The circle a peak sits on

Four readings of one distance took a single pole off the frequency axis and found the four readings of a response — reactance, phase, corner and roll-off — to be one geometric fact: the magnitude is the pole’s distance from the origin over its distance from the point $j\omega$ on the imaginary axis, and the phase is the angle that separation subtends. It then described, in words, what a second pole would add. A pair near the axis makes a peak, because as $j\omega$ passes the pair one of the two distances becomes small; how small is how close the pair is to the axis, which is the damping. And it left the pair undrawn, with a remark that a figure of it would make the resonance essays’ quality factor a geometric quantity, which no algebra makes as obvious.

This essay draws it. The circuit is the simplest pair a netlist has — a series resistor, inductor and capacitor, taken as a low-pass across the capacitor — and its poles are solved from the netlist rather than written down. What comes out is more than a restatement: the frequency of the peak is the crossing of a particular circle with the axis, the quality factor is an angle with a forty-five degree threshold in it, the bandwidth is a length that can be seen, and the same two points read by projection instead of by distance give the ringing in time.

A product of two distances #

The network is 10 mH and 100 nF with a resistor chosen for the quality factor on the slider; its undamped frequency is $\omega_0 = 1/\sqrt{LC}$, 5.03 kHz. The transfer function is built from the netlist by sampling its determinant, and its two poles come from the roots. At $Q = 2$ they sit at $-0.2500 \pm j0.9682$ in units of $\omega_0$: on the circle of radius $\omega_0$, which is what $\omega_0$ means, and at a distance $1/2Q$ of that radius from the imaginary axis, which is what $Q$ means.

With two poles the single-pole construction multiplies. The response at $j\omega$ is

$$|H(j\omega)| = \frac{\omega_0^2}{d_1\,d_2},$$

with $d_1$ and $d_2$ the distances from $j\omega$ to the upper and lower pole, and $\omega_0^2$ the product of the poles’ own distances from the origin, which sets the response to one at direct current. Checked against the nodal solve at 241 frequencies over four decades, the product of distances agrees to a part in $10^{15}$ — the solve’s own rounding. Every statement below is a statement about two lengths.

For a point low on the axis both legs are about $\omega_0$ long and the response is about one. For a point far up the axis both legs are about $\omega$ and the response falls as $\omega_0^2/\omega^2$, forty decibels a decade. In between, the point passes the upper pole, the upper leg becomes short, and if it becomes short enough the product falls below $\omega_0^2$ and the response rises above one. The lower leg is then nearly twice $\omega_0$ and changes slowly. The peak is a story about the short leg.

The peak is where a circle meets the axis #

Where exactly is the product least? Write the pole pair as $-\sigma \pm j\omega_d$. The point $j\omega$ on the axis makes a triangle with the two poles, and the product of the two sides from $j\omega$ is least where

$$\omega_{pk}^2 = \omega_d^2 - \sigma^2,$$

which has a construction. Draw the circle centred at $-\sigma$ on the real axis — directly between the two poles — with radius $\omega_d$, so that it passes through both. It meets the imaginary axis at $\pm\sqrt{\omega_d^2 - \sigma^2}$: exactly the peak. At $Q = 2$ that crossing is at 0.9354 of $\omega_0$, and a golden-section search on the solved response finds its maximum at 0.935414 of $\omega_0$ with a height of 2.0656, which is $\omega_0^2/(2\sigma\omega_d)$ to nine figures.

The figure’s panel reads the two legs at the peak: 0.2521 and 1.9200 of $\omega_0$, whose product’s reciprocal is 2.0656. The short leg is barely longer than the pair’s distance from the axis, 0.2500; that closeness is the peak.

Move the slider down to $Q = 0.6$ and the construction answers a different question. The pair is now further from the axis than it is high — $\sigma = 0.833$ against $\omega_d = 0.553$ — the circle centred on its real part is too small to reach the axis, and the response has no peak: it falls from direct current. The search on the solve agrees. Whether a response peaks at all is whether that circle reaches the axis.

Q is an angle #

The circle’s reach depends only on the ratio of $\omega_d$ to $\sigma$, which is an angle, and so is the quality factor.

As the resistance falls the pair slides round its circle towards the imaginary axis. Measured from the negative real axis, the angle $\theta$ of the upper pole satisfies

$$Q = \frac{1}{2\cos\theta},$$

because $\sigma = \omega_0\cos\theta$ and $\sigma = \omega_0/2Q$. Where the behaviour is written down put it the other way round — the cosine of the pair’s angle is the damping ratio — and the two statements are one, since the damping ratio is $1/2Q$. The twelve solved pairs sit at 0° for $Q = 0.5$ — both poles meeting on the real axis, critical damping — 24.6° at 0.55, 45.0° at 0.707, 60° at 1, 75.5° at 2, 84.3° at 5 and 87.1° at 10, each to six figures of the formula.

