On a Method for Studying Transient Regimes in a Four-Terminal Network with Non-Simultaneous Switching of Terminals
Abstract Generated abstract
This paper develops a matrix operational method for calculating transient regimes in linear multi-terminal electrical networks when terminals are switched at different times and in arbitrary order. The approach represents the network state through resistance, inductance, and capacitance matrices, incorporates initial capacitor voltages and inductor currents produced by earlier switchings, and derives complete and reduced impedance and admittance matrices for a four-terminal network, including cases with singular impedance matrices or a disconnected common terminal. The method is then specialized to through four-terminal networks and illustrated by a simple circuit in which three switches close successively, yielding current transforms for each time interval. The paper argues that the procedure is comparatively simple and useful for complex circuits such as communication equipment and phase-by-phase switching in three-phase systems.
Full Text
On the Methodology for Studying Transient Conditions in a Four-Terminal Network with Non-Simultaneous Switching of the Terminals
A. N. ZHILIN
Introduction
The methodology for studying transient conditions in electrical circuits with simultaneous switching of the terminals has been developed in considerable detail \[1;2\]. However, in modern electrical engineering, circuits are used in a number of cases whose terminals are switched on non-simultaneously and in any order of sequence. The time intervals between switchings may then be such that the transient processes are superimposed on one another. The transient processes that occur when terminals are switched on non-simultaneously have not been sufficiently investigated. The methods of Kron, Pen-Tung-Sah, and others are hardly applicable here, since they do not yield simplifications even when ordinary steady-state conditions are considered. The use of the delay theorem in a circuit of general form is also difficult, while the Thevenin–Helmholtz method leads to cumbersome calculations that are practically inadvisable.
The absence of material on this question prompted us to attempt to provide a method for calculating transient processes in a linear \(n\)-terminal network when its terminals are switched on non-simultaneously \[3\].
According to this method, the electrical state of a linear multi-terminal network can be written for any instant of time by the equation
\[ [R]\vec{i} + [L]\frac{d\vec{i}}{dt} + [K]\int \vec{i}\,dt \overset{m}{=} \vec{U}_{e}. \tag{1} \]
Here
\[
\vec{U}_{e}
\]
is an \(m\)-dimensional voltage vector whose components are the external voltages applied to terminals \(1,2,\ldots,m \le n\).
\[
\vec{i}
\]
is an \(m\)-dimensional current vector whose components are the currents flowing through terminals \(1,2,\ldots,m \le n\).
\([R], [L], [K] = [C]^{-1}\) are matrices that are the generalized parameters of the \(n\)-terminal network when \(m\) terminals are switched on.
The solution of matrix equation (1) in operational form has the form:
\[ \vec{I}(p) \overset{m}{=} [Y(p)]\left(\vec{U}_{e}(p) - \vec{U}_{i}(p)\right), \tag{2} \]
where
\[ [Y(p)] \overset{m}{=} [Z(p)]^{-1} \overset{m}{=} \left\{[R] + p[L] + \frac{1}{p}[K]\right\}^{-1}. \tag{3} \]
the admittance matrix of the circuit with \(m\) terminals switched on, and
\[ \overrightarrow{U_i}(p) = \overrightarrow{U_c}(0) - p[L]\overrightarrow{i(0)} \tag{4} \]
is the voltage appearing in equation (2) as a result of the nonsimultaneous switching of the terminals of the \(n\)-terminal network when \(m\) terminals are switched on.
With simultaneous switching of \(m\) terminals, in view of the assumed passivity of the \(n\)-terminal network, the voltage \(U_i(p)\) will be equal to zero.
The transition from the operational relation (2) to time dependences is carried out by the usual methods, for example, by means of integral residues or by means of the expansion theorem.
