Question

Hi all... have a bit of a challengeing question if anyone has some time to help please...?

i'm totally lost.


Find the inverse Laplace transform of the following function.

\[\large F(s) = \frac {10}{s(s+2)(s+3)^2}\]

Answer 1

\[\large F(s) = \frac {10}{s(s+2)(s+3)^2}\]

Answer 2

@ganeshie8 or @skullpatrol ...if u guys have a spare couple of minutes, wood love a hand please?

Answer 3

This looks nothing more than an incredibly tedious partial fraction decomposition problem :3

Answer 4

ok... but u lost me at decomposition

Answer 5

partial fractions.
Just split the bloody bugger into partial fractions.

Crud... this is going to be tedious XD

Answer 6

\[\Large F(s) = \frac{A}{s}+ \frac{B}{s+2}+ \frac{C}{s+3}+ \frac{D }{(s+3)^2}\]

Answer 7

there's no way i can just split it somehow and use the laplace transform tables to reverse it somehow?

Answer 8

This... IS splitting it somehow :D

Answer 9

or something

Answer 10

I haven't done Laplace transforms in about 18 months XD

Answer 11

\[\Large F(s) = \frac{A}{s}+ \frac{B}{s+2}+ \frac{C}{s+3}+ \frac{D }{(s+3)^2}\]
is this one too many? should the C/s+3 disappear?

Answer 12

Let's try. Maybe I did overdo things XD

Answer 13

It is rather daunting, isn't it? :D
\[\Large \frac{A(s+2)(s+3)^2+ Bs(s+3)^2 + Cs(s+2)}{s(s+2)(s+3)^2}\]

Answer 14

Wonder what this tells me...
A+B = 0

Answer 15

whoa...whoa... that's ... ummm... huge

Answer 16

wait... why does A + B = 0
sorry man, i'm missing some logic thing there

Answer 17

Because I said so.
Isn't that logic enough? :)

Answer 18

LOL kidding.
Hang on...

Answer 19

lawl ;P

Answer 20

When you evaluate all that out, the coefficient of \(x^3\) is going to be A+B.
Try it.

Answer 21

Convinced?

Answer 22

Actually, I don't think this works at all. We need that D.

Because this'll probably give us an overdefined system.

Answer 23

no, sorry, haven't dont partial fractions in ...11 months... where does x come in again?

Answer 24

Oi... you better brush up on partial fractions, mate :P
Laplace transforms crawl with them XD

Answer 25

\[\Large F(s) = \frac{A}{s}+ \frac{B}{s+2}+ \frac{C}{s+3}+ \frac{D }{(s+3)^2}\]

Answer 26

Don't worry.. it's not THAT hard... hopefully XD
\[\large F(s) = \frac{A(s+2)(s+3)^2 + Bs(s+3)^2 + Cs(s+2)(s+3) + Ds(s+2)}{s(s+2)(s+3)^2}\]

Answer 27

crud...
\[F(s) = \frac{A(s+2)(s+3)^2 + Bs(s+3)^2 + Cs(s+2)(s+3) + Ds(s+2)}{s(s+2)(s+3)^2}\]

Answer 28

hang on... u've lost me again...
s(s+2)(s+3)(S+3)^2 doesnt equal s(s+2)(s+3)^2

Answer 29

Don't be silly... true enough, if you multiply the denominators, it doesn't give you
s(s+2)(s+3)^2

However, when adding fractions, it's not always the product of the denominators you get, but their LEAST COMMON DENOMINATOR.

Which happens to be s(s+2)(s+3)^2

If you don't believe me, try separating
\[F(s) = \frac{A(s+2)(s+3)^2 + Bs(s+3)^2 + Cs(s+2)(s+3) + Ds(s+2)}{s(s+2)(s+3)^2}\]

And you'll definitely arrive at
\[\Large F(s) = \frac{A}{s}+ \frac{B}{s+2}+ \frac{C}{s+3}+ \frac{D }{(s+3)^2}\]
after some cancellations.

Answer 30

ahhh. k with u now, cheers for clearing that up tj

Answer 31

Okay, now to the hard part... I hope you're ready for this :D

\[F(s) = \frac{A(s+2)(s+3)^2 + Bs(s+3)^2 + Cs(s+2)(s+3) + Ds(s+2)}{s(s+2)(s+3)^2}\]

Answer 32

I can deduce
A+B+C = 0

Any objections? :D

Answer 33

But then again, my charge isn't really to solve this problem myself, than to ensure that you solve it.
So... do you want to evaluate that long and scary polynomial? :D

Answer 34

some... just the deduction.
A+B+C = 0 because...?

Answer 35

Okay, good question.
We need to find the coefficients of s^3, s^2, s, and the constant.
So, to find the coefficient of s^3,
say, in the first term
we have
(s+2)(s+3)^2
This'll be long and hard, but it'll inevitably start with an s^3 right?
(s+2)(s+3)^2 = s^3 +...(some stuff)

catch me so far?

Answer 36

yep, following.

Answer 37

s^3 + 8s^2 + 21s + 18

Answer 38

So, the first term is
A(s^3 + <some other stuff> )

Using similar logic, the second and third terms respectively become
B(s^3 + <some other stuff> )
C(s^3 + <some other stuff> )

Don't worry about the other stuff just yet, for now, just concern yourself with the coefficient of s^3.

