Laplace Transform help!

cunnins4

Hi guys, i'm having a bit of trouble with the following question, if someone could shed some light i'd appreciate it:

here's the question:


it's the right hand side that's bothering me

I keep getting the laplace transform of the RHS to be 1/s - 2*(exp^-3*s)/s but the solution says it's 1/(s+3)??? There's no mention of exp in the solution, which seems really odd!

Michael Collins

Don't see how the answer can avoid having e^s terms in it. I get pretty much the same as you for the RHS, i.e.

The RHS translates to

u(t) + u(t-2) - 2u(t-3), taking the LT of this we get:

[1+e^(-2s) - 2e^(-3s)]/s

The answer 1/(s+3) is just a decaying exponential in the time domain, which is certaintly not what the RHS is! I'd say your answer is wrong...

joe_chicken

Agreed, I got the same answer as above.
[1+e^(-2s) - 2e^(-3s)]/s

The right hand side contains the sum of 2 step functions (one delayed), and from what little I remember in Laplace, a delayed step function is given by an expression with exponentials in it

Maybe 1/(s+3) has something to do with the factorising of the LHS?
(i.e. with f'(0) = 0 and f(0) = 0 => L[f'' + 5f' + 6] = F(s)(s^s +5s + 6) = F(s)(s+3)(s+2))

cunnins4

I actually posted wrong earlier: i got the same as you guys, i just forgot to type the other exponentials. Thanks for clearing that up for me. It was wrecking my head, and the other lads in college were confused too.

I'd imagine when he was doing the solution he just wrote one of the factors of the lhs or something by accident.

Sure not to worry.

Cheers again lads.