Inverse Laplace Transformation


by playboy
Tags: inverse, laplace, transformation
playboy
#1
Mar28-06, 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?
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HallsofIvy
HallsofIvy is offline
#2
Mar28-06, 08:01 PM
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PF Gold
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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
#3
Mar28-06, 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]

seang
seang is offline
#4
Mar28-06, 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.
playboy
#5
Mar28-06, 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?
seang
seang is offline
#6
Mar28-06, 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?
playboy
#7
Mar28-06, 10:17 PM
P: n/a
Im so lost right now..... :(
playboy
#8
Mar28-06, 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 :(
HallsofIvy
HallsofIvy is offline
#9
Mar29-06, 06:01 AM
Math
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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.
topsquark
topsquark is offline
#10
Mar29-06, 07:09 AM
P: 335
Quote Quote by HallsofIvy
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.
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
#11
Mar29-06, 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....
playboy
#12
Mar29-06, 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?
playboy
#13
Mar29-06, 01:34 PM
P: n/a
Quote 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...

[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?
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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