New Method of State-Space Formulation for Degenerate Circuit and Coupling Circuit

Abstract

The state-space formulation overcomes many limitations of traditional differential equation approach and is utilized as alternative to many traditional approaches in the modern electrical field. This paper proposes a new method of finding the state equation for degenerate circuit and coupling circuit that have not been systematically solved now. This paper also introduces some sound improvements to solve complicated dependent-source circuits. Four comparative examples are demonstrated to show the significant merits that our method owns over the traditional approaches.

This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

The state-space formulation plays a significant part to probe the significant properties including robustness and stability in the electrical related theories [1, 2]. The state-space formulation overcomes many nonsystematic limitations of traditional differential equation approach in the modern electrical field [3, 4].

The work proposed in [4] has focused on reducing the computational complexity of the state equation by substituting capacitors with voltage generators and inductors with current generators for the electrical circuits. At this point, the circuit contains only generators and resistors. The claim cannot work for the electrical circuit with coupling circuit due to the appearance of both inductor and mutual differential item $Md{i}_L/dt$. Apply the substitution theorem such that the capacitors (inductors) are changed with the current sources (the voltage sources), respectively, in our previous result [5]. The substitution theorem prevents the derivative elements $Ld{i}_L/dt$ and $Cd{v}_C/dt$ from being in the electrical system. However, [5] cannot derive the state equation for electrical circuit with coupling elements due to the coupling term $Md{i}_L/dt$. In order to overcome this shortcoming, we propose a new method of substituting the coupling term with $Md{i}_{L\mathrm{2}}/dt=(M/L_{\mathrm{2}})v_{L\mathrm{2}}$, $Md{i}_{L\mathrm{1}}/dt=(M/L_{\mathrm{1}})v_{L\mathrm{1}}$, and hence the problem of the coupling differential item $Md{i}_L/dt$ can be solved.

References [4] and [5] have claimed that the proposed method of finding the state equation cannot to be applied to degenerate circuits in which there are fewer state variables than energy-storage elements (capacitors and inductors), where the excess components are called the redundancy elements. The study of degenerate networks with the redundancy element will need more mathematical work for advanced electrical networks theories. On the other hand, a new method of finding the state equations for degenerate circuit is successfully presented in this paper. In order to solve the difficulty for degenerate circuits, we propose a new method of putting the redundancy element $R_e(t)$ into evidence in proper current or voltage terms $i_{C\mathrm{2}}=\left(\pm C_{\mathrm{2}}/C_{\mathrm{1}}\right)i_{C\mathrm{1}}\pm C_{\mathrm{2}}{\dot{R}}_e(t)$ and $v_{L\mathrm{2}}=\left(\pm L_{\mathrm{2}}/L_{\mathrm{1}}\right)v_{L\mathrm{1}}\pm L_{\mathrm{2}}{\dot{R}}_e$ and hence the problem of the redundancy element can be solved.

Reference [3, page131] had claimed that the problem will yield increased difficulties in using network analysis to obtain the state equation for the electrical network with a dependent source. The existing studies [6, page683] [7, page521] [8] show that the designs of the state equation become very complex for dependent-source circuit. The main characteristic of this paper is exploited in its systematic and unified structure that solves the limitations of those traditional approaches.

Up to now, the existing conventional approaches for the derivation of state-space formulation are summarized as follows.

Method 1.

Apply the Thevenin equivalent theorem to get the state equation [9]. The conventional standard methods of deriving the Thevenin equivalent circuit need two steps as follows: (Step 1) Let the load device be open-circuited and the Thevenin-voltage parameter is calculated as the open-circuit voltage across the load device. From the existent literatures [10, 11], it is obvious to see that the calculation of the open-circuit voltage is tedious when the network has the complicated devices consisting the dependent sources and coupling devices based on using the Kirchhoff’s voltage law (Kirchhoff’s current law). (Step 2) The left work of finding Thevenin equivalent circuit is the derivation of the Thevenin-resistance parameter. There are mainly three techniques for the derivation of Thevenin-resistance in existent researches as follows: (Technique 1) Set all independent sources to be zero with the load device being disconnected and then the equivalent resistance of the zero-energy circuit at the load device is the desired Thevenin-resistance parameter via the conventional node-voltage matrix method [12]. Technique 1 cannot be utilized when some voltage (current) source is not in series (parallel) with a resistor based on the intrinsical properties of the conventional node-voltage (mesh-current) matrix method. (Technique 2) Find the open-circuit voltage $V_{OC}$ and the short-circuit current $I_{SC}$ across the load device, then the Thevenin-resistance parameter is equal to $Z_{TH}=V_{OC}/I_{SC}$ [10, 13]. Technique 2 is not only impractical but also at the risk of encountering the interminate case 0/0. For some circuits consisting only dependent sources and resistors, one needs to apply the alternative method because the ratio of the open-circuit voltage $V_{OC}$ to the short-circuit current $I_{SC}$ is interminate form 0/0. (Technique 3) Firstly set all independent sources to be zero with the load device being disconnected. Then, use additionally the independent current (voltage) source $I_t$ ($V_t$) to the load device and thus find the load voltage (current) voltage (current) $V_k$ ($I_k$) by $Z_{TH}=V_k/I_t$ ($Z_{TH}=V_t/I_k$) [11]. Technique 3 is very restricted to certain conditions due to the requirement of setting all independent sources to be zero. Moreover, it also is obvious to see that using conventional Technique 3 is very messy based on using the Kirchhoff’s voltage law (Kirchhoff’s current law).

Method 2.

Obtain the state-space formulation for electrical circuits based on the superposition theorem [14]. In particular, it formulates an n-order circuit and splits this into n first-order circuit, each of which is solved separately. However, the splitting method is not easily be applied to get the state-space formulation when the network has the complicated devices consisting the bridging feedback element or dependent sources devices due to the intrinsical properties.

Method 3.

Formulate the state equation for electrical via the normal tree theory [15]. The method defines a complex normal tree in order to select the state variables. Moreover, it builds a set of equations via the nonsystematic Kirchhoff’s law.

Method 4.

Get the state-space formulation for electrical circuits based on using filling matrices [16, 17]. The method is only limited to nondegenerate circuits. Moreover, the operations of inverting the matrices make the method be impractical.

To solve all above shortcomings of the conventional approaches, we have proposed an efficient method for the calculation of the state-space formulation in a straightforward way. The significant novelties of this proposed approach include the following: (1) it can easily be applied to the electrical circuits that voltage (current) source is not in series (parallel) with a resistor, mutual-inductance element, and the dependent-source element without using tedious Kirchhoff’s law; (2) it does not formulate an n-order circuit and split this into n first-order circuit; (3) it is conceptually simpler since it does not need the use of normal tree theory; (4) it is applicable to both nondegenerate and degenerate circuits; (5) it does not need impractical operations of inverting the matrices.

2. New Method of Finding the State Equations

View the capacitor voltages and the inductor currents as state variables. The main structure of our proposed new simple unifying approach is given below.

Step 1.

Apply the substitution theorem to replace the capacitors with the current sources and the inductors with the voltage sources. Using the source-transformation theorem changes all voltage elements to be current elements and all current elements to be voltage elements, respectively.

If the voltage element is not in series with a resistor and the current element is not in parallel with a resistor, we add a zero resistor to be in series with the voltage element and an infinite resistor to be in parallel with the current element.

If the desired circuit contains the coupling elements, the coupling element can be replaced with with $Md{i}_{L\mathrm{2}}/dt=(M/L_{\mathrm{2}})v_{L\mathrm{2}}$, $Md{i}_{L\mathrm{1}}/dt=(M/L_{\mathrm{1}})v_{L\mathrm{1}}$. If the considered circuit is degenerate with redundancy voltage v(t) or current i(t), then we elaborate in such a way to show the redundancy element into equation $i_{C\mathrm{2}}=\left(\pm C_{\mathrm{2}}/C_{\mathrm{1}}\right)i_{C\mathrm{1}}\pm C_{\mathrm{2}}\dot{v}(t)$ and/or $v_{L\mathrm{2}}=\left(\pm L_{\mathrm{2}}/L_{\mathrm{1}}\right)v_{L\mathrm{1}}\pm L_{\mathrm{2}}\dot{i}(t)$ where current generator $i_{C\mathrm{2}}$ substitutes capacitor C2 and voltage source $v_{L\mathrm{2}}$ substitutes inductor L2 on a case by case basis.

