I have this circuit shown below:
simulate this circuit – Schematic created using CircuitLab
Basically this circuit has two inputs and two outputs. The voltage and current at the inputs are defined as: $$V_{1}^{in},I_{1}^{in},V_{3}^{in},I_{3}^{in}$$
and at the output:
$$V_{2}^{out},I_{2}^{out},V_{4}^{out},I_{4}^{out}$$
I know, I can represent these four elements (if you combine RLC together) by using transfer matrix as:
$$\begin{bmatrix} V_{1}^{out}\\ I_{1}^{out} \end{bmatrix} = T_{1}\begin{bmatrix} V_{1}^{in}\\ I_{1}^{in} \end{bmatrix}$$
$$\begin{bmatrix} V_{2}^{out}\\ I_{2}^{out} \end{bmatrix} = T_{2}\begin{bmatrix} V_{2}^{in}\\ I_{2}^{in} \end{bmatrix}$$
$$\begin{bmatrix} V_{3}^{out}\\ I_{3}^{out} \end{bmatrix} = T_{3}\begin{bmatrix} V_{3}^{in}\\ I_{3}^{in} \end{bmatrix}$$
$$\begin{bmatrix} V_{4}^{out}\\ I_{4}^{out} \end{bmatrix} = T_{4}\begin{bmatrix} V_{4}^{in}\\ I_{4}^{in} \end{bmatrix}$$
On the other hand we know that: $$I_{1}^{out} - I_{2}^{in} = I_{b}$$ and $$I_{3}^{out} - I_{4}^{in} = -I_{b}$$ where $$I_{b}$$ is defined as:
$$I_{b} = \frac{V_{1}^{out}-V_{3}^{out}}{R} = \frac{V_{2}^{in}-V_{4}^{in}}{R}$$
Also we know:
$$V_{1}^{out} = V_{2}^{in}$$
$$V_{3}^{out} = V_{4}^{in}$$
As a result finally we have this matrix equation for this circuit:
$$\begin{bmatrix} V_{2}^{out}\\ I_{2}^{out}\\ V_{4}^{out}\\ I_{4}^{out} \end{bmatrix} = \begin{bmatrix} T_{2}T_{1} & 0 \\ 0 & T_{4}T_{3} \end{bmatrix} \begin{bmatrix} V_{1}^{in}\\ I_{1}^{in}\\ V_{3}^{in}\\ I_{3}^{in} \end{bmatrix} + \begin{bmatrix} -T_{2}\begin{bmatrix} 0\\ I_{b} \end{bmatrix}\\ T_{4}\begin{bmatrix} 0\\ I_{b} \end{bmatrix} \end{bmatrix}$$
The above matrix equation gives me 4 equations but, I have 5 unknowns. My known parameters are:
$$I_{1}^{in},I_{3}^{in},V_{2}^{out},V_{4}^{out}$$
and my unknown parameters are:
$$V_{1}^{in},V_{3}^{in},I_{2}^{out},I_{4}^{out},I_{b}$$
Is there any way to find $I_{b}$ here without further assumption?
