# Proof;Only Reciprocal Circuit are passive circuit

Consider a two-port reciprocal network : $$\ Y= \left[ {\begin{array}{cc} 0 & -1/2 \\ -1/2 & 0 \\ \end{array} } \right] \$$
then how can we proof that it is a passive network or circuit?

I have no idea how to prove it ....any hints or suggestions please...?

• Hi, welcome to eesx. To me your question is not clear, namely I do not understand what $M$ represents. Two-port networks are usually represented with Z, Y, H, G, ABCD, S, T... Parameters, but I don't know about M parameters. Feb 18 '18 at 9:32
• Convert it to a circuit. Feb 18 '18 at 10:43

Prove that the circuit only dissipates energy, never creates it. (=the total input power via the ports to the circuit must be greater or equal to zero), no matter what thevenin sources are connected to it.

• Since the network is represented with the impedance matrix, it's better to use Norton, current sources, rather than Thévenin's. Feb 18 '18 at 10:54
• @MassimoOrtolano the question has evolved. Maybe this comment needs something. Feb 18 '18 at 21:34
• You're right, when I commented the matrix was the impedance one. If we wait enough, the OP will end up with the scattering matrix... Feb 18 '18 at 21:36
• @MassimoOrtolano A couple of times I have included a copy of the question to the answer "As a protection against edits which can render the answers nonsense" Feb 18 '18 at 21:41

Remark: When I first wrote this answer, the matrix in the question was the impedance matrix, whereas in the edited question it's the admittance one. The passivity conditions reported below carry over to the admittance matrix ($\boldsymbol{Y}(s)+\boldsymbol{Y^*}(s)$ should be non-negative definite etc.).

First, a remark on the title, "Only Reciprocal Circuit are passive circuit": this implication is wrong, and, actually, there are non-reciprocal circuits that are passive too (e.g. a Hall device).

In general, it can be proved that an electrical network represented by an impedance matrix $\boldsymbol{Z}(s)$ is passive if and only if (for a review, see e.g. P. Triverio et al., "Stability, Causality, and Passivity in Electrical Interconnect Models"):

1. each element of $\boldsymbol{Z}(s)$ is defined and analytic in $\operatorname{Re} s > 0$ (this is indeed verified for your matrix);
2. $\boldsymbol{Z}(s)+\boldsymbol{Z^*}(s)$, where $\boldsymbol{Z^*}(s)$ is the Hermitian conjugate of $\boldsymbol{Z}(s)$, is a non-negative definite (or, synonymously, positive semidefinite) matrix for all $s$ such that $\operatorname{Re} s > 0$ (this is easy to check for your matrix, since it's a constant matrix);
3. $\boldsymbol{Z}(s^*)=\boldsymbol{Z^*}(s)$ (this is indeed verified for your matrix).

The second condition is related to the fact that in the complex exponential regime the average power entering the network is given by (I'm not giving a complete proof, just a few hints to get the idea: the complete proof is much more involved)

$$P = \operatorname{Re} \boldsymbol{V}^*\boldsymbol{I} = \frac{1}{2}(\boldsymbol{V}^*\boldsymbol{I}+\boldsymbol{I}^*\boldsymbol{V})$$

where $\boldsymbol{V}$ and $\boldsymbol{I}$ are respectively the vectors of the port voltages and of the port currents. Given that $\boldsymbol{V} = \boldsymbol{Z}\boldsymbol{I}$, the above equation yields

$$P = \frac{1}{2}\boldsymbol{I}^*(\boldsymbol{Z^*}+\boldsymbol{Z})\boldsymbol{I}.$$

Since for a passive device the average power entering the network should be positive for all admissible voltages and currents, the above equation implies that $\boldsymbol{Z}(s)+\boldsymbol{Z^*}(s)$ should be non-negative definite.

$$\boldsymbol{Z}+\boldsymbol{Z^*} = \pmatrix{0 & -1 \\ -1 & 0}$$

What are its eigenvalues? Is this matrix positive definite?

• First of all I apologize for the inconvenience ; secondly its Eigen Values will be $-1$ and $1$ ,so the matrix is non-positive definite, hence it is an active network ; thirdly i want to know whether a $2 \times 2$ Matrix is positive-definite if its Eigen values are $0$ and $1$ ? Feb 21 '18 at 6:05
• @user9198116 Exactly, it's an active network. If, instead, the eigenvalues (don't capitalize "eigen": it's not a name and it means "auto") were 0 and 1, the matrix would be non-negative definite (or, positive semidefinite) and, yes, it would be a passive network. Feb 21 '18 at 7:57
• lastly if the eigenvalues are $0$ and $0$ of a $2 \times 2$ matrix ,then whether it will be a passive network? Feb 21 '18 at 13:13