Step 2.

Draw up the matrix node-voltage equations or matrix mesh-current equations by the following matrix equations:$$\begin{array}{}\text{(1)}& \left[\begin{array}{cccc}{Y}_{\mathrm{11}}& -Y_{\mathrm{12}}& \cdots & -Y_{\mathrm{1}N}\\ -Y_{\mathrm{21}}& Y_{\mathrm{22}}& \cdots & -Y_{\mathrm{2}N}\\ ⋮& ⋮& & ⋮\\ -Y_{N\mathrm{1}}& -Y_{N\mathrm{2}}& \cdots & Y_{NN}\end{array}\right]\left[\begin{array}{c}{V}_{\mathrm{1}}\\ V_{\mathrm{2}}\\ ⋮\\ V_N\end{array}\right]=\left[\begin{array}{c}\pm I_{S\mathrm{11}}\pm I_{S\mathrm{12}}\pm \cdots \\ \pm I_{S\mathrm{21}}\pm I_{S\mathrm{22}}\pm \cdots \\ ⋮\\ \pm I_{SN\mathrm{1}}\pm I_{SN\mathrm{2}}\pm \cdots \end{array}\right]\text{(2)}& \left[\begin{array}{cccc}{Z}_{\mathrm{11}}& -Z_{\mathrm{12}}& \cdots & -Z_{\mathrm{1}N}\\ -Z_{\mathrm{21}}& Z_{\mathrm{22}}& \cdots & -Z_{\mathrm{2}N}\\ ⋮& ⋮& & ⋮\\ -Z_{N\mathrm{1}}& -Z_{N\mathrm{2}}& \cdots & Z_{NN}\end{array}\right]\left[\begin{array}{c}{I}_{\mathrm{1}}\\ I_{\mathrm{2}}\\ ⋮\\ I_N\end{array}\right]=\left[\begin{array}{c}\pm V_{S\mathrm{11}}\pm V_{S\mathrm{12}}\pm \cdots \\ \pm V_{S\mathrm{21}}\pm V_{S\mathrm{22}}\pm \cdots \\ ⋮\\ \pm V_{SN\mathrm{1}}\pm V_{SN\mathrm{2}}\pm \cdots \end{array}\right]\end{array}$$where $V_{sij}$ and $I_{sij}$ are denoted as the additional voltage sources and the current sources respectively, $Z_{ij}$ and $Y_{ij}$ are denoted as the impedance and the admittances, respectively, and $V_i$ and $I_i$ are denoted as the node-voltage variables and the mesh-current variables, respectively. Moreover, $I_{ii}$, $i=\mathrm{1,2},\cdots ,N$ are in clockwise direction.

If the dependent current elements and voltage elements are concerned in the listed matrix, then the dependent elements must be changed to be node-voltage elements or mesh-current elements and then the easy moving-term processes are performed.

Step 3.

Applying our invented easy matrix operations changes the node-voltage elements or mesh-current elements to the state elements $v_{Ck},i_{Lk}$. Some easy and significant matrix operations are outlined as follows:

(rule i)$$\begin{array}{c}\text{(3)}& \left[\begin{array}{ccc}A& B& C\\ D& E& F\\ G& H& I\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}J\\ K\\ L\end{array}\right]\Rightarrow \left[\begin{array}{ccc}A& B·k& C\\ D& E·k& F\\ G& H·k& I\end{array}\right]\left[\begin{array}{c}x\\ \frac{y}{k}\\ z\end{array}\right]=\left[\begin{array}{c}J\\ K\\ L\end{array}\right]\end{array}$$ (rule ii)$$\begin{array}{c}\text{(4)}& \left[\begin{array}{ccc}A& B& C\\ D& E& F\\ G& H& I\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}J\\ K\\ L\end{array}\right]\Rightarrow \left[\begin{array}{ccc}A& B& C\\ D& E& F\\ G& H& I\end{array}\right]\left[\begin{array}{c}x-k\\ y\\ z\end{array}\right]=\left[\begin{array}{c}J-Ak\\ K-Dk\\ L-Gk\end{array}\right]\end{array}$$ (rule iii)$$\begin{array}{c}\text{(5)}& \left[\begin{array}{ccc}A& B& C\\ D& E& F\\ G& H& I\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}J\\ K\\ L\end{array}\right]\Rightarrow \left[\begin{array}{ccc}A& B+A& C\\ D& E+D& F\\ G& H+G& I\end{array}\right]\left[\begin{array}{c}x-y\\ y\\ z\end{array}\right]=\left[\begin{array}{c}J\\ K\\ L\end{array}\right]\end{array}$$ (rule iv)$$\begin{array}{c}\text{(6)}& \left[\begin{array}{ccc}A& B& C\\ D& E& F\\ G& H& I\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}J\\ K\\ L\end{array}\right]\Rightarrow \left[\begin{array}{ccc}B& A& C\\ E& D& F\\ H& G& I\end{array}\right]\left[\begin{array}{c}y\\ x\\ z\end{array}\right]=\left[\begin{array}{c}J\\ K\\ L\end{array}\right]\end{array}$$ (rule v) $$\begin{array}{c}\text{(7)}& \left[\begin{array}{ccc}A& B& C\\ D& \mathrm{0}& F\\ G& \mathrm{0}& I\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}J\\ K\\ L\end{array}\right]\Rightarrow \left[\begin{array}{cc}D& F\\ G& I\end{array}\right]\left[\begin{array}{c}x\\ z\end{array}\right]=\left[\begin{array}{c}K\\ L\end{array}\right]\end{array}$$ (rule vi) (8) A B C D E F G H I x y z = J K L ⇒ - J B C - K E F - L H I x x 2 y x z x = - A - D - G (rule vii)$$\begin{array}{c}\text{(9)}& \left[\begin{array}{ccc}A& B& C\\ D& E& F\\ G& H& I\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}J\\ K\\ L\end{array}\right]\Rightarrow \left[\begin{array}{ccc}A& B& C\\ D& E& F\\ G+D& H+E& I+F\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}J\\ K\\ L+K\end{array}\right]\end{array}$$

Step 4.

Reorganize the elements $i_{Ck},v_{Lk}$ to build a new matrix equation from the right-hand side of original listed matrix equation. Use the Cramer’s rule to obtain the elements $i_{C_k}$ and $v_{L_k}$. Then substitute the elements $i_{C_k}$ with $C_k\left(d{v}_{C_k}/dt\right)$ and the elements $v_{L_k}$ with $L_k\left(d{i}_{L_k}/dt\right)$, respectively, to get the state equation.

3. Demonstrated Examples

In order to show the significant merits that our method owns over the traditional approaches [3, 4, 18], four comparative examples are proposed in this section.

Example 1.

Consider the electrical network [4] shown in Figure 1. The presented approach will be compared to the state-space formulation proposed in [4]. The state equation is easily determined by using only four-step approach in unified process.

[figure omitted; refer to PDF]

Step 1.

Applying the substitution theorem and source-transformation theorem yields Figures 2 and 3.

[figure omitted; refer to PDF] [figure omitted; refer to PDF]

Step 2.

Draw up the matrix node-voltage equations and solve the dependent element by easy moving-term processes.$$\begin{array}{}\text{(10)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R}+\frac{\mathrm{1}}{\mathrm{2}R}& \mathrm{0}& -\frac{\mathrm{1}}{\mathrm{2}R}\\ \mathrm{0}& \frac{\mathrm{1}}{R_{CCC}}& \mathrm{0}\\ -\frac{\mathrm{1}}{\mathrm{2}R}& \mathrm{0}& \frac{\mathrm{1}}{\mathrm{3}R}+\frac{\mathrm{1}}{\mathrm{2}R}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{11}}\\ v_{\mathrm{22}}\\ v_{\mathrm{33}}\end{array}\right]=\left[\begin{array}{c}+\frac{v_s}{R}-i_{C\mathrm{1}}\\ i_{C\mathrm{1}}-i_{C\mathrm{2}}+\frac{v_L}{R_{CCC}}\\ i_{C\mathrm{2}}\end{array}\right]\end{array}$$

Step 3.