The resistance and admittance matrices \(\left[{}^{m}Z(p)\right]\), \(\left[{}^{m}Y(p)\right]\) for \(m\) switched-on terminals are found from the complete matrices \(\left[{}^{n}Z(p)\right]\), \(\left[{}^{n}Y(p)\right]\) for \(n\) switched-on terminals of the \(n\)-terminal network. If the resistance matrices are nonsingular, then the complete and reduced admittance matrices are found as their inverses:
\[ \left[{}^{n}Y(p)\right] = \left[{}^{n}Z(p)\right]^{-1} \quad \text{and} \quad \left[{}^{m}Y(p)\right] = \left[{}^{m}Z(p)\right]^{-1}. \tag{5} \]
The reduced resistance matrices \(\left[{}^{m}Z(p)\right]\) are determined from the complete resistance matrix \(\left[{}^{n}Z(p)\right]\) by simply deleting the outer columns and lower rows up to the \(m\)-th column and \(m\)-th row according to the scheme:
\[ \left[{}^{n}Z(p)\right] = \left[ \begin{array}{ccccc} Z_{11}, Z_{12}, \ldots, Z_{1m}, \ldots, Z_{1n} \\ Z_{21}, Z_{22}, \ldots, Z_{2m}, \ldots, Z_{2n} \\ \ldots\ldots\ldots\ldots\ldots\ldots \\ Z_{m1}, Z_{m2}, \ldots, Z_{mm}, \ldots, Z_{mn} \\ \ldots\ldots\ldots\ldots\ldots\ldots \\ Z_{n1}, Z_{n2}, \ldots, Z_{nm}, \ldots, Z_{nn} \end{array} \right] ; \quad \left[{}^{m}Z(p)\right] = \left[ \begin{array}{cccc} Z_{11}, Z_{12}, \ldots, Z_{1m} \\ Z_{21}, Z_{22}, \ldots, Z_{2m} \\ \ldots\ldots\ldots\ldots \\ Z_{m1}, Z_{m2}, \ldots, Z_{mm} \end{array} \right] \tag{6} \]
The elements of the complete resistance matrix may be calculated on the basis of the relation
\[ \left\{{}^{n}Z\right\}_{ik} = \left(\frac{U_i}{I_k}\right)_{{}^{n}I = I_k}. \tag{7} \]
In expression (7) the vector \({}^{n}I\) has the form
\[ I = (0,0,\ldots,I_k,\ldots,0), \quad k = 1,2,\ldots,n. \tag{8} \]
In those cases when the resistance matrix is singular, the values of the elements of the complete admittance matrix can be determined from the dependence
\[ \left\{{}^{n}Y\right\}_{ik} = \left(\frac{\dot I_i}{\dot U_k}\right)_{{}^{n}\dot U = \dot U_k}, \tag{9} \]
in which
\[ {}^{n}\dot U = (0,0,\ldots,\dot U_k,\ldots,0), \quad k = 1,2,\ldots,n. \tag{10} \]
Knowing the elements of the complete conductance matrix, the elements of the reduced matrices are found from the expression:
\[ \left\{Y^{m-1}\right\}_{ik} = \left\{Y^m\right\}_{ik} - \frac{ \left\{Y^m\right\}_{im}\left\{Y^m\right\}_{mk} }{ \left\{Y^m\right\}_{mm} }. \tag{11} \]
Equation (11) makes it possible to express the matrix \(\left\{Y^{m-1}\right\}\) through the matrix of the next higher order \(\left\{Y^m\right\}\).
In the course of the calculation, in addition to the conductance and resistance matrices, it is necessary to know the matrices \([R^m]\), \([L^m]\), and \([K^m]\). The element of the matrix \([Z^m]\) located at the intersection of the \(i\)-th row and the \(k\)-th column has the form:
\[ \left\{Z^m(p)\right\}_{ik}=r_{ik}+pL_{ik}+\frac{1}{pC_{ik}}, \tag{12} \]
therefore the elements of the matrices \([R^m]\), \([L^m]\), and \([K^m]\) are determined by the formulas:
\[ \left\{R^m\right\}_{ik}=r_{ik},\qquad \left\{L^m\right\}_{ik}=L_{ik} \quad\text{and}\quad \left\{K^m\right\}_{ik}=\frac{1}{C_{ik}}. \tag{13} \]
The advantages of the method set forth are especially apparent when studying transient processes in complex networks. In the present article this general method is applied to the calculation of transient processes in a four-terminal network when its terminals are switched on non-simultaneously.
Basic equations
Let us suppose that in a four-terminal network (Fig. 1) one of its terminals is permanently connected to sources of electrical energy. We shall further agree to neglect the influence of the internal parameters of the sources and not to take into account the action of the electric arc when connecting the terminals.
Fig. 1
The equations of the electrical state of the four-terminal network under these conditions, for a different number of connected terminals, may be written, according to expression (2), in the following form.
