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
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.
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
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