## ragman find inverse laplace transform of F(s)= (s^2+2s+5/(s^2(s+1)) one year ago one year ago

1. Skaematik

note

2. ragman

'note' sorry this is my first time using this what do you mean by that?

3. Skaematik

Im just bookmarking this question so that I can return to it later. :D Id want to know the answer too

4. ragman

oh ok sounds good thanks

5. lgbasallote

your question is $\huge \mathcal L ^{-1} \left \{ s^2 + 2s + \frac 5{s^2(s+1)}\right \}$

6. lgbasallote

??

7. lgbasallote

@ragman you there?

8. ragman

hey sorry about the late reply but my question is actually (s^2+22+5) / (s^2(s+1) sorry i forgot the other )

9. ragman

@lgbasallote

10. lgbasallote

uhh s^2 + 22 + 5??

11. ragman

dang i keep doing typos 2s not 22 so yeah its (s^2+2s+5)

12. lgbasallote

so.. $\huge \mathcal L ^{-1} \left \{ \frac{s^2 + 2s + 5}{s^2(s+1)} \right \}$

13. ragman

yup perfect

14. hartnn

can u do partial fractions ??

15. ragman

@hartnn yeah thats what i was doing but for some reason i could figure out the other two variable maybe im solving it incorrectly but i got -3 for A and 5 for B

16. lgbasallote

where's C?

17. hartnn

those are correct, so u need only C right ? where u took C above s+1

18. lgbasallote

your P.F. should be $\huge \frac As + \frac B{s^2} + \frac C{s+1}$

19. lgbasallote

so you have $\implies As(s+1) + B(s+1) + Cs^2 = s^2 + 2s + 5$

20. ragman

oh ok so thats where i messed up i thought that where C is it should be (Cs+D) so i was trying to solve for both of those variables

21. hartnn

nopes only C, if it were s^2+1 u would have taken Cx+D

22. hartnn

so i think u can get C now, right?

23. ragman

oh i see now and yeah i believe i can get C now trying it right now

24. lgbasallote

i think it will change your A value too

25. lgbasallote

so you best check

26. hartnn

great tell us what u get C as....we all will verify that. A is correct.

27. lgbasallote

nope nevermind..it stays the same

28. ragman

ok sounds good got B=5 finding A now

29. ragman

A=-3 and C=4 is that what you guys got?

30. hartnn

yup, same.Good work.

31. lgbasallote

A =-3?

32. lgbasallote

yeah that's right

33. ragman

ok cool i was about to write out how i got it lol

34. lgbasallote

everything's right

35. ragman

and now i just plug back all the variables right

36. hartnn

yup, and use inverse laplace formulas.

37. lgbasallote

yep find $\huge \mathcal L ^{-1} \left \{ -\frac 3s + \frac 5s + \frac 4{s+1} \right \}$

38. lgbasallote

uhh that should be 5/s^2

39. ragman

ok sorry im just learning this but how do you go about doing that im trying to find the L^-1 formula but my notes arent really helping

40. lgbasallote

here's a hint: $\huge \mathcal L \{ a \} = \frac as$ $\huge \mathcal L \{ t^n\} = \frac{n!}{s^{n+1}}$ $\huge \mathcal L \{e^{at} \} = \frac{a}{s -a}$ does that help?

41. lgbasallote

uhhh wait... wrong formula for last one

42. lgbasallote

$\huge \mathcal L \{e^{at} \} = \frac{1}{s-a}$ better

43. ragman

yes thats very helpful but im getting confused with what n should be in the middle equations

44. lgbasallote

n is the exponent

45. ragman

so it would be 2?

46. lgbasallote

for example: $\huge \mathcal L \{ t^2 \} = \frac{2!}{s^{2+1}} \implies \frac{2}{s^3}$

47. ragman

thats whats a lil confusing to me because on the denominator it would have to be n=1 so that it would equal s^2 but on the numerator 1! equals 1 right ?

48. hartnn

yes thats true, so u will have 5t^1= 5t ....isn't it ? u can verify that u should get final answer as 5t-3+4e^(-t) did u get that?

49. ragman

yup thats what i got thanks

50. hartnn

welcome :)