Using our invented easy matrix operations changes the node-voltage elements to the state elements $v_{Ck},i_{Lk}$ based on the following equations:$$\begin{array}{}\text{(11)}& v_{C\mathrm{1}}=v_{\mathrm{11}}-v_{\mathrm{22}}\text{(12)}& v_{C\mathrm{2}}=v_{\mathrm{22}}-v_{\mathrm{33}}=-\left(v_{\mathrm{33}}-v_{\mathrm{22}}\right)\text{(13)}& i_L=\frac{v_{\mathrm{22}}-v_L}{R_{CCC}}\end{array}$$Applying (rule iii) to (10) yields$$\begin{array}{}\text{(14)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R}+\frac{\mathrm{1}}{\mathrm{2}R}& \mathrm{0}+\frac{\mathrm{1}}{R}+\frac{\mathrm{1}}{\mathrm{2}R}& -\frac{\mathrm{1}}{\mathrm{2}R}\\ \mathrm{0}& \frac{\mathrm{1}}{R_{CCC}}& \mathrm{0}\\ -\frac{\mathrm{1}}{\mathrm{2}R}& \mathrm{0}-\frac{\mathrm{1}}{\mathrm{2}R}& \frac{\mathrm{1}}{\mathrm{3}R}+\frac{\mathrm{1}}{\mathrm{2}R}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{11}}-v_{\mathrm{22}}\\ v_{\mathrm{22}}\\ v_{\mathrm{33}}\end{array}\right]=\left[\begin{array}{c}+\frac{v_s}{R}-i_{C\mathrm{1}}\\ i_{C\mathrm{1}}-i_{C\mathrm{2}}+\frac{v_L}{R_{CCC}}\\ i_{C\mathrm{2}}\end{array}\right]\end{array}$$From (rule iii), (14) is changed to be$$\begin{array}{}\text{(15)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R}+\frac{\mathrm{1}}{\mathrm{2}R}& \mathrm{0}+\frac{\mathrm{1}}{R}+\frac{\mathrm{1}}{\mathrm{2}R}-\frac{\mathrm{1}}{\mathrm{2}R}& -\frac{\mathrm{1}}{\mathrm{2}R}\\ \mathrm{0}& \frac{\mathrm{1}}{R_{CCC}}& 0\\ -\frac{\mathrm{1}}{\mathrm{2}R}& \mathrm{0}-\frac{\mathrm{1}}{\mathrm{2}R}+\frac{\mathrm{1}}{\mathrm{3}R}+\frac{\mathrm{1}}{\mathrm{2}R}& \frac{\mathrm{1}}{\mathrm{3}R}+\frac{\mathrm{1}}{\mathrm{2}R}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{11}}-v_{\mathrm{22}}\\ v_{\mathrm{22}}\\ v_{\mathrm{33}}-v_{\mathrm{22}}\end{array}\right]=\left[\begin{array}{c}+\frac{v_s}{R}-i_{C\mathrm{1}}\\ i_{C\mathrm{1}}-i_{C\mathrm{2}}+\frac{v_L}{R_{CCC}}\\ i_{C\mathrm{2}}\end{array}\right]\end{array}$$Using (rule ii) and (rule iii) to (15) gets$$\begin{array}{}\text{(16)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R}+\frac{\mathrm{1}}{\mathrm{2}R}& R_{CCC}\left(\mathrm{0}+\frac{\mathrm{1}}{R}+\frac{\mathrm{1}}{\mathrm{2}R}-\frac{\mathrm{1}}{\mathrm{2}R}\right)& -\frac{\mathrm{1}}{\mathrm{2}R}\\ \mathrm{0}& \frac{\mathrm{1}}{R_{CCC}}{R}_{CCC}& \mathrm{0}\\ -\frac{\mathrm{1}}{\mathrm{2}R}& R_{CCC}\left(\mathrm{0}-\frac{\mathrm{1}}{\mathrm{2}R}+\frac{\mathrm{1}}{\mathrm{3}R}+\frac{\mathrm{1}}{\mathrm{2}R}\right)& \frac{\mathrm{1}}{\mathrm{3}R}+\frac{\mathrm{1}}{\mathrm{2}R}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{11}}-v_{\mathrm{22}}\\ \frac{v_{\mathrm{22}}-v_L}{R_{CCC}}\\ v_{\mathrm{33}}-v_{\mathrm{22}}\end{array}\right]=\left[\begin{array}{c}+\frac{v_s}{R}-i_{C\mathrm{1}}-\frac{\mathrm{1}}{R}{v}_L\\ i_{C\mathrm{1}}-i_{C\mathrm{2}}+\frac{v_L}{R_{CCC}}-\frac{v_L}{R_{CCC}}\\ i_{C\mathrm{2}}-\frac{\mathrm{1}}{\mathrm{3}R}{v}_L\end{array}\right]\end{array}$$i.e.,$$\begin{array}{}\text{(17)}& \left[\begin{array}{ccc}\frac{\mathrm{3}}{\mathrm{2}R}& \mathrm{0}& \frac{\mathrm{1}}{\mathrm{2}R}\\ \mathrm{0}& \mathrm{1}& \mathrm{0}\\ -\frac{\mathrm{1}}{\mathrm{2}R}& \mathrm{0}& -\frac{\mathrm{5}}{\mathrm{6}R}\end{array}\right]\left[\begin{array}{c}{v}_{C\mathrm{1}}\\ i_L\\ v_{C\mathrm{2}}\end{array}\right]=\left[\begin{array}{c}\frac{v_S}{R}-i_{C\mathrm{1}}-\frac{\mathrm{1}}{R}{v}_L\\ i_{C\mathrm{1}}-i_{C\mathrm{2}}\\ i_{C\mathrm{2}}-\frac{\mathrm{1}}{\mathrm{3}R}{v}_L\end{array}\right]\end{array}$$

Step 4.

Reorganize the elements $i_{Ck},v_{Lk}$ to build a new matrix equation from the right-hand side of original listed matrix equation. $$\begin{array}{}\text{(18)}& \left[\begin{array}{ccc}-\mathrm{1}& \mathrm{0}& -\frac{\mathrm{1}}{R}\\ \mathrm{1}& -\mathrm{1}& \mathrm{0}\\ \mathrm{0}& \mathrm{1}& -\frac{\mathrm{1}}{\mathrm{3}R}\end{array}\right]\left[\begin{array}{c}{i}_{C\mathrm{1}}\\ i_{C\mathrm{2}}\\ v_L\end{array}\right]=\left[\begin{array}{ccc}\frac{\mathrm{3}}{\mathrm{2}R}& \frac{\mathrm{1}}{\mathrm{2}R}& \mathrm{0}\\ \mathrm{0}& \mathrm{0}& \mathrm{1}\\ -\frac{\mathrm{1}}{\mathrm{2}R}& -\frac{\mathrm{5}}{\mathrm{6}R}& \mathrm{0}\end{array}\right]\left[\begin{array}{c}{v}_{C\mathrm{1}}\\ v_{C\mathrm{2}}\\ i_L\end{array}\right]+\left[\begin{array}{c}-\frac{\mathrm{1}}{R}\\ \mathrm{0}\\ \mathrm{0}\end{array}\right]v_S\end{array}$$Use the Cramer’s rule to obtain the elements $i_{C_k}$ ($v_{L_k}$) and then get the desired state equation. $$\begin{array}{}\text{(19)}& \left[\begin{array}{c}{C}_{\mathrm{1}}\frac{d{v}_{C\mathrm{1}}}{dt}\left(i.e\text{.}{i}_{C\mathrm{1}}\right)\\ C_{\mathrm{2}}\frac{d{v}_{C\mathrm{2}}}{dt}\left(i.e\text{.}{i}_{C\mathrm{2}}\right)\\ L\frac{d{i}_L}{dt}\left(i.e\text{.}{v}_L\right)\end{array}\right]=\left[\begin{array}{ccc}-\frac{\mathrm{3}}{\mathrm{4}R}& -\frac{\mathrm{3}}{\mathrm{4}R}& \frac{\mathrm{3}}{\mathrm{4}}\\ -\frac{\mathrm{3}}{\mathrm{4}R}& -\frac{\mathrm{3}}{\mathrm{4}R}& -\frac{\mathrm{1}}{\mathrm{4}}\\ -\frac{\mathrm{3}}{\mathrm{4}}& \frac{\mathrm{1}}{\mathrm{4}}& -\frac{\mathrm{3}R}{\mathrm{4}}\end{array}\right]\left[\begin{array}{c}{v}_{C\mathrm{1}}\\ v_{C\mathrm{2}}\\ i_L\end{array}\right]+\left[\begin{array}{c}\frac{\mathrm{1}}{\mathrm{4}R}\\ \frac{\mathrm{1}}{\mathrm{4}R}\\ \frac{\mathrm{3}}{\mathrm{4}}\end{array}\right]v_S\end{array}$$

Example 2.

