## Inverse Laplace Transformation

Okay I really really need somebody to help me

Find the Inverse Laplace Transformation of $\{ 1/(s^3 + 1)\}(t)$

(for those of you who don't know, you look it up in a table. the closest thing that I can find is $\{ 1/(s^2 + 1)\}(t)$ which is sin(t) )

Well, I started of by breaking up $s^3 + 1$ into $(s+1)(s^2 - s + 1)$

After this, im so lost because $s^2 - s + 1$ cannot be broken up any further....

anybody have any ideas?

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 Recognitions: Gold Member Science Advisor Staff Emeritus COMPLETE THE SQUARE!! s2-s+1 can be written as a perfect square plus something: s2-s+ 1/4+ 1-1/4= s2-s+ 1/4+ 3/4 = (s- 1/2)2 + 3/4
 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 $\{ 1/(s^3 + 1)\}(t)$ =$\{ 1/(s+1)((s- 1/2)^2 + 3/4)\}(t)$ =$\{ 1/(s+1)(s- 1/2)^2\}[/tex] + [itex]\{ 1/(s+1)(3/4))\}(t)$ and this inverse leplace transformation is too messy: $\{ 1/(s+1)(s- 1/2)^2\}$

## Inverse Laplace Transformation

no that's not too messy man! You can do it! You mean:

$$\frac{1}{(s+1)(s-(1/2)^2)}$$

yeah?
In fact you don't don't even have to bother finding any constants, these should be right in your table.

 you made an error, im looking for the Inverse Laplace Transformation of: $$\frac{1}{(s+1)(s-(1/2))^2}$$ 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?
 $$\frac{t^{n}}{n!}e^{-\alpha t} \cdot u(t) =\frac{1}{(s+\alpha)^{n+1}}$$ 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?
 Im so lost right now..... :(
 I want to break $$\frac{1}{(s+1)(s-(1/2))^2}$$ = $$\frac{A}{(s+1)} + \frac{B}{(s-(1/2)} + \frac{C}{(s-(1/2))^2}$$ = $$\frac{A}{(s+1)} + \frac{B((s-(1/2)) + C}{(s-(1/2))^2}$$ I tried solving for A, B and C like 3 times and got different answers :(
 Recognitions: Gold Member Science Advisor Staff Emeritus Obviously you aren't going to find an inverse transform for $$\frac{1}{(s+1)(s-\frac{1}{2})^2}$$ in a table- reduce it by partial fractions. As you say, you want $$\frac{1}{(s+1)(s-\frac{1}{2})^2}= \frac{A}{s+1}+ \frac{B}{(s-\frac{1}{2})}+ \frac{C}{(s-\frac{1}{2})^2}$$ Multiply on both sides by the common denominator and you get $$1= A(s-\frac{1}{2})^2+ B(s+1)(s-\frac{1}{2})+ C(s+1)$$ 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.

 Quote by HallsofIvy Obviously you aren't going to find an inverse transform for $$\frac{1}{(s+1)(s-\frac{1}{2})^2}$$ in a table- reduce it by partial fractions. As you say, you want $$\frac{1}{(s+1)(s-\frac{1}{2})^2}= \frac{A}{s+1}+ \frac{B}{(s-\frac{1}{2})}+ \frac{C}{(s-\frac{1}{2})^2}$$ Multiply on both sides by the common denominator and you get $$1= A(s-\frac{1}{2})^2+ B(s+1)(s-\frac{1}{2})+ C(s+1)$$ 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.
You forgot to square the factor in your A equation, and you solved for B incorrectly.
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

 omg, i was actually expanding this: $$1= A(s-\frac{1}{2})^2+ B(s+1)(s-\frac{1}{2})+ C(s+1)$$ and trying to equate cooeffiecnts... which got soooo messy....
 okay, this is getting ridiculus. After solving for A, B, C, and substituting the values back into the original equation, we get... $$\frac{1}{(s+1)(s-\frac{1}{2})^2}= \frac{(4/9)}{s+1}+ \frac{(-4/9)}{(s-\frac{1}{2})}+ \frac{(2/3)}{(s-\frac{1}{2})^2}$$ 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: $$\frac{(2/3)}{(s-\frac{1}{2})^2}$$ do you complete the square again?

 Quote by playboy okay, this is getting ridiculus. After solving for A, B, C, and substituting the values back into the original equation, we get... $$\frac{1}{(s+1)(s-\frac{1}{2})^2}= \frac{(4/9)}{s+1}+ \frac{(-4/9)}{(s-\frac{1}{2})}+ \frac{(2/3)}{(s-\frac{1}{2})^2}$$ 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: $$\frac{(2/3)}{(s-\frac{1}{2})^2}$$ do you complete the square again?
Never mind, I figured this one out.

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