One linear operator, three coordinate systems — and the matrix properties that decide whether a measured network is reciprocal, passive, lossless, and safe to simulate.
Take a system with $N$ ports — a filter, a length of cable, an amplifier, or simply a box with $N$ coax connectors on it. Because such a system is linear, its behaviour at any one frequency is captured entirely by an $N \times N$ matrix. Which matrix you get depends only on what you choose to call the input.
Drive the ports with currents, read the voltages that appear.
$$\mathbf{V} = \mathbf{Z}\,\mathbf{I}$$
Entries in ohms.
Drive with voltages, read the currents. The exact reverse.
$$\mathbf{I} = \mathbf{Y}\,\mathbf{V}, \qquad \mathbf{Y} = \mathbf{Z}^{-1}$$
Entries in siemens — amps per volt.
Work with the travelling waves a transmission line carries.
$$\mathbf{b} = \mathbf{S}\,\mathbf{a}$$
Entries dimensionless.
The three matrices are three coordinate systems laid over the same underlying linear operator, and slide 14 gives the formulae for moving between them. All three are functions of frequency: writing $\mathbf{Z}$ rather than $\mathbf{Z}(f)$ is only a shorthand.
$\mathbf{Z}$ is current in, voltage out. It is $\mathbf{Y}$ that answers "voltage in, current out", which is why its units are amps per volt. Mixing the two up is the single most common slip in this material.
Column $j$ of the impedance matrix has a direct experimental meaning. Inject a known current into port $j$, leave every other port open so no current flows there, and record the voltage at each port:
$$Z_{ij} = \left.\frac{V_i}{I_j}\right|_{I_k = 0,\ k \neq j}, \qquad Y_{ij} = \left.\frac{I_i}{V_j}\right|_{V_k = 0,\ k \neq j}.$$
The admittance matrix is the mirror image, with every other port shorted instead.
Harder than at audio, but the difficulty is routinely overstated. An open has to be an open at the reference plane, and a physical open has fringing capacitance and may radiate, while a physical short has series inductance. Yet open, short and load standards are exactly what a conventional SOLT calibration uses, accurately, at 5 GHz and a good deal higher — the standards are not assumed ideal but characterised, their fringing capacitance and residual inductance captured in a model the analyser applies. Higher still, and on wafer, TRL takes over because it leans on transmission-line sections rather than lumped terminations.
Realising ideal terminations is challenging, not hopeless, and the inconvenience is real but secondary. The actual reason waves win is on slide 04.
Build the network from ordinary R, L, C and PCB copper and transmission from port 1 to port 2 equals transmission from 2 to 1. The matrix records that as symmetry about the diagonal:
$$\mathbf{Z} = \mathbf{Z}^{\mathsf{T}}, \qquad \mathbf{Y} = \mathbf{Y}^{\mathsf{T}}, \qquad \mathbf{S} = \mathbf{S}^{\mathsf{T}}.$$
Note the plain transpose, not the conjugate transpose: $\mathbf{Z}$ stays complex, so this is weaker than Hermitian symmetry. Ferrite isolators, circulators and active devices break reciprocity on purpose, and in measured data the size of the antisymmetric part $\mathbf{Z} - \mathbf{Z}^{\mathsf{T}}$ is a direct measure of error.
A passive network can absorb energy and store it, but it cannot on average deliver more than it receives, however you drive it. In port variables the average power delivered into the network by a set of phasor drives is
$$P_{\text{avg}} = \tfrac12 \operatorname{Re}\!\left(\mathbf{V}^{\mathsf{H}}\mathbf{I}\right) = \tfrac12\, \mathbf{I}^{\mathsf{H}} \mathbf{Z}_{\mathsf{H}} \mathbf{I}, \qquad \mathbf{Z}_{\mathsf{H}} = \tfrac12\left(\mathbf{Z} + \mathbf{Z}^{\mathsf{H}}\right).$$
Requiring this to be non-negative for every drive vector is exactly the statement that the Hermitian part is positive semidefinite, so passivity collapses to a real eigenvalue test:
$$\mathbf{Z}_{\mathsf{H}} \succeq 0 \iff \lambda_i(\mathbf{Z}_{\mathsf{H}}) \ge 0 \ \ \forall i.$$
Split $\mathbf{Z} = \mathbf{R} + j\mathbf{X}$ and the physics is obvious: only $\mathbf{R}$ absorbs average power, $\mathbf{X}$ merely stores and returns it, and a lossless network is the boundary case $\mathbf{R} = 0$.
