Reciprocity, passivity and scattering energy
Derive S†S passivity and lossless unitarity, test reciprocal symmetry, coherent eigenchannels, a dissipative series resistor, an ideal gyrator and an isolator, with hidden physical power channels.
- Derive real absorbed power from peak incident and outgoing wave amplitudes.
- Prove passivity is equivalent to I−S†S being positive semidefinite.
- Test both diagonal and determinant conditions in a 2×2 Hermitian matrix.
- Derive the lossless unitarity S†S=I criterion.
- Distinguish reciprocal Sᵀ=S from power conservation.
- Calculate coherent eigenchannels of a series resistor.
- Demonstrate why individual column-power checks are insufficient.
- Test ideal gyrator nonreciprocal losslessness.
- Test ideal isolator passivity with reverse absorption.
- Identify unmeasured radiation/modes and calibration uncertainty in experiments.
A scattering matrix must conserve or absorb power consistently for every incident-wave combination, not merely for one VNA excitation. We now derive the full Hermitian passivity criterion, distinguish it from reciprocity and losslessness, test coherent eigenchannels of a series resistor, and construct lossless nonreciprocal as well as absorptive one-way examples.
0104.1 — From port waves to net average absorbed power
For peak-phasor amplitudes normalized to positive real port references, incident wave vector a and outgoing vector b both have units √W. RF-04.03 defined b=Sa. Because a and b are based on peak voltages and currents, real average power is half the difference of squared Euclidean norms.
Substitute b=Sa to obtain P_abs=(1/2)a†(I−S†S)a. A negative absorbed power for an admissible incident-wave excitation would mean the chosen two-port delivers net real power. The test requires consistent port and mode normalization, and excludes stored initial energy or unmodeled active pumping.
Figure 4.1 — 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.
0204.2 — Passivity is a positive-semidefinite matrix condition
Since P_abs must be nonnegative for every complex incident vector a, the Hermitian matrix H=I−S†S must be positive semidefinite. For a 2×2 Hermitian H, the necessary and sufficient principal-minor tests are H11≥0, H22≥0 and H11H22−|H12|²≥0.
Checking only |S11|²+|S21|²≤1 and |S12|²+|S22|²≤1 is insufficient: these are tests of the two diagonal entries of H. Coherent excitation may reveal an eigenvector with negative absorbed power if the cross terms of S†S are too large.
0304.3 — Losslessness means unitarity, not reciprocity
A lossless two-port that includes all its propagating power channels has P_abs=0 for every excitation. Consequently H=0 and S†S=I. The matrix S is unitary, so its columns are orthonormal in complex power-wave coordinates and S⁻¹=S†.
The unitarity condition is independent of matrix symmetry. A reciprocal lossy network can have S12=S21 while failing S†S=I. Conversely, a nonreciprocal lossless network can be unitary without being symmetric. Excluded radiative or higher-order channels make the measured reduced S appear lossy even if the complete physical system conserves energy.
0404.4 — Reciprocity is a transpose condition under real references
With reciprocal linear media, correctly defined traveling modal ports and positive real power normalization, reciprocity gives Sᵀ=S, or S12=S21 for the two-port. The two diagonal reflections are allowed to differ when the network is geometrically asymmetric.
Reciprocity neither proves passivity nor unitarity; energy conditions instead involve a conjugate transpose S†. Confusing ordinary transpose with conjugate transpose is especially damaging when phases are nonzero. For complex reference impedances, the wave definition and reciprocity transformation must be reconsidered.
0504.5 — A resistive two-port has lossless and lossy coherent channels
Reuse the 25 Ω series resistor with two 50 Ω references: S=[[0.2,0.8],[0.8,0.2]]. Its columns each have squared norm 0.68, so exciting port 1 alone dissipates 32% of incident power. Compute S†S=[[0.68,0.32],[0.32,0.68]], whose eigenvalues are 1 and 0.36.
The normalized common-mode excitation a∝(1,1) has S eigenvalue +1 and no voltage drop across the series resistor, so it is lossless. The differential excitation a∝(1,−1) has eigenvalue −0.6 and absorbs 1−0.36=64% of its incident power. Passivity must hold for both coherent combinations.
0604.6 — Why checking two single-port excitations is not enough
Consider the formal matrix S=[[0.8,0.8],[0,0]]. Driving port 1 alone gives total outgoing power fraction 0.64, and so does driving port 2 alone. These two column tests look passive, but they do not test interference between coherent input waves.
For a=(1,1), outgoing b=(1.6,0). The incoming squared norm is 2 while the outgoing squared norm is 2.56: net average absorbed power is (2−2.56)/2=−0.28 W for wave amplitudes in √W. S†S has eigenvalues 1.28 and 0, so the full passivity test correctly rejects this hypothetical passive device.
