Circuit Response
every model has an edge, and the figure carries it
Theme

The thread: Two routes to a number

A phase margin from the loop gain and an overshoot from the step response. A transfer function from the matrix and the same one from the poles it was factored into. A ladder solved by nodal analysis and by a chain-matrix product. Neither route in any of those pairs can confirm itself, and they share no arithmetic.
a bridge, which no series-parallel reduction reachesnode a7.5566 Vnode b4.7993 Vcurrent law, rebuilt from the element laws2.71e-16 of the largest branch currentpower delivered against power dissipated4.33e-16 apart · 48.07 mWsolved, then checked — 6 elementsa linear network has no edge: this one is exact Networks, and how a solve is checked

What a network answers, and how the answer is checked

A circuit has exactly one answer and a matrix finds it. The part that matters is not that the answer exists but that it can be checked twice, by routes that share no arithmetic — and that a circuit with no answer is refused by name rather than returned as a large plausible number.

 -60-40-200gain (decibels)the sketch: flat, then −20 dB/decade−3.01 dB-90-450101001k10k100kfrequency (hertz)phase (degrees)sketch within 0.1 dB below 152 Hzsolved, then checked — checked against a chain-matrix productthe sketch is 3.01 dB wrong at 995 Hz Frequency, which is the same solve

One solve, read four ways

Reactance, phase, the corner frequency and the roll-off are not four ideas. They are four readings of one complex number, obtained from the same matrix that answers direct-current questions — and the straight-line sketch every engineer draws of them is itself a model, three decibels wrong exactly where it is read.

 00.50011.50024output (volts), for a 1 V step inthe final valuesolid: from the poles · dashed: stepped forwardtime (milliseconds) above · the same span as a fraction, belowgap between the two routes (volts)1e-71e-61e-51e-41.0m10m1.0e+2m1solved, then checked — residues against 500 trapezoidal stepsthe numerical route is out by 1.7e-3 V Before the steady state

One step, computed twice

A step response from the poles is exact. The same step walked forward in time is not, and the difference between them is the trapezoidal rule's own error rather than anything about the circuit. It falls by a factor of four every time the step is halved, which is a claim about a method and can be watched.

 -90-60-3001001k10kfrequency (hertz)gain (decibels)ButterworthChebyshevBesselhalf power1.00 kHzthe passband, magnified-1-0.500000.2000.4000.6000.8001solved, then checked — three networks, 133 frequencies eachall normalised to a measured −3 dB at 1.00 kHz Filters, measured not tabulated

Three families, one corner

Butterworth is flat, Chebyshev is steep, Bessel has good delay. None of those is a number, so the table they appear in cannot answer the question anybody has. Here each family's poles are computed from its definition, built as an actual network, and then measured — starting with the step every comparison skips.

 -40-200204060loop gain (decibels)unity loop gaincrossover 5.73 kHz46.1 dB of gain margin-180-135-901101001k10k100k1M10M100Mfrequency (hertz)loop phase (degrees)−180°34.9° of marginsolved, then checked — the loop cut and injected34.9° of phase margin at 5.73 kHz Feedback, and the margin

What is left at crossover

A feedback loop is stable or not according to one number read at one frequency — how much phase remains before −180° at the point where the loop gain passes unity. The loop gain here is obtained the way it is obtained on a bench: cut the loop, drive one side of the cut, and measure what comes back to the other.

 02461001k10k100k1M10Mload resistance across the output (ohms)output voltage, solved with the load in place6.0 V with nothing connected1% low at 495 kΩthe circuit12 VR₁R₂R_Lsolved, then checked — the load swept over six decadesthe ratio is 1% wrong below 495 kΩ Networks, and how a solve is checked

The divider, and the thing it does not know about

A two-resistor divider's output is set by the ratio of its resistances — with nothing connected. Connect anything at all and what decides the answer is the quantity the ratio was built to discard: the magnitude. Two dividers of identical ratio give six volts and one volt into the same load.

