Simplifying Laplace transform

**ArTiCK** · October 9th 2009, 03:42 PM

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)

Thank in advance,
ArTiCk

**mr fantastic** · October 9th 2009, 08:38 PM

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)

Thank in advance,
ArTiCk

Are you sure this is the correct expression?

**chisigma** · October 9th 2009, 08:46 PM

If You write the LT as...

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

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

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

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

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

... where...

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

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

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

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

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

Kind regards

$\chi$ $\sigma$

**mr fantastic** · October 9th 2009, 08:50 PM

Originally Posted by chisigma

If You write the LT as...

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

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

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

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

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

... where...

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

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

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

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

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

Kind regards

$\chi$ $\sigma$

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.

**CaptainBlack** · October 10th 2009, 12:22 AM

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)

Thank in advance,
ArTiCk

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

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

or may-be:

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

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

CB

**ArTiCK** · October 10th 2009, 02:49 AM

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

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