Hybrid h and inverse-hybrid g parameters
Choose mixed port variables, derive the h and g matrices from Z/Y by elimination, preserve both entering-current signs, and verify a reciprocal π-network with explicit port power.
- Define hybrid independent and dependent port variables.
- Preserve two inward port currents and the output-current minus sign.
- Derive all h entries by eliminating I₂ from Z.
- Extract h entries under explicit open/short experiments.
- Derive h from Y using the appropriate nonzero pivot.
- Deduce h₁₂=−h₂₁ under reciprocity.
- Check mixed units Ω, S and dimensionless.
- Construct inverse hybrid g from Z and verify its matrix relation.
- Confirm port power is nonnegative for a passive π network.
- Audit h and g on one numerical example.
Start from the same entering-current two-port references of RF-03.01 and the Z/Y descriptions derived in RF-03.02. Mixed variables are a rearrangement of port equations, not a new law of physics.
0103.1 — Why choose mixed port variables?
At a selected frequency, different choices of independent port variables lead to different but equivalent two-port representations, provided the required inverse exists. The hybrid form uses an input current and an output voltage as independent variables; it is useful in models of amplifiers and transistor stages.
The output current is still defined entering the device, as in RF-03.01. Hybrid does not mean changing voltage polarity, reversing an arrow or assuming that the device is active.
0203.2 — Define the h matrix and its units
Write V₁=h₁₁I₁+h₁₂V₂ and I₂=h₂₁I₁+h₂₂V₂. Dimensional consistency gives h₁₁ in ohms, h₂₂ in siemens and both transfer ratios dimensionless. The four entries have different units, so the matrix is not an ordinary homogeneous impedance matrix.
The h representation depends on the solvability of its mixed-variable boundary conditions. Avoid treating entries as universal constants independent of frequency, bias, loading or linearization assumptions.
Figure 3.1 — Signed electrical references of a two-port
Both I₁ and I₂ are defined ENTERING the network at their positive terminals; each voltage arrow points from minus to plus.
I₁ enters the left port pointing right, while I₂ enters the right port pointing left. V₁ and V₂ are positive on the upper terminals. The conventional output current toward an attached load equals −I₂. These arrows define references, not actual solved current flow.
0303.3 — Derive h from Z without guessing signs
From V₂=Z₂₁I₁+Z₂₂I₂, solve I₂=(V₂−Z₂₁I₁)/Z₂₂ when Z₂₂≠0. Substituting into V₁=Z₁₁I₁+Z₁₂I₂ yields the first row of h. The second row follows directly from the solved current.
The minus sign of h₂₁=−Z₂₁/Z₂₂ is crucial: port 2 current is entering, so a passive through-current leaving the output is negative in the reference convention. The determinant term in h₁₁ arises from elimination, not a new approximation.
0403.4 — Experimental short/open conditions for h
To measure h₁₁ and h₂₁, impose V₂=0 by short-circuiting port 2 and excite nonzero I₁. Then V₁/I₁=h₁₁ and I₂/I₁=h₂₁. Notice that I₂ generally does not vanish under a short.
To measure h₁₂ and h₂₂, open the input so I₁=0, then apply nonzero V₂. Ratios V₁/V₂ and I₂/V₂ provide the other column. The physical open and short are ideal definitions; at microwave frequencies fixtures and reference planes must be included.
0503.5 — Derive h from Y and identify the singular pivot
Start instead from I₁=Y₁₁V₁+Y₁₂V₂. If Y₁₁≠0, isolate V₁=(I₁−Y₁₂V₂)/Y₁₁. Inserting this result in I₂=Y₂₁V₁+Y₂₂V₂ gives all four hybrid coefficients.
The Z-based transformation requires Z₂₂≠0; the Y-based version requires Y₁₁≠0. A valid physical two-port may possess an h description even when a particular Z or Y chart is unavailable. Conditions must be checked before division.
0603.6 — Reciprocal hybrid transfer signs and power
Under the established sign convention a reciprocal Z matrix obeys Z₁₂=Z₂₁, which gives h₁₂=−h₂₁. Forgetting the minus sign is a frequent error when importing h-parameter conventions from a drawing with output current pointing outward.
Power remains P_abs=(1/2)Re(V₁I₁*+V₂I₂*) for peak phasors. A reciprocal dissipative π network has h₁₂=−h₂₁ but positive internal losses, so the reciprocity identity must never be used as a losslessness test.
0703.7 — Inverse-hybrid g and matrix transformations
The inverse-hybrid form takes V₁ and I₂ as independent variables and returns I₁ and V₂. Solve V₁=Z₁₁I₁+Z₁₂I₂ for I₁ when Z₁₁≠0; substitute into V₂ to identify the g entries.