 realimaginaryacross Racross Lacross Cthe source, 1 Vmagnitudes|v_R| = 1.000 V|v_L| = 2.128 V|v_C| = 2.128 Vsum 5.255 Vvector sum 1.000 Vsolved, then checked — one solve at 1.59 kHzsteady state only: 3 cycles to settle Frequency, which is the same solve

Three voltages that close on one, and the steady state they assume

Kirchhoff's voltage law drawn as a polygon in the complex plane. The three element voltages of a series circuit add head to tail to the source exactly — while their magnitudes add to five times it. And the whole picture is a statement about a settled circuit, which takes a computable number of cycles to arrive.

 the step this produces00.50011.5001234σjωζ = 0.3000ω₀ = 1592 Hzsolved, then checked — poles by rooting the determinantnatural frequency recovered to 6 digits Before the steady state

Where the behaviour is written down

Two numbers in the complex plane contain everything a second-order circuit will ever do. Their distance from the origin is the natural frequency, the cosine of their angle is the damping — and the fastest-settling circuit is not the critically damped one, which is the case the textbooks name.

 passband deviationdecibels, peak to trough below 0.8 f_cButterworth0.443 dBChebyshev0.500 dBBessel1.882 dBattenuation at three times the cornerdecibels downButterworth47.7 dBChebyshev64.0 dBBessel28.3 dBgroup-delay variation across the passbandper cent, slowest against fastestButterworth48.0%Chebyshev49.0%Bessel0.1%solved, then checked — nine measurements, three networksevery number here moves with the order Filters, measured not tabulated

What a steep skirt costs

A filter's order buys attenuation at a known rate — twenty decibels per decade per pole, and no arrangement of components changes it. What varies between families is how quickly the slope is reached, and the currency it is paid for in is delay: the steepest of the three distorts delay eight hundred times more than the gentlest.

 0501001500200400600closed-loop output (volts) for a 1 V stepthe 100× the divider asks for35.1% overphase margin, measured two waysfrom the loop gain34.9°from the overshoot35.0°apart by 0.1° — the relation assumes two poles and this loop has threesolved, then checked — margin against overshootthe second-order relation is 0.1° out here Feedback, and the margin

Two measurements of one margin

A phase margin is computed from the loop gain in the frequency domain, without ever looking at a step. An overshoot is measured from the closed-loop step response in the time domain, without ever looking at a Bode plot. Inverting the standard relation on the second returns 34.9° against the first's 34.9°, and the residue is the third pole.

 1101001k10k100k1M10M100M1Gfrequency (hertz)impedance looking into 10.0 cm of track (ohms)the lumped model: one L, one C1° long at 3.97 MHza tenth of a wavelength at 143 MHzthe 200 Ω at the far endsolved, then checked — the line against a two-element modelKirchhoff's laws run out at 143 MHz Where the models stop

Kirchhoff's own frequency

The current law says the current entering a node equals the current leaving it at the same instant, which assumes the signal crosses the circuit in no time. It crosses at about two-thirds the speed of light, so the law has a frequency of its own — set by nothing but the physical size of the board.

 00.200.400.600.8011001k10kfrequency (hertz)fraction of the source across the resistorhalf the power198.9 Hz measuredresonance 1.59 kHzsolved, then checked — half-power points by bisectionf₀/Q predicts 198.9 Hz — exactly Frequency, which is the same solve

Resonance, and the bandwidth it sets exactly

The half-power bandwidth of a resonant circuit is f₀/Q — not approximately, but to every digit the arithmetic has, which is rare enough to be worth checking. What is not exact, and is drawn as though it were, is the idea that the band sits centred on the resonance. At a quality factor of one its middle is twelve per cent above.

 00.5011.521001kfrequency (hertz)group delay (milliseconds)ButterworthChebyshevBesselthe corner, 1.00 kHzsolved, then checked — −dφ/dω on the unwrapped phaseflat magnitude is not flat delay Filters, measured not tabulated

Flat magnitude, unflat delay

A filter that passes every frequency in its band at the right amplitude and the wrong time has not passed the signal. Group delay is the measurement that says so, it is absent from the classical comparison, and it varies by fifty per cent across the passband of the two families everybody uses.

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