For the connection of the first terminal \(K_1\), with \(K_2\) and \(K_3\) disconnected:
\[ \vec I(p)=\left[Y^1(p)\right]\left\{ \vec U_\ell-\left\{\vec U_\ell(0)-p\left[L^1\right]\vec i(0)\right\} \right\}, \tag{14} \]
where, under zero initial conditions,
\[ \overset{1}{\vec I}(p)=(I_1^{1}(p)) \text{—current through terminal 1;} \]
\[ [\overset{1}{Y}(p)] \text{—conductance matrix of the circuit with terminal 1 switched on;} \]
\[ \overset{1}{\vec U_e}(p)=(U_1^{1}(p)) \text{—voltage with terminal 1 switched on;} \]
\[ \overset{1}{\vec U_c}(0)=(U_{c1}^{1}(0))=0 \text{—initial voltage on the capacitance;} \]
\[ p[\overset{1}{L}]\,\overset{1}{\vec i}(0)=p[L_{11}^{1}](i_1^{1}(0))=0 \text{—initial voltage on the inductance.} \]
For switching on the second terminal \(K_2\), with \(K_1\) already switched on and \(K_3\) switched off:
\[ \overset{2}{\vec I}(p)=[\overset{2}{Y}(p)] \left\{\overset{2}{\vec U_e}(p)-\left(\overset{2}{\vec U_c}(0)-p[\overset{2}{L}]\,\overset{2}{\vec i}(0)\right)\right\}, \tag{15} \]
where
\[ \overset{2}{\vec I}(p)=(i_1^{2}(p),i_2^{2}(p)) \text{—currents through terminals 1,2;} \]
\[ [\overset{2}{Y}(p)] \text{—conductance matrix of the circuit with terminals 1,2 switched on;} \]
\[ \overset{2}{\vec U_e}(p)=(U_1^{2}(p),U_2^{2}(p)) \text{—voltage with terminals 1,2 switched on;} \]
\[ \overset{2}{\vec U_c}(0)=(U_{c1}^{2}(0),U_{c2}^{2}(0)) \text{—initial voltage on the capacitance;} \]
\[ p[\overset{2}{L}]\,\overset{2}{\vec i}(0) = p \begin{bmatrix} L_{11},L_{12}\\ L_{21},L_{22} \end{bmatrix} (i_1^{2}(0),i_2^{2}(0)) \text{—initial voltage on the inductance.} \]
For switching on the third terminal \(K_3\), with \(K_1\) and \(K_2\) already switched on:
\[ \overset{3}{\vec I}(p)=[\overset{3}{Y}(p)] \left\{\overset{3}{\vec U_e}(p)-\left(\overset{3}{\vec U_c}(0)-p[\overset{3}{L}]\,\overset{3}{\vec i}(0)\right)\right\}, \tag{16} \]
where
\[ \overset{3}{\vec I}(p)=(i_1^{3}(p),i_2^{3}(p),i_3^{3}(p)) \text{—current through terminals 1,2,3;} \]
\[ [\overset{3}{Y}(p)] \text{—conductance matrix of the circuit with terminals 1,2,3 switched on;} \]
\[ \overset{3}{\vec U_e}(p)=(U_1^{3}(p),U_2^{3}(p),U_3^{3}(p)) \text{—voltage with terminals 1,2,3 switched on;} \]
\[ \overset{3}{\vec U_c}(0)=(U_{c1}^{3}(0),U_{c2}^{3}(0),U_{c3}^{3}(0)) \text{—initial voltage on the capacitance;} \]
\[ p[\overset{3}{L}]\,\overset{3}{\vec i}(0) = p \begin{bmatrix} L_{11},L_{12},L_{13}\\ L_{21},L_{22},L_{23}\\ L_{31},L_{32},L_{33} \end{bmatrix} (i_1^{3}(0),i_2^{3}(0),i_3^{3}(0)) \text{—initial voltage on the inductance.} \]
Resistance and Conductance Matrices of a Four-Terminal Network
In order to use the calculation equations (14), (15), and (16), it is necessary to be able to find the complete resistance and conductance matrices \(\overset{3}{[Z(p)]}\), \(\overset{3}{[Y(p)]}\), and, from them, the reduced matrices, for example, \(\overset{2}{[Z(p)]}\), \(\overset{2}{[Y(p)]}\). The complete and reduced resistance matrices are determined in accordance with the circuit (Fig. 2), according to rule (6).
Fig. 2
The resistance matrices of the four-terminal network may be written as follows:
\[ \overset{3}{[Z(p)]}= \begin{bmatrix} Z_{11}, Z_{12}, Z_{13}\\ Z_{21}, Z_{22}, Z_{23}\\ Z_{31}, Z_{32}, Z_{33} \end{bmatrix}, \tag{17} \]
\[ \overset{2}{[Z(p)]}= \begin{bmatrix} Z_{11}, Z_{12}\\ Z_{21}, Z_{22} \end{bmatrix} \quad \text{and} \quad \overset{1}{[Z(p)]}=[Z_{11}], \tag{18} \]
where
\[ \overset{3}{[Z(p)]} \]
is the complete resistance matrix, and
\[ \overset{2}{[Z(p)]},\ \overset{1}{[Z(p)]} \]
are reduced resistance matrices.
