# Bounds of impedance matrix of a passive two port network

If we consider the impedance matrix of a linear, reciprocal passive network (Z12 =Z21 because of the reciprocity), $$\begin{bmatrix} Z_{11} & Z_{12} \\ Z_{12} & Z_{22} \end{bmatrix} = \begin{bmatrix} r_{11} + i x_{11} & r_{12} + i x_{12}\\ r_{12} + i x_{12} & r_{22} + i x_{22} \end{bmatrix}$$ will the following condition be always true? $$r_{12}^2 \leq (r_{11}r_{22})$$ If true, what is the rationale behind it? If not, what would an example circuit?

• First an error must have crept into the relation you wrote, can you possibly compare ohms versus squared ohms? Once fixed it is however true for any passive bouble bipole . It just comes out from power dissipated inside to be greater or equal to zero for any port currents. Try to workout this power as function of I1 and I2 first.... – carloc Aug 16 '18 at 7:51

A necessary condition for an impedance matrix to be passive is that its Hermitian part must be nonnegative-definite along the imaginary axis (of Laplace transform domain). See this paper for detail. The Hermitian part is defined as $$\Z_H(s) = (Z(s) + Z^H(s))/2\$$. For reciprocal networks, this corresponds to the real part of the impedance matrix. Next, for a 2-by-2 real symmetric matrix to be positive semidefinite, a necessary condition is that its determinant be greater than zero. This then gives the condition you asked. A simple way to appreciate the presence of the Hermitian part has been posted here.