If h is nonsingular, g=h⁻¹ in the stated ordering of mixed variables. Both g₁₁ and h₂₂ have siemens units, whereas g₂₂ and h₁₁ are in ohms. Reciprocity gives g₁₂=−g₂₁ when the representations exist.
0803.8 — Numeric π-network audit and domain limits
For the passive π-network example from RF-03.02, Y₁₁=0.030 S, Y₁₂=Y₂₁=−0.020 S and Y₂₂=0.025 S. Using the Y-to-h formulas gives h₁₁=33.3333 Ω, h₁₂=2/3, h₂₁=−2/3 and h₂₂≈0.0116667 S.
Excite the hybrid independent variables I₁=0.020 A and V₂=0.4 V. Then V₁≈0.933333 V and I₂≈−0.00866667 A; verify directly against I=YV. The sign of I₂ is negative because the device delivers current out of the output terminal for these conditions.
Practice — 10 worked problems on h and g
- Derive h11,h12,h21,h22 from Z by eliminating I2.
- State the physical experiment for h21 and explain its sign.
- State the physical experiment for h12.
- Derive the dimensions of all four h entries.
- Derive h from Y by eliminating V1.
- For reciprocal Z, prove h12=−h21.
- Compute h for the reciprocal π network Y=[0.030 −0.020;−0.020 0.025] S.
- For I1=0.020 A and V2=0.4 V, calculate V1 and I2 from the preceding h matrix.
- Derive g from Z by eliminating I1 and give the required pivot.
- Explain why equal h12 and h21 are generally inconsistent with the two-entering-current reciprocal convention.
Worked solutions
1. From V₂=Z₂₁I₁+Z₂₂I₂ solve I₂=(V₂−Z₂₁I₁)/Z₂₂. Substitute into V₁=Z₁₁I₁+Z₁₂I₂: h₁₁=Z₁₁−Z₁₂Z₂₁/Z₂₂, h₁₂=Z₁₂/Z₂₂, h₂₁=−Z₂₁/Z₂₂, h₂₂=1/Z₂₂.
2. Set V₂=0 by shorting port 2. Then I₂=h₂₁I₁, hence h₂₁=(I₂/I₁)|V₂=0. A through-current supplied toward a short exits port 2, so the entering-current I₂ may be negative.
3. Set I₁=0 by opening port 1. Then V₁=h₁₂V₂, hence h₁₂=(V₁/V₂)|I₁=0 for nonzero V₂.
4. V₁=h₁₁I₁+h₁₂V₂ gives h₁₁ Ω and h₁₂ dimensionless. I₂=h₂₁I₁+h₂₂V₂ gives h₂₁ dimensionless and h₂₂ S.
5. From I₁=Y₁₁V₁+Y₁₂V₂, V₁=I₁/Y₁₁−(Y₁₂/Y₁₁)V₂. Substitute in I₂=Y₂₁V₁+Y₂₂V₂: h₁₁=1/Y₁₁, h₁₂=−Y₁₂/Y₁₁, h₂₁=Y₂₁/Y₁₁, h₂₂=Y₂₂−Y₂₁Y₁₂/Y₁₁.
6. Reciprocity means Z₁₂=Z₂₁. Therefore h₁₂=Z₁₂/Z₂₂ and h₂₁=−Z₂₁/Z₂₂ give h₁₂=−h₂₁, provided Z₂₂≠0.
7. Use Y₁₁=.030,Y₁₂=Y₂₁=−.020,Y₂₂=.025 S. The formulas give h₁₁=33.3333 Ω, h₁₂=+2/3, h₂₁=−2/3 and h₂₂=.025−.0004/.030=.0116667 S.
8. V₁=(33.3333)(.020)+(2/3)(.4)=.666667+.266667=.933333 V. I₂=(−2/3)(.020)+(.0116667)(.4)=−.0133333+.00466667=−.00866667 A.
9. With Z₁₁≠0, I₁=(V₁−Z₁₂I₂)/Z₁₁. Then V₂=Z₂₁I₁+Z₂₂I₂. Thus g₁₁=1/Z₁₁, g₁₂=−Z₁₂/Z₁₁, g₂₁=Z₂₁/Z₁₁, g₂₂=Z₂₂−Z₂₁Z₁₂/Z₁₁.
10. For reciprocal Z we have h₁₂=−h₂₁. Equal signs would require h₁₂=h₂₁=0 (zero coupling) in the same entering-current convention. A typical sign-equal formula silently uses an outward output current.
Next derive ABCD chain matrices with the explicit −I₂ convention and cascade from Kirchhoff.