The complete resistance matrix may be found from Table 1 or from formula (7).
Table 1
Determination of the complete resistance matrix \([Z]\)
| \(N/n\) | [[circuit diagram for determining column 1]] | [[circuit diagram for determining column 2]] | [[circuit diagram for determining column 3]] |
|---|---|---|---|
| 1 | \(Z_{11}=\dfrac{\dot{U}_1}{\dot{I}_1}\) | \(Z_{12}=\dfrac{\dot{U}_1}{\dot{I}_2}\) | \(Z_{13}=\dfrac{\dot{U}_1}{\dot{I}_3}\) |
| 2 | \(Z_{21}=\dfrac{\dot{U}_2}{\dot{I}_1}\) | \(Z_{22}=\dfrac{\dot{U}_2}{\dot{I}_2}\) | \(Z_{23}=\dfrac{\dot{U}_2}{\dot{I}_3}\) |
| 3 | \(Z_{31}=\dfrac{\dot{U}_3}{\dot{I}_1}\) | \(Z_{32}=\dfrac{\dot{U}_3}{\dot{I}_2}\) | \(Z_{33}=\dfrac{\dot{U}_3}{\dot{I}_3}\) |
The complete admittance matrix is determined as the inverse impedance matrix by formula (5)
\[ \overset{3}{[Y(p)]} = [\overset{3}{Z}(p)]^{-1} = \left[ \begin{array}{ccc} Z_{11}, & Z_{12}, & Z_{13} \\ Z_{21}, & Z_{22}, & Z_{23} \\ Z_{31}, & Z_{32}, & Z_{33} \end{array} \right]^{-1}, \tag{19} \]
where
\[ \overset{3}{[Y(p)]} \]
is the complete admittance matrix, and
\[ \overset{2}{[Y(p)]} = [\overset{2}{Z}(p)]^{-1} = \left[ \begin{array}{cc} Z_{11}, & Z_{12} \\ Z_{21}, & Z_{22} \end{array} \right]^{-1}; \qquad \overset{1}{[Y(p)]} = [\overset{1}{Z}(p)]^{-1} = \left[ \frac{1}{Z_{11}} \right] \tag{20} \]
are the reduced admittance matrices.
If the complete impedance matrix is singular, then the complete admittance matrix can be determined from Table 2 or by formula (9).
Table 2
Determination of the complete admittance matrix \([Y]\)
| \(N/n\) | Excitation at terminal 1 | Excitation at terminal 2 | Excitation at terminal 3 |
|---|---|---|---|
| Diagram: voltage \(\dot u_1\) applied; currents \(\dot I_1,\dot I_2,\dot I_3\) enter the network \(Y\). | Diagram: voltage \(\dot u_2\) applied; currents \(\dot I_1,\dot I_2,\dot I_3\) enter the network \(Y\). | Diagram: voltage \(\dot u_3\) applied; currents \(\dot I_1,\dot I_2,\dot I_3\) enter the network \(Y\). | |
| 1 | \(Y_{11}=\dfrac{\dot I_1}{\dot u_1}\) | \(Y_{12}=\dfrac{\dot I_1}{\dot u_2}\) | \(Y_{13}=\dfrac{\dot I_1}{\dot u_3}\) |
| 2 | \(Y_{21}=\dfrac{\dot I_2}{\dot u_1}\) | \(Y_{22}=\dfrac{\dot I_2}{\dot u_2}\) | \(Y_{23}=\dfrac{\dot I_2}{\dot u_3}\) |
| 3 | \(Y_{31}=\dfrac{\dot I_3}{\dot u_1}\) | \(Y_{32}=\dfrac{\dot I_3}{\dot u_2}\) | \(Y_{33}=\dfrac{\dot I_3}{\dot u_3}\) |
The reduced admittance matrices can be obtained through the elements of the complete matrix by formula (11):
\[ \overset{2}{[Y(p)]} = \left[ \begin{array}{cc} Y_{11}-\dfrac{Y_{13}Y_{31}}{Y_{33}}, & Y_{12}-\dfrac{Y_{13}Y_{32}}{Y_{33}} \\[1.2em] Y_{21}-\dfrac{Y_{23}Y_{31}}{Y_{33}}, & Y_{22}-\dfrac{Y_{23}Y_{32}}{Y_{33}} \end{array} \right] \tag{21} \]
