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I have this circuit shown below:

schematic

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?

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  • \$\begingroup\$ What do the capacitors connect to? \$\endgroup\$ – Andy aka Aug 19 at 6:57
  • \$\begingroup\$ @Andyaka All the capacitors are connected to the ground. \$\endgroup\$ – Alone Programmer Aug 19 at 12:46
  • \$\begingroup\$ Is input on right? Or left? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 19 at 13:52
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 The 1 and 3 on the left are the inputs and 2 and 4 are the outputs on the right. \$\endgroup\$ – Alone Programmer Aug 19 at 13:53
  • \$\begingroup\$ What is the input waveform f and V. Zc=1/(2Pi *fC] then Ic(f)=V(f)/Zc. Then for 3kV spikes 4us rise time f=0.35/Tr \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 19 at 13:58
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You forgot to include one of these equations (that you already gave):

$$I_b = \frac{V_1^{out}-V_3^{out}}{R}$$

or

$$I_b = \frac{V_2^{out}-V_4^{out}}{R}$$

This should give you 5 equations and 5 unknowns.

[edit] I did not redo your calculations. Don't hold it against me if there was an error in there somewhere.

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