The phase a decibel buys

Every reader of a Bode plot carries an exchange rate between its two halves: twenty decibels a decade of slope goes with ninety degrees of phase. The phase the magnitude already knows showed where that rate comes from. For a network with no zeros in the right half-plane, the phase at any frequency is a weighted average of the magnitude’s slope over every frequency, and the rule of thumb is that average taken over a slope that never changes.

A rule stated at one frequency is awkward to use, because real slopes do change, and a designer rarely wants the phase at a single frequency as much as the answer to a blunter question: how much phase can a network supply at all, and what does it cost? The same integral answers that more cleanly than it answers the pointwise one. Integrate the phase over every frequency and the weighting drops out completely. What is left depends on one number, which is how far the gain moves from the bottom of the spectrum to the top.

Ninety degree-decades for every twenty decibels #

The unit needs saying once. A Bode plot draws phase in degrees against frequency on a logarithmic axis, so the natural measure of area on it is degrees multiplied by decades. Ninety degrees held for one decade encloses ninety degree-decades, and so does forty-five degrees held for two.

The network in the first figure is the lead from the phase the magnitude already knows: nine kilohms with a capacitor across them, feeding one kilohm to ground. Its gain climbs from −20 dB at low frequency to 0 dB at high frequency, and its phase rises from nothing to a peak of 54.903° at the geometric mean of its two corners and falls back to nothing again. That peak is $\arcsin(9/11)$ to every digit printed, on the solved network.

The area under the hump, integrated on the solved phase out to seven decades beyond each corner, is 90.000 degree-decades. The statement being measured is

$$\int \varphi \; d(\log_{10} f) \;=\; 90^\circ \times \frac{\Delta G}{20\ \text{dB}}$$

where $\Delta G$ is the gain at the top of the spectrum minus the gain at the bottom. It holds for any network whose phase is fixed by its magnitude and whose gain is finite and non-zero at both ends. Twenty decibels of rise buys ninety degree-decades of lead; forty buys a hundred and eighty. A network whose gain falls instead of rising has the same area on the other side of zero.

Nothing about a lead of ten to one is special. Pulled down to a ratio of two, the same section rises 6.00 dB, its phase peaks at 19.41°, and the area under it is 27.000 degree-decades. At 3.16 to one it rises 10.00 dB, peaks at 31.30° and encloses 45.000. The peaks are not in proportion to the gain — a lead of twice the decibels does not reach twice the phase — and the areas are, to every digit shown.

Why the area forgets the shape #

The reason is in the integral that ties the phase to the magnitude. With $M$ the natural logarithm of the magnitude and $u$ the natural logarithm of frequency, Bode’s relation writes the phase at $u_0$ as

$$\varphi(u_0) = \frac{1}{\pi}\int \frac{dM}{du}\,\ln\coth\frac{|u-u_0|}{2}\,du .$$

The weight depends only on the distance between the frequency being asked about and the frequency whose slope is being counted. Integrate the phase over every $u_0$ and each piece of slope is counted exactly once against the whole of the weight, whose total over both sides is $\pi^2/2$ — twice the $\pi^2/4$ that makes a constant slope worth exactly ninety degrees. So

$$\int \varphi\,du_0 = \frac{1}{\pi}\cdot\frac{\pi^2}{2}\int\frac{dM}{du}\,du = \frac{\pi}{2}\,\bigl(M(\infty)-M(0)\bigr).$$

The integral of a slope is the change in the thing that slopes. Every detail of where the slope sat — sharp or gentle, in one place or scattered over ten decades — integrates away, and only the two ends survive. Converted from radians to degrees and from natural logarithms to decades, $\pi/2$ per neper becomes ninety degree-decades per decade of magnitude, which is ninety per twenty decibels.

So the famous rule and this one are two readings of a single number. The rule of thumb is the weight’s total applied at one frequency to a slope that never changes. The area theorem is the same total applied to the whole curve, and it needs no assumption about the slope at all. The rule is an approximation everywhere except on an idealised straight line; the area is exact for every minimum-phase network there is.

The peak is free, and the arrangement decides it #

Fix the gain change at forty decibels and ask what can be done with it.

As a single section of a hundred to one, the forty decibels produce a broad, low hump. The phase climbs for two decades, peaks at 78.579°, and needs another two decades to come down. The same forty decibels as two sections of ten to one, sitting on top of each other, produce something quite different.

Stacked, the two leads peak at 109.806°, which is exactly twice the 54.903° of one of them, because their peaks coincide. The hump is taller and narrower than the single section’s, and its area is 180.000 degree-decades — the same, to the third decimal place.

