# Simplifying Laplace transform

• Oct 9th 2009, 03:42 PM
ArTiCK
Simplifying Laplace transform
Hi,

I need help simplifying the following so that i can use the Laplace transforms table

LT[y] = (2/3s^2 -5s -2)/ (s^3 -3s^2 -1)

ArTiCk
• Oct 9th 2009, 08:38 PM
mr fantastic
Quote:

Originally Posted by ArTiCK
Hi,

I need help simplifying the following so that i can use the Laplace transforms table

LT[y] = (2/3s^2 -5s -2)/ (s^3 -3s^2 -1)

ArTiCk

Are you sure this is the correct expression?
• Oct 9th 2009, 08:46 PM
chisigma
If You write the LT as...

$\displaystyle \mathcal {L} \{y(t)\} = L(s)=\frac{N(s)}{D(s)}$ (1)

... where $\displaystyle N(s)$ and $\displaystyle D(s)$ are both polynomial in s, and You 'factorize' $\displaystyle D(s)$ as follows ...

$\displaystyle D(s) = (s-\alpha_{1})\cdot (s-\alpha_{2})\dots (s-\alpha_{n})$ (2)

... the LT can be written as...

$\displaystyle L(s) = \frac{\rho_{1}}{s-\alpha_{1}} + \frac{\rho_{2}}{s-\alpha_{2}} + \dots + \frac{\rho_{n}}{s-\alpha_{n}}$ (3)

... where...

$\displaystyle \rho_{n} = \lim_{s \rightarrow \alpha_{n}} (s-\alpha_{n})\cdot L(s)$ (4)

In your case $\displaystyle D(s)$ is a polynomial of order 3 and its roots are...

$\displaystyle \alpha_{1} = 3.1038034...$

$\displaystyle \alpha_{2} = -.051901701... - i \cdot .56523585...$

$\displaystyle \alpha_{2} = -.051901701... + i \cdot .56523585...$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Oct 9th 2009, 08:50 PM
mr fantastic
Quote:

Originally Posted by chisigma
If You write the LT as...

$\displaystyle \mathcal {L} \{y(t)\} = L(s)=\frac{N(s)}{D(s)}$ (1)

... where $\displaystyle N(s)$ and $\displaystyle D(s)$ are both polynomial in s, and You 'factorize' $\displaystyle D(s)$ as follows ...

$\displaystyle D(s) = (s-\alpha_{1})\cdot (s-\alpha_{2})\dots (s-\alpha_{n})$ (2)

... the LT can be written as...

$\displaystyle L(s) = \frac{\rho_{1}}{s-\alpha_{1}} + \frac{\rho_{2}}{s-\alpha_{2}} + \dots + \frac{\rho_{n}}{s-\alpha_{n}}$ (3)

... where...

$\displaystyle \rho_{n} = \lim_{s \rightarrow \alpha_{n}} (s-\alpha_{n})\cdot L(s)$ (4)

In your case $\displaystyle D(s)$ is a polynomial of order 3 and its roots are...

$\displaystyle \alpha_{1} = 3.1038034...$

$\displaystyle \alpha_{2} = -.051901701... - i \cdot .56523585...$

$\displaystyle \alpha_{2} = -.051901701... + i \cdot .56523585...$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

IF the given expression is correct (and I doubt that it is) then the technique would be to factorise the denominator in a linear factor and an irreducible quadratic factor and then use a partical fraction decomposition. The difficulty of doing this convinces me that the given expression, wherever it has come from, is wrong.
• Oct 10th 2009, 12:22 AM
CaptainBlack
Quote:

Originally Posted by ArTiCK
Hi,

I need help simplifying the following so that i can use the Laplace transforms table

LT[y] = (2/3s^2 -5s -2)/ (s^3 -3s^2 -1)

ArTiCk

Please repost with brackets so deployed to make the meaning of this explicit. As it stands you have (probably) posted:

$\displaystyle \mathcal{L}[y] = \frac{\frac{2}{3}s^2 -5s -2}{s^3 -3s^2 -1}$

or may-be:

$\displaystyle \mathcal{L}[y] = \frac{\frac{2}{3s^2} -5s -2}{s^3 -3s^2 -1}$

If you do not make yourself clear you are wasting our time

CB
• Oct 10th 2009, 02:49 AM
ArTiCK
Hi all,

I would like to apologise for causing so much trouble. I tried to delete this post but i couldn't find a button to do so. As Mr. F has mentioned i have made a mistake earlier on in my calculations. I appreciate all the help that all you guys have given.

Sorry once again,
ArTiCk