# 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.... Commented Aug 16, 2018 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.