Circuit Response
every model has an edge, and the figure carries it
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The thread: The constants decide

A divider's ratio does not predict what happens when something is connected to it — its magnitude does. A filter's family does not say how much delay distortion it costs — a measurement does. The quantity that gets discarded to make a rule memorable is repeatedly the one that decides the outcome.
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.

 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.

 00.2500.5000.750105101520time (microseconds)output, divided by the size of its own step20 mV step1.0e+2 mV step5.0e+2 mV step2 V step8 V steplinear below 79.6 mVsolved, then checked — integrated with the rate limitscaling fails above a 79.6 mV step Before the steady state

The step that is too big

A linear circuit scales — double the input and the output doubles, exactly. A real amplifier does not, because its output can only move at a fixed rate, and the amplitude at which the two stop agreeing is about eighty millivolts for an ordinary part. No transfer function contains that number, because no transfer function can.

 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.

 010203040501101001k10k100k1Mfrequency (hertz)closed-loop gain (decibels)the ideal amplifier: two resistors, no frequencythe circuit+−1% low at 1.35 kHz3 dB down at 10.0 kHzsolved, then checked — a nullor against a real devicethe ideal answer is 1% wrong above 1.35 kHz Feedback, and the margin

The ideal amplifier, and where it stops being one

An ideal operational amplifier's closed-loop gain is set by two resistors and nothing else — a horizontal line at every frequency. The real one is already a tenth of a per cent low at direct current, one per cent low by 1.35 kHz, and above 10 kHz has no loop gain left, at which point the ideal answer is not an approximation to anything.

 1.0m10m1.0e+2m110100100m110100drive amplitude (millivolts)how much the linear model understates the gain (per cent)1% understated10% understated1% at 7.3 mVV_T = 25.9 mVsolved, then checked — the Bessel ratio from its seriesthe tangent is 1% wrong above 7.3 mV Where the models stop

How small is small signal

Linearising an exponential replaces a curve by its tangent, which is exact at a point and progressively wrong away from it. The amplitude at which it is one per cent wrong is 7.3 millivolts at room temperature — 28 per cent of the thermal voltage, not a small fraction of it, and a good deal smaller than "small signal" suggests.

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