\[ \overset{1}{[Y(p)]} = \left[ \overset{2}{Y}_{11} - \dfrac{\overset{2}{Y}_{12}\overset{2}{Y}_{21}}{\overset{2}{Y}_{22}} \right] = \]
\[ =\left[ Y_{11}-\frac{Y_{13}Y_{31}}{Y_{33}}- \frac{\left(Y_{12}-\dfrac{Y_{13}Y_{32}}{Y_{33}}\right)\left(Y_{21}-\dfrac{Y_{23}Y_{31}}{Y_{33}}\right)} {Y_{22}-\dfrac{Y_{23}Y_{32}}{Y_{33}}} \right]. \tag{22} \]
The complete conductance matrix, expressed in terms of the elements of the complete resistance matrix, is represented in the following form:
\[ {}^{3}[Y(p)]= \left[ \begin{array}{ccc} \dfrac{1}{Z_{11}}+\dfrac{1}{Z_{12}}+\dfrac{1}{Z_{13}}; & -\dfrac{1}{Z_{12}}; & -\dfrac{1}{Z_{13}} \\[1.2em] -\dfrac{1}{Z_{12}}; & \dfrac{1}{Z_{22}}+\dfrac{1}{Z_{12}}+\dfrac{1}{Z_{23}}; & -\dfrac{1}{Z_{23}} \\[1.2em] -\dfrac{1}{Z_{13}}; & -\dfrac{1}{Z_{23}}; & \dfrac{1}{Z_{33}}+\dfrac{1}{Z_{13}}+\dfrac{1}{Z_{23}} \end{array} \right] \tag{23} \]
The reduced conductance matrices in terms of the elements of the complete resistance matrix are written as:
\[ {}^{2}[Y(p)]= \left[ \begin{array}{cc} \dfrac{1}{Z_{11}}+\dfrac{1}{Z_{12}}+\dfrac{1}{Z_{13}} -\dfrac{\left(\dfrac{1}{Z_{13}}\right)^2} {\dfrac{1}{Z_{33}}+\dfrac{1}{Z_{13}}+\dfrac{1}{Z_{23}}}; & -\dfrac{1}{Z_{12}} -\dfrac{\dfrac{1}{Z_{13}\cdot Z_{23}}} {\dfrac{1}{Z_{33}}+\dfrac{1}{Z_{13}}+\dfrac{1}{Z_{23}}} \\[1.8em] -\dfrac{1}{Z_{12}} -\dfrac{\dfrac{1}{Z_{13}\cdot Z_{23}}} {\dfrac{1}{Z_{33}}+\dfrac{1}{Z_{13}}+\dfrac{1}{Z_{23}}}; & \dfrac{1}{Z_{22}}+\dfrac{1}{Z_{12}}+\dfrac{1}{Z_{23}} -\dfrac{\left(\dfrac{1}{Z_{23}}\right)^2} {\dfrac{1}{Z_{33}}+\dfrac{1}{Z_{13}}+\dfrac{1}{Z_{23}}} \end{array} \right] \tag{24} \]
\[ {}^{1}[Y(p)]= \left[ \dfrac{1}{Z_{11}}+\dfrac{1}{Z_{12}}+\dfrac{1}{Z_{13}} -\dfrac{\left(\dfrac{1}{Z_{13}}\right)^2} {\dfrac{1}{Z_{33}}+\dfrac{1}{Z_{13}}+\dfrac{1}{Z_{23}}} - \dfrac{ \left( -\dfrac{1}{Z_{12}} -\dfrac{\dfrac{1}{Z_{13}Z_{23}}} {\dfrac{1}{Z_{33}}+\dfrac{1}{Z_{13}}+\dfrac{1}{Z_{23}}} \right)^2 } { \dfrac{1}{Z_{22}}+\dfrac{1}{Z_{12}}+\dfrac{1}{Z_{23}} -\dfrac{\left(\dfrac{1}{Z_{23}}\right)^2} {\dfrac{1}{Z_{33}}+\dfrac{1}{Z_{13}}+\dfrac{1}{Z_{23}}} } \right] \tag{25} \]
In a number of cases it is necessary to calculate transients in a four-terminal network whose common terminal is disconnected from the external circuit. Using here, as well, the relations derived earlier, one can obtain the complete and reduced conductance matrices.
The complete conductance matrix, for example, for a circuit without a common terminal has the form:
\[ {}^{3}[Y(p)]= \left[ \begin{array}{ccc} Y_{11}, & Y_{12}, & Y_{13} \\ Y_{21}, & Y_{22}, & Y_{23} \\ Y_{31}, & Y_{32}, & Y_{33} \end{array} \right], \tag{26} \]
where
\(\overset{3}{[Y(p)]}\) is the complete conductance matrix of the four-terminal network whose terminal 4 is disconnected (Fig. 3).