Reference [4] claimed that the state equation was mainly obtained by substituting capacitors with voltage generators and inductors with current generators for the electrical circuit. At this point, the circuit contains only generators and resistors. The claim cannot work for the electrical circuit with coupling circuit due to the appearance of both inductor and mutual differential item. Consider the electrical network with coupling elements shown in Figure 4, where the mutual inductance is denoted as $M$.

[figure omitted; refer to PDF]

From the general circuit theory, the related equivalent circuit of coupling circuit is given by Figure 5.

[figure omitted; refer to PDF]

Since$$\begin{array}{}\text{(20)}& v_{L\mathrm{2}}=L_{\mathrm{2}}\frac{d{i}_{L\mathrm{2}}}{dt}\end{array}$$and$$\begin{array}{}\text{(21)}& v_{L\mathrm{1}}=L_{\mathrm{1}}\frac{d{i}_{L\mathrm{1}}}{dt}\end{array}$$the dependent coupling element can be replaced with$$\begin{array}{}\text{(22)}& \frac{d{i}_{L\mathrm{2}}}{dt}=\frac{\mathrm{1}}{L_{\mathrm{2}}}{v}_{L\mathrm{2}}\end{array}$$and$$\begin{array}{}\text{(23)}& \frac{d{i}_{L\mathrm{1}}}{dt}=\frac{\mathrm{1}}{L_{\mathrm{1}}}{v}_{L\mathrm{1}}\end{array}$$Apply the new approach to obtain the state equations as follows.

Step 1.

Applying the substitution theorem and source-transformation theorem gets Figure 6.

[figure omitted; refer to PDF]

Step 2.

Draw up the matrix node-voltage equations and solve the dependent source by simple moving-term processes. $$\begin{array}{}\text{(24)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R_{CCC}}+\frac{\mathrm{1}}{R_{\mathrm{1}}}& \mathrm{0}& -\frac{\mathrm{1}}{R_{CCC}}\\ \mathrm{0}& \frac{\mathrm{1}}{R_{CCC}}+\frac{\mathrm{1}}{R_{\mathrm{2}}}& -\frac{\mathrm{1}}{R_{CCC}}\\ -\frac{\mathrm{1}}{R_{CCC}}& -\frac{\mathrm{1}}{R_{CCC}}& \frac{\mathrm{2}}{R_{CCC}}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{1}}\\ v_{\mathrm{2}}\\ v_{\mathrm{3}}\end{array}\right]=\left[\begin{array}{c}\frac{v_S}{R_{\mathrm{1}}}-\frac{v_{L\mathrm{1}}+\left(M/L_{\mathrm{2}}\right)v_{L\mathrm{2}}}{R_{CCC}}\\ -\frac{\left(M/L_{\mathrm{1}}\right)v_{L\mathrm{1}}+v_{L\mathrm{2}}}{R_{CCC}}\\ \frac{v_{L\mathrm{1}}+\left(M/L_{\mathrm{2}}\right)v_{L\mathrm{2}}}{R_{CCC}}+\frac{\left(M/L_{\mathrm{1}}\right)v_{L\mathrm{1}}+v_{L\mathrm{2}}}{R_{CCC}}-i_C\end{array}\right]\end{array}$$

Step 3.

Use our invented easy matrix operations changes the node-voltage elements to the state elements $v_{Ck},i_{Lk}$ based on the following equations:$$\begin{array}{}\text{(25)}& i_{L\mathrm{1}}=\frac{v_{\mathrm{1}}-v_s}{R_{\mathrm{1}}}\text{(26)}& i_{L2}=\frac{v_2}{R_2}\text{(27)}& v_C=v_3\end{array}$$From (rule ii), we get$$\begin{array}{}\text{(28)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R_{CCC}}+\frac{\mathrm{1}}{R_{\mathrm{1}}}& \mathrm{0}& -\frac{\mathrm{1}}{R_{CCC}}\\ \mathrm{0}& \frac{\mathrm{1}}{R_{CCC}}+\frac{\mathrm{1}}{R_{\mathrm{2}}}& -\frac{\mathrm{1}}{R_{CCC}}\\ -\frac{\mathrm{1}}{R_{CCC}}& -\frac{\mathrm{1}}{R_{CCC}}& \frac{\mathrm{2}}{R_{CCC}}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{1}}-v_S\\ v_{\mathrm{2}}\\ v_{\mathrm{3}}\end{array}\right]=\left[\begin{array}{c}\frac{v_S}{R_{\mathrm{1}}}-\frac{v_{L\mathrm{1}}+\left(M/L_{\mathrm{2}}\right)v_{L\mathrm{2}}}{R_{CCC}}-\frac{v_S}{R_{CCC}}-\frac{v_S}{R_{\mathrm{1}}}\\ -\frac{\left(M/L_{\mathrm{1}}\right)v_{L\mathrm{1}}+v_{L\mathrm{2}}}{R_{CCC}}\\ \frac{v_{L\mathrm{1}}+\left(M/L_{\mathrm{2}}\right)v_{L\mathrm{2}}}{R_{CCC}}+\frac{\left(M/L_{\mathrm{1}}\right)v_{L\mathrm{1}}+v_{L\mathrm{2}}}{R_{CCC}}-i_C+\frac{v_S}{R_{CCC}}\end{array}\right]\end{array}$$Adding first row and second row to third row yields$$\begin{array}{}\text{(29)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R_{CCC}}+\frac{\mathrm{1}}{R_{\mathrm{1}}}& \mathrm{0}& -\frac{\mathrm{1}}{R_{CCC}}\\ \mathrm{0}& \frac{\mathrm{1}}{R_{CCC}}+\frac{\mathrm{1}}{R_{\mathrm{2}}}& -\frac{\mathrm{1}}{R_{CCC}}\\ \frac{\mathrm{1}}{R_{\mathrm{1}}}& \frac{\mathrm{1}}{R_{\mathrm{2}}}& 0\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{1}}-v_S\\ v_{\mathrm{2}}\\ v_{\mathrm{3}}\end{array}\right]=\left[\begin{array}{c}-\frac{v_{L\mathrm{1}}+\left(M/L_{\mathrm{2}}\right)v_{L\mathrm{2}}}{R_{CCC}}-\frac{v_S}{R_{CCC}}\\ -\frac{\left(M/L_{\mathrm{1}}\right)v_{L\mathrm{1}}+v_{L\mathrm{2}}}{R_{CCC}}\\ -i_C\end{array}\right]\end{array}$$Multiplying first row and second row with $R_{CCC}$ yields $$\begin{array}{}\text{(30)}& \left[\begin{array}{ccc}\mathrm{1}& \mathrm{0}& -\mathrm{1}\\ \mathrm{0}& \mathrm{1}& -\mathrm{1}\\ \frac{\mathrm{1}}{R_{\mathrm{1}}}& \frac{\mathrm{1}}{R_{\mathrm{2}}}& \mathrm{0}\end{array}\right]\left[\begin{array}{c}{i}_{L\mathrm{1}}\\ i_{L\mathrm{2}}\\ v_C\end{array}\right]=\left[\begin{array}{c}-v_{L\mathrm{1}}-\frac{M}{L_{\mathrm{2}}}{v}_{L\mathrm{2}}-v_S\\ -\frac{M}{L_{\mathrm{1}}}{v}_{L\mathrm{1}}-v_{L\mathrm{2}}\\ -i_C\end{array}\right]\end{array}$$

Step 4.