0704.7 — Nonreciprocal does not mean active or dissipative
For an ideal gyrator oriented so that S=[[0,1],[−1,0]], Sᵀ≠S but S†S=I. The two ports exchange power without net loss, and reversing the propagation direction flips an output phase. One realization can have a real antisymmetric impedance matrix Z=[[0,R0],[−R0,0]].
An ideal matched isolator with S=[[0,0],[1,0]] instead transmits wave power from port 1 to 2 but absorbs power entering from port 2. It remains passive because S†S=diag(1,0)≼I, yet it is not lossless. Real isolators require nonreciprocal bias and a dissipation path or additional physical ports.
0804.8 — What a measured two-port can and cannot prove
A measured two-port S usually describes only two calibrated propagating channels. Free-space radiation, substrate modes, cable losses, heat, and fixture absorption may carry power not represented by those two output channels. Reduced S can then be contractive even when a larger modeled device is otherwise energy conserving.
Before declaring a fabricated DUT active or passive from its reported S matrix, check that both ports use compatible real power-wave normalization, that the calibration planes match, and that uncertainty is quantified. An apparent eigenvalue slightly above unity can reflect measurement error; a repeatable violation warrants a controlled energy and calibration audit.
Practice — passivity matrices, eigenvalues and energy
- Derive the average absorbed peak-wave power for b=Sa.
- State all three scalar conditions for a 2×2 Hermitian H=I−S†S to be positive semidefinite.
- Explain why losslessness requires S†S=I, not merely S12=S21.
- Give a reciprocal but asymmetric S condition, and distinguish transpose from conjugate transpose.
- For a 25 Ω resistor in series between 50 Ω ports, compute S†S and its eigenvalues.
- Find the absorbed fractions for in-phase and opposite-phase coherent excitations of that resistor.
- Show why S=[[0.8,0.8],[0,0]] fails passivity despite each column norm being less than one.
- Check whether the ideal gyrator S=[[0,1],[−1,0]] is passive, reciprocal and lossless.
- Check whether the ideal matched isolator S=[[0,0],[1,0]] is passive and lossless.
- Explain why a measured S†S≺I does not alone identify microscopic heat loss.
Worked solutions
1. From peak power P=(a†a−b†b)/2 and b=Sa, we have b†b=a†S†Sa. Factor out a to obtain P_abs=(1/2)a†(I−S†S)a. All ports and modes must use a compatible power normalization.
2. Write H=[[h11,h12],[h12*,h22]]. For H≽0, require h11≥0, h22≥0 and det H=h11h22−|h12|²≥0. The diagonal inequalities are not sufficient without the determinant.
3. If a passive network absorbs zero real power for every incident vector, then a†(I−S†S)a=0 for all a. The Hermitian quadratic form vanishes only if I−S†S=0, so S†S=I. Reciprocity separately requires Sᵀ=S under our real-reference convention.
4. A reciprocal asymmetric two-port may have S12=S21 but S11≠S22, as in the π network with S11=−1/37 and S22=3/37. Ordinary transpose Sᵀ interchanges port indices; conjugate transpose S† also conjugates phase and enters power conservation.
5. For S=[[0.2,0.8],[0.8,0.2]], multiplication gives S†S=[[0.68,0.32],[0.32,0.68]]. Symmetric eigenvectors (1,1)/√2 and (1,−1)/√2 have eigenvalues 0.68+0.32=1 and 0.68−0.32=0.36.
6. For in-phase a=(1,1)/√2, b=Sa=a, so outgoing-to-incident power ratio is one and absorption is zero. For opposite-phase a=(1,−1)/√2, b=−0.6a, so 0.36 of power emerges and 0.64 is absorbed. The resistor heats only for a nonzero differential mode.
7. Each isolated input column has squared norm 0.8²=0.64. But S†S=[[0.64,0.64],[0.64,0.64]] has eigenvalues 1.28 and 0. For a=(1,1), input norm²=2 and output norm²=1.6²=2.56, so P_abs=(2−2.56)/2=−0.28 W in peak-wave units; impossible for a passive two-port.
8. For S_g=[[0,1],[−1,0]], S_g†S_g=I and thus net absorbed power is zero. But S_gᵀ=−S_g≠S_g, so it is nonreciprocal, passive and lossless under the idealized complete two-port model.
9. For S_iso=[[0,0],[1,0]], S_iso†S_iso=diag(1,0)≤I, hence it is passive. A wave incident at port 1 transmits with unit power, but a wave entering at port 2 is absorbed; therefore it is not lossless. It is also nonreciprocal because S12=0≠S21=1.
10. The difference between measured incident and outgoing channel powers can include radiation, substrate or higher-order modes, fixtures, and heating. A contractive reduced S only proves net loss from the included channels; additional measurements are needed to attribute the destination of that power.
Next: voltage, pseudo and power-wave definitions, complex port references and transformations between V/I and S.