Fig. 3
\[ Y_{11}=\frac{1}{Z_{12}}+\frac{1}{Z_{13}}+ \frac{Z_{22}+Z_{33}}{Z_{11}Z_{22}+Z_{22}Z_{33}+Z_{11}Z_{33}}, \tag{27} \]
\[ Y_{12}=-\left( \frac{1}{Z_{12}}+\frac{1}{Z_{11}}- \frac{(Z_{11}+Z_{33})Z_{22}} {(Z_{11}Z_{22}+Z_{22}Z_{33}+Z_{11}Z_{33})Z_{11}} \right), \tag{28} \]
\[ Y_{13}=-\left( \frac{1}{Z_{13}}+\frac{1}{Z_{11}}- \frac{(Z_{11}+Z_{22})Z_{33}} {(Z_{11}Z_{22}+Z_{22}Z_{33}+Z_{11}Z_{33})Z_{11}} \right), \tag{29} \]
\[ Y_{21}=-\left( \frac{1}{Z_{12}}+\frac{1}{Z_{22}}- \frac{(Z_{22}+Z_{33})Z_{11}} {(Z_{11}Z_{22}+Z_{22}Z_{33}+Z_{11}Z_{33})Z_{22}} \right), \tag{30} \]
\[ Y_{22}=\frac{1}{Z_{12}}+\frac{1}{Z_{23}}+ \frac{Z_{11}+Z_{33}}{Z_{11}Z_{22}+Z_{22}Z_{33}+Z_{11}Z_{33}}, \tag{31} \]
\[ Y_{23}=-\left( \frac{1}{Z_{23}}+\frac{1}{Z_{22}}- \frac{(Z_{11}+Z_{22})Z_{33}} {(Z_{11}Z_{22}+Z_{22}Z_{33}+Z_{11}Z_{33})Z_{22}} \right), \tag{32} \]
\[ Y_{31}=-\left( \frac{1}{Z_{13}}+\frac{1}{Z_{33}}- \frac{(Z_{22}+Z_{33})Z_{11}} {(Z_{11}Z_{22}+Z_{22}Z_{33}+Z_{11}Z_{33})Z_{33}} \right), \tag{33} \]
\[ Y_{32}=-\left( \frac{1}{Z_{23}}+\frac{1}{Z_{33}}- \frac{(Z_{11}+Z_{33})Z_{22}} {(Z_{11}Z_{22}+Z_{22}Z_{33}+Z_{11}Z_{33})Z_{33}} \right), \tag{34} \]
\[ Y_{33}=\frac{1}{Z_{13}}+\frac{1}{Z_{23}}+ \frac{Z_{11}+Z_{22}}{Z_{11}Z_{22}+Z_{22}Z_{33}+Z_{11}Z_{33}}. \tag{35} \]
If the parameters of the four-terminal network \(Z_{11}, Z_{22}, Z_{33}\) are coupled to one another only by electric and magnetic fields, then the general matrices of resistance and conductance given in the article cannot be used. In this case one must find the complete matrices from Tables 1 and 2, and the reduced matrices from the elements of the complete ones.
Calculation Procedure
The calculation, according to the equations given above, is carried out in the following order:
-
For the specified circuit, using formulas 17, 18 or Table 1, the complete and reduced resistance matrices are determined.
-
From the resistance matrices, the inductance and capacitance matrices are determined. The elements of these matrices are found from formulas 12, 13.
-
From the resistance matrices, if they are nonsingular, the conductance matrices are determined from formulas 19 and 20.
If the resistance matrices are singular, the complete conductance matrix may be determined from Table 2, and the reduced matrices from the elements of the complete matrix according to formulas 21 and 22.
Finally, the complete and reduced conductance matrices may be determined from the elements of the complete resistance matrix by formulas 23, 24, and 25.
If one of the terminals of the four-terminal network is not connected to the common point of the sources, then the complete conductance matrix is found from formula 26 from the known elements of the complete resistance matrix.
-
From equations 14, 15, and 16 (in accordance with the specified sequence of switching on the individual terminals), after substituting into them the complete and reduced matrices, the desired currents at various instants of time are determined.
-
The resulting operational expressions for the currents are transformed into time functions by means of integral residues or by the expansion theorem. The voltages are determined from the known current functions by integration and differentiation of the latter.
Often in radio-engineering circuits not the general, but the through four-terminal network is considered. A through four-terminal network is one¹ in which the currents at the output and input are pairwise equal.