Reorganize the elements $i_{Ck},v_{Lk}$ to build a new matrix equation from the right-hand side of original listed matrix equation.$$\begin{array}{}\text{(31)}& \left[\begin{array}{ccc}-\mathrm{1}& -\frac{M}{L_{\mathrm{2}}}& \mathrm{0}\\ -\frac{M}{L_{\mathrm{1}}}& -\mathrm{1}& \mathrm{0}\\ \mathrm{0}& \mathrm{0}& -\mathrm{1}\end{array}\right]\left[\begin{array}{c}{v}_{L\mathrm{1}}\\ v_{L\mathrm{2}}\\ i_C\end{array}\right]=\left[\begin{array}{ccc}\mathrm{1}& \mathrm{0}& -\mathrm{1}\\ \mathrm{0}& \mathrm{1}& -\mathrm{1}\\ \frac{\mathrm{1}}{R_{\mathrm{1}}}& \frac{\mathrm{1}}{R_{\mathrm{2}}}& \mathrm{0}\end{array}\right]\left[\begin{array}{c}{i}_{L\mathrm{1}}\\ i_{L\mathrm{2}}\\ v_C\end{array}\right]+\left[\begin{array}{c}\mathrm{1}\\ \mathrm{0}\\ \mathrm{0}\end{array}\right]v_S\end{array}$$Use the Cramer’s rule to obtain the elements $i_{C_k}$ ($v_{L_k}$) and then get the desired state equation.$$\begin{array}{}\text{(32)}& \left[\begin{array}{c}{L}_{\mathrm{1}}\frac{d{i}_{L\mathrm{1}}}{dt}\left(i.e\text{.}{v}_{L\mathrm{1}}\right)\\ L_{\mathrm{2}}\frac{d{i}_{L\mathrm{2}}}{dt}\left(i.e\text{.}{v}_{L\mathrm{2}}\right)\\ C\frac{d{v}_C}{dt}\left(i.e\text{.}{i}_C\right)\end{array}\right]=\left[\begin{array}{ccc}\frac{\mathrm{1}}{\Delta }& -\frac{M}{\Delta L_{\mathrm{2}}}& \left(\frac{M}{L_{\mathrm{2}}}-\mathrm{1}\right)\frac{\mathrm{1}}{\Delta }\\ -\frac{M}{\Delta L_{\mathrm{1}}}& \frac{\mathrm{1}}{\Delta }& \left(\frac{M}{L_{\mathrm{1}}}-\mathrm{1}\right)\frac{\mathrm{1}}{\Delta }\\ -\frac{\mathrm{1}}{R_{\mathrm{1}}}& -\frac{\mathrm{1}}{R_{\mathrm{2}}}& 0\end{array}\right]\left[\begin{array}{c}{i}_{L\mathrm{1}}\\ i_{L\mathrm{2}}\\ v_C\end{array}\right]+\left[\begin{array}{c}\frac{\mathrm{1}}{\Delta }\\ -\frac{M}{\Delta L_{\mathrm{1}}}\\ 0\end{array}\right]v_S\end{array}$$where $\Delta \equiv M^{\mathrm{2}}/L_{\mathrm{1}}{L}_{\mathrm{2}}-\mathrm{1}$.

Example 3.

Reference [4] has claimed that the proposed method of finding the state equation is only applicable to nondegenerate circuits. An example for degenerate circuit is presented by using only four-step approach in unified process. Consider the degenerate circuit shown in Figure 7 consisting of an inductive cutest and a capacitive loop [18, page784]. For the sake of simplicity, let all numerical values be equal to one.

[figure omitted; refer to PDF]

Since$$\begin{array}{}\text{(33)}& i_{C\mathrm{2}}=C_{\mathrm{2}}\frac{d{v}_{C\mathrm{2}}}{dt}=C_{\mathrm{2}}\frac{d}{dt}\left(v_{C\mathrm{1}}-v_{S\mathrm{2}}\right)=\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}{C}_{\mathrm{1}}\frac{d{v}_{C\mathrm{1}}}{dt}-C_{\mathrm{2}}\frac{d{v}_{S\mathrm{2}}}{dt}=\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}{i}_{C\mathrm{1}}-C_{\mathrm{2}}{\dot{v}}_{S\mathrm{2}}\end{array}$$and$$\begin{array}{}\text{(34)}& v_{L\mathrm{2}}=L_{\mathrm{2}}\frac{d{i}_{L\mathrm{2}}}{dt}=L_{\mathrm{2}}\frac{d}{dt}\left(-i_{L\mathrm{1}}-i_S\right)=\frac{-L_{\mathrm{2}}}{L_{\mathrm{1}}}{v}_{L\mathrm{1}}-L_{\mathrm{2}}{\dot{i}}_S\end{array}$$We will substitute the inductor $L_2$ and capacitor $C_2$ with voltage source $\left(-L_2/L_1\right)v_{L1}-L_2{\dot{i}}_S$ and current source $\left(C_2/C_1\right)i_{C1}-C_2{\dot{v}}_{S2}$, respectively, as follows.

Step 1.

Applying the substitution theorem and source-transformation theorem obtains Figure 8.

[figure omitted; refer to PDF]

Step 2.

Draw up the matrix node-voltage equations and solve the dependent source by simple moving-term processes. $$\begin{array}{}\text{(35)}& \left[\begin{array}{ccc}\frac{\mathrm{2}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{1}}}& -\frac{\mathrm{1}}{R_{ccc}}& -\frac{\mathrm{1}}{R_{ccc}}\\ -\frac{\mathrm{1}}{R_{ccc}}& \frac{\mathrm{1}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{2}}}& 0\\ -\frac{\mathrm{1}}{R_{ccc}}& 0& \frac{\mathrm{1}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{3}}}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{1}}\\ v_{\mathrm{2}}\\ v_{\mathrm{3}}\end{array}\right]=\left[\begin{array}{c}\frac{v_{S\mathrm{1}}}{R_{\mathrm{1}}}-i_{C\mathrm{1}}+\frac{v_{L\mathrm{1}}}{R_{ccc}}+\frac{v_{S\mathrm{2}}}{R_{ccc}}\\ \frac{-v_{L\mathrm{1}}}{R_{ccc}}+i_S+\frac{L_{\mathrm{2}}}{L_{\mathrm{1}}{R}_{\mathrm{2}}}{v}_{L\mathrm{1}}+\frac{L_{\mathrm{2}}}{R_{\mathrm{2}}}{\dot{i}}_S\\ \frac{-v_{S\mathrm{2}}}{R_{ccc}}-i_S-\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}{i}_{C\mathrm{1}}+C_{\mathrm{2}}{\dot{v}}_{S\mathrm{2}}\end{array}\right]\end{array}$$

Step 3.