Moving the second lead upward in frequency, with the slider, shows the trade without anything else changing. At 3.16 times apart the peak is 101.13° at 5.62 kHz. At ten times apart it is 78.58° at 10.0 kHz, which is the single hundred-to-one lead exactly, and not by coincidence: the first lead’s pole now sits on the second lead’s zero, the two cancel, and what remains is one zero at 1 kHz and one pole at 100 kHz. At 31.6 times apart the peak is 60.65°; at a hundred, 56.59°. The area is 180.000 degree-decades at every setting.

Three decades apart the two sections hardly know about each other. There are two humps, each close to a single lead’s shape, and the taller peaks at 55.067°, lifted above 54.903° by the other lead’s distant tail. The area between them is still 180.000 degree-decades.

That is the whole of what the arrangement of a network’s corners decides about its phase: the height of the phase against its width. A lead meant to hold up the phase at one crossover frequency wants its area concentrated, and a network meant to hold a phase over a band wants it spread. Neither choice changes what is spent. The area is decided by the gain the network gives up, and it was decided before any corner was placed.

Even the straight-line phase sketch honours this, which is a small vindication of a construction the straight lines, and where they are not the curve spends its time finding fault with. Each corner’s sketch is a ramp of forty-five degrees a decade, antisymmetric about the corner exactly as the true curve is, so the sketch of a ten-to-one lead encloses ninety degree-decades too. What it gets wrong is the peak: its two ramps overlap for only one decade, and at 3.16 kHz they give 45° against a solved 54.903°.

A ceiling for real leads, and the pair that walks through it #

Stacking more coincident leads raises the peak further. One section of a hundred to one peaks at 78.579°, two of ten to one at 109.806°. The same forty decibels shared between eight sections pushes it further still.

Eight leads of 1.78 to one peak at 130.144°, and that is close to as far as real corners can go. A single lead of ratio $r$ peaks at $\arcsin\bigl((r-1)/(r+1)\bigr)$, and that is always less than $\tfrac12\ln r$ radians. When several leads are combined their peaks add at best, and only when they coincide, while the logarithms of their ratios add to the logarithm of the total gain change. So real leads sharing a rise of $\Delta G$ decibels can never peak above $\Delta G \cdot \ln 10/40$ radians, however they are arranged. For forty decibels that is 131.93°. Every figure stays under it, and eight sections get within 1.8° of it.

That ceiling is a property of real corners, not of the area. A zero pair and a pole pair, each with a quality factor of twenty and placed half a decade apart, rise by the same twenty decibels as the single ten-to-one lead of the first figure — a pair falls or rises forty decibels a decade, so half a decade is all twenty decibels needs.

Its phase peaks at 175.291°, more than three times what the real lead reached and far through the 65.96° ceiling that binds every real lead with the same rise. And the area is 90.000 degree-decades, the same as the lead’s.

Both statements hold because the shape is a rectangle. The zero pair turns the phase through nearly 180° in a narrow band around 1 kHz, the pole pair turns it back around 3.16 kHz, and between them the phase sits close to 180° for half a decade. A rectangle 180° tall and half a decade wide is ninety degree-decades. The high quality factor has not bought extra phase; it has bought square corners, and the gain changes needed to put those corners in place are the same twenty decibels as before. The gain panel shows what that costs elsewhere, since a pair at a quality factor of twenty is a deep notch and a tall resonance rather than a gentle step, and no network meant to add phase to a loop would be built that way. That is a judgement about practicality, not about the theorem, which is exact for both.

Phase that has to be given back #

If the area is fixed by the difference between the two ends of the gain, a network whose gain ends where it started has none.

The lead of the first figure followed by a lag of the same ratio, two decades higher, gives the gain back. The phase still rises — to 53.320°, a little short of the lead’s own peak because the lag’s tail already reaches down into it — and then it goes negative, and the two lobes are 76.00 and −76.00 degree-decades. Their sum is 0.000. Each lobe is short of ninety because the two overlap and each cancels part of the other; their total cannot be anything but zero.

This is the general form of a fact every designer of compensated networks meets in a particular form. Phase lead at one frequency without a net change of gain is phase lag of exactly equal area at another. A lead-lag network moves phase from one band into another. It cannot add any.

