Kindly simplify this equation with partial fraction and tell me procedure as well
3/[s(s+1)(s+3)]
Thanks
$\displaystyle \frac{3}{s(s+1)(s+3) } $
$\displaystyle = \frac{3}{ (s+1) ( s^2 + 3s ) } $
$\displaystyle = \frac{3}{ (s+1) ( (s+1)(s+2) - 2 )} $
$\displaystyle = \frac{1}{2} \left( 3~\frac{s+2 }{ (s+1)(s+2) - 2 } - \frac{3}{ s+1} \right) $
$\displaystyle = \frac{1}{2} \left( 3 ~ \frac{s+2}{ s(s+3) } - \frac{3}{ s+1} \right) $
$\displaystyle = \frac{1}{2} \left( ( \frac{2}{s} + \frac{1}{s+3} ) - \frac{3}{ s+1} \right) $
$\displaystyle = \frac{1}{s} + \frac{1}{2 } \frac{1}{s+3} - \frac{3}{2} ~ \frac{1}{s+1} $
Or, since all the poles are simple, you could use the Heaviside cover-up method.
I regard $\displaystyle s+2 $ in $\displaystyle (s+1)(s+2) - 2 $ as a constant $\displaystyle a$ and let $\displaystyle X = s+1$ so we have
$\displaystyle \frac{2}{X(aX-2 )} = \frac{a}{aX-2} - \frac{1}{X} $
$\displaystyle \frac{1}{X(aX-2)} = \frac{1}{2}( \frac{a}{aX-2} - \frac{1}{X} )
$