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Laplace Transforms

by popo902
Tags: laplace, transforms
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popo902
#1
May15-11, 11:51 PM
P: 60
1. The problem statement, all variables and given/known data

Im having trouble finding ways to manipulate equations to fit something from the table

The two i'm stuck on are these

1. [itex]\frac{1}{s^{2}- 2s + 3} (\frac{1+(s^{2}+1)e^{-3\Pi S}}{(s^{2}+1)})[/itex] = Y(s)

2.[itex]\frac{1}{s^{2}- 2s + 2} (\frac{s}{s^{2}+1} + s - 2)[/itex] = Y(s)



2. Relevant equations

These are the IVPs i got them from
1. y" - 2y' + 3y = sint + [tex]\delta[/tex](t - 3*pi)
y(0) = 0
y'(0) = 0

2. y'' - 2y' + 2y = cost
y(0) = 1
y'(o) = 0

3. The attempt at a solution

I tried all sorts of things like multiplying the equations out
i still can't seem to find a way to comfortably manipulate it to match anything on the laplace table
can some one help or give me a tip?
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hunt_mat
#2
May16-11, 03:27 AM
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The first thing I would recommend to do is write:

[tex]
s^{2}-2s+3=(s-1)^2+2,\quad s^{2}-2s+2=(s-1)^2+1
[/tex]

Then I think the transform looks like a convolution doesn't it?
vela
#3
May16-11, 05:04 AM
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Use partial fractions to break them up.

hunt_mat
#4
May16-11, 05:15 AM
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Laplace Transforms

You can't use partial fractions here.
Ray Vickson
#5
May16-11, 01:39 PM
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Quote Quote by hunt_mat View Post
You can't use partial fractions here.
Yes, you can. Some of the denominators remain quadratic in s if you restrict yourself to reals, but can be fully expanded out to linear factors if you use complex roots.

RGV
popo902
#6
May16-11, 05:49 PM
P: 60
I see i see
but how did you get the equations to look like that?

and could you get the inverse laplace transforms with complex numbers??
vela
#7
May16-11, 06:34 PM
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Quote Quote by popo902 View Post
I see i see
but how did you get the equations to look like that?
Do you mean what hunt_mat did? If so, he just completed the square.
and could you get the inverse laplace transforms with complex numbers??
Yes, you can, but it's usually best to avoid that if possible. Sign mistakes are bad enough, but when you start throwing factors of i around, you increase your chance of making a mistake immensely. You can definitely invert the transform without resorting to using complex algebra.
popo902
#8
May16-11, 08:09 PM
P: 60
yeh i saw that after about 30 minutes of just staring at the problem haha
however, when you do carry j's into your laplace and get the inverse, they will still be complex right?
so technically you could have a laplace shifted by a complex value using the e^at rule, where a = some j?

thank you all though, it really helped
vela
#9
May16-11, 08:17 PM
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If you do everything correctly using complex algebra, it will simplify down to a purely real result. In other words, the math is consistent. If it doesn't come out real, you made a mistake somewhere.


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