Use our invented easy matrix operations changes the node-voltage elements to the state elements $v_{Ck},i_{Lk}$ based on the following equations:$$\begin{array}{}\text{(36)}& v_{C\mathrm{1}}=v_{\mathrm{1}}\text{(37)}& i_{L1}=\frac{v_1-v_2-v_{L1}}{R_{ccc}}\end{array}$$Applying (rule i) to (35) yields$$\begin{array}{}\text{(38)}& \left[\begin{array}{ccc}\frac{\mathrm{2}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{1}}}& \frac{\mathrm{1}}{R_{ccc}}& -\frac{\mathrm{1}}{R_{ccc}}\\ -\frac{\mathrm{1}}{R_{ccc}}& -\frac{\mathrm{1}}{R_{ccc}}-\frac{\mathrm{1}}{R_{\mathrm{2}}}& \mathrm{0}\\ -\frac{\mathrm{1}}{R_{ccc}}& \mathrm{0}& \frac{\mathrm{1}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{3}}}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{1}}\\ -v_{\mathrm{2}}\\ v_{\mathrm{3}}\end{array}\right]=\left[\begin{array}{c}\frac{v_{S\mathrm{1}}}{R_{\mathrm{1}}}-i_{C\mathrm{1}}+\frac{v_{L\mathrm{1}}}{R_{ccc}}+\frac{v_{S\mathrm{2}}}{R_{ccc}}\\ \frac{-v_{L\mathrm{1}}}{R_{ccc}}+i_S+\frac{L_{\mathrm{2}}}{L_{\mathrm{1}}{R}_{\mathrm{2}}}{v}_{L\mathrm{1}}+\frac{L_{\mathrm{2}}}{R_{\mathrm{2}}}{\dot{i}}_S\\ \frac{-v_{S\mathrm{2}}}{R_{ccc}}-i_S-\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}{i}_{C\mathrm{1}}+C_{\mathrm{2}}{\dot{v}}_{S\mathrm{2}}\end{array}\right]\end{array}$$From (rule iii), we get$$\begin{array}{}\text{(39)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{1}}}& \frac{\mathrm{1}}{R_{ccc}}& -\frac{\mathrm{1}}{R_{ccc}}\\ \frac{\mathrm{1}}{R_{\mathrm{2}}}& -\frac{\mathrm{1}}{R_{ccc}}-\frac{\mathrm{1}}{R_{\mathrm{2}}}& 0\\ -\frac{\mathrm{1}}{R_{ccc}}& 0& \frac{\mathrm{1}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{3}}}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{1}}\\ -v_{\mathrm{2}}+v_{\mathrm{1}}\\ v_{\mathrm{3}}\end{array}\right]=\left[\begin{array}{c}\frac{v_{S\mathrm{1}}}{R_{\mathrm{1}}}-i_{C\mathrm{1}}+\frac{v_{L\mathrm{1}}}{R_{ccc}}+\frac{v_{S\mathrm{2}}}{R_{ccc}}\\ \frac{-v_{L\mathrm{1}}}{R_{ccc}}+i_S+\frac{L_{\mathrm{2}}}{L_{\mathrm{1}}{R}_{\mathrm{2}}}{v}_{L\mathrm{1}}+\frac{L_{\mathrm{2}}}{R_{\mathrm{2}}}{\dot{i}}_S\\ \frac{-v_{S\mathrm{2}}}{R_{ccc}}-i_S-\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}{i}_{C\mathrm{1}}+C_{\mathrm{2}}{\dot{v}}_{S\mathrm{2}}\end{array}\right]\end{array}$$From (rule ii), (39) is changed to be$$\begin{array}{}\text{(40)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{1}}}& \frac{\mathrm{1}}{R_{ccc}}& -\frac{\mathrm{1}}{R_{ccc}}\\ \frac{\mathrm{1}}{R_{\mathrm{2}}}& -\frac{\mathrm{1}}{R_{ccc}}-\frac{\mathrm{1}}{R_{\mathrm{2}}}& 0\\ -\frac{\mathrm{1}}{R_{ccc}}& 0& \frac{\mathrm{1}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{3}}}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{1}}\\ -v_{\mathrm{2}}+v_{\mathrm{1}}-v_{L\mathrm{1}}\\ v_{\mathrm{3}}\end{array}\right]=\left[\begin{array}{c}\frac{v_{S\mathrm{1}}}{R_{\mathrm{1}}}-i_{C\mathrm{1}}+\frac{v_{S\mathrm{2}}}{R_{ccc}}\\ +i_S+\frac{L_{\mathrm{1}}+L_{\mathrm{2}}}{L_{\mathrm{1}}{R}_{\mathrm{2}}}{v}_{L\mathrm{1}}+\frac{L_{\mathrm{2}}}{R_{\mathrm{2}}}{\dot{i}}_S\\ \frac{-v_{S\mathrm{2}}}{R_{ccc}}-i_S-\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}{i}_{C\mathrm{1}}+C_{\mathrm{2}}{\dot{v}}_{S\mathrm{2}}\end{array}\right]\end{array}$$Applying (rule i) to (40) yields$$\begin{array}{}\text{(41)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{1}}}& 1& -\frac{\mathrm{1}}{R_{ccc}}\\ \frac{\mathrm{1}}{R_{\mathrm{2}}}& -1& 0\\ -\frac{\mathrm{1}}{R_{ccc}}& 0& \frac{\mathrm{1}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{3}}}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{1}}\\ \frac{-v_{\mathrm{2}}+v_{\mathrm{1}}-v_{L\mathrm{1}}}{R_{ccc}}\\ v_{\mathrm{3}}\end{array}\right]=\left[\begin{array}{c}\frac{v_{S\mathrm{1}}}{R_{\mathrm{1}}}-i_{C\mathrm{1}}+\frac{v_{S\mathrm{2}}}{R_{ccc}}\\ +i_S+\frac{L_{\mathrm{1}}+L_{\mathrm{2}}}{L_{\mathrm{1}}{R}_{\mathrm{2}}}{v}_{L\mathrm{1}}+\frac{L_{\mathrm{2}}}{R_{\mathrm{2}}}{\dot{i}}_S\\ \frac{-v_{S\mathrm{2}}}{R_{ccc}}-i_S-\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}{i}_{C\mathrm{1}}+C_{\mathrm{2}}{\dot{v}}_{S\mathrm{2}}\end{array}\right]\end{array}$$i.e.,$$\begin{array}{}\text{(42)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{1}}}& \mathrm{1}& -\frac{\mathrm{1}}{R_{ccc}}\\ \frac{\mathrm{1}}{R_{\mathrm{2}}}& -\mathrm{1}& 0\\ -\frac{\mathrm{1}}{R_{ccc}}& 0& \frac{\mathrm{1}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{3}}}\end{array}\right]\left[\begin{array}{c}{v}_{C\mathrm{1}}\\ i_{L\mathrm{1}}\\ v_{\mathrm{3}}\end{array}\right]=\left[\begin{array}{c}\frac{v_{S\mathrm{1}}}{R_{\mathrm{1}}}-i_{C\mathrm{1}}+\frac{v_{S\mathrm{2}}}{R_{ccc}}\\ +i_S+\frac{L_{\mathrm{1}}+L_{\mathrm{2}}}{L_{\mathrm{1}}{R}_{\mathrm{2}}}{v}_{L\mathrm{1}}+\frac{L_{\mathrm{2}}}{R_{\mathrm{2}}}{\dot{i}}_S\\ \frac{-v_{S\mathrm{2}}}{R_{ccc}}-i_S-\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}{i}_{C\mathrm{1}}+C_{\mathrm{2}}{\dot{v}}_{S\mathrm{2}}\end{array}\right]\end{array}$$Adding third row to first row yields$$\begin{array}{}\text{(43)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R_{\mathrm{1}}}& \mathrm{1}& \frac{\mathrm{1}}{R_{\mathrm{3}}}\\ \frac{\mathrm{1}}{R_{\mathrm{2}}}& -\mathrm{1}& 0\\ -\frac{\mathrm{1}}{R_{ccc}}& 0& \frac{\mathrm{1}}{R_{ccc}}+\frac{\mathrm{1}}{R_{\mathrm{3}}}\end{array}\right]\left[\begin{array}{c}{v}_{C\mathrm{1}}\\ i_{L\mathrm{1}}\\ v_{\mathrm{3}}\end{array}\right]=\left[\begin{array}{c}\frac{v_{S\mathrm{1}}}{R_{\mathrm{1}}}-i_{C\mathrm{1}}-i_S-\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}{i}_{C\mathrm{1}}+C_{\mathrm{2}}{\dot{v}}_{S\mathrm{2}}\\ +i_S+\frac{L_{\mathrm{1}}+L_{\mathrm{2}}}{L_{\mathrm{1}}{R}_{\mathrm{2}}}{v}_{L\mathrm{1}}+\frac{L_{\mathrm{2}}}{R_{\mathrm{2}}}{\dot{i}}_S\\ \frac{-v_{S\mathrm{2}}}{R_{ccc}}-i_S-\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}{i}_{C\mathrm{1}}+C_{\mathrm{2}}{\dot{v}}_{S\mathrm{2}}\end{array}\right]\end{array}$$Multiplying third row with $R_{CCC}$ yields $$\begin{array}{}\text{(44)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R_{\mathrm{1}}}& 1& \frac{\mathrm{1}}{R_{\mathrm{3}}}\\ \frac{\mathrm{1}}{R_{\mathrm{2}}}& -\mathrm{1}& 0\\ -\mathrm{1}& 0& 1\end{array}\right]\left[\begin{array}{c}{v}_{C\mathrm{1}}\\ i_{L\mathrm{1}}\\ v_{\mathrm{3}}\end{array}\right]=\left[\begin{array}{c}\frac{v_{S\mathrm{1}}}{R_{\mathrm{1}}}-i_{C\mathrm{1}}-i_S-\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}{i}_{C\mathrm{1}}+C_{\mathrm{2}}{\dot{v}}_{S\mathrm{2}}\\ +i_S+\frac{L_{\mathrm{1}}+L_{\mathrm{2}}}{L_{\mathrm{1}}{R}_{\mathrm{2}}}{v}_{L\mathrm{1}}+\frac{L_{\mathrm{2}}}{R_{\mathrm{2}}}{\dot{i}}_S\\ -v_{S\mathrm{2}}\end{array}\right]\end{array}$$Multiplying third row with $-1/R_3$ to first row obtains $$\begin{array}{}\text{(45)}& \left[\begin{array}{ccc}\frac{\mathrm{1}}{R_{\mathrm{1}}}+\frac{\mathrm{1}}{R_{\mathrm{3}}}& 1& 0\\ \frac{\mathrm{1}}{R_{\mathrm{2}}}& -\mathrm{1}& 0\\ -\mathrm{1}& 0& 1\end{array}\right]\left[\begin{array}{c}{v}_{C\mathrm{1}}\\ i_{L\mathrm{1}}\\ v_{\mathrm{3}}\end{array}\right]=\left[\begin{array}{c}\frac{v_{S\mathrm{1}}}{R_{\mathrm{1}}}-\left(\mathrm{1}+\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}\right)i_{C\mathrm{1}}-i_S+C_{\mathrm{2}}{\dot{v}}_{S\mathrm{2}}+\frac{v_{S\mathrm{2}}}{R_{\mathrm{3}}}\\ +i_S+\frac{L_{\mathrm{1}}+L_{\mathrm{2}}}{L_{\mathrm{1}}{R}_{\mathrm{2}}}{v}_{L\mathrm{1}}+\frac{L_{\mathrm{2}}}{R_{\mathrm{2}}}{\dot{i}}_S\\ -v_{S\mathrm{2}}\end{array}\right]\end{array}$$From (rule v), we get$$\begin{array}{}\text{(46)}& \left[\begin{array}{cc}\frac{\mathrm{1}}{R_{\mathrm{1}}}+\frac{\mathrm{1}}{R_{\mathrm{3}}}& 1\\ \frac{\mathrm{1}}{R_{\mathrm{2}}}& -1\end{array}\right]\left[\begin{array}{c}{v}_{C\mathrm{1}}\\ i_{L\mathrm{1}}\end{array}\right]=\left[\begin{array}{c}\frac{v_{S\mathrm{1}}}{R_{\mathrm{1}}}-\left(\mathrm{1}+\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}\right)i_{C\mathrm{1}}-i_S+C_{\mathrm{2}}{\dot{v}}_{S\mathrm{2}}+\frac{v_{S\mathrm{2}}}{R_{\mathrm{3}}}\\ +i_S+\frac{L_{\mathrm{1}}+L_{\mathrm{2}}}{L_{\mathrm{1}}{R}_{\mathrm{2}}}{v}_{L\mathrm{1}}+\frac{L_{\mathrm{2}}}{R_{\mathrm{2}}}{\dot{i}}_S\end{array}\right]\end{array}$$

