Question

Inverse Laplace transform hint needed

Answer 1

\[L(y)=\frac{e^{-s}}{s(s^2+1)}-\frac{e^{-2s}}{s(s^2+1)}\]

Answer 2

partial

Answer 3

you think partial fractions will work here? if so, how can I do it for an exponential?

Answer 4

oh true

Answer 5

yep that's my issue too

Answer 6

no

Answer 7

is it just do it for 1/ s(s^2+1) then multiply by the exp?

Answer 8

oh is it umm shift?

Answer 9

i think heaveside

Answer 10

expect step functions as there there e^-s and e^-2s terms

Answer 11

yep

Answer 12

and the s on the bottom

Answer 13

L(U_c(t) f(t-c))=e^-cs * L(f(t))

Answer 14

that doesn't help here I don't think

Answer 15

\[\mathcal{L}^{-1}\left\{e^{-cs}F(s)\right\} = u_c(t) f(t-c)\]

Answer 16

yes, I know both of you are right about when there is a function of t

Answer 17

make it t-c

Answer 18

notice that your F(s) is same for both the pieces

Answer 19

we don't have to make it f-c

Answer 20

\[F(s) = \dfrac{1}{s(s^2+1)}\]

Answer 21

yea, I distributed the s^2+1

Answer 22

find the inverse of that first ^

Answer 23

but that is my issue

Answer 24

do I do the partial fraction or no?

Answer 25

should I use a convolution integral?

Answer 26

p frac works

Answer 27

whats the convolution integral for

Answer 28

where we do \[f(x)=\frac{1}{s^2+1}~~and~~g(x)=\frac{e^{-s}}{s}-\frac{e^{-2s}}{s}\]

Answer 29

f(s)*g(S) = f(t)*g(t)
u wanna extra f(t) and g(t) from there?? thats hardd woorkk

Answer 30

s*

Answer 31

you can use convolution, but p fracs is much sumpler here : \[\dfrac{1}{s(s^2+1)} = \dfrac{(s^2+1)-s^2}{s(s^2+1)} = \dfrac{1}{s} - \dfrac{s}{s^2+1}\]

Answer 32

I guess I'm asking is that legal?

Answer 33

yes convolution is an alternative legal method for finding inverses

Answer 34

you want to work it using p fracs or convolution ?

Answer 35

i wanna see convolution

Answer 36

I'm gonna do p fracs cause they're a heck of a lot simpler

Answer 37

=[

Answer 38

btw where did that minus come from ganeshie?

Answer 39

dan :
http://tutorial.math.lamar.edu/Classes/DE/ConvolutionIntegrals.aspx

Answer 40

nvm

Answer 41

\[\dfrac{1}{s(s^2+1)} = \dfrac{(s^2+1)\color{red}{-}s^2}{s(s^2+1)} = \dfrac{1}{s} \color{red}{-} \dfrac{s}{s^2+1}\]

Answer 42

oh so i can do L^-1(F(s)) convolution L^-1(G(S)) ?

Answer 43

but how do \[\frac{se^{-2s}}{s^2+1}\]

Answer 44

another partial frac?

Answer 45

nope we are done, let me write the last line

Answer 46

but how?

Answer 47

wait huh?

Answer 48

that doesn't make sense

Answer 49

hmm wait

Answer 50

u might also have to put it in f(t-c) form if u have e^-cs multiplying

Answer 51

that equation works only for this doesnt it
http://prntscr.com/5h1k4s

Answer 52

yeah shifting is missing

Answer 53

it makes sure we have unique solutions then :)

Answer 54

I have \[\frac{e^{-s}}{s}-\frac{se^{-s}}{s^2+1}-\frac{e^{-2s}}{s}+\frac{se^{-2s}}{s^2+1}\]

Answer 55

otherwise the tails for the previous f(t) will play a role when shifted

Answer 56

http://www.wolframalpha.com/input/?i=inverse+laplace+of+%28se%5E%28-s%29%29%2F%28s%5E2%2B1%29

Answer 57

\[F(s)=\frac{e^{-s}}{s(s^2+1)}-\frac{e^{-2s}}{s(s^2+1)}\]

say \(g(s) = \frac{1}{s(s^2+1)}\), the inverse would be \(g(t) = 1-\cos(t)\)

\[f(t) =u_1(t) g(t-1) - u_2(t) g(t-2)\]

Answer 58

but we should distribute ya?

Answer 59

i would leave it like that
plugging in 1-cost for g(t) makes it look messy

Answer 60

I'm getting different answer

Answer 61

http://www.wolframalpha.com/input/?i=inverse+laplace+of+%5B%28e%5E%28-s%29%29%2F%28s%29-%28se%5E%28-s%29%29%2F%28s%5E2%2B1%29-%28e%5E%28-2s%29%29%2F%28s%29%2B%28se%5E%28-2s%29%29%2F%28s%5E2%2B1%29%5D

Answer 62

ok so I still have to shift these no matter what

Answer 63

http://www.wolframalpha.com/input/?i=inverse+laplace+%5Cfrac%7Be%5E%7B-s%7D%7D%7Bs%28s%5E2%2B1%29%7D-%5Cfrac%7Be%5E%7B-2s%7D%7D%7Bs%28s%5E2%2B1%29%7D

Answer 64

wolfram agrees wid our answer, it just uses a different notation for step function :
\[u_c(t) \sim \theta(t-c)\]

Answer 65

no it has extra

Answer 66

your missing the -1s

Answer 67

are you refering to this
http://prntscr.com/5h1n53

Answer 68

yea

Answer 69

hmm i dont see anything missing yet

Answer 70

f(t) =u_1(t) [g(t-1)-1] - u_2(t) [g(t-2)-1]

Answer 71

we have defined g(t) = 1-cos(t)

Answer 72

....I'm still confused... I didn't do it that way

Answer 73

thats how paul does
maybe plugin g(t) into the final function and see what you get

Answer 74

now I see, but how do I shift it to make it work?

Answer 75

it is already shifted.

Answer 76

before that

Answer 77

I don't see the shift

Answer 78

wouldn't it be s-2 on top then?

Answer 79

ohhhhhhhhhhh

Answer 80

ok i get it

Answer 81

:) this is the final inverse
http://prntscr.com/5h1our

Answer 82

yea, that's what I got

Answer 83

thank you

Answer 84

thank you both

Answer 85

all that for 6 steps woo!

Answer 86

2 steps.

Answer 87

6 steps on my paper from start to finish