A tolerance is an arrow

The circle a peak sits on drew a series RLC low-pass’s pole pair in the complex plane and read the response off it as $\omega_0^2$ over the product of two distances — from the point on the imaginary axis to each pole. The peak was where a circle through the pair met the axis, the quality factor was the pair’s angle, and the half-power width was twice the pair’s distance from the axis. Every part in that essay was exact.

No part is exact, and the question a designer actually asks of a resonant network is what its parts’ tolerances do to the response. The usual answer is a table of sensitivities computed by differentiating a transfer function, one derivative per part, and it is correct and opaque. The plane offers a different reading. A change in a part moves the pair; the response at each frequency changes by how much that movement lengthens or shortens the two legs; and the two things that decide the answer — which direction the part moves the pair, and how long the legs are — can be seen. This essay moves each part by one per cent on the netlist, solves the moved pair, and checks the picture against the solve.

Three parts, three directions #

The network is the earlier essay’s at $Q = 5$: 10 mH, 100 nF and a resistor of 63.2 Ω, with the undamped frequency $\omega_0 = 1/\sqrt{LC}$ at 5.03 kHz and the pair at $-0.1 \pm j0.995$ of $\omega_0$. Each part is raised by one per cent in turn, the transfer function is rebuilt from the moved netlist, and its poles are found again. Nothing about how the poles should move is assumed; the movement is the difference between two sets of roots.

The capacitor moves the pair 49.88 × $10^{-4}$ $\omega_0$ straight down, and not at all sideways. That is the arithmetic of the two poles’ coordinates: the distance from the axis is $\sigma = R/2L$, which has no capacitance in it, so a change in $C$ leaves $\sigma$ alone and changes only $\omega_0$ — down by 0.496 per cent, half the capacitor’s change — and with it the pole’s height. The pair slides parallel to the imaginary axis.

The resistor moves it 10.00 × $10^{-4}$ $\omega_0$ away from the axis, and 1.01 × $10^{-4}$ down: round the circle. $\omega_0$ has no resistance in it and does not move at all, so the pair keeps its distance from the origin and changes only its angle, towards the real axis. Near the top of the circle, where a $Q = 5$ pair sits, round the circle is nearly horizontal.

The inductor does both: it is in $\omega_0$ and in $\sigma$, and it moves the pair 48.88 down and 9.90 towards the axis. Raising $L$ lowers the frequency like raising $C$ and raises the quality factor like lowering $R$.

The response change is the arrow projected on the legs #

The response is $\omega_0^2/(d_1 d_2)$, so its logarithm changes by the change in $\ln\omega_0^2$ less the fractional changes in the two legs. A leg’s length changes, to first order, by the component of the pole’s displacement along the leg, and its fractional change is that component over its length. The prediction needs no transfer function: the displacement, the two legs at each frequency, and a dot product.

For the capacitor at $Q = 5$ the solved change — the ratio of the two responses — and the projected one agree to within 3.9 per cent of the largest change, over 401 frequencies from half of $\omega_0$ to one and a half. The largest change is −3.20 per cent at 1.093 of $\omega_0$, just above the peak. For the resistor the largest is −0.99 per cent, at the peak itself, and for the inductor −2.73 per cent at 1.115. The residual disagreement is the second-order term that a one per cent change leaves: a one per cent move is small against the long leg and not quite small against the short one.

A one per cent part has become a three per cent change in response, and where in the band it appears depends on which part it was. The two features of the picture that decide that are the next two figures.

Drawn together, the three are two shapes. The capacitor and inductor make an S: +1.78 per cent at 0.893 of $\omega_0$ and −3.20 at 1.093 for the capacitor, +2.29 and −2.73 for the inductor, crossing zero near the peak where the response is only lowered by the overall one per cent of $\omega_0^2$. That is a frequency shift: the response has slid down the axis by half a per cent, so the side it slid towards rises and the side it slid away from falls, and on a resonance the slopes are steep. The resistor makes a dip centred on the peak and nothing else, because it changes the peak’s height and leaves its position.

Where each direction is felt #

The shapes come from the angle of the short leg, and that part of the argument needs no solve at all.