Step 4.

Reorganize the elements $i_{Ck},v_{Lk}$ to build a new matrix equation from the right-hand side of original listed matrix equation.$$\begin{array}{}\text{(47)}& \left[\begin{array}{cc}-\left(\mathrm{1}+\frac{C_{\mathrm{2}}}{C_{\mathrm{1}}}\right)& \mathrm{0}\\ \mathrm{0}& \frac{L_{\mathrm{1}}+L_{\mathrm{2}}}{L_{\mathrm{1}}{R}_{\mathrm{2}}}\end{array}\right]\left[\begin{array}{c}{i}_{C\mathrm{1}}\\ v_{L\mathrm{1}}\end{array}\right]=\left[\begin{array}{cc}\frac{\mathrm{1}}{R_{\mathrm{1}}}+\frac{\mathrm{1}}{R_{\mathrm{3}}}& \mathrm{1}\\ \frac{\mathrm{1}}{R_{\mathrm{2}}}& -\mathrm{1}\end{array}\right]\left[\begin{array}{c}{v}_{C\mathrm{1}}\\ i_{L\mathrm{1}}\end{array}\right]+\left[\begin{array}{ccc}-\frac{\mathrm{1}}{R_{\mathrm{1}}}& -\frac{\mathrm{1}}{R_{\mathrm{3}}}& \mathrm{1}\\ \mathrm{0}& \mathrm{0}& -\mathrm{1}\end{array}\right]\left[\begin{array}{c}{v}_{S\mathrm{1}}\\ v_{S\mathrm{2}}\\ i_S\end{array}\right]+\left[\begin{array}{ccc}\mathrm{0}& -C_{\mathrm{2}}& \mathrm{0}\\ \mathrm{0}& \mathrm{0}& -\frac{L_{\mathrm{2}}}{R_{\mathrm{2}}}\end{array}\right]\left[\begin{array}{c}{\dot{v}}_{S\mathrm{1}}\\ {\dot{v}}_{S\mathrm{2}}\\ {\dot{i}}_S\end{array}\right]\end{array}$$Use the Cramer’s rule to obtain the elements $i_{C_k}$ ($v_{L_k}$) and then get the desired state equation. $$\begin{array}{}\text{(48)}& \left[\begin{array}{c}{C}_{\mathrm{1}}\frac{d{v}_{C\mathrm{1}}}{dt}\left(i.e\text{.}{i}_{C\mathrm{1}}\right)\\ L_{\mathrm{1}}\frac{d{i}_{L\mathrm{1}}}{dt}\left(i.e\text{.}{v}_{L\mathrm{1}}\right)\end{array}\right]=\left[\begin{array}{cc}-\frac{C_{\mathrm{1}}}{C_{\mathrm{1}}+C_{\mathrm{2}}}\left(\frac{\mathrm{1}}{R_{\mathrm{1}}}+\frac{\mathrm{1}}{R_{\mathrm{3}}}\right)& -\frac{C_{\mathrm{1}}}{C_{\mathrm{1}}+C_{\mathrm{2}}}\\ \frac{L_{\mathrm{1}}}{L_{\mathrm{1}}+L_{\mathrm{2}}}& -\frac{L_{\mathrm{1}}{R}_{\mathrm{2}}}{L_{\mathrm{1}}+L_{\mathrm{2}}}\end{array}\right]\left[\begin{array}{c}{v}_{C\mathrm{1}}\\ i_{L\mathrm{1}}\end{array}\right]+\left[\begin{array}{ccc}\frac{C_{\mathrm{1}}}{C_{\mathrm{1}}+C_{\mathrm{2}}}\frac{\mathrm{1}}{R_{\mathrm{1}}}& \frac{C_{\mathrm{1}}}{C_{\mathrm{1}}+C_{\mathrm{2}}}\frac{\mathrm{1}}{R_{\mathrm{3}}}& -\frac{C_{\mathrm{1}}}{C_{\mathrm{1}}+C_{\mathrm{2}}}\\ 0& 0& -\frac{L_{\mathrm{1}}{R}_{\mathrm{2}}}{L_{\mathrm{1}}+L_{\mathrm{2}}}\end{array}\right]\left[\begin{array}{c}{v}_{S\mathrm{1}}\\ v_{S\mathrm{2}}\\ i_S\end{array}\right]+\left[\begin{array}{ccc}\mathrm{0}& \frac{C_{\mathrm{1}}{C}_{\mathrm{2}}}{C_{\mathrm{1}}+C_{\mathrm{2}}}& \mathrm{0}\\ \mathrm{0}& \mathrm{0}& -\frac{L_{\mathrm{1}}{L}_{\mathrm{2}}}{L_{\mathrm{1}}+L_{\mathrm{2}}}\end{array}\right]\left[\begin{array}{c}{\dot{v}}_{S\mathrm{1}}\\ {\dot{v}}_{S\mathrm{2}}\\ {\dot{i}}_S\end{array}\right]\end{array}$$

Example 4.

It has been stated in [3, page131] that the problem will yield increased difficulties in using network analysis to obtain the state equation for the electrical circuit with a dependent element. Consider the electrical circuit with dependent element demonstrated in Figure 9 [3, page130].

[figure omitted; refer to PDF]

Step 1.

Applying the substitution theorem and source-transformation theorem yields Figures 10 and 11.

[figure omitted; refer to PDF] [figure omitted; refer to PDF]

Step 2.

