How can I rewrite a transfer function in terms of resonance frequency \$\omega\_0\$ and damping factor Q? Referred to as "standard form" in the university materials.

I'm still at it, trying to understand LCL filters, and found a gap in the university material. They always let us calculate the transfer function, then the standard form was given, so we just had to fill in the blanks and use the given function to draw a Bode plot. Now that I have a real circuit, I'm stuck. The university book only contains this section on the matter

[![Circuit analysis book](https://i.stack.imgur.com/zvgcq.png)](https://i.stack.imgur.com/zvgcq.png)

Nilsson & Riedel has a section devoted to Bode diagrams in the appendix. It says all you need to do is divide away the poles and zeros and factor the result. Poles and zeroes seems to refer to the coefficients of the highest exponents in the numerator and denominator.

None of this is very revealing to me. Say I have the following transfer function. This is indeed in the general form, but how on earth do you factorise that? Getting rid of the poles and zeros is not very helpful either.

![schematic](https://i.stack.imgur.com/Mo4wr.png)

[simulate this circuit](/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fMo4wr.png) – Schematic created using [CircuitLab](https://www.circuitlab.com/)

$$
H(j\omega)=\frac{j\omega C\_fR\_f+1}{j\omega(L\_1+L\_2) + (j\omega)^2C\_fR\_f(L\_1+L\_2)+ (j\omega)^3 L\_1L\_2C\_f}\\
H(j\omega)=\frac{C\_fR\_f}{L\_1L\_2C\_4}\frac{j\omega+\frac{1}{C\_fR\_f}}{j\omega\frac{L\_1+L\_2}{L\_1L\_2C\_4} + (j\omega)^2\frac{R\_f(L\_1+L\_2)}{L\_1L\_2}+ (j\omega)^3 }\\
H(j\omega)=\frac{\omega C\_fR\_f-j}{\omega(L\_1+L\_2) + j\omega^2C\_fR\_f(L\_1+L\_2)+ j^2\omega^3 L\_1L\_2C\_f}\\
$$

I put that in Wolfram Alpha, and it gave the following roots for the denominator. Besides being humongous, I don't feel they bring me much closer to a solution.
[![first root](https://i.stack.imgur.com/VU8Oh.gif)](https://i.stack.imgur.com/VU8Oh.gif)
[![second root](https://i.stack.imgur.com/xH5BU.gif)](https://i.stack.imgur.com/xH5BU.gif)

[update]

The factorization finally clicked, and I came up with the following for the undamped case:
$$
\begin{align}
H(j\omega)&=\frac{1}{(j\omega-0)((L\_1+L\_2) + (j\omega)^2L\_1L\_2C\_4)} \\
j\omega&=\frac{\pm j \sqrt{4L\_1L\_2C\_4(L\_1+L\_2)}}{2L\_1L\_2C\_4} \\
H(j\omega)&=\frac{1}{(j\omega-0)(j\omega-j\frac{ \sqrt{4L\_1L\_2C\_4(L\_1+L\_2)}}{2L\_1L\_2C\_4})(j\omega+j\frac{ \sqrt{4L\_1L\_2C\_4(L\_1+L\_2)}}{2L\_1L\_2C\_4})} \\
&=\frac{1}{(j\omega)(\frac{L\_1+L\_2}{L\_1L\_2C\_4}+(j\omega)^2)} \\
&=\frac{\frac{L\_1L\_2C\_4}{L\_1+L\_2}}{(j\omega)(1+(j\omega)^2\frac{L\_1L\_2C\_4}{L\_1+L\_2})}
\end{align}
$$
Putting this in standard form gives
$$
\begin{align}
H(j\omega)&=\frac{1}{(j\frac{\omega}{\omega\_0})(1+j\frac{\omega}{\omega\_1 Q}+(j\frac{\omega}{\omega\_1})^2)} \\
Q&=0 \\
\omega\_0&=1 \\
\omega\_1&=\frac{L\_1+L\_2}{L\_1L\_2C\_4}
\end{align}
$$

I hope that's not terribly wrong.