At the peak the short leg is horizontal — the point on the axis is level with the pole — so a pole moving horizontally moves straight along it and changes its length in full, while a pole moving vertically moves across it and changes its length not at all. A quarter of the way round, at the half-power points $\omega_d \pm \sigma$, the leg is at forty-five degrees and a vertical move has half its effect, with opposite signs on the two sides because the leg points up on one side and down on the other. The effect per unit of movement is $1/\sigma$ for the horizontal move at the peak and $1/2\sigma$ for the vertical move at the band edges: 10 and 5 per unit of $\omega_0$ for this pair, whose distance from the axis is a tenth of $\omega_0$.

That is the whole reason a capacitor’s tolerance and a resistor’s are felt in different places. The resistor moves the pair horizontally and is felt where the short leg is horizontal — at the peak. The capacitor moves it vertically and is felt where the short leg has a vertical component — at the band edges — and not at the peak, where its effect is only the one per cent it takes off $\omega_0^2$. A filter whose specification is a peak height is a filter whose resistors matter; one whose specification is the frequencies of its band edges is one whose reactive parts matter, and the plane says why in a single glance.

Proportional to Q #

The legs’ lengths decide how large the effect is, and the short one’s length is set by the quality factor.

A one per cent capacitor moves the pair by half a per cent of $\omega_0$ whatever the quality factor, because $\omega_0$ goes as the inverse square root of $C$. The short leg that displacement is measured against is $\sigma = \omega_0/2Q$ long. So the largest change grows in proportion to $Q$: 1.32 per cent at $Q = 1$, 3.20 at 5, 5.56 at 10 and 10.11 at 20 — rising by 0.46 points for each unit of $Q$ between 10 and 20, close to the half point the geometry predicts, the rest being the long leg and the overall change in $\omega_0^2$. A $Q = 20$ resonator built from one per cent capacitors has a ten per cent tolerance on its response at the band edges, with no part out of specification.

The resistor behaves differently, and the plane says why too. It sets $\sigma$ itself, so a one per cent change in the resistor is a one per cent change in the short leg’s length at the peak, whatever that length is. Its largest effect is 0.99 per cent at every quality factor in the sweep. The resistor’s tolerance is transmitted, not amplified; the capacitor’s is amplified by $Q$.

The tolerance that can only take away found a doubly terminated ladder filter’s response stationary at its ripple peaks, so that no small component change could raise it there. The single pair has no such protection: at its peak the capacitor’s first-order effect is small only because the displacement is perpendicular to the short leg, not because the network is designed to be stationary. It is the same geometric fact — a displacement perpendicular to the leg changes its length only to second order — occurring here by accident of direction rather than by construction.

Two hundred builds #

The single-part changes add up to a population, and the picture predicts its spread.

Two hundred builds with each part drawn uniformly within one per cent give a spread in the response — its standard deviation at each frequency — that peaks at ±2.46 per cent at 1.100 of $\omega_0$, beside the peak, and dips to 0.83 per cent at $\omega_0$. The prediction from the three single-part changes, each scaled to a uniform part’s standard deviation of $1/\sqrt3$ per cent and added in quadrature, lies on the measured spread to within two per cent of its largest value. The spread is the displacements, added.

Its shape is the S’s shape folded: largest at the band edges, where the capacitor’s and inductor’s vertical movements are felt most, and least at the peak, where only the resistor’s horizontal movement and the small change in $\omega_0^2$ remain. A test that measures such a resonator’s response at its peak will find it within a per cent from build to build, and one that measures at the band edge will find it within five — from the same parts. The tolerance that is not on any part found a divider’s measured spread well inside its worst case; here the spread’s size depends on where along the axis it is looked for, which is a second way the tolerance is not on any part.

What the arrow says to buy #

Read as a specification, the capacitor’s law is a rule for choosing parts. The largest change at the band edges is close to half the capacitor’s tolerance times $Q$, so a response held to one per cent at its band edges needs capacitors to about $2/Q$ per cent: two per cent at $Q = 1$, 0.4 at $Q = 5$, 0.1 at $Q = 20$. The resistor’s rule is $Q$-independent — its tolerance arrives at the peak undivided and unamplified — so at high $Q$ the reactive parts are the whole of the problem and the resistor is a detail, which is the opposite of the intuition that a resonance’s height is its fragile feature.

