Z and Y matrices from open/short experiments

At a fixed frequency, a linear source-free network can admit a matrix relationship between terminal phasors. The impedance representation treats currents as independent variables and expresses both port voltages as their linear combinations. Because each term must have voltage units, every Z coefficient has units of ohms.

This representation is only valid when the chosen currents define the response uniquely. Ideal constraints can make Z unavailable or singular; the existence of a physical network does not require every imaginable parameter chart to exist. Throughout, I₁ and I₂ follow the same entering-current convention used in RF-03.01.

Setting I₂=0 opens port 2 in the ideal terminal model, but it does not force V₂=0. The first matrix column follows V₁=Z₁₁I₁ and V₂=Z₂₁I₁; divide by nonzero I₁ to measure Z₁₁ and Z₂₁. This is the open-circuit driving-point impedance and the forward transfer impedance.

To obtain the other column, open port 1 by setting I₁=0 and excite port 2. Then V₁=Z₁₂I₂ and V₂=Z₂₂I₂. Do not confuse Z₁₂ with an output-to-input voltage gain: its denominator is a current. In practice an RF open fixture has fringing capacitance and requires a reference-plane model.

An admittance description instead imposes port voltages and gives terminal currents. Thus Y has units of siemens, and a zero-voltage boundary condition means an ideal short. With V₂=0 the first column reads I₁=Y₁₁V₁ and I₂=Y₂₁V₁.

With V₁=0, excite port 2 to obtain Y₁₂=I₁/V₂ and Y₂₂=I₂/V₂. A short circuit does not mean that its port current vanishes: the current through the short can be substantial. The signs of the off-diagonal terms depend on keeping both currents referenced into the two-port.

Write V=ZI. If det Z is nonzero, multiply by Z⁻¹ on the left to obtain I=Z⁻¹V, hence Y=Z⁻¹. For the two-by-two matrix the determinant is Z₁₁Z₂₂−Z₁₂Z₂₁ and the inverse swaps the diagonal terms and negates both off-diagonal terms.

If det Z=0, this inverse does not exist. It is incorrect to divide by zero or to infer that the physical device itself disappears. Some ideal devices admit one representation but not another; numerical inversion also becomes unreliable near a poorly conditioned matrix.

Model the actual π network with a shunt conductance G₁ at input node V₁, a shunt G₂ at output node V₂, and a bridge conductance G_b between those nodes. The lower conductors share a reference node. Each branch has positive conductance, so the network is linear, reciprocal and dissipative.

Kirchhoff gives I₁=G₁V₁+G_b(V₁−V₂) and I₂=G₂V₂+G_b(V₂−V₁). Collecting coefficients produces Y₁₁=G₁+G_b, Y₂₂=G₂+G_b and Y₁₂=Y₂₁=−G_b. The negative transfer terms are a consequence of entering currents, not evidence of active gain.

The π network is reciprocal because its transfer admittances are equal. It is not port-interchange symmetric unless G₁=G₂; reciprocity does not require identical input and output driving-point admittances. A passive device can also have loss, so reciprocity alone cannot establish energy conservation without dissipation.

For peak phasors, substitute the signed KCL currents into P_abs=(1/2)Re(V₁I₁*+V₂I₂*). Rearrangement yields the nonnegative sum (G₁|V₁|²+G₂|V₂|²+G_b|V₁−V₂|²)/2. For a general valid Z model, passivity at a real frequency implies its Hermitian part is positive semidefinite; broadband positive-realness additionally demands analytic conditions.

Attach an external shunt load Y_L to the output port. The current I₂ is defined entering the two-port, whereas load current Y_LV₂ leaves it. Kirchhoff at the connection therefore gives I₂=−Y_LV₂, not +Y_LV₂.

Inserting this condition in the second π-network equation yields (G₂+G_b+Y_L)V₂=G_bV₁. The loaded voltage ratio is G_b/(G₂+G_b+Y_L), and input admittance follows from I₁/V₁=G₁+G_b−G_b²/(G₂+G_b+Y_L). These identities make load/reference signs directly testable.

Take G₁=0.010 S, G₂=0.005 S and G_b=0.020 S. Then Y=[[0.030,−0.020],[−0.020,0.025]] S, det Y=0.00035 S², and its inverse Z has entries 71.4286, 57.1429, 57.1429 and 85.7143 Ω. Multiply YZ to confirm the identity matrix.

With V₁=1 V and V₂=0.4 V using real peak phasors, I₁=0.022 A and I₂=−0.010 A; the network absorbs P=0.009 W. Independent resistive-branch powers sum to the same result. Adding a 0.010 S output load gives V₂/V₁=4/7 and Z_in≈53.846 Ω. Short/open tests in a real RF fixture need de-embedding and calibrated reference planes.

1. Starting with V=ZI, derive Z₁₁ and Z₂₁ under the ideal open-port-2 condition.
2. Which physical boundary condition is imposed by I₁=0, and how is Z₁₂ extracted?
3. Derive Y₁₁ and Y₂₁ when port 2 is short-circuited. State the unit of Y₂₁.
4. Why does V₂=0 not imply I₂=0? Give a circuit example.
5. Compute the determinant and inverse of Z=[[4,1],[1,3]] Ω.
6. Derive the signed admittance matrix of the three-resistor π network with shunts G₁,G₂ and bridge G_b.
7. Is a π network with G₁≠G₂ reciprocal? Is it symmetric under exchanging the two ports?
8. For G₁=0.010 S, G₂=0.005 S, G_b=0.020 S, V₁=1 V, V₂=0.4 V, compute I₁,I₂ and absorbed mean power.
9. Verify the preceding power by summing the three resistive-branch dissipations.
10. With an output shunt load Y_L=0.010 S, derive I₂ sign, V₂/V₁ and Z_in. Why is using I₂=+Y_LV₂ wrong?

Having obtained both Z and Y from defined terminal experiments, chapter 03 will derive hybrid h parameters from the same signed equations and explain why different matrix representations require different nonzero pivots.