Question

Find the inverse laplace transform of e^(-10s)/(s(s^2+3s+2)).

Answer 1

hey, for this problem partial fraction decompose is what I shud say, hehe

Answer 2

So it's e^(-10s)/s+e^(-10s)/(s^2+3s+2)?

Answer 3

the last term can be fraction again, girl

Answer 4

find the inverse of:

\[\frac{ 1 }{ s(s+1)(s+2) }\]

the e^(as) makes you modify each term to F(s-a)

Answer 5

So you do A/s+B/(s+1)+C/(s+2) and you find A, B, C?

Answer 6

yes

Answer 7

don't forget time e^(-10s) to each term

Answer 8

Wait a minute.

Answer 9

nope, hehehe

Answer 10

Mean woman...

Answer 11

sorry made a mistake

Answer 12

hahaha....

Answer 13

the e^(as) ALSO multiplies each terms by:

\[U(t - a)\] where U is the unit step function

Answer 14

and i meant turn the inverse into f(t-a)

Answer 15

after finding f(t) of everything except e^(as) in F(s)

Answer 16

But how do I find A, B, C?

Answer 17

\[\frac{ 1 }{ s(s+1)(s+2) } = \frac{ A }{ s} + \frac{ B }{ s+1 } + \frac{ C }{ s+2 }\]

The partial fraction decomposition is another way of writing the fraction.

If you've ever added fractions, you know that they need to have the same denominator. so make them all the same.

\[\frac{ 1 }{ s(s+1)(s+2) } = \frac{ A(s+1)(s+2) }{ s(s+1)(s+2) } + \frac{ B(s)(s+2) }{s(s+1)(s+2) } + \frac{ C(s)(s+1) }{ s(s+1)(s+2) }\]

\[\frac{ 1 }{ s(s+1)(s+2) } = \frac{ expand }{ s(s+1)(s+2) }\]

Answer 18

yes, thank a lot Euler271

Answer 19

and we are saying that 1 is equal to the expanded numerator which is in terms of A, B and C

Answer 20

let me know