The angle is also the cleanest way to see why a quality factor below a half is not a quality factor at all. Below $Q = 0.5$ the formula would need a cosine above one: the two poles have left the circle and split along the real axis, one towards the origin and one away from it, and the response is two first-order sections in cascade rather than a pair. The circle of radius $\omega_0$ is where complex pairs live, and the real axis is where it ends.

The forty-five-degree line is the threshold the circle construction found. A peak exists when $\omega_d > \sigma$, which is when the pair is further round than forty-five degrees, which is when $Q > 1/\sqrt2$. So the familiar statement that a second-order low-pass is maximally flat at $Q = 0.707$ is a statement that its poles sit exactly on the diagonal: any less damped and the pair is closer to the imaginary axis than to the real one, and the response rises before it falls. The best damping is not the one to build arrived at a damping choice through a step’s settling time; the plane says what the choice is, as a direction.

The height, and the width that is a length #

The peak’s height and its width both come out of the two legs.

The height is $Q/\sqrt{1 - 1/4Q^2}$: 1.155 at $Q = 1$, 2.066 at 2, 5.025 at 5, 20.006 at 20 — to eight figures of the searched maxima at all eleven quality factors from 0.75 to 20. For a sharp pair it is just $Q$, the familiar statement, and the correction is the long leg: it is not quite $2\omega_0$ but $2\omega_d$, which the square root accounts for.

The width is where the geometry is most direct. At the peak the short leg has its least length, which for a sharp pair is $\sigma$, the pair’s distance from the axis. The response has fallen by $\sqrt2$ where the short leg has grown by $\sqrt2$ — where it has become the hypotenuse of a right isosceles triangle with the pair’s distance from the axis as one side — which is at $\omega_d \pm \sigma$. So the half-power width is $2\sigma$, and since $\sigma = \omega_0/2Q$ it is $\omega_0/Q$. The bisected widths approach it: 0.2021 at $Q = 5$, 0.1003 at 10, 0.0500 at 20, in units of $\omega_0$. Resonance, and the bandwidth it sets exactly found the width exactly $f_0/Q$ on a band-pass network; on this low-pass it is only asymptotically so, and the plane says why.

At $Q = 5$ the short leg’s least length is 0.100 of $\omega_0$ at $\omega_d$, and at $\omega_d \pm \sigma$ it has grown by exactly $\sqrt2$ — to twelve figures, since that is Pythagoras. Across the same band the long leg changes by 10.6 per cent. That is why the width is $2\sigma$ only asymptotically: the long leg is not constant, and at a quality factor of a few its change is a noticeable part of the product. At $Q = 1.5$ the bisected width is 0.80 of $\omega_0$ against $1/Q = 0.67$, a fifth wider than the short leg alone would make it.

The same pair, read by projection #

The step response is not a reading along the imaginary axis at all, and the plane gives it by a different operation.

The marched step into the $Q = 5$ network matches $1 - e^{-\sigma t}(\cos\omega_d t + (\sigma/\omega_d)\sin\omega_d t)$ to nine parts in a million. Its ringing is at 5.01 kHz — the pole’s height up the imaginary axis, $\omega_d$, not $\omega_0$ — and its envelope decays as $e^{-\sigma t}$, at the pole’s distance from the axis. Successive overshoots are 0.7292, 0.3878, 0.2062 and 0.1097 of the step: each 0.5318 of the one before, which is $e^{-2\pi\sigma/\omega_d}$ to five figures.

So the frequency response reads the pole’s distances from points on the axis, and the time response reads its projections onto the two axes. The two readings of one diagram were never drawn together in the earlier essays, and they are worth holding together, because the quality factor means the same thing in both: the ratio of the pole’s height to its distance from the axis is, in the frequency domain, how tall and narrow the peak is, and in time, how many cycles the ringing lasts. The decay per cycle, $e^{-2\pi\sigma/\omega_d}$, is about $e^{-\pi/Q}$; the ringing loses a factor of $e$ in about $Q/\pi$ cycles, 1.6 at $Q = 5$. Where the behaviour is written down found the step’s shape fixed by the poles’ positions; the projection is the geometric form of that sentence.