So, we deduce (because in the fourth term, involving D, you don't get a s^3)
that the coefficient of s^3 is
A+B+C

right?

Answer 39

ah, ok, understand the logic now... soory, u took some leaps before i didn't intuitively follow, with u now tj, cheers

Answer 40

Now the harder bits... the coefficients of s^2 and s...
This'll be a bloody nightmare XD

We already know that in the first term (the A-bit) the coefficient of s^2 is 8 (as you already graciously figured :P )

Answer 41

In the second bit, thankfully, it IS simpler.
We have a square of a binomial, (s+3)^2 so we simply have to find the coefficient of the s in this binomial, which is 6.

And that's the coefficient of s^2 in the second part (note the extra s outside)

Answer 42

\[s^3+6 s^2+9 s ~~is ~~B\]

Answer 43

Well, I suppose I could leave the hard parts to you :D

Answer 44

But I figured out it's 6 first >:)
#OverlyCompetitive

Answer 45

5 in the third.

Answer 46

aww... dont b like that... r u still hung up on the malaysian/phillipino confusion from earlier ? ;P

Answer 47

Of course not. You know I respect you way too much to be worked up over a trivial error like that :D

Answer 48

sweeeet, and ditto dude

Answer 49

Now, back to work.
No objections to 5 being the coefficient of s^2 in the third term, I suppose?

Answer 50

nah, all good, its 5

Answer 51

And what about the fourth?
You didn't forget the fourth, did you? :)

Answer 52

1

Answer 53

1s^2 + 2s

Answer 54

winning ;D

Answer 55

there we go.
So we can deduce that
8A + 6B + 5C + D = 0

So far, so good?

Answer 56

yep, with u

Answer 57

No you're not, you're like, miles away XD

Anyway, now find the coefficients of the lone s.
:P

Answer 58

21a, 9b, 2d and...

Answer 59

6c?

Answer 60

Bingo :P
21A + 9B + 6C + 2D = 0

What about the remaining terms... the constants? :)

Answer 61

18,0,0,0.

Answer 62

Yup.
So we now have four unknowns, A,B,C,D
and four equations.

Now solve for A, B, C, and D.

Answer 63

Hey, you were right, why delete that? XD

Answer 64

18A = 10

Answer 65

shame and low self confidence...?

Answer 66

A = 5/9, B = -5, C = 40/9, D = 10/3 ...?

Answer 67

that was fast :D

Answer 68

it was simultaneous eqn, used a website, dont need to relaern that.

Answer 69

Okay... unfair. >:D

Anyway, we now have values for A,B,C,D that should pretty much solve all your problems XD

Answer 70

wait... not really

Answer 71

if a + b + c = 0
and A + B + C + D = 10
then d = 10
but d = 10/3...?

Answer 72

no, nvmd, understood the glaring flaws in that logic, sorry

Answer 73

I was going to say "Do I have to dignify that with a response" but decided against it :D

Answer 74

also glad u held back ;D

Answer 75

Technically, I didn't, since you knew what I was going to say.

Anyway, I take it you'll have no further problems in getting the inverse Laplace transform from here on in?

Answer 76

[\Large F(s) = \frac{A}{s}+ \frac{B}{s+2}+ \frac{C}{s+3}+ \frac{D }{(s+3)^2}]

[\Large F(s) = \frac{5/9}{s}+ \frac{-5}{s+2}+ \frac{40/9}{s+3}+ \frac{10/3}{(s+3)^2}]

then pick and choose from the table whatever suits this?

Answer 77

[\Large F(s) = \frac{A}{s}+ \frac{B}{s+2}+ \frac{C}{s+3}+ \frac{D }{(s+3)^2} \]

[\Large F(s) = \frac{5/9}{s}+ \frac{-5}{s+2}+ \frac{40/9}{s+3}+ \frac{10/3}{(s+3)^2}\]

then pick and choose from the table whatever suits this?...whay didn't that latexx properly...?

Answer 78

I suggest you get the inverse Laplace first in terms of A,B,C, and D first, and then replace only AFTERWARDS.

Less confusing, in my opinion.

Answer 79

And you forgot the backslash \ in the beginning of those equations...

Answer 80

\[Large F(s) = \frac{A}{s}+ \frac{B}{s+2}+ \frac{C}{s+3}+ \frac{D }{(s+3)^2} \]

\[Large F(s) = \frac{5/9}{s}+ \frac{-5}{s+2}+ \frac{40/9}{s+3}+ \frac{10/3}{(s+3)^2}\]

then pick and choose from the table whatever suits this?... much better

Answer 81

Well, I suppose you could work with this too.
Yeah, you do that :D

Answer 82

cheers for all the help tj, really appreciate it hey!

Answer 83

Of course you do >:)

Answer 84

5/9 ?

\[\large -5e^{-2t}\]
\[\Large \frac{40e^{-3t}}{9}\]
\[\Large \frac{10e^{-3t}}{3} t\]

Answer 85

did i choose the right ones?

Answer 86

Yup.
I sense a little doubt with your first term, however... :P

Is this what you call confidence? ;)

Answer 87

no, that's the shame and self loathing again... :'(

Answer 88

;P

Answer 89

so i just add them together into 1 equation from here then, yeah?

Answer 90

one expression, and that's your answer, yes :)

Answer 91

back now.... sweeeeeeeeeet!