(s) = -1/(s+2) + 1/(s+1)
inverse laplace transform of H(s) = -e^(-2t) + e^(-t)
applying inverse laplace trasform of e^(-5s)/s gives the heavy-side function u_{s}(t) = u(t-c) where u_{s}(t) = t-5
plugging in we get:
-e^(-2(t-5))+e^(-(t-5))
I think this is correct.