Reciprocity of electrical networks is a special case of
Lorentz reciprocity, but it can also be proven more directly from network theorems. This proof shows reciprocity for a two-node network in terms of its
admittance matrix, and then shows reciprocity for a network with an arbitrary number of
nodes by an
induction argument. A linear network can be represented as a set of linear equations through
nodal analysis. For a network consisting of
n+1 nodes (one being a reference node) where, in general, an admittance is connected between each pair of nodes and where a current is injected in each node (provided by an ideal current source connected between the node and the reference node), these equations can be expressed in the form of an admittance matrix, : \begin{bmatrix} I_1\\ I_2\\ \vdots\\ I_n \end{bmatrix}= \begin{bmatrix} Y_{11} & Y_{12} & \cdots & Y_{1n} \\ Y_{21} & Y_{22} & \cdots & Y_{2n} \\ \vdots & \vdots & \ddots & \vdots\\ Y_{n1} & Y_{n2} & \cdots & Y_{nn} \end{bmatrix} \begin{bmatrix} V_1\\ V_2\\ \vdots\\ V_n \end{bmatrix} where :I_k is the current injected into node
k by a generator (which amounts to zero if no current source is connected to node
k) :V_k is the voltage at node
k with respect to the reference node (one could also say, it is the electric potential at node
k) :Y_{jk} (
j ≠
k) is the negative of the admittance directly connecting nodes
j and
k (if any) :Y_{kk} is the sum of the admittances connected to node
k (regardless of the other node the admittance is connected to). This representation corresponds to the one obtained by
nodal analysis. If we further require that network is made up of passive, bilateral elements, then :Y_{jk} = Y_{kj} since the admittance connected between nodes
j and
k is the same element as the admittance connected between nodes
k and
j. The matrix is therefore symmetrical. For the case where n = 2 the matrix reduces to, : \begin{bmatrix} I_1 \\ I_2 \end{bmatrix}= \begin{bmatrix} Y_{11} & Y_{12} \\ Y_{21} & Y_{22} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \end{bmatrix} . From which it can be seen that, : Y_{12} = \left . \frac {I_1}{V_2} \right |_{V_1=0} and : Y_{21} = \left . \frac {I_2}{V_1} \right |_{V_2=0} \ . But since Y_{12} = Y_{21} then, : \left . \frac {I_1}{V_2} \right |_{V_1=0} = \left . \frac {I_2}{V_1} \right |_{V_2=0} which is synonymous with the condition for reciprocity. In words, the ratio of the current at one port to the voltage at another is the same ratio if the ports being driven and measured are interchanged. Thus reciprocity is proven for the case of n = 2. For the case of a matrix of arbitrary size, the order of the matrix can be reduced through
node elimination. After eliminating the
sth node, the new admittance matrix will have the form, : \begin{bmatrix} (Y_{11}Y_{ss} - Y_{s1}Y_{1s}) & (Y_{12}Y_{ss} - Y_{s2}Y_{1s}) & (Y_{13}Y_{ss} - Y_{s3}Y_{1s}) & \cdots \\ (Y_{21}Y_{ss} - Y_{s1}Y_{2s}) & (Y_{22}Y_{ss} - Y_{s2}Y_{2s}) & (Y_{23}Y_{ss} - Y_{s3}Y_{2s}) & \cdots \\ (Y_{31}Y_{ss} - Y_{s1}Y_{3s}) & (Y_{32}Y_{ss} - Y_{s2}Y_{3s}) & (Y_{33}Y_{ss} - Y_{s3}Y_{3s}) & \cdots \\ \cdots & \cdots & \cdots & \cdots \end{bmatrix} It can be seen that this new matrix is also symmetrical. Nodes can continue to be eliminated in this way until only a 2×2 symmetrical matrix remains involving the two nodes of interest. Since this matrix is symmetrical it is proved that reciprocity applies to a matrix of arbitrary size when one node is driven by a voltage and current measured at another. A similar process using the impedance matrix from
mesh analysis demonstrates reciprocity where one node is driven by a current and voltage is measured at another. == References ==