Im trying to do partial fraction but cant seem to get it right, and the examiner have not showed how he did the partial fraction, just the answer.
I want to partial fraction;
$$\frac{1}{((s+\frac{1}{2})^2+\frac{3}{4})(s+\frac{1}{2})}$$
My solution so far:
$\frac{1}{((s+\frac{1}{2})^2+\frac{3}{4})(s+\frac{1}{2})} = \frac{A}{s+\frac{1}{2}}+\frac{Bs+D}{(s+\frac{1}{2})^2+\frac{3}{4}} = \frac{A((s+\frac{1}{2})^2+\frac{3}{4})+(Bs+D)(s+\frac{1}{2})}{((s+\frac{1}{2})^2+\frac{3}{4})(s+\frac{1}{2})}$
$\Rightarrow As^2+As+A+Bs^2+\frac{1}{2}Bs+Ds+\frac{1}{2}D=1$
Comparing coefficients gives me
$s^2:A+B=0$
$s^1:A+\frac{1}{2}B+\frac{1}{2}D=0$
$s^0:A+\frac{1}{2}D=1$
Which gives me $A=2, B=-2, C=-2$ which is wrong.
Can someone tell me what I'm doing wrong? Something tells me that its the $Bs+D$ that is wrong.
Thanks!
There are two approaches here - complete the square later, or as Steven Stadnicki suggests, do a change of variable to make your life easier. I'll show my intended solution of completing squares later.
Let $$F(s) = \frac{1}{((s+\frac{1}{2})^2+\frac{3}{4})(s+\frac{1}{2})}.$$
We can show that after re-expanding and clearing fractions, we can write $$F(s) = \frac{2}{(2s+1)(s^2 + s + 1)}$$ which makes our lives easier. Now do the usual expansion:
$$\frac{2}{(2s+1)(s^2 + s + 1)} = \frac{A}{2s+1} + \frac{Bs + C}{s^2 + s + 1}$$
and clearing fractions gives
$$2 = A(s^2 + s + 1) + (Bs + C)(2s+1).$$
One can equate coefficients here but I will opt to use the Heaviside cover-up method. With $s = -\frac{1}{2}$ we get $A = \frac{8}{3}.$ For $s = 0$ we obtain the equation $2 = \frac{8}{3} + C$ which gives $C = -\frac{2}{3}.$ Finally, as we have run out of "nice" $s$ values, we'll choose $s = 1$ which will give $B = -\frac{4}{3}$ (check this!). Thus our expansion is
$$F(s) = \frac{\frac{8}{3}}{2s+1} + \frac{-\frac{4}{3}s - \frac{2}{3}}{s^2 + s + 1}$$
and, after re-completing the square and normalizing,
$$F(s) = \frac{\frac{4}{3}}{s+\frac{1}{2}} + \frac{-\frac{4}{3}s - \frac{2}{3}}{\left(s + \frac{1}{2} \right)^2 + \frac{3}{4}}.$$
The inverse Laplace transform should then be easy to compute (an exponential, a sine, and a cosine).