In the N-port theory, it says that the total admittance matrix for the parallel connection of two 2-ports is the sum of individual admittance matrices "if it is possible to connect them in parallel". \begin{equation} \hat{Y}_\text{tot} = \hat{Y}_1 + \hat{Y}_2 \end{equation}

Is there an example of two 2-ports with well-defined admittance matrices that do not yield a total admittance matrix according to the above sum rule when connected in parallel?


I don't have an exact answer to your question, but I think that the quoted phrase refers to the fact that not all pairs of two-ports can be connected in parallel. I quote somewhat extensively from Seshu and Balabanian's Linear Network Analysis (pages 321-322; bolds are mine):

Let us next turn our attention to the parallel combination of two-ports. Two two-ports are said to be connected in parallel if the corresponding terminals (1,1', 2,2') of the two are connected together as in Fig. 27 [note: can't include it, sorry]. This condition forces the equality of the terminal voltages of the two networks. If we can assume that the relationships among the voltages and currents of the individual networks Na and Nb remain unaltered when the two are connected in parallel, then we can write

Total admittance

Thus the short-circuit admittance matrix of the composite network is the sum of the short-circuit admittance matrices of the individual networks. We must now inquire into the conditions under which the two networks can be connected in this way without causing the voltage-current relationships at the terminals of each to be modified in any way. For example, we can see that if there is a straight-through connection between terminals 1' and 2' of network b but not in network a, then the branch between terminals 1' and 2' of network a will be shorted when the parallel connection is made. Equation (79) will not be valid in such a case.

In order for the voltage-current relationships of the individual networks to remain unaltered under interconnection, the following condition (due to O. Brune) is necessary and sufficient. When the two two-ports are interconnected at either of the two ends and the other ends are short-circuited as in Fig. 28 the voltage marked V must be zero.

Brune's test for parallel-connected two-ports

If this condition is not satisfied, the matrix addition will not give the correct answer for the parameters of the composite network, unless isolating ideal transformers are introduced at one of the two ends. It is a simple matter to prove this result by calculating the voltage between terminals 1' and 2' of both networks. The details will be left to you.

This is a rather old book. I've searched for more information online and Brune's test might not be as simple as the authors claim. See, for example:

  • A. M. Sommariva, "On Brune's Tests", in IEEE Transactions on Circuits and Systems II: Express Briefs, vol. 61, no. 4, pp. 249-253, April 2014.

  • E. W. Zelyakh, "Application of brune's tests to a class of network interconnections", in International Journal of Circuit Theory and Applications, Vol. 6, Issue 1, December 2006 (sorry, can't post more links).*

There's also a google book that shows how the isolation transformer can be used: Electrical Circuit Analysis, by A.V.Bakshi and U.A.Bakshi, page 12-52.


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