Fig. 4
The problem of non-simultaneous switching of the terminals of a through four-terminal network is solved in the same way as for a non-through one, since the through four-terminal network is a special case of the non-through one.
Let us assume, under the same assumptions as were adopted earlier, that we have a through four-terminal network (Fig. 4), whose terminals are switched on non-simultaneously. In this case, as is known, the physical state of the circuit is described by two equations, i.e., the four-terminal network is reduced to a three-terminal network having a superconducting side or a common terminal (Fig. 5). The voltage \(U'\) in this case does not correspond to the voltage drop in the upper element of the through four-terminal network, but this circumstance does not cause any complications, since the calculated
¹ The term “through four-terminal network” was introduced by Yu. T. Velichko.
quantities in radio-engineering circuits, in the majority of cases, are the input and output voltages of the four-terminal network, while the lateral one is usually not calculated.
The circuit (Fig. 5) can then be reduced to our general computational circuit (Fig. 6).
Fig. 5
The complete and reduced resistance matrices are found for the circuit in this case by the methods indicated earlier:
\[ {}^{2}[Z'(p)] = \begin{bmatrix} Z_{11}, & Z_{12} \\ Z_{21}, & Z_{22} \end{bmatrix}; \qquad {}^{1}[Z'(p)] = [Z_{11}]. \tag{36} \]
The conductance matrices are determined either as the inverses of the resistance matrices, or from Tables 1 and 2, from the open-circuit and short-circuit experiments.
Fig. 6
The equations of the transient process under non-simultaneous switching-on of the terminals for the through four-terminal network will be written as follows.
For switching on switch \(K_1\) with switch \(K_2\) switched off, in the time interval \(0 \leq t \leq \tau_1\),
\[ {}^{1}\vec{I}(p) = \bigl({}^{1}I_1(p)\bigr) = {}^{1}[Y(p)] \left\{ {}^{1}U_1(p) - \left[ {}^{1}\vec{U}_c(0) - p\,{}^{1}[L]\,{}^{1}\vec{i}(0) \right] \right\}. \tag{37} \]
For switching on switch \(K_2\) onto the previously switched-on switch \(K_1\), in the time interval \(\tau_1 \leq t \leq \infty\),
\[ {}^{2}\vec{I}(p) = \bigl({}^{2}I_1(p),\,{}^{2}I_2(p)\bigr) = {}^{2}[Y(p)] \left\{ {}^{2}U_1(p),\,{}^{2}U_2(p) - \left[ {}^{2}\vec{U}_c(0) - p\,{}^{2}[L]\,{}^{2}\vec{i}(0) \right] \right\}. \tag{38} \]
Example of a calculation
Let us show, by means of a simple example, the procedure for calculating the transient process in a four-terminal network with non-simultaneous switching of the terminals.
We shall calculate the circuit shown in Fig. 7. Let the time intervals of non-simultaneous switching be specified as follows: switch \(K_1\) is closed at \(t=0\), \(K_2\) at \(\tau_1\), and \(K_3\) at \(\tau_2\).
Fig. 7
We determine the full and reduced resistance matrices
According to the given circuit, the matrices are written as:
\[ \overset{3}{[Z(p)]}= \begin{bmatrix} r+pL, & 0, & 0\\ 0, & r, & r\\ 0, & 0, & r \end{bmatrix}, \tag{39} \]
\[ \overset{2}{[Z(p)]}= \begin{bmatrix} r+pL, & 0\\ 0, & r \end{bmatrix}, \tag{40} \]
\[ \overset{1}{[Z(p)]}=[r+pL] \tag{41} \]
We determine the inductance and capacitance matrices
\[ \overset{3}{[L]}= \begin{bmatrix} L, & 0, & 0\\ 0, & 0, & 0\\ 0, & 0, & 0 \end{bmatrix}; \qquad \overset{2}{[L]}= \begin{bmatrix} L, & 0\\ 0, & 0 \end{bmatrix}; \qquad \overset{1}{[L]}=[L]. \tag{42} \]
