How to calculate the transfer function from a group of equations below?

Question

The group of equations below describe the relationship of variables from a circuit(C stands for capicator, L is for inductor etc.). And the equation at the bottom shows how the transfer function is defined.

My question is how can the transfer function be determined from the group of equations above. And can you enunciate on that? A step by step enunciation would be really appreciated.

\begin{equation} \left\{ \begin{aligned} &V_A=V_{IN}\cdot \frac{\left(\frac{1}{j\omega C_A}\mid\mid\left(R_{A1}+R_{A2}\right)\right)}{\left(\frac{1}{j\omega C_A}\mid\mid\left(R_{A1}+R_{A2}\right)\right) + R_A + j\omega L_A}\cdot\frac{R_{A2}}{R_{A1}+R_{A2}}\\ &V_B=V_{IN}\cdot \frac{\left(\frac{1}{j(\omega+\pi) C_B}\mid\mid\left(R_{B1}+R_{B2}\right)\right)}{\left(\frac{1}{j(\omega+\pi) C_B}\mid\mid\left(R_{B1}+R_{B2}\right)\right) + R_B + j(\omega+\pi) L_B}\cdot\frac{R_{B2}}{R_{B1}+R_{B2}}\\ &V_{IN}=\left[V_{DC}-\left(V_A+V_B\right)\right]\cdot\frac{\Re_A\mid\mid\Re_B}{R_s+\left(\Re_A\mid\mid\Re_B\right)}\\ &\Re_A=\frac{1}{j\omega C_A}\mid\mid\left(R_{A1}+R_{A2}\right) + R_A + j\omega L_A\\ &\Re_B=\frac{1}{j\omega C_B}\mid\mid\left(R_{B1}+R_{B2}\right) + R_B + j\omega L_B\\ \end{aligned} \right. \end{equation}

\begin{equation} G(\omega)=\frac{V_A-V_B}{V_{DC}} \end{equation}

P.S. In the fig above, double slanted bar stands for parallel connection in a circuit, i.e. $R_A\mid\mid R_B=\frac{R_AR_B}{R_A+R_B}$

Answer 1

I think about your first $3$ equations as $$\begin{align}V_A & =V_{IN}\cdot ugly_1\\ V_B & =V_{IN}\cdot ugly_2\\ V_{IN} & = \left(V_{DC}-(V_A+V_B)\right)\cdot ugly3\end{align}$$ If we take the sum and difference of the first two equtions, we get $$\begin{align}V_A+V_B & =V_{IN}(ugly_1+ugly_2)\\ V_A-V_B & =V_{IN}(ugly_1-ugly_2)\end{align}$$ The sum equation may be substituted into the third of your equations to get $$V_{IN}=\left\{V_{DC}-V_{IN}(ugly_1+ugly_2)\right\}ugly_3$$ $$V_{IN}=\frac{V_{DC}\cdot ugly_3}{1+(ugly_1+ugly_2)ugly_3}$$ And that result can go into the difference equation as $$V_A-V_B=\frac{V_{DC}(ugly_1-ugly_2)ugly_3}{1+(ugly_1+ugly_2)ugly_3}$$Now just divide by $V_{DC}$, plug in your expressions for $ugly_1$, $ugly_2$, and $ugly_3$, simplify, and Bob's your uncle!