#1
Mar2806, 06:56 PM

P: n/a

Okay I really really need somebody to help me
Find the Inverse Laplace Transformation of [itex]\{ 1/(s^3 + 1)\}(t)[/itex] (for those of you who don't know, you look it up in a table. the closest thing that I can find is [itex]\{ 1/(s^2 + 1)\}(t)[/itex] which is sin(t) ) Well, I started of by breaking up [itex]s^3 + 1[/itex] into [itex](s+1)(s^2  s + 1)[/itex] After this, im so lost because [itex]s^2  s + 1[/itex] cannot be broken up any further.... anybody have any ideas? 



#2
Mar2806, 08:01 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,890

COMPLETE THE SQUARE!!
s^{2}s+1 can be written as a perfect square plus something: s^{2}s+ 1/4+ 11/4= s^{2}s+ 1/4+ 3/4 = (s 1/2)^{2} + 3/4 


#3
Mar2806, 08:38 PM

P: n/a

HallsofIvy, thank you for that.
I was thinking of completeing the square, but it made things even more messy. The question no becomes: Take the inverse leplace transformation of [itex]\{ 1/(s^3 + 1)\}(t)[/itex] =[itex]\{ 1/(s+1)((s 1/2)^2 + 3/4)\}(t)[/itex] =[itex]\{ 1/(s+1)(s 1/2)^2\}[/tex] + [itex]\{ 1/(s+1)(3/4))\}(t)[/itex] and this inverse leplace transformation is too messy: [itex]\{ 1/(s+1)(s 1/2)^2\}[/itex] 



#4
Mar2806, 09:13 PM

P: 185

Inverse Laplace Transformation
no that's not too messy man! You can do it! You mean:
[tex] \frac{1}{(s+1)(s(1/2)^2)} [/tex] yeah? In fact you don't don't even have to bother finding any constants, these should be right in your table. 


#5
Mar2806, 09:37 PM

P: n/a

you made an error, im looking for the Inverse Laplace Transformation of:
[tex] \frac{1}{(s+1)(s(1/2))^2} [/tex] I looked that up in my text book, and no, its not in the tabel :( Then i tried partial fractions to break it up and boy, that too isn't any easier... So I have absolutly no idea what to do with this? 



#6
Mar2806, 10:02 PM

P: 185

[tex]\frac{t^{n}}{n!}e^{\alpha t} \cdot u(t) =\frac{1}{(s+\alpha)^{n+1}}[/tex]
Don't be upset if that one's not in your book, I had to fish around to find it my first time too. From there I think you can do a partial fraction expantion right? maybe? 


#7
Mar2806, 10:17 PM

P: n/a

Im so lost right now..... :(



#8
Mar2806, 10:19 PM

P: n/a

I want to break [tex] \frac{1}{(s+1)(s(1/2))^2} [/tex] = [tex] \frac{A}{(s+1)} + \frac{B}{(s(1/2)} + \frac{C}{(s(1/2))^2} [/tex] = [tex] \frac{A}{(s+1)} + \frac{B((s(1/2)) + C}{(s(1/2))^2} [/tex]
I tried solving for A, B and C like 3 times and got different answers :( 



#9
Mar2906, 06:01 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,890

Obviously you aren't going to find an inverse transform for
[tex]\frac{1}{(s+1)(s\frac{1}{2})^2}[/tex] in a table reduce it by partial fractions. As you say, you want [tex]\frac{1}{(s+1)(s\frac{1}{2})^2}= \frac{A}{s+1}+ \frac{B}{(s\frac{1}{2})}+ \frac{C}{(s\frac{1}{2})^2}[/tex] Multiply on both sides by the common denominator and you get [tex]1= A(s\frac{1}{2})^2+ B(s+1)(s\frac{1}{2})+ C(s+1)[/tex] Taking x= 1 gives 1= A(3/2) so A= 2/3. Taking x= 1/2 gives 1= c(3/2) so C= 2/3. Taking x= 0 gives 1= A/2 B/2+ C or 1= 1/3 B/2+ 2/3 so B/2= 2/3 1/3 1= 2/3 and B= 1/3. 



#10
Mar2906, 07:09 AM

P: 335

s=1 gives 1 = A(3/2)^2 so A = 4/9. s=1/2 gives 1 = C(3/2) so C = 2/3 s=0 gives 1 = A(1/2)^2 + B(1/2) + C or 1 = A/4  B/2 + C so B/2 = 1 + 1/9 + 2/3 = 2/9. So B = 4/9. Dan 


#11
Mar2906, 07:47 AM

P: n/a

omg, i was actually expanding this:
[tex]1= A(s\frac{1}{2})^2+ B(s+1)(s\frac{1}{2})+ C(s+1)[/tex] and trying to equate cooeffiecnts... which got soooo messy.... 


#12
Mar2906, 09:26 AM

P: n/a

okay, this is getting ridiculus.
After solving for A, B, C, and substituting the values back into the original equation, we get... [tex]\frac{1}{(s+1)(s\frac{1}{2})^2}= \frac{(4/9)}{s+1}+ \frac{(4/9)}{(s\frac{1}{2})}+ \frac{(2/3)}{(s\frac{1}{2})^2}[/tex] The first two terms for an Inverse Leplace Transform are easy, however, the thrid term is messy again. How on earth do you take the inverse leplace transform of this: [tex]\frac{(2/3)}{(s\frac{1}{2})^2}[/tex] do you complete the square again? 


#13
Mar2906, 01:34 PM

P: n/a

You have to play around with one of the transforms in the tabel in the book, then it works. 


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