Reading a measured response backwards #

The construction runs in both directions, and the backwards direction is the useful one on a bench. A measured response with a peak of height $h$ at frequency $\omega_{pk}$ fixes the pair completely: the height gives $Q$ through $Q/\sqrt{1 - 1/4Q^2}$, the angle gives the pair’s position round its circle, and the peak frequency gives the circle’s radius through $\omega_{pk} = \omega_0\sqrt{1 - 1/2Q^2}$. At $Q = 2$ the peak sits at 0.935 of $\omega_0$ — six and a half per cent below the undamped frequency — and a designer reading the peak off an analyser and calling it the resonance has mis-stated $\omega_0$ by that much. At $Q = 5$ the peak is at 0.990 of $\omega_0$, one per cent low; at $Q = 20$ it is indistinguishable.

The ringing gives a second, independent reading. Its frequency is $\omega_d$, which at $Q = 2$ is 0.968 of $\omega_0$ — between the peak and $\omega_0$ — and its decay per cycle gives $\sigma$. So three frequencies are commonly called “the resonance” of one pair: the peak of the magnitude at $\sqrt{\omega_d^2 - \sigma^2}$, the ringing at $\omega_d$, and the undamped $\omega_0$ that the components define. On the plane they are the circle’s crossing with the axis, the pole’s height, and the pole’s distance from the origin — three different lengths of one triangle, which coincide only as the pair approaches the axis. Two parasitics, and the resonance neither has measured a resonance made of stray elements; which of the three it measured is the question every such measurement should state.

Why the geometry is worth having #

None of this is new algebra; every closed form above is in any text on second-order systems. What the plane adds is a single picture from which all of them can be read without the algebra, and a way of asking questions the algebra does not suggest.

It says which quantity is sensitive to what. The peak’s height depends on the short leg, which depends on the pair’s distance from the axis; so anything that moves the pair towards or away from the axis moves the peak’s height in proportion, and something that moves it along the axis moves the peak’s frequency without changing its height. The first is damping and the second is tuning, and the plane shows them to be perpendicular.

It says where approximations fail. The width is $\omega_0/Q$ only while the long leg is constant, and the long leg is about $2\omega_0$ long while the width is $\omega_0/Q$ wide, so the approximation is good to a fraction of order $1/2Q$ — a statement about two lengths, readable off the figure.

And it generalises. The phase the magnitude already knows showed that for networks with nothing in the right half plane the phase is fixed by the magnitude; on the plane that is the statement that the legs’ angles and lengths are not independent. A zero is a leg in the numerator. A notch is a zero on the axis, and the corner error a filter hides in its sections is a set of pairs each close enough to the axis to matter.

How the numbers were obtained #

The network is a netlist: a unit source, a resistor of $\sqrt{L/C}/Q$, a 10 mH inductor and a 100 nF capacitor, with the output across the capacitor. The transfer function is built by sampling the circuit’s determinant and its source-substituted determinant on a circle of radius $\omega_0$ and fitting the polynomials; the poles are the denominator’s roots, found numerically, and nothing about their positions is assumed. The magnitude is evaluated from that transfer function and compared with $\omega_0^2/(d_1 d_2)$ at 241 frequencies. The peak is found by golden-section search on the magnitude and the half-power points by bisection on it. The step is marched by the trapezoidal rule at 12,000 steps over ten ringing periods and compared with the closed form; the overshoot ratios are read from the marched trace.

What it leaves out #

Zeros. The low-pass has none, so every leg is in the denominator. A band-pass across the resistor has a zero at the origin, whose leg is $\omega$ itself, and that changes the peak’s position — to exactly $\omega_0$ — and makes the width exactly $\omega_0/Q$; a high-pass has two.

More than one pair. A higher-order filter’s response is the product over all its pairs, and the pair nearest the axis near a given frequency dominates there. The construction applies pair by pair, and which pair owns which part of the band is itself a geometric question.

Anything that moves the poles. Every part here is exact.

Still open: a tolerance as a displacement, the zero’s own circle, and pairs in a chain #

A component’s tolerance, drawn as an arrow. A change in any of the three parts moves the pair, and the response changes by the displacement projected onto the two legs. Since the short leg is $\sigma = \omega_0/2Q$ long, a displacement that is a fixed fraction of $\omega_0$ changes the response near the peak by an amount that grows with $Q$ — and whether a resistor’s tolerance and a capacitor’s are felt in the same place depends on the directions they move the pair. That is the plane’s reading of sensitivity, and it would put the tolerance that is not on any part in the same picture.

The circle for a zero. A pair of zeros near the axis makes a notch, and the notch’s depth and width are the same construction inverted. Where a zero pair and a pole pair share a frequency — an elliptic section — the response is a ratio of two short legs, and its sensitivity is the difference of two displacements.

Pairs in a chain. A cascade of sections is a product of pairs, each with its own circle and its own peak. The overall peak is not at any one section’s crossing, and where it is, is a question the plane can answer by multiplying lengths.