Question

find the inverse laplace of 2s exp^-s/(s-1)^2

Answer 1

Could you rewrite that using the equation tool?

Answer 2

y u dont understand?

Answer 3

\[(2s \times e^{-s})/(s-1)^2\]
?

Answer 4

yes this is right.

Answer 5

Have you tried anything so far?

Answer 6

yes but im little bit confuse.

Answer 7

Show us what you've tried so far and we'll see if we can help

Answer 8

ok my ans is 2exp^t-1.H(t-1) but the book ans is different.

Answer 9

What's the book answer?

Answer 10

2t.exp^t-1H(t-1).

Answer 11

Did you do this with partial fractions?

Answer 12

no.i use unit step function.

Answer 13

After applying time shifting, you'll need to take inverse laplace of
2s/(s-1)^2

how did you find it?
you say you got 2e^(t) for this?

Answer 14

lets see :)
Applying frequency shifting,

\(\Large L^{-1}[ 2s/ (s-1)^2] = e^t L^{-1} [2(s+1)/s^2 ]\)

clear till here ?

Answer 15

because next steps are quite trivial !

Answer 16

the book ans is 2t.exp^t-1H(t-1).

Answer 17

i need this answer.

Answer 18

lets see how we can get there :)

you understood the properties i applied?
Have the inverse laplace transform table with you ?

Answer 19

yes i have.

Answer 20

so from 2s exp^-s/(s-1)^2
we came to 2 (s+1)/s^2 by applying 2 properties.

you understood both ?

Answer 21

for better clarity, do you want me to give you the steps till that step?

Answer 22

yes give me details plz

Answer 23

give me a moment to type,
by that time, please think of next steps :)

Answer 24

ok.thank u.

Answer 25

Since \(\Large L^{-1} [e^{-as}F(s)] = f(t-a) H(t-a) \)

we first find
\(\Large L^{-1} [\dfrac{2s }{(s-1)^2}] \)
and then use the above property

Now
\(\Large L^{-1}f(s-a) = e^{at}f(t) \)
so
\(\Large L^{-1} [\dfrac{2s }{(s-1)^2}] = e^t L^{-1} [\dfrac{2(s+1) }{(s)^2}] \)

see whether you get these steps first :)

Answer 26

\(\Large L^{-1}[f(s-a)] = e^{at}L^{-1} [f(s)]\)

Answer 27

ok thank u so much.and which softwaer u used to write this things clearly.

Answer 28

last thing is tell me the the laplace of exp^t-1 is what??

Answer 29

used \(Latex \), which is supported by this site.
you can use it too :)

Answer 30

can i download it?/

Answer 31

why would you need that ?
we're here :
\(e^t L^{-1} [\dfrac{2(s+1) }{(s)^2}]\)
you'll get the same answer which is in you book, be patient lol
we're just half way done

Answer 32

and for the latex
you can just start typing in `\( ..\)`

try it
type `\(Hi\)`
and see what u get

Answer 33

ok

Answer 34

so we need to solve
\(e^t L^{-1} [\dfrac{2(s+1) }{(s)^2}]\)

`\(e^t L^{-1} [\dfrac{2(s+1) }{(s)^2}]\)`

any ideas how to proceed?

Answer 35

sorry exp ^t-x laplace??

Answer 36

sorry, didn't get your question.

we're solving an inverse laplace question,
and I just asked you whether you know how we can proceed with
\( L^{-1} [\dfrac{2(s+1) }{(s)^2}]\)

are we on a same page?

Answer 37

i dont understand why u right 2(s+1) in numerater?

Answer 38

glad that you ask doubts :)

\(\Large L^{-1}[f(s-a)] = e^{at}L^{-1} [f(s)]\)
This is a standard property

what does it say? If you want to remove e^at from inverse laplace ,
replace 's' by 's+1' in your function <<Note how s-1 on left became s>>

Answer 39

so applying that property to your function,
the numerator s became (s+1)

Answer 40

correction :
If you want to remove e^at from inverse laplace ,
replace 's' by 's+a' in your function <<Note how s-a on left became s>>

Answer 41

can u right the correct ans which i post in above comments??

Answer 42

if I just post all the steps, you'll understand none of them.
I don't suppose you want that...
thats why I am posting few steps and asking you whether you got them...

if this approach doesn't suit you, I am really not the right person to help you...

Answer 43

ok thank u so much to give me precious time.nice to meet u.

Answer 44

ah, you too :)

Answer 45

here
\[F(s)=\frac{ 2s.e^{-s} }{ (s-1)^2 }\\s~In ~laplace~domain~means~\frac{ d }{ dt }~In ~time~domain\\f(t)=2.\frac{ d }{ dt }L^{-1}[\frac{ e^{-s} }{ (s-1)^2 }]=2.\frac{ d }{ dt }[(t-1)e^{t-1}H(t-1)]\\used~frequency~shifting~In~last~step\\f(t)=2[(t-1)e^{t-1}+e^{t-1}]H(t-1)=2t.e^{t-1}H(t-1)\]
easy this way :) @Yasirist