Inverse Laplace Transformation

[playboy]

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, I am so lost because s^2 - s + 1 cannot be broken up any further...

anybody have any ideas?

 

[HallsofIvy]

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

 

[playboy]

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] + \{ 1/(s+1)(3/4))\}(t)<br /> <br /> and this inverse leplace transformation is too messy: \{ 1/(s+1)(s- 1/2)^2\}

 

[seang]

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.

 

[playboy]

you made an error, I am looking for the Inverse Laplace Transformation of:
\frac{1}{(s+1)(s-(1/2))^2}

I looked that up in my textbook, 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?

 

[seang]

\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 expansion right? maybe?

 

[playboy]

Im so lost right now... :(

 

[playboy]

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 :(

 

[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.

 

[topsquark]

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

 

[playboy]

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...

 

[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?

 

[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.