How many measurements are needed to determine a black box with four terminals?

Question

I am having a black box with four terminals and just resistors inside. I can model it like this:

^{simulate this circuit – Schematic created using CircuitLab}

If I have four different currents and four different potentials at the four different terminals, that seems not enough to determine the circuit inside, as the determinant of my matrix to solve that seems 0, when I create equations from Kirchhoff rules. Either I created them wrong, or I must make multiple measurements with, e.g., one or two terminals open. What is the minimum amount of measurements I need to do to determine the resistors in such a black box?

Answer 1

You have 5 unknowns. You will need at least 5 measurements to determine them.

If I have 4 different currents and 4 different potentials at the 4 different terminals, that seems not enough to determine the circuit inside, as the determinant of my matrix to solve that seems 0, when I create equations from Kirchhoff rules.

If you were thinking these 4 currents and 4 voltages were a total of 8 measurements, that's not correct.

In simple terms, 4 of these values depend on the sources you applied externally and not on the internal structure of the unknown network. If you did the measurements by, for example, applying ideal voltage sources to each terminal, then this should be obvious.

Even if you actually used a more complex source (for example a Thevenin source with an internal resistance), it will turn out that you added 4 degrees of freedom from the sources, and measured 4 responses; rather than obtained 8 independent measurements.

Another way to think of it is, you are trying to determine 5 resistances and what you measured was 4 resistances at 4 terminals, not 8 independent characteristics of the circuit.

Answer 2

You do need 5 measurements to get 5 variables out. Say:

• R1+R3 (left arm)
• R2+R4 (right arm)
• R1+R6+R2 (upper branch)
• R3+R6+R4 (lower branch)
• R1+R6+R4 or R2+R6+R3 (cross branch)

It is possible to do them all at once by using floating AC sources - a different frequency for each measurement. The sources have to act like open circuits outside of their tuned frequency. In practice that means gyrators or digitally controlled sources that synthesize this floating-like behavior.

Answer 3

For the standpoint of a network with n accessible nodes, measurement of steady voltages and currents on/in the nodes can be summarized as n-1 voltages (one node serving as reference) and n-1 currents (the last current is the negative of the sum of the others), thus 2n-2 quantities characterize the network state. Further, in a network made of resistors, scaling the n-1 voltages scales the n-1 currents linearly. Without proof: this means we need 2n-3 resistors to model an arbitrary resistors network.

For n = 4 in the question, a network of 5 resistors indeed is enough. However no all resistor networks with 4 external nodes can be represented by the question's H construction when we want non-negative resistances and force the order of nodes. For example this:

Proof: in the above circuit the resistance measured between nodes A and B (with C and D open) is R7 ∥ (R8+R10+R9) < 1Ω, when in the question it's R1+R6+R2. It follows we need R1 < 1Ω. Similarly, we need R3 < 1Ω. But the resistance measured between A and C in the above circuit is R8 ∥ (R7+R9+R10) > 50Ω, and is R1+R3 in the question's circuit, leading to the contradiction 50Ω < 2Ω.

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What is the minimum amount of measurements I need to do to determine the resistors in such a black box?

If we count as measurement a reading of a multimeter, 2n-3 = 5 measurements at least, since we have as many unknowns. And exactly so if we are cautious in how we measure.

It's convenient to put a multimeter in resistor mode with autoranging, and measure 2n-3 = 5 apparent resistances across node pairs with the other n-2 = 2 disconnected. For the question, labeling accessible nodes of the H model ABCD in reading order:

1. R_AC = R1+R3
2. R_BD = R2+R4
3. R_AB = R1+R6+R2
4. R_CD = R3+R6+R4
5. R_AD = R1+R6+R4

We then compute

• R1 = (R_AC-R_CD+R_AD)/2
• R2 = (R_BD+R_AB-R_AD)/2
• R3 = (R_AC+R_CD-R_AD)/2
• R4 = (R_BD-R_AB+R_AD)/2
• R6 = (-R_AC-R_BD+R_AB+R_CD)/2

That works except in some degenerate cases where measurements return infinite. E.g. when R1 is infinite, in which case measurements 1 3 5 return infinite, and measurements 2 4 do not allow to make a valid model (of R2, R3+R6, and R4). This can be worked around by dynamically deciding what measurements we make according to earlier measurements: e.g. if measurements 1 and 3 return infinite, we want to change measurement 5 to R_BC instead of R_AD. More generally, I conjecture that a reasonable heuristic to minimize tolerance in the end model for arbitrary n is to measure across node pairs not already explored, within this favor nodes the least explored, and within this favor nodes with the lower sum of the resistances measured previously for this node.

Among many, an alternative measurement method is to ground one node, inject known currents in the other nodes and measure the voltages, inject known voltages in these others nodes and measure all but one current. There are special cases with open nodes and shorts, and in practice that's less convenient.

As explained above, we might obtain negative resistor values in the model. For networks made of actual non-negative resistors, that can be avoided by permutation of the external nodes of the model. It's not necessary to redo the measurements themselves.