Question

Inverse laplace trasform of
s^2+2s+3/(s^2+2s+2) (s^2+2s+5)

Answer 1

i'd say use partial fractions first

Answer 2

when i factorise the bottom half i get complex number?

Answer 3

i have attempted partial fractions .. still no look

Answer 4

for the first part of the fraction i got (-1-i)(-1+i) , second part i got (-1-2i) (-1+2i)

Answer 5

then i did a/(-1-i)+b(-1+i)+c/-1-2i+b/(-1+2i)

Answer 6

see this\[\frac{s^2+2s+3}{(s^2+2s+2) (s^2+2s+5)}\]\[\frac{1}{2}\frac{2(s^2+2s+3)}{(s^2+2s+2) (s^2+2s+5)}\]\[\frac{1}{2}\frac{(s^2+2s+2)+(s^2+2s+5)-1}{(s^2+2s+2) (s^2+2s+5)}\]

Answer 7

is that the corrrect way ?

Answer 8

whats with the 1/2 and -1 where is that come from?

Answer 9

yeah see\[\frac{1}{2}\frac{(s^2+2s+2)+(s^2+2s+5)-1}{(s^2+2s+2) (s^2+2s+5)}\]\[\frac{1}{2}({\frac{1}{s^2+2s+5}+\frac{1}{s^2+2s+2}}-\frac{1}{(s^2+2s+2) (s^2+2s+5)})\]

Answer 10

right from this step you factorise ?