Transpose symmetry $\mathbf{Z} = \mathbf{Z}^{\mathsf{T}}$ expresses reciprocity and needs a reciprocal medium; it assumes nothing about signals being real. The Hermitian condition $\mathbf{Z}_{\mathsf{H}} \succeq 0$ expresses passivity and likewise needs no reality, because average power is a Hermitian quadratic form even for complex phasors. What does need real time-domain signals is conjugate symmetry, $H(-j\omega) = H^*(j\omega)$. A VNA measuring real voltages obeys it; a complex-baseband IQ model generally does not. Trust the Hermitian passivity test for complex signals; distrust any negative-frequency mirror argument unless $h(t)$ has been stated to be real.
Once a connection is long enough that the signal's phase changes appreciably along it, the natural solution of the equations on that connection is not a single voltage but a pair of waves — one heading towards the load, one heading back. Transmission-line theory is built on that pair from the start, and the quantities an RF engineer uses daily are all relationships between them: return loss, insertion loss, VSWR, group delay, the action of a matching network, the effect of a length of line.
Scattering parameters carry that description into network theory, and that is why they exist. When a signal arrives at one port and leaves by another, the useful thing to know is how its magnitude and phase were altered on the way — and an S-parameter is precisely that number. Cascade two components and the phases add; insert a quarter-wave line and you can read off what it does. The description composes the way the hardware does.
Waves are the natural currency of transmission-line and network theory, and an S-parameter is the natural bookkeeping for magnitude and phase from port to port.
Unused ports sit in $Z_0$ rather than open or short, so the device stays in the broadband-matched state it meets in the real system — and a good 50 Ω load holds over a wide sweep more easily than an ideal open. Real, but a consequence of the choice of variables rather than the motivation for it.
Start with what is physically on the line. At a port on a transmission line of characteristic impedance $Z_0$, the total voltage and current are the superposition of a wave heading into the network and one heading back out:
$$V = V^{+} + V^{-}, \qquad I = \frac{V^{+} - V^{-}}{Z_0},$$
$$V^{+} = \frac{V + Z_0 I}{2}, \qquad V^{-} = \frac{V - Z_0 I}{2}.$$
These are voltages, in volts. They are what a TDR displays and what superposes at an impedance discontinuity. A scattering parameter is a ratio of one of them to another:
$$S_{ij} = \left.\frac{V_i^{-}}{V_j^{+}}\right|_{V_k^{+} = 0,\ k \neq j}$$
— that is, with every other port terminated in $Z_0$ so nothing is incident there. In words: $S_{ij}$ is the complex ratio of the wave leaving port $i$ to the wave arriving at port $j$, and being complex it carries a magnitude and a phase.
The magnitude and phase of the signal emerging at port 2 relative to the signal driven into port 1, at whichever frequency you are looking at. Sweep the source and that collection of numbers is the transmission response.
The wave that comes back out of port 1 relative to the one you sent in: the input reflection coefficient.
In general the diagonal entries are reflections and the off-diagonal entries are transmissions, and the second subscript is always the port you drove.
It is often convenient to rescale as $a_i = V_i^{+}/\sqrt{Z_0}$ and $b_i = V_i^{-}/\sqrt{Z_0}$, so that $|a_i|^2$ is the power into port $i$ and $|b_i|^2$ the power out of it, and the matrix becomes $\mathbf{b} = \mathbf{S}\mathbf{a}$. The rescaling is by a constant common to every port, so it cancels from every ratio and leaves each $S_{ij}$ untouched; its only job is to let the power bookkeeping of slide 07 be written without dragging factors of $1/2Z_0$ around. With $Z_0$ real these are also Kurokawa's power waves — the two descriptions part company only at a complex reference, which slide 14 returns to.
Send 1 W into port 1 with the other port matched. A fraction $|S_{11}|^2$ comes straight back and $|S_{21}|^2$ arrives at port 2, so $|S_{11}| = -10$ dB means a tenth of the power reflected and $|S_{21}| = -3$ dB means about half gets through. One trap: cascading is not matrix multiplication of $\mathbf{S}$, because the wave leaving the first component is partly reflected by the second and returns to be re-reflected. Convert to ABCD or T, multiply there, convert back. A second trap is the decibel convention: $S_{21}$ is a ratio of wave amplitudes, so $|S_{21}|$ is a voltage ratio while $|S_{21}|^2$ is the power ratio, and the figure on every datasheet is $20\log|S_{21}|$, not $10\log$. Halving the voltage is −6 dB and quarters the power; −3 dB is $|S_{21}| = 0.707$, which halves it.
