## anonymous one year ago find the inverse laplace of 2s exp^-s/(s-1)^2

1. Luigi0210

Could you rewrite that using the equation tool?

2. anonymous

y u dont understand?

3. anonymous

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

4. anonymous

yes this is right.

5. Luigi0210

Have you tried anything so far?

6. anonymous

yes but im little bit confuse.

7. Luigi0210

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

8. anonymous

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

9. Luigi0210

10. anonymous

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

11. Luigi0210

Did you do this with partial fractions?

12. anonymous

no.i use unit step function.

13. hartnn

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?

14. hartnn

lets see :) Applying frequency shifting, $$\Large L^{-1}[ 2s/ (s-1)^2] = e^t L^{-1} [2(s+1)/s^2 ]$$ clear till here ?

15. hartnn

because next steps are quite trivial !

16. anonymous

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

17. anonymous

18. hartnn

lets see how we can get there :) you understood the properties i applied? Have the inverse laplace transform table with you ?

19. anonymous

yes i have.

20. hartnn

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

21. hartnn

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

22. anonymous

yes give me details plz

23. hartnn

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

24. anonymous

ok.thank u.

25. hartnn

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 :)

26. hartnn

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

27. anonymous

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

28. anonymous

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

29. hartnn

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

30. anonymous

31. hartnn

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

32. hartnn

and for the latex you can just start typing in $$..$$ try it type $$Hi$$ and see what u get

33. anonymous

ok

34. hartnn

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?

35. anonymous

sorry exp ^t-x laplace??

36. hartnn

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?

37. anonymous

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

38. hartnn

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>>

39. hartnn

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

40. hartnn

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>>

41. anonymous

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

42. hartnn

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...

43. anonymous

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

44. hartnn

ah, you too :)

45. sidsiddhartha

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