The rule also says where to spend a trim. A single adjustable capacitor moves the pair vertically and so can put the band back where it belongs, cancelling the drift of every reactive part at once, because they all move the pair in the same direction; it cannot restore the height, which the resistor and the parts’ losses set. The q the components allow found the height capped by the worst component’s own quality factor — a horizontal displacement no trimmer reaches, since a capacitor’s loss moves the pair away from the axis rather than along it. A tuned circuit is therefore adjusted in one direction and specified in the other, and the plane shows both directions on one drawing. Resonance, and the bandwidth it sets exactly measured the width as $f_0/Q$; the arrow says the width is set by the horizontal coordinate and survives any vertical move, so a tuning trim changes where a band is and not how wide.

Two derivatives, one arrow #

The derivative the plane is using is the pole’s, and a response’s derivative is built from it. The derivative of a root distinguished the two and found them disagreeing about which of two realisations of one filter was better — by eight orders of magnitude on the response and a factor of 2.17 on the poles. The arrow shows how both can be right. The pole’s derivative is the arrow’s length and direction. The response’s derivative is that arrow projected onto legs whose lengths vary along the axis: short near a pole and long elsewhere. A realisation whose poles move little can still have a response that moves a great deal, if its poles are close to the axis and the arrow happens to point along a short leg.

So pole sensitivity alone cannot rank realisations, and response sensitivity alone cannot say why one ranks above another. The arrow and the legs together do both, and for a single pair they can be drawn.

How the numbers were obtained #

Each network is a netlist: a unit source, a resistor of $\sqrt{L/C}/Q$, 10 mH and 100 nF, output across the capacitor, with one part multiplied by 1.01. Each transfer function is built by sampling the circuit’s determinants, and its poles are the numerical roots of its denominator, so each displacement is the difference between two sets of computed roots. The solved change in response is the ratio of the two transfer functions’ magnitudes at 401 frequencies from 0.5 to 1.5 of $\omega_0$. The predicted change is $2\,\delta\omega_0/\omega_0$ less each leg’s fractional change, each leg’s change being the dot product of the pole’s displacement with the leg over the leg’s length squared, the lower pole moving as the mirror image of the upper. The where-it-is-felt figure uses the geometry alone, with no solve. The population draws each part uniformly within one per cent from a seeded generator and builds two hundred networks.

What it leaves out #

Large changes. Everything is first order in the displacement. A five per cent capacitor at $Q = 20$ moves the pair by more than its distance from the axis, and then the short leg changes direction as well as length, and the projection is no longer a good picture.

Parts that move together. The population draws the three parts independently. Parts from one reel, or one temperature coefficient shared by two, move the pair in a correlated direction, and the spread is then the projection of a correlated cloud of arrows — which, for a capacitor and inductor tracking in opposite senses, can cancel the frequency shift entirely.

Zeros. The low-pass has no zeros to displace. A notch’s sensitivity is the same construction with a leg in the numerator, and a zero that moves off the axis fills the notch in, which is a displacement felt at exactly one frequency.

Still open: tracking parts, the notch’s zero, and a ladder filter’s many arrows #

Parts that track. A capacitor and an inductor whose values move in opposite senses with temperature keep $LC$ constant and move the pair horizontally only — round the circle — trading a frequency tolerance for a height tolerance. How much tracking converts one into the other, and whether a filter can be designed so its parts’ dominant drift moves every pole along the direction each is least felt in, is the arrow picture used as a design rule.

A zero’s arrow. At the bottom of a notch the response is a short leg in the numerator. What actually fills a null found a ten per cent error in a notch’s arm leaving the null three hundred decibels deep — moved, not filled — and loss filling it. On the plane that is the difference between a zero’s arrow along the axis, which moves the null and keeps its leg reaching zero, and an arrow off it, which leaves a floor as deep as the arrow is long. Drawing the arrows for each part of a real notch would say which of its tolerances are harmless and which are not, in one picture.

A ladder filter’s arrows. A fifth-order filter has five poles and five arrows per part. The LC ladder realisation’s famous insensitivity should appear as arrows that lie along directions the nearby legs do not feel, which is a picture of the three tolerances that do nothing found as eigenvectors.