Electronic – Under what conditions is the star-mesh transform invertible

circuit analysis

We all know and love the Δ-Y (delta-wye) and Y-Δ (wye-delta) transforms for simplifying three-resistor networks:

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The Δ-Y and Y-Δ transforms have the nice property that a Δ can always be turned into a Y, and a Y can always be turned into a Δ, no matter the value of the resistances involved.

There's a generalised version of the Y-Δ transform called the star-mesh transform. This converts a "star" of \$ N \$ resistors into a "mesh" of \$ ^{N}C_{2} \$resistors.

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Wikipedia suggests that the star-to-mesh transform will always exist – but that the inverse transform, mesh-to-star, may not exist. To wit:

The transform replaces N resistors with \$^{N}C_{2}\$ resistors. For N > 3, the result is an increase in the number of resistors, so the transform has no general inverse without additional constraints.

What are the constraints that must be satisfied in order for the inverse to exist?

I am particularly interested in converting a 4-node mesh network into a 4-resistor star network.

Motivation for the question: I have an industrial power systems model (really just a very large network of constant-voltage sources and impedances) containing ~ 2,000 nodes. I am attempting to reduce it to just four nodes of interest.


There are some published papers on this topic.

  • Versfeld, L., "Remarks on star-mesh transformation of electrical networks," Electronics Letters , vol.6, no.19, pp.597,599, September 17 1970

    Two new aspects of the well known star-mesh transformation are studied: (a) the necessary and sufficient conditions for the transformation of a given general mesh network into an equivalent star network; (b) an extension to networks containing sources.

  • Bapeswara Rao, V.V.; Aatre, V.K., "Mesh-star transformation," Electronics Letters , vol.10, no.6, pp.73,74, March 21 1974

    An equivalent star network exists for a given mesh network if the latter satisfies the Wheatstone relaflonship. Using this fact, it is shown that all the offdiagonal cofactors of the datum-node admittance matrix of such a mesh network are equal. From this property, a simple relationship between the elements of the two networks is derived.

I don't have IEEE Xplore access so I can't read them.

Best Answer

For the mesh-star conversion the problem is that you have more equations than variables, so the number of bond \$N_b\$ is: \$N_b=N_e-N_v\$, where \$N_e\$ is the number of equations, equal also to the number of the resistance in the mesh, \$N_v\$ is the number of variables, equal to the number of the resistance in the star. In the 4-case, I've demonstrated that the bonds for the transformation are \$G_{AB}G_{CD}=G_{AC}G_{BD}=G_{AD}G_{BC}\$, in other words the products between the resistance without node in common must be the same.

Ps: The "demonstration" is: The formula for the star-mesh transformation is \$G_{XY}=\frac{G_XG_Y}{G_{TOT}}\$ with \$G_{TOT}=\sum_{i=1}^nG_i\$. So, assuming \$G_{XY}\ne0\$, we can divide two of those equation and obtain \$\frac{G_X}{G_Y}=\frac{G_{XZ}}{G_{YZ}}\$ for every Z different from X or Y. In the 4-case this means 6 equations, one is the following: \$\frac{G_A}{G_B}=\frac{G_{AC}}{G_{BC}}=\frac{G_{AD}}{G_{BD}}\Rightarrow G_{AC}G_{BD}=G_{AD}G_{BC}\$. We get the same result from: \$\frac{G_C}{G_D}=\frac{G_{AC}}{G_{AD}}=\frac{G_{BC}}{G_{BD}}\$. From the last 4 equations we obtain \$G_{AB}G_{CD}=G_{AD}G_{BC}\$ and \$G_{AB}G_{CD}=G_{AC}G_{BD}\$ and so we finally have the \$G_{AB}G_{CD}=G_{AC}G_{BD}=G_{AD}G_{BC}\$ condition. So this is a necessary condition. But if the ratio between any two conductances of the mesh is known ,we can express the \$G_{TOT}\$ depending on only one of those, like \$G_{TOT}=G_{A}+G_B+G_C+G_D=G_A(1+\beta+\gamma+\delta)\$, where \$\beta=\frac{G_B}{G_A}=\frac{G_{BC}}{G_{AC}}=\frac{G_{BD}}{G_{AD}}\$, and so on..\$\Rightarrow G_{AB}=\frac{G_AG_B}{G_{TOT}}=\frac{G_AG_B}{G_A(1+\beta+\gamma+\delta)}=\frac{G_B}{(1+\beta+\gamma+\delta)}\Rightarrow G_B=G_{AB}(1+\beta+\gamma+\delta)\$. With similar calculations we can find all the 4 conductances(resistances) of the star.

I suppose all of this means the condition is also a sufficient condition.

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