S-parameters account for power that leaves through a port. They say nothing about where the rest of it went.
From a complete $N$-port set, the total power that failed to emerge:
$$1 - \sum_i |S_{i1}|^2.$$
It lumps dissipation and radiation together. A structure that reflects less than everything may be heating copper and dielectric, or radiating into the room as an antenna. No scattering measurement separates the two.
A perfect open reflects everything and does so in phase, sitting at $S_{11} = +1$. But the moment the structure radiates, part of what would have returned is gone, and the S-parameters record only that it did not come back. Distinguishing the mechanisms needs a different experiment — a calorimetric one, or a pattern measurement in an anechoic chamber.
The condition on the next slide certifies that the network is not supplying energy. It says nothing about which mechanism is consuming it.
Count the power. The difference between what goes in and what comes out is
$$\tfrac12\left(\mathbf{a}^{\mathsf{H}}\mathbf{a} - \mathbf{b}^{\mathsf{H}}\mathbf{b}\right) = \tfrac12\,\mathbf{a}^{\mathsf{H}} \left(\mathbf{I} - \mathbf{S}^{\mathsf{H}}\mathbf{S}\right)\mathbf{a}.$$
If the network absorbs nothing whatever you send it, that must vanish for every drive vector, which forces
$$\mathbf{S}^{\mathsf{H}}\mathbf{S} = \mathbf{I}.$$
A lossless network has a unitary scattering matrix: it rotates the vector of wave amplitudes without changing its length. The columns are orthonormal, which is the algebraic way of saying that power sent into one port must emerge from some port.
Allow loss and the equality relaxes to an inequality,
$$\mathbf{I} - \mathbf{S}^{\mathsf{H}}\mathbf{S} \succeq 0 \iff \sigma_i(\mathbf{S}) \le 1 \ \ \forall i,$$
so a passive network is a contraction on wave space — it can shrink the wave vector but never stretch it — and losslessness is the boundary case in which nothing shrinks at all.
Once there is loss, $\mathbf{S}$ need not be normal and its eigenvectors need not stay orthogonal. The singular-value form remains the correct test; the eigenvalues do still satisfy $|\lambda_k| \le 1$, but they are the weaker statement.
A single resistor in series between two 50 Ω ports. Small enough to solve in your head, big enough to show what an eigenvector of $\mathbf{S}$ means:
$$\mathbf{S} = \begin{pmatrix} s_{11} & s_{21} \\ s_{21} & s_{11}\end{pmatrix}, \quad s_{11} = \frac{R}{R + 2Z_0}, \quad s_{21} = \frac{2Z_0}{R + 2Z_0}.$$
Because the matrix is symmetric with equal diagonals, its eigenvectors are fixed: the even drive $(1,1)/\sqrt2$ and the odd drive $(1,-1)/\sqrt2$, with eigenvalues $s_{11} + s_{21} = 1$ and $s_{11} - s_{21}$.
Drive both sides identically and both ends of the resistor sit at the same potential, so no current flows through it and nothing is dissipated — the pattern passes through untouched, $|\lambda| = 1$. Drive them in antiphase and the full difference lands across the resistor, which warms up, so that pattern comes out smaller. That is all an eigenvector of $\mathbf{S}$ amounts to: the drive pattern that keeps its shape, with $|\lambda|$ telling you how much of it survives. Set $R = 0$ and both eigenvalues reach the unit circle — a lossless through-connection.
It is tempting to present "the matrix at one frequency" and "the swept response" as two different concepts. They are not. The matrix is a function of frequency, and a swept measurement is that function sampled at many points rather than one.
$\mathbf{S}(f_0)$ is an $N \times N$ matrix of complex numbers. Ask about reciprocity, passivity, losslessness, eigenvectors — all answered from that matrix.
Exactly the same questions, answered once per frequency. Plot one entry against $f$ and you have a frequency response. No new concept has entered.
Everything on slides 01–08 therefore applies unchanged across a sweep. What genuinely deserves to be called a second view is the time domain, and that is the next slide.
Calling the single-frequency picture "spatial" invites a comparison with beam-steering and phased arrays, where sources separated in space play the role that taps separated in time play in a filter. That is a real and interesting analogy — and a different article. Here, the honest split is time against frequency.
