# Inverse Laplace Transformation

by playboy
Tags: inverse, laplace, transformation
 P: n/a 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?
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,890 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
 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 $\{ 1/(s^3 + 1)\}(t)$ =$\{ 1/(s+1)((s- 1/2)^2 + 3/4)\}(t)$ =$\{ 1/(s+1)(s- 1/2)^2\}[/tex] + [itex]\{ 1/(s+1)(3/4))\}(t)$ and this inverse leplace transformation is too messy: $\{ 1/(s+1)(s- 1/2)^2\}$
P: 185

## 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.
 P: n/a 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?
 P: 185 $$\frac{t^{n}}{n!}e^{-\alpha t} \cdot u(t) =\frac{1}{(s+\alpha)^{n+1}}$$ 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?
 P: n/a Im so lost right now..... :(
 P: n/a 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 :(
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,890 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.
P: 335
 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
 P: n/a 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....
 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... $$\frac{1}{(s+1)(s-\frac{1}{2})^2}= \frac{(4/9)}{s+1}+ \frac{(-4/9)}{(s-\frac{1}{2})}+ \frac{(2/3)}{(s-\frac{1}{2})^2}$$ 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?
P: n/a
 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.

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