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Laplace Transforms 
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#1
May1511, 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? 


#2
May1611, 03:27 AM

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P: 1,583

The first thing I would recommend to do is write:
[tex] s^{2}2s+3=(s1)^2+2,\quad s^{2}2s+2=(s1)^2+1 [/tex] Then I think the transform looks like a convolution doesn't it? 


#3
May1611, 05:04 AM

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PF Gold
P: 11,673

Use partial fractions to break them up.



#4
May1611, 05:15 AM

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Laplace Transforms
You can't use partial fractions here.



#5
May1611, 01:39 PM

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RGV 


#6
May1611, 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?? 


#7
May1611, 06:34 PM

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#8
May1611, 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 


#9
May1611, 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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