Plotting an entry of $\mathbf{S}$ against frequency raises a new question: why does the curve have the shape it has? Answering that means looking inside the box, and the natural language inside is time.
For a network built from R, L and C, every entry is a ratio of polynomials, $S_{21}(s) = N(s)/D(s)$. Collect every capacitor voltage and inductor current into a state vector $\mathbf{q}$ and the dynamics become
$$\frac{d\mathbf{q}}{dt} = \mathbf{A}_c \mathbf{q} + \mathbf{b}\,x(t),$$
and for a minimal model — one with no hidden states that cancel out of the output — the poles are exactly the eigenvalues of $\mathbf{A}_c$. Stable and passive means all of them sit strictly left of the $j\omega$-axis, an axis which is both the fence they must stay behind and the road along which you measure.
$S_{21}(j\omega)$ against $\omega$. A pole near the axis gives a tall narrow peak; a zero on the axis gives a null.
$h(t)$ against $t$. The same pole gives a slowly decaying ring at the same frequency. One Laplace transform apart.
(1) Eigenvalues of $\mathbf{S}(f)$. At a fixed frequency, lossless means unitary means every $|\lambda| = 1$. A statement about how ports exchange energy; nothing about stability. (2) Poles in the $s$-plane. Eigenvalues of $\mathbf{A}_c$; stable when strictly left of $\operatorname{Re}(s) = 0$, which is also the evaluation axis. (3) Poles in the $z$-plane — the discrete-time world of the companion deck, where the boundary is $|z| = 1$ and the response is evaluated on the circle itself. The pair worth keeping apart is (1) and (3): both are unit circles in a complex plane, and neither has anything to do with the other.
A notched resonance, the workhorse shape of filter design:
$$H(s) = \frac{s^2 + \omega_z^2}{s^2 + 2\sigma s + \omega_0^2}.$$
Move the pole pair towards the $j\omega$-axis and watch the peak grow and the ringing lengthen — two readings of one change. Move the zero and watch the null follow it.
Small pushes at the right rhythm build a large swing; grab the ropes — add loss, move the pole away from the axis — and it dies quickly. A zero sitting on the axis is the opposite trick: the numerator vanishes exactly there, so the response nulls out. Notch filters put zeros on the axis deliberately.
What the laboratory hands you is samples of $\mathbf{S}(j\omega)$ — discrete, band-limited, noisy. Turning that into something a circuit simulator can use means fitting a rational model, and the fit has to respect two physical constraints the raw data only approximately encodes.
A causal impulse response, $h(t) = 0$ for $t < 0$, corresponds to a transfer function analytic in the right half-plane, and Cauchy's theorem then ties the real and imaginary parts of $H(j\omega)$ together as a Hilbert pair:
$$\operatorname{Re} H(j\omega) = \frac{1}{\pi}\,\mathcal{P}\!\!\int_{-\infty}^{\infty} \frac{\operatorname{Im} H(j\omega')}{\omega' - \omega}\, d\omega'.$$
These are the Kramers–Kronig relations, known in the network literature as Bode's attenuation–phase relations. Their practical content: magnitude and phase are not independent quantities you may fit separately.
Kramers–Kronig itself requires only causality. The additional mirror rule — real part even, imaginary part odd, $H(-j\omega) = H^*(j\omega)$ — additionally requires $h(t)$ to be real. True of VNA data taken on real voltages; false in general for complex-envelope models.
The reliable way to enforce causality is structural rather than integral: fit a strictly proper rational function with all poles in the open left half-plane, as vector fitting does, and causality follows by construction.
Causality restricts where the poles may sit — a condition on a finite set of numbers, easy to impose. Passivity is a condition at every frequency:
$$\mathbf{S}^{\mathsf{H}}(j\omega)\mathbf{S}(j\omega) \preceq \mathbf{I} \quad \text{for all } \omega,$$
and it must hold not merely at the frequencies you happened to sample but everywhere in between. An unconstrained fit will sail past 1 in the gaps, and a model that does so manufactures energy and makes a transient simulation diverge — often long after anyone was still suspicious of the model.
Only a model that is both causal and passive is safe to hand to a time-domain simulator. (Grivet-Talocia & Gustavsen, Passive Macromodeling, 2016.)
