The two resistors a ladder was designed between

A filter design table gives element values. An order-five Butterworth low-pass is 0.618, 1.618, 2.000, 1.618, 0.618, in normalised units, and every textbook prints them.

What the table does not print in the same typeface is that those numbers are values for a two-port between two stated resistances. The five reactances alone are not a filter. They become one when a source of the right resistance drives them into a load of the right resistance, and the response in the table is the response of the seven components together.

That is not a pedantic distinction. Getting it wrong by twenty per cent puts a decibel on the passband.

Six decibels that are not a loss #

The first number a mismatched ladder draws attention to is the one at match: a doubly-terminated filter loses 6.021 dB at direct current, which is exactly half the voltage, and it does so by design.

The reason is the divider. At low frequency the inductors are wires and the capacitors are open, so the source resistance and the load resistance are in series across the source and the load gets half. Nothing about the filter can avoid it, because the filter is not there yet.

Halving a signal deliberately looks like waste, and this is the place to say what it buys. At a passband maximum, a doubly-terminated ladder is delivering all the power the source is able to deliver into any load — it is matched, in the maximum-power-transfer sense. A quantity at a maximum is stationary: perturb any element and the response can only move down, and it moves down as the square of the perturbation rather than in proportion to it.

That is the whole of why a ladder is insensitive. The filters field measures it: one per cent on one component moves a five-pole cascade fifty-four times further at the ripple peaks than it moves the ladder realising the same response. The six decibels are the price of that, and they are not recoverable — a singly-terminated ladder has no insertion loss and no stationarity either.

What a wrong source resistance does #

The figure sweeps the source resistance from a twentieth of the design value to twenty times it, keeping the five reactances and the load exactly as designed, and measures the worst departure of the normalised passband shape.

Three things are going on in that table and they are worth separating, because two of them are usually merged into “the filter is loaded”.

The level moves, and that is the least of it. A source resistance twenty times too high drops twenty-six decibels rather than six, which is obvious and correctable with gain elsewhere.

The shape moves, and that is not correctable. The normalised passband is 2.34 dB out at twice the design resistance and 5.33 dB out at twenty times. No amount of gain in front of or behind the filter recovers a passband that has grown a ripple.

The corner moves, and it moves the opposite way from the intuition. Too high a source resistance lowers the corner, from 1000 Hz to 902; too low a one raises it, to 1300. A ladder driven from a very stiff source is a faster filter than the one that was designed, which is why the mismatch cannot be compensated by rescaling the components either.

The window, and how narrow it is #

Bisecting on the departure gives the band inside which the passband shape survives to half a decibel:

$$0.8857 \le \frac{R_{\mathrm{source}}}{R_{\mathrm{design}}} \le 1.1371$$

That is −11.4 per cent to +13.7 per cent, and it is a startling requirement when it is written next to the way source resistances are usually arrived at. A “50 Ω” generator is 50 Ω to a few per cent and a “600 Ω” line is a nominal figure with no tolerance attached at all. An amplifier’s output impedance is a fraction of an ohm at low frequency and rises with frequency; a preceding filter’s output impedance is its own function of frequency and is nothing like a resistance.

The window is also not symmetric, and the asymmetry runs the wrong way. Going high, the departure settles: from ten times to twenty times it moves only from 4.94 to 5.33 dB, because the ladder is approaching a current-driven limit and stops noticing further increases. Going low, it keeps growing: 15.51 dB at a twentieth, with 7.21 dB of peaking on the passband.

A source that is too stiff is worse than a source that is too soft, which is the opposite of the advice usually given about driving filters. The reason is that the reflections a doubly-terminated ladder is designed to absorb are absorbed in the source resistance, and a source resistance of nearly zero absorbs nothing: the energy that the ladder rejects at a passband maximum has nowhere to go but back into the passband.

Why the reflection is the designed quantity #

The synthesis makes this explicit and it is worth following, because it says why the terminations are not an afterthought.