Draw up the matrix node-voltage equations and solve the dependent source by simple moving-term processes.$$\begin{array}{}\text{(49)}& \left[\begin{array}{cc}\frac{\mathrm{1}}{R_{\mathrm{1}}}+\frac{\mathrm{1}}{R_{CCC}}& \mathrm{0}\\ \mathrm{0}& \frac{\mathrm{1}}{R_{\mathrm{2}}}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{1}}\\ v_{\mathrm{2}}\end{array}\right]=\left[\begin{array}{c}{i}_S-i_C+\frac{v_L}{R_{CCC}}\\ i_C+g_m{v}_L\end{array}\right]\end{array}$$

Step 3.

Using our invented easy matrix operations changes the node-voltage elements to the state elements $v_{Ck},i_{Lk}$ based on the following equations:$$\begin{array}{}\text{(50)}& v_c=-v_{\mathrm{2}}+v_{\mathrm{1}}\text{(51)}& i_L=\frac{v_1-v_L}{R_{CCC}}\end{array}$$From (rule i), we get$$\begin{array}{}\text{(52)}& \left[\begin{array}{cc}\frac{\mathrm{1}}{R_{\mathrm{1}}}+\frac{\mathrm{1}}{R_{CCC}}& \mathrm{0}\\ \mathrm{0}& -\frac{\mathrm{1}}{R_{\mathrm{2}}}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{1}}\\ -v_{\mathrm{2}}\end{array}\right]=\left[\begin{array}{c}{i}_S-i_C+\frac{v_L}{R_{CCC}}\\ i_C+g_m{v}_L\end{array}\right]\end{array}$$Applying (rule iii) to (52) yields$$\begin{array}{}\text{(53)}& \left[\begin{array}{cc}\frac{\mathrm{1}}{R_{\mathrm{1}}}+\frac{\mathrm{1}}{R_{CCC}}& \mathrm{0}\\ \frac{\mathrm{1}}{R_{\mathrm{2}}}& -\frac{\mathrm{1}}{R_{\mathrm{2}}}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{1}}\\ -v_{\mathrm{2}}+v_{\mathrm{1}}\end{array}\right]=\left[\begin{array}{c}{i}_S-i_C+\frac{v_L}{R_{CCC}}\\ i_C+g_m{v}_L\end{array}\right]\end{array}$$i.e.,$$\begin{array}{}\text{(54)}& \left[\begin{array}{cc}\frac{\mathrm{1}}{R_{\mathrm{1}}}+\frac{\mathrm{1}}{R_{CCC}}& \mathrm{0}\\ \frac{\mathrm{1}}{R_{\mathrm{2}}}& -\frac{\mathrm{1}}{R_{\mathrm{2}}}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{1}}\\ v_C\end{array}\right]=\left[\begin{array}{c}{i}_S-i_C+\frac{v_L}{R_{CCC}}\\ i_C+g_m{v}_L\end{array}\right]\end{array}$$From (rule ii), (54) is changed to be$$\begin{array}{}\text{(55)}& \left[\begin{array}{cc}\frac{\mathrm{1}}{R_{\mathrm{1}}}+\frac{\mathrm{1}}{R_{CCC}}& \mathrm{0}\\ \frac{\mathrm{1}}{R_{\mathrm{2}}}& -\frac{\mathrm{1}}{R_{\mathrm{2}}}\end{array}\right]\left[\begin{array}{c}{v}_{\mathrm{1}}-v_L\\ v_C\end{array}\right]=\left[\begin{array}{c}{i}_S-i_C-\frac{v_L}{R_{\mathrm{1}}}\\ i_C+v_L\left(g_m-\frac{\mathrm{1}}{R_{\mathrm{2}}}\right)\end{array}\right]\end{array}$$From (rule i), we get$$\begin{array}{}\text{(56)}& \left[\begin{array}{cc}\left(\frac{\mathrm{1}}{R_{\mathrm{1}}}+\frac{\mathrm{1}}{R_{CCC}}\right)\times R_{CCC}& \mathrm{0}\\ \left(\frac{\mathrm{1}}{R_{\mathrm{2}}}\right)\times R_{CCC}& -\frac{\mathrm{1}}{R_{\mathrm{2}}}\end{array}\right]\left[\begin{array}{c}\frac{v_{\mathrm{1}}-v_L}{R_{CCC}}\left(i.e\text{.}{i}_L\right)\\ v_C\end{array}\right]=\left[\begin{array}{c}{i}_S-i_C-\frac{v_L}{R_{\mathrm{1}}}\\ i_C+v_L\left(g_m-\frac{\mathrm{1}}{R_{\mathrm{2}}}\right)\end{array}\right]\end{array}$$i.e.,$$\begin{array}{}\text{(57)}& \left[\begin{array}{cc}\mathrm{0}& \mathrm{1}\\ -\frac{\mathrm{1}}{R_{\mathrm{2}}}& \mathrm{0}\end{array}\right]\left[\begin{array}{c}{v}_C\\ i_L\end{array}\right]=\left[\begin{array}{c}{i}_S-i_C-\frac{v_L}{R_{\mathrm{1}}}\\ i_C+v_L\left(g_m-\frac{\mathrm{1}}{R_{\mathrm{2}}}\right)\end{array}\right]\end{array}$$

Step 4.

Reorganize the elements $i_{Ck},v_{Lk}$ to build a new matrix equation from the right-hand side of original listed matrix equation.$$\begin{array}{}\text{(58)}& \left[\begin{array}{cc}-\mathrm{1}& -\frac{\mathrm{1}}{R_{\mathrm{1}}}\\ \mathrm{1}& g_m-\frac{\mathrm{1}}{R_{\mathrm{2}}}\end{array}\right]\left[\begin{array}{c}{i}_C\\ v_L\end{array}\right]=\left[\begin{array}{cc}\mathrm{0}& \mathrm{1}\\ -\frac{\mathrm{1}}{R_{\mathrm{2}}}& \mathrm{0}\end{array}\right]\left[\begin{array}{c}{v}_C\\ i_L\end{array}\right]+\left[\begin{array}{c}-\mathrm{1}\\ \mathrm{0}\end{array}\right]i_S\end{array}$$Use the Cramer’s rule to obtain the elements $i_{C_k}$ ($v_{L_k}$) and then get the desired state equation.$$\begin{array}{}\text{(59)}& \left[\begin{array}{c}C\frac{d{v}_{C\mathrm{1}}}{dt}\left(i.e\text{.}{i}_C\right)\\ L\frac{d{i}_{L\mathrm{1}}}{dt}\left(i.e.v_L\right)\end{array}\right]=\left[\begin{array}{cc}-\frac{\mathrm{1}}{R_{\mathrm{1}}{R}_{\mathrm{2}}\Delta }& \frac{\mathrm{1}}{\Delta }\left(g_m-\frac{\mathrm{1}}{R_{\mathrm{2}}}\right)\\ \frac{\mathrm{1}}{R_{\mathrm{2}}\Delta }& -\frac{\mathrm{1}}{\Delta }\end{array}\right]\left[\begin{array}{c}{v}_C\\ i_L\end{array}\right]+\left[\begin{array}{c}\frac{\mathrm{1}}{\Delta }\left(-g_m+\frac{\mathrm{1}}{R_{\mathrm{2}}}\right)\\ \frac{\mathrm{1}}{\Delta }\end{array}\right]i_S\end{array}$$where $\Delta \equiv -g_m+\mathrm{1}/R_{\mathrm{1}}+\mathrm{1}/R_{\mathrm{2}}$.

4. Conclusion

Modern circuit analysis approaches model the electrical circuit by integral-differential equation with several nonsystematic limitations that are overcome by the state-space formulation in the modern electrical field. The main contribution of this paper is to present a new method of finding the state equation for degenerate circuit and coupling circuit that have not been systematically solved now. Moreover, this paper also proposes some strong improvements to solve the dependent-source circuit via four demonstrated examples. We will extend the proposed result to construct the low-pass, band-pass and high-pass filter in state-space environment in near future.

Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

This work is supported by the Institute-Level Major Scientific Research and Training Projects (Lead Doctor special projects) for City College of Dongguan University of Technology, Grant no. 2017YZDYB01Z, Grant Title: Create a new theory and technology of electric circuit and electronics science based on Chen’s Electrical Unifying Approach. This work was also supported by City College of Dongguan University of Technology, Grant nos. 201713844031, 2019QJY006Z, and 201813844009.

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