With a single real reference impedance common to all ports the conversions are compact:
$$\mathbf{Y} = \mathbf{Z}^{-1}, \qquad \mathbf{S} = (\mathbf{Z} + Z_0\mathbf{I})^{-1}(\mathbf{Z} - Z_0\mathbf{I}), \qquad \mathbf{Z} = Z_0(\mathbf{I} + \mathbf{S})(\mathbf{I} - \mathbf{S})^{-1}.$$
This is a matrix Möbius transformation, and it carries the half-plane $\operatorname{Re}(\mathbf{Z}) \succeq 0$ onto the unit ball $\|\mathbf{S}\| \le 1$: passive networks become contractions, purely reactive ones land on the boundary. Different real reference impedances at different ports need only the appropriate per-port $\sqrt{Z_0}$ factors, and nothing conceptual changes. A complex reference is a different matter.
The literature carries three definitions of the incident and reflected amplitudes. They agree everywhere except when the reference impedance is complex — and there the difference is physical, not cosmetic.
Same reference in both: $V^{\pm} = (V \pm Z_0 I)/2$.
$$\Gamma = \frac{Z - Z_0}{Z + Z_0}$$
$\Gamma = 0$ means reflectionless — no backward wave on the line. The signal-integrity convention, and the one used here.
Conjugate in the outgoing term: $b = (V - Z_0^{*}I)/2\sqrt{\operatorname{Re}Z_0}$.
$$\Gamma = \frac{Z - Z_0^{*}}{Z + Z_0}$$
$\Gamma = 0$ means conjugate matched — maximum power transfer. A different termination.
Marks & Williams' generalisation, for a reference impedance that is itself complex and frequency-dependent — lossy line, waveguide, on wafer.
With a real $Z_0$ the conjugate does nothing and all three coincide, which is why the distinction can be set aside for most 50 Ω coaxial and PCB work. The conjugate is not a normalisation factor you can absorb, so state which convention you are using whenever $Z_0$ is not real.
The matrices become scalars and the map reduces to
$$\Gamma = \frac{z - 1}{z + 1}, \qquad z = Z/Z_0.$$
The familiar grid of circles is nothing more than the image of the constant-resistance and constant-reactance lines under this bilinear map. The rim $|\Gamma| = 1$ is the one-port case of the unit circle from slide 07, carrying the purely reactive terminations, with everything passive inside it. There is no useful $N$-port generalisation of the picture, which is why multiport work uses the matrix formulae above.
Drag the normalised resistance and reactance and watch the termination move on the chart. $Z_0 = 50\,\Omega$ throughout.
Left of centre is low impedance, right of centre is high, the rim is purely reactive, and the centre is a perfect match. A short sits at $\Gamma = -1$, an open at $\Gamma = +1$, and 25 Ω at $\Gamma = -1/3$ — a VSWR of 2.
A differential pair on a VNA is a four-port: two single-ended ports at the near end, one per conductor, and two at the far end. That 4×4 matrix says nothing directly about differential behaviour, because differential and common mode are combinations of the single-ended variables:
$$v_d = v_P - v_N, \qquad v_c = \tfrac{1}{2}(v_P + v_N).$$
Order the single-ended ports 1, 2 as the near-end pair and 3, 4 as the far-end pair, and the mixed-mode vector as $(d_1, d_2, c_1, c_2)$. Then the change of variables is one matrix:
$$\mathbf{M} = \frac{1}{\sqrt2}\begin{pmatrix} 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & -1 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix}, \qquad \mathbf{S}_{mm} = \mathbf{M}\,\mathbf{S}\,\mathbf{M}^{\mathsf{T}}.$$
The $1/\sqrt2$ makes $\mathbf{M}$ orthogonal, so $\mathbf{M}^{-1} = \mathbf{M}^{\mathsf{T}}$ and this is a similarity transform — the same linear operator in a different basis, exactly as on slide 01. Nothing is measured twice.
An orthogonal $\mathbf{M}$ is a rotation of wave space, and rotations preserve eigenvalues and singular values. Every test on slides 02, 03, 07 and 13 — reciprocity, passivity, losslessness, the Hamiltonian search — returns the same answer in either basis. Mixed mode reorganises the information into the combinations an engineer cares about; it adds none and destroys none.