A ladder is derived from its reflection rather than from its transmission. Writing the filter’s own polynomial as $E$ and its reflection zeros as $F$, the element values are the successive quotients of a continued-fraction expansion of $(E+F)/(E-F)$. The load resistance falls out of the same expansion as its last quotient — which is why an even-order equiripple design cannot be realised between equal terminations at all, and returns a load of something other than one.

So the design procedure’s input is “how much shall be reflected, at which frequencies”, and its outputs are the elements and the two resistances together. Changing one of the resistances afterwards is changing an answer while keeping the rest, which is why the response that comes back is not any filter anybody designed.

The source that is not a resistance at all #

The window above is stated in ohms, which quietly assumes the thing at the left-hand end is a resistance. Usually it is not, and that is a second failure mode rather than a refinement of the first.

An amplifier’s output impedance is a fraction of an ohm at direct current, rises a decade per decade once the loop gain starts falling, and settles at the open-loop output resistance somewhere above the crossover. The node that is at ground for a while measures that curve directly on the complementary node and finds the same three regions — a tenth of an ohm at direct current, ten ohms at a kilohertz, and 909 ohms above a megahertz, which is the two feedback resistors in parallel with the amplifier contributing nothing at all. So an “amplifier-driven” ladder is driven from something that is 0.1 Ω across the first part of its passband and tens of ohms across the last part of it, and the transition sits wherever that particular part’s loop gain runs out.

There is a worse case than a rising impedance, and it is one this collection has measured on a regulator rather than on an amplifier. A source below a frequency finds the impedance not merely rising but peaking — 1.95 ohms at ten kilohertz, a third above the same circuit’s own open-loop value, because near crossover $1+T$ is smaller than one and dividing by it is multiplying. A source whose impedance has a peak in it puts a bump in a ladder’s passband at the frequency of the peak, which is a shape no mismatch in this essay’s sweep produces, and which would be attributed to the filter by anyone measuring the filter.

A preceding filter is worse. Its output impedance is a function with peaks and nulls in it, and there is no ohm value to compare with the window at all.

The consequence is that “matched” and “mismatched” are not the two cases. A ladder driven from a frequency-dependent impedance has a response that is neither the designed one nor any single mismatched one, and the departure is largest wherever the driving impedance is furthest from the design value — which is generally the top of the passband, where the shape matters most.

The repair is the same as before and is worth stating as a rule: put the design resistance in as a real component, so that the frequency-dependent part of what is driving the ladder is small compared with a resistor that is not. A kilohm in series with an amplifier whose output impedance runs from 0.1 Ω to 30 Ω leaves the source inside a three per cent window across the whole band, which is comfortably inside the eleven the shape needs.

What the ladder is doing to the source #

The reason the window has the width it has is visible from the other side: at a passband maximum the ladder’s input impedance is the design resistance, and between the maxima it is not.

An order-five Butterworth has its input impedance equal to one kilohm at direct current and at each of its passband maxima, and swings either side of that in between. The source resistance is therefore right at some frequencies and wrong at others no matter what value it takes, and the design’s whole achievement is to arrange the frequencies where it is right so that the response between them is the one asked for.

Mismatching the source moves every one of those crossings at once. That is why the departure is not a small ripple added to the design response but a different response — and why the corner frequency moves, since the corner is decided by where the last maximum sits relative to the load.

It is also why the effect is first order rather than second. A component error moves the response at a maximum, where the response is stationary; a termination error moves where the maxima are, and nothing about that is stationary.

Where the same idea is already in the collection #

Two other fields have met the reflection-as-the-designed-quantity idea, and putting them beside each other is the useful part.

A quarter-wave transformer matches one resistance to another over a band, and the band is set by how much reflection is tolerated. Its single section is exactly the narrow-band case; several sections whose reflections cancel over a band are a filter design with the reflection as the shaped quantity, which is what a ladder is.

A transmission line terminated in its characteristic impedance has no reflection and therefore no frequency dependence whatever. Terminate it in something else and the mismatch produces exactly the staircase the lines field draws — the same energy-with-nowhere-to-go that puts peaking on the passband above.

