- #1

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If someone could point me to the next step that'd be great!

Thanks a lot guys and girls.

What I have is in the link below, I'd embed it but it's too big and will mess with the formatting of the page.

http://i46.tinypic.com/2z4aaup.png

- Thread starter chief10
- Start date

- #1

- 78

- 0

If someone could point me to the next step that'd be great!

Thanks a lot guys and girls.

What I have is in the link below, I'd embed it but it's too big and will mess with the formatting of the page.

http://i46.tinypic.com/2z4aaup.png

- #2

FOIWATER

Gold Member

- 434

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I believe I haven't checked it more than once may have made a mistake but it definitely seems that it would be easiest to handle each term on its own. Maybe the first one can also be forced to resemble either sin or cosine without using partial fractions

- #3

- 798

- 34

Hi !

If someone could point me to the next step that'd be great!

Thanks a lot guys and girls.

What I have is in the link below, I'd embed it but it's too big and will mess with the formatting of the page.

http://i46.tinypic.com/2z4aaup.png

The Laplace transform of exp(-at)sin(bt) is b/((s+a)²+b²)

The Laplace transform of exp(-at)cos(bt) is (s+a)/((s+a)²+b²)

So, you have to rewrite Y(s) on the form :

Y(s) = C1/s + C2* b/((s+a)²+b²)+ C3* (s+a)/((s+a)²+b²)

First, compute a and b, then C1, C2 and C3

- #4

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Do you mean [5(e^-2) x tsint]? you're not raising the sine function are you, just multiplying it by the exp?

I believe I haven't checked it more than once may have made a mistake but it definitely seems that it would be easiest to handle each term on its own. Maybe the first one can also be forced to resemble either sin or cosine without using partial fractions

The second component appears to be solvable in that sense, however your completion of the square I believe is incorrect. You can't simplify the denominator to my knowledge.

That specific Laplace transform wasn't on my given formula sheet, that's a bummer.

Hmmm the first term is a little trickier! any ideas on that one?

10/s(s^2+2s+5)

transforming that is quite a task..

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- #5

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in the C3 function, could I ask why you put (s+a) in the numerator? thanks mateHi !

The Laplace transform of exp(-at)sin(bt) is b/((s+a)²+b²)

The Laplace transform of exp(-at)cos(bt) is (s+a)/((s+a)²+b²)

So, you have to rewrite Y(s) on the form :

Y(s) = C1/s + C2* b/((s+a)²+b²)+ C3* (s+a)/((s+a)²+b²)

First, compute a and b, then C1, C2 and C3

- #6

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I mean s²+2s+5 = (s+1)²+2² = (s+a)²+b², then a=1 and b=2

- #7

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how does s²+2s+5 = (s+1)²+2²?I mean s²+2s+5 = (s+1)²+2² = (s+a)²+b², then a=1 and b=2

that's not right is it?

while you're here though, if you could take a look at my other thread involving Laplace that'd be great!!!

https://www.physicsforums.com/showthread.php?t=639980

- #8

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Can you develop (s+1)²+2² = ?how does s²+2s+5 = (s+1)²+2²?

that's not right is it?

- #9

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excuse me it's 3am here in Aus, lol.. my bad.. got itCan you develop (s+1)²+2² = ?

do you mind elaborating on the partial fraction a little if you don't mind?

i've drawn up the initial equation you gave me below, how does the completed square factor in here?

http://i49.tinypic.com/w39cj.jpg

BTW, I just ran it through Matlab and this is the answer it gave me to the ODE, seems like we're on the right track.

(3*sin(2*t))/(2*exp(t)) - (2*cos(2*t))/exp(t) + 2

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- #10

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That's OK. Then you have to compute C1, C2, C3i've drawn up the initial equation you gave me below, how does the completed square factor in here?

http://i49.tinypic.com/w39cj.jpg

Rewrite all with only one denominator s((s+a)²+b²) and compare to Y(s) also rewriten with the same denominator.

- #11

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Alright I think I get what you're saying, i posted a link below.That's OK. Then you have to compute C1, C2, C3

Rewrite all with only one denominator s((s+a)²+b²) and compare to Y(s) also rewriten with the same denominator.

http://i48.tinypic.com/2zits9l.jpg

I won't post back for a few hours now though because it's 4:15am here so i'll head off to bed now. I really appreciate all of your help. Hopefully you'll be here tomorrow to help me too mate :)

I await your reply! thanks bud.

- #12

FOIWATER

Gold Member

- 434

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And yeah as you said it's not raised to the power it's a multiplication

- #13

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any ideas on which step I should take next?

- #14

- 798

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Do you forget that you already have writen Y(s) on another form whitout C1, C2, C3 : http://i46.tinypic.com/2z4aaup.png

Compare both in order to compute C1,C2, C3

- #15

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wait so are you saying to equate both of the Y(s) equations?

- #16

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Of course, obviously, YES ! :zzz:

- #17

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lol alright, i'll do that and post back in a bit

- #18

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i tried it out but i've got brackets and fractions everywhere? sorry to seem a little out of touch here.

- #19

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Do you have the same denominator for ALL fractions ?

- #20

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nopeDo you have the same denominator for ALL fractions ?

in this one: http://i46.tinypic.com/2z4aaup.png - The Y(s) function has two different denominators so you can't cancel them out on both sides if you equated the Y(s) equations.

i mean even if you did equate them and cancel them out, does one side just become 10+5? (15)

- #21

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Oh my God ! Can't you make all fractions with the same denominator s((s+1)²+4) just like you did before for the three other fractions ?

- #22

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yeah I thought that's what you meant, no issue

Oh my God ! Can't you make all fractions with the same denominator s((s+1)²+4) just like you did before for the three other fractions ?

lol relax man, it's confusing over the net with just text and stuff, i appreciate the help, i'll post back in a bit.

- #23

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thanks a lot mate I think i'm pretty sure I worked it out! with your help!

Oh my God ! Can't you make all fractions with the same denominator s((s+1)²+4) just like you did before for the three other fractions ?

much appreciated dude, i've attached it below

http://i48.tinypic.com/2ep3t3c.jpg

- #24

- 798

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You are right to be pretty sure !thanks a lot mate I think i'm pretty sure I worked it out! with your help!

much appreciated dude, i've attached it below

http://i48.tinypic.com/2ep3t3c.jpg

- #25

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haha :)

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