| Block | What it describes | Why you look at it |
|---|---|---|
| $\mathbf{S}_{dd}$ | Differential in, differential out | The actual signal path. $S_{dd21}$ is the insertion loss that sets how much equalisation a link needs; $S_{dd11}$ the differential return loss |
| $\mathbf{S}_{cc}$ | Common in, common out | How common-mode energy propagates — whether a choke or ferrite will achieve anything, and how much common-mode current reaches a cable |
| $\mathbf{S}_{cd}$ | Differential in, common out | Your own signal leaking into common mode. Common-mode current on a cable radiates efficiently — this is the quadrant that fails an EMC scan |
| $\mathbf{S}_{dc}$ | Common in, differential out | External common-mode noise arriving as differential noise at the receiver. Lands straight on the eye |
If the two halves of the pair are genuinely identical, the off-diagonal blocks vanish and $\mathbf{S}_{mm}$ is block diagonal. Every non-zero entry in $\mathbf{S}_{cd}$ or $\mathbf{S}_{dc}$ is a calibrated measure of how asymmetric the pair really is — a far more useful thing to hand a layout engineer than a vague complaint about EMC.
Touchstone records the numbers but not what the ports were connected to, and two conventions are in use: 1–2 / 3–4 as near and far pairs, or 1–3 / 2–4 as the two conductors. Apply the wrong $\mathbf{M}$ and the blocks permute — most visibly $\mathbf{S}_{dd}$ swaps with $\mathbf{S}_{cc}$ — giving plots that look like a catastrophic channel. Sanity check: $S_{dd21}$ should be the least lossy trace on the chart. If it looks like rubbish while $S_{cc21}$ looks pristine, you have the permutation backwards, not a broken board.
The cleanest asymmetry to work through is intra-pair skew. Take two conductors identical in every respect except that one is electrically longer by $\Delta t$. Split the delay symmetrically, drive differentially, and the far-end modes come out as
$$|S_{dd21}| = |H|\cos(\pi f \Delta t), \qquad |S_{cd21}| = |H|\sin(\pi f \Delta t).$$
Energy the differential mode loses turns up in the common mode, exactly and without waste. Note how differently the two respond.
$S_{dd21}$ is almost blind to skew, so a channel can look perfectly healthy on the plot everyone shows in review while radiating badly. $S_{cd21}$ is the sensitive instrument, and it is the one to put a limit line on. Usual causes: unequal path lengths where a pair turns a corner or dodges a via; asymmetric barrels, antipads and connector footprints; and — most awkwardly — fibre weave, where one conductor runs over glass bundles and the other over resin, giving them different effective permittivity. Weave skew accumulates with length, which is why long pairs are routed at a slight angle to the weave or on spread-glass laminate.
Mode conversion is not loss. The energy changes mode rather than disappearing, and the four-port stays perfectly passive — the singular values are untouched, as the orthogonality of $\mathbf{M}$ guarantees. The trouble is that the energy is now in a mode the receiver does not read and an enclosure does not contain.
.s2p / .sNp.| Claim | Test | Which boundary |
|---|---|---|
| Reciprocity | $\mathbf{Z} = \mathbf{Z}^{\mathsf{T}}$, or $\mathbf{S} = \mathbf{S}^{\mathsf{T}}$ | Symmetry; no boundary |
| Passivity | $\mathbf{Z}_{\mathsf{H}} \succeq 0$, equivalently $\sigma(\mathbf{S}) \le 1$ at every $\omega$ | PSD and contraction |
| Losslessness | $\mathbf{S}^{\mathsf{H}}\mathbf{S} = \mathbf{I}$ | $|\lambda(\mathbf{S})| = 1$ — the eigenvalue plane |
| Resonance or null | Pole near the $j\omega$-axis, or zero on it; $Q \approx \omega_0/2\sigma$ | Distance from the $j\omega$-axis |
| Pair is symmetric | $S_{cd21}$ small across the band | Off-diagonal mixed-mode blocks vanish for a symmetric pair |
| Safe to simulate | LHP poles plus Hamiltonian passivity enforcement | No RHP poles; $\sigma \le 1$ everywhere |
1. Reciprocal? Unless it contains ferrite or active circuitry, $\mathbf{S}$ should equal its transpose — the antisymmetric part measures your error. 2. Causal? Fitted poles all in the left half-plane, plus a Kramers–Kronig spot check. 3. Passive everywhere? Use the all-frequency Hamiltonian test, not a sweep through the sample points — that is exactly where violations hide. 4. Resonances physical? Every peak should match an LHP pole pair of the right $Q$, every null a nearby zero. 5. Reference consistent? One real $Z_0$, usually 50 Ω, before applying any conversion formula.
Deck 01 of the Matrix Methods in Engineering series. Companions: Digital Filter Design and Equalisation in High-Speed Serial Links.
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