I've got a problem with factorising the denominator of an LTI system. The system is a simple boost converter and my current transfer function is
$$ U_O = \frac{\lambda' \cdot U_I – I_O \cdot L \cdot s}{C \cdot L \cdot s^2 + \lambda'^2}. $$
I want to solve for the systems poles next and and bring them into time constant representation. I thus use the common approach
$$ C \cdot L \cdot s^2 + \lambda'^2 \overset{!}{=} 0 $$
which can be solved with the quadratic formula
$$ s = \pm \frac{\sqrt{-4 \cdot C \cdot L \cdot \lambda'^2}}{2 \cdot C \cdot L} = \pm \frac{\lambda'}{\underbrace{\sqrt{C \cdot L}}_{T_{CL}}} j = \frac{\lambda'}{T_{CL}} j $$
and leads to the factorised form
$$ (s + \frac{\lambda'}{T_{CL}} j) \cdot (s – \frac{\lambda'}{T_{CL}} j). $$
This result is obviously wrong since expanding the term doesn't give the same result and I'm not able to wrap my head around why this is the case.
Factorising the denominator with the third binomial formula gives the correct factors instead
$$ (T_{CL} \cdot s + \lambda' j) \cdot (T_{CL} \cdot s – \lambda' j) $$
and these are already in time constant representation, which leads to the fully factorised transfer function in time constant representation
$$ U_O = \frac{\lambda' \cdot U_I – I_O \cdot L \cdot s}{\lambda'^2 \cdot \underbrace{(\frac{T_{CL}}{\lambda'} \cdot s + j)}_p \cdot \underbrace{(\frac{T_{CL}}{\lambda'} \cdot s – j)}_{p^*}}. $$
Where did I mess up the quadratic formula?
My only approach is that the coefficients aren't
$$
\begin{split}
a &= C \cdot L \\
b &= 0 \\
c &= \lambda'^2
\end{split}
$$
and I'm not solving for $s$, but that the coefficients are
$$
\begin{split}
a &= 1 \\
b &= 0 \\
c &= \lambda'^2
\end{split}
$$
and I'm solving for $T_{CL} \cdot s$ with respect to the quadratic equation
$$ {\underbrace{(T_{CL} \cdot s)}_x}^2 + \lambda'^2. $$
Please give some advice on how to deal with such situations, I'm seriously confused and couldn't find an answer browsing through the material about quadratic equations. Different problems are far to over represented.
Best Answer
I think you mean that you don't get back the original denominator. But this is ok. Remember that, when you have a quadratic equation, the quadratic formula gives you the roots of the equation. In your case, the quadratic equation is $$ CLs^2 + \lambda^2 = 0. $$
Note that we can divide and multiply both sides of the equation by any number that we want. Let's go ahead and divide both sides of the equation by $CL$: $$ s^2 + \frac{\lambda^2}{CL} = 0. $$ And this is exactly what you get when you multiply out the following: $$ (s + \frac{\lambda'}{T_{CL}} j)(s - \frac{\lambda'}{T_{CL}} j) = 0. $$
This shows you that you can multiply quadratic equation by a scalar and you will get the same roots.