\[ \overset{3}{[K]}= \begin{bmatrix} 0, & 0, & 0\\ 0, & 0, & 0\\ 0, & 0, & 0 \end{bmatrix}; \qquad \overset{2}{[K]}= \begin{bmatrix} 0, & 0\\ 0, & 0 \end{bmatrix}; \qquad \overset{1}{[K]}=[0]. \tag{43} \]
We determine the full and reduced conductance matrices
Since the resistance matrices are nonsingular, the conductance matrices are found as their inverses:
\[ \overset{3}{[Y(p)]} = \overset{3}{[Z(p)]}^{-1} = \begin{bmatrix} \dfrac{1}{r+pL}; & 0; & 0\\[6pt] 0; & \dfrac{1}{r}; & -\dfrac{1}{r}\\[6pt] 0; & 0; & \dfrac{1}{r} \end{bmatrix} \tag{44} \]
\[ \overset{2}{|Y(p)|}=\overset{2}{[Z(p)]^{-1}}= \begin{bmatrix} \dfrac{1}{r+pL} & 0\\[4pt] 0 & \dfrac{1}{r} \end{bmatrix}, \tag{45} \]
\[ \overset{1}{|Y(p)|}=\overset{1}{[Z(p)]^{-1}}= \left[\dfrac{1}{r+pL}\right]. \tag{46} \]
Determining the transforms of the currents
For switching on switch \(K_1\) in the time interval \(0 \leq t \leq \tau_1\)
\[ \overset{1}{\vec I}(p)=(\overset{1}{i_1}(p)) =\overset{1}{|Y(p)|}\{(E_1)-\overset{1}{\vec u_c}(0)-p\overset{1}{[L]}\overset{1}{\vec i}(0)\}, \quad \text{since } \overset{1}{\vec u_c}(0)=0 \text{ and } \overset{1}{\vec i}(0)=0 \]
\[ \overset{1}{I_1}(p)=\frac{E_1}{r+pL}. \tag{47} \]
For switching on switch \(K_2\) onto the previously switched-on switch \(K_1\) in the time interval \(\tau_1 \leq t \leq \tau_2\)
\[ \overset{2}{\vec I}(p)=(\overset{2}{I_1}(p),\overset{2}{I_2}(p)) =\overset{2}{|Y(p)|}\{(E_1,E_2)+p\overset{2}{[L]}(\overset{1}{i_1}(\tau_1),0)\}, \quad \text{since } \overset{2}{\vec u_c}(\tau_1)=0 \]
\[ \overset{2}{\vec I}(p)=(\overset{2}{I_1}(p),\overset{2}{I_2}(p)) = \begin{bmatrix} \dfrac{1}{r+pL} & 0\\[4pt] 0 & \dfrac{1}{r} \end{bmatrix} \left\{(E_1,E_2)+p \begin{bmatrix} L & 0\\ 0 & 0 \end{bmatrix} (\overset{1}{i_1}(\tau_1),0)\right\} \]
\[ \overset{2}{\vec I}(p)=(\overset{2}{I_1}(p),\overset{2}{I_2}(p)) = \left( \frac{E_1+pL\overset{1}{i_1}(\tau_1)}{r+pL}, \frac{E_2}{r} \right). \tag{48} \]
For switching on switch \(K_3\) onto the previously switched-on switches \(K_1\) and \(K_2\) in the time interval \(\tau_2 \leq t \leq \infty\)
\[ \overset{3}{\vec I}(p)=(\overset{3}{I_1}(p),\overset{3}{I_2}(p),\overset{3}{I_3}(p)) =\overset{3}{|Y(p)|} \left\{ (E_1,E_2,E_3)+p \begin{pmatrix} L,0,0\\ 0,0,0\\ 0,0,0 \end{pmatrix} (\overset{2}{i_1}(\tau_2),\overset{2}{i_2}(\tau_2),0) \right\} \]
\[ \overset{3}{\vec I}(p)=(\overset{3}{I_1}(p),\overset{3}{I_2}(p),\overset{3}{I_3}(p)) = \left( \frac{E_1+pL\overset{2}{i_1}(\tau_2)}{r+pL}; \frac{E_2-E_3}{r}; \frac{E_3}{r} \right). \tag{49} \]
Determining the originals of the currents
The originals of the currents are determined from the transforms 47, 48, 49 in the usual way, for example, by means of integral residues or by applying the expansion theorem.
The method set forth above is simple in application. Its advantages become especially apparent in the calculation of complex circuits.
Such circuits occur in communication devices, as well as in three-phase current systems with phase-by-phase switching.
The author takes the opportunity to note that a number of valuable comments on this work were received by him from his teacher, Professor R. A. Voronov.
REFERENCES
- Bulgakov B. V., Oscillations, vol. 1, 1949.
- Kalantarov I. L. and Neiman L. R. Theoretical Foundations of Electrical Engineering, part II, 1949.
- Pukhov G. E. and Zhizhin A. N. Calculation equations of a multipole network with nonsimultaneous switching of terminals. Scientific Notes of LPI, issue V, Lviv, 1949.