Question

\[\mathcal L^{-1} \left \{\frac{2s^2 + 1}{(s+1)^2}\right\} = 1 + e^{-t} - 3te^{-t}?\]

Answer 1

take laplace of right hand side and check .are you are getting the original Question.?

Answer 2

\[\frac{1}{s} + \frac{1}{s-1} -\frac{3}{(s-1)^2}??\]

Answer 3

it means something is wrong :P you are not getting original question. are you?

Answer 4

well that's what i got when i turned it into partial fractions

Answer 5

Before Partial fraction make sure it is proper fraction ?

Answer 6

sounds proper to me

Answer 7

uhmm wait...that should be s+1 not s-1

Answer 8

if not long division . i do not think it is proper. long division is required.
Because the degree of both numerator and denominator polynomial is same ??

Answer 9

...well i still think it will come out same :/

Answer 10

wolfram also admits it is improper and do long division first.
http://www.wolframalpha.com/input/?i=Partial+fractions++%282s%5E2%2B1%29%2F%28s%2B1%29%5E2%29

Answer 11

seems im wrong >.<

Answer 12

http://www.wolframalpha.com/input/?i=%282s%5E2+%2B+1%29%2F%28s%2B1%29%5E2

Answer 13

yes it seems :(

Answer 14

so you should first perform long division.
right ?

Answer 15

ugh i dont know anymore =_= i think i'll retire for now....this is the second to the last topic anyway @_@

Answer 16

if you want .
but an interesting function is going to come out of this inverse Laplace;
wanna know ?

Answer 17

sure

Answer 18

ok
have you ever heard of inverse Laplace of 1??.
it is here
\[\Large L^{-1}(1)=\delta(t)\]
where\[\Large \delta(t)\] it is delta function.
when you will perform long division and partial fraction you will have.
\[\Large +2-\frac{4}{s+1}+\frac{3}{(s+1)^2}\]

i hope you can find the inverse of second and third term
for the first term you will have
\[\Large 2L^{-1}(1)=2\delta(t)\]

Answer 19

i have no idea what you just said..

Answer 20

but since it's \[\frac{-4}{s+1} + \frac{3}{(s+1)^2} +2\]
i assume inverse laplace would be \[-4e^{-t} + 3te^{-t} + \frac2s\]
correct?

Answer 21

the first two terms are correct but third one is not.
how an s can be there when you have moved to the t domain ?
always remember when you take Laplace all t should be converted to s
and when you take inverse all s should be back into t.

Answer 22

hmm..im pretty sure it's 2/s though :/

Answer 23

ok lets go from start!!!!
what is Laplace of 1 ?

Answer 24

wolfram saidd it was \[2\delta t\]
o.O wth does that even mean

Answer 25

it mean i was right :)

Answer 26

i think i got meself mixed up..lpalace of 2 is 2/s but it's not the inverse

Answer 27

good .

Answer 28

here we goes to delta function!
inverse Laplace of 1 is delta function.as i pointed out earlier.

Answer 29

always?

Answer 30

always !!!!!!

Answer 31

thanks. i learn something new every hour today

Answer 32

too bad i'll forget them in the morning haha

Answer 33

just one thing more i know you will need this in your questions.
what is the Laplace of t ?

Answer 34

@lgbasallote

Answer 35

1/s^2 i suppose

Answer 36

good . since it is transformation i wanted to tell you.
if you can move from t to s
how to come back from s to t
what is inverse Laplace of s?
\[\Large L^{-1}(s)=??\]

Answer 37

@lgbasallote

Answer 38

ok just m going to answer it.
the inverse Laplace of s is derivative of delta function.
\[\Large L^{-1}(s)=\delta'(t)\]
i just wanted to tell you this because you will need this in very very near future (may be in your right next question) . so that if you encounter it you will be able to do it !!!:)