The unifying statement is that a passive two-port cannot dispose of energy it does not accept, so every design that shapes what it accepts is a design about its terminations.

What to do about it #

The window is narrow, so the practical question is what to put at each end of a ladder in a circuit that has no natural 1 kΩ anywhere.

Use a real resistor for the source. If the driving stage’s output impedance is much lower than the design value, a series resistor of the design value makes the source correct — at the cost of the signal it drops, which is the six decibels showing up as a real loss rather than as an inherent one. This is what a well-built passive filter does and it is why they look wasteful on a schematic.

Terminate the load in a real resistor too, for the same reason, and take the output at high impedance across it. A ladder feeding an amplifier’s input directly is loaded by whatever that input is, which is usually far too high.

Or do not use a ladder. A cascade of active sections has no terminations at all: each section drives the next from a low impedance into a high one, so there is nothing to match. It pays for that with the sensitivity a ladder is not a cascade measures at fifty-four times — and the number is not the claim there either. Fitted over two decades of tolerance the cascade’s passband error grows as the 0.99 power and the ladder’s as the 2.00, because at maximum power transfer the response is stationary in every element it contains. So the same matched condition that makes this essay’s window narrow is what makes the ladder insensitive inside it, and the two results are one theorem read in opposite directions.

Two further essays say what the cascade costs beyond that exponent, and both are reasons the trade is less even than it looks. The band that does not close finds a cascade of active sections buildable at three decades of impedance level at second order, one at sixth, and none at all at eighth, because fifty ohms of amplifier output resistance and two picofarads of stray bind it from opposite ends and every added section brings three more nodes and one more amplifier — while a doubly terminated ladder still has four decades at order nine, an inductor’s loss being a fixed quality factor rather than a fixed resistance. And the Q the amplifier decides finds a section built with an amplifier a hundred times its corner coming out with its quality factor two per cent high and its pole two per cent low, which on a fifth-order half-decibel Chebyshev is 2.1 decibels of ripple.

Set against those, a source resistance that has to sit inside 0.886 to 1.137 of its design value is a requirement on one component that a designer can meet by fitting it.

What is checked #

The figure asserts four things, and three of them would have caught a wrong netlist.

That the matched ladder loses exactly $20\log_{10} 2$ decibels, to a part in a million: if the load resistance came out of the expansion wrong, or the source were stamped between the wrong nodes, this is the first number to move.

That the matched passband has no peak anywhere in it — every point at or below the low-frequency value, to $10^{-9}$ — which is what a maximally flat response between correct terminations means, and what a Chebyshev would fail.

That the half-decibel window is narrower than a factor of one and a half, which is the claim the essay is for.

And that the departure at a twentieth of the design resistance exceeds the departure at twenty times it, which is the asymmetry stated as an inequality rather than as an impression.

What the window is not #

Two things this window is easy to confuse with, and both are measured elsewhere in this field with different answers.

It is not the impedance level. Scaling every resistance up and every capacitance down leaves the response bit-for-bit identical — the same filter a thousand times larger measures that invariance to a part in $10^{15}$ — and the terminations scale with everything else, so a scaled design is still matched. The window here is about the ratio between the source and the design value, which scaling does not move. A designer has a free parameter in the level and no freedom at all in the ratio.

It is not a tolerance on the ladder’s own elements either. Those are the quantity a ladder is not a cascade measures, and the answer there is the opposite of the one here: at the passband maxima the response is stationary in every element, so a one per cent inductor moves it by 0.003 dB where the same part in a cascade moves it by 0.162. The ladder is the insensitive structure and the one with an eleven per cent window on a component that is not part of the filter. Both statements come from the same theorem — stationarity at maximum power transfer holds only when maximum power transfer holds — which is why the insensitivity and the termination sensitivity cannot be had one without the other.

The practical form of that is worth one sentence, because it inverts the usual reading. The reason to fit a real resistor at each end is not that a ladder is fragile; it is that a ladder is exceptionally robust inside a condition, and the resistor is what buys the condition.