A divider with two ratios is the most familiar instance. A resistive divider with capacitance across both arms divides by its resistances at low frequency and by its capacitances at high frequency, and when the two ratios differ the divider is a lead or a lag of exactly that mismatch, with a phase area of ninety degree-decades for every twenty decibels of it. When the trimmer makes the ratios agree, the magnitude is flat. For a minimum-phase network a flat magnitude has no slope anywhere for the phase to be built from, so the phase is zero at every frequency too. The square wave on the front panel of an oscilloscope is flat when both are, and the adjustment that flattens the gain has flattened the phase along with it.

What a lead compensator is paying for #

Stable and unstable with less gain states the price of a lead section plainly: it attenuates at direct current by exactly the ratio of its pole to its zero, so two sections of ten to one give up a factor of a hundred before the loop does anything at all. The area theorem turns that sentence into an exchange rate. A factor of a hundred is forty decibels, and forty decibels is 180 degree-decades of phase. That is all the phase those two sections can supply, and where it lands is the only thing the designer chooses.

A phase margin is read at one frequency, so the useful part of the purchase is only what lands near crossover. The two sections of the hero figure put 109.806° at the frequency where their peaks coincide and the remaining area in the tails, where it props up a loop that does not need it. Spread across three decades they would put 55° at each of two places. The loop decides which is better, while the total spent is fixed.

The other side of the account is the poles. What is left at crossover measures how far one reaches: about −6° a full decade below itself and −84° a decade above. A pole’s gain falls without limit, so its phase area has no bound. Every loop has more of them than it wants, and every lead is a finite loan of phase set against an unbounded debt.

Sometimes the loan arrives unasked for. Two requirements pulling one capacitor finds a regulator whose margin exists only because the output capacitor’s series resistance puts a zero into the loop gain. That zero is the rising half of a lead whose own pole comes later, and the phase it contributes near crossover is paid for, at the same rate, by the gain it adds above that frequency until the pole takes it back.

The networks that have no area to measure #

Everything above assumed that the phase is fixed by the magnitude. The first network drawn to show where that assumption fails is the one that shows what the area theorem becomes without it.

Behind an all-pass, the gain is untouched: it still rises by 20.000 dB. The phase goes down to −180° and does not come back — −179.37° by 1.00 MHz — so the area under it depends only on where the sweep is stopped, and each decade more adds −180.0 degree-decades. There is no number for it to be compared with. That is the other face of the minimum-phase result: the phase a magnitude cannot see is also a phase with no fixed total, and excess phase is exactly the part of a network’s phase for which the area statement has nothing to say.

The same is true of every way out of the minimum-phase class, and each is ordinary. Flat delay, bought with more delay uses all-pass sections on purpose, because leaving the class is the only way to change a filter’s delay without touching its magnitude. The staircase in time is a metre of cable, 4.83 nanoseconds of pure delay, whose phase grows in proportion to frequency; there the area of each decade is ten times the decade before. And the frequency a device sets for itself finds a right-half-plane zero in a common-emitter stage — gain that rises like a zero’s, with a phase that falls like a pole’s.

That gives the area a second use besides accounting. For a network measured over a wide enough band, comparing the phase area with ninety degree-decades per twenty decibels of measured gain change is a test of whether the network is minimum phase. The comparison is not exact on a finite sweep, since a truncated area is short by whatever its tails carried, and that shortfall shrinks as the span widens. But a network hiding an all-pass or a delay does not disagree by a tail. It disagrees by an amount that grows with every decade added, and that is a signature a truncated measurement cannot fake.

Still open: the level the phase cannot know, and the energy that arrives first #

Two continuations follow from the subject, and each has an argument of its own.

The magnitude from the phase, up to one constant. The integral in this essay runs from magnitude to phase, and a companion runs the other way: given the phase at every frequency, the log-magnitude of a minimum-phase network is fixed everywhere except for one additive constant. The area theorem already says half of this — the phase knows the gain change, ninety degree-decades per twenty decibels — and the companion says the other half, that the phase cannot know where the gain starts. Measuring that recovery on the lead of the first figure would find the shape of the gain right to the precision the sweep allows, and the level wrong by exactly the one number no phase measurement contains.

Minimum phase as minimum delay, in time. A minimum-phase network is named for lagging least among all networks that share its magnitude, and that has a time-domain face that no frequency plot shows. Two networks with identical magnitude and therefore identical total impulse-response energy do not deliver that energy on the same schedule. The one with its zeros in the left half-plane has delivered more of it by every instant, and mirroring a zero into the right half-plane makes its step response start in the wrong direction. Both can be measured on two solved networks that share a magnitude to the last bit — the delivered fraction of energy against time, and the undershoot — and that is an argument about when a network responds rather than about how much phase it has.