Question

Help: inverse laplace transform

Answer 1

\[\frac{ -3/10S - 11/10 }{ S^2+2S+5 }\]

Answer 2

\[\frac{ -3/10S }{ S^2+2S+5 } + \frac{ -11/10 }{ S^2+2S+5 }\]

Answer 3

\[-3/10*\frac{ S }{ (S+1)^2+4 }\]

Answer 4

My trouble is the algebraic manipulation at this point

Answer 5

\[-3/10\frac{ (S+1)-1 }{ (s+1)^2 +4}\]

Answer 6

You have formula10 on the table, it says
\[\dfrac{s-a}{(s-a)^2 +b^2}\]
so you need convert s+1=s-(-1)

Answer 7

yes, you are good.

Answer 8

ok , so now observe this :
\[\frac{ -1 }{ (s+1)^2 +4 }\]

Answer 9

@loser66

Answer 10

Yes,

Answer 11

I cant seem to use any laplace transform from my table for that one

Answer 12

So far you have 3 terms, right?
\[(-3/10)\dfrac{s+1}{(s+1)^2+2^2}+(3/10)\dfrac{1}{(s+1)^2+2^2}-(11/10)\dfrac{1}{(s+1)^2+2^2}\]

the last 2 terms gives you \((-4/5)\dfrac{1}{(s+1)^2+2^2}\)
right?

Answer 13

yes.

Answer 14

ok, (4/5) = 2*(2/5) , take (2/5) out and apply formula 9

Answer 15

BRILLIANT !!

Answer 16

:)

Answer 17

Thank you Very Much !