The residue that falls with its slope

The residue that has only one sign found that a bridged-T in a selective amplifier’s feedback path trades the twin-T’s unlimited Q for a ceiling: the loop’s quality factor rises with the amplifier’s gain and stops near $\sqrt{R_1/R_2}/2$, because the bridged-T’s residue at its null, $2/(2 + n)$ with $n = R_1/R_2$, is real, positive and never zero. The dual that is the same function confirmed that the dual network has exactly the same residue and ceiling, and that what differs between the two is only which parts carry the ratio.

Both essays left the same question standing. If the ceiling exists because the residue cannot reach zero, then a small change that moves the residue towards zero from the positive side should raise the ceiling. The earlier essay named two candidates — a resistance in series with the shunt element, or a fraction of the output returned to the junction — and asked how close the residue could come to zero before the parts’ tolerances made its sign uncertain, at which point the bridged-T would have become a twin-T with a twin-T’s problem — a network whose null, as the null that is stationary in nothing measured, every imperfection fills at the same rate and none moves.

The answer is that the residue can be made as small as desired, and the ceiling does not move.

The shunt resistor returned to the output #

The trim is the simplest one that acts on the residue: instead of taking the shunt resistor $R_2$ from the junction to ground, take it to a node driven at $k$ times the network’s own output. With $k = 0$ it is the earthed bridged-T. With $k$ negative — an inverting buffer driving the resistor’s far end — the resistor carries more current than its value alone would pass, and it behaves as a smaller resistor; with $k$ between nought and one it behaves as a larger one.

Solving the junction’s node equation gives the effective ratio at once: the network behaves, at its null frequency, as a bridged-T of ratio $n(1 - k)$, and the residue is

$$\rho(k) = \frac{2}{2 + n(1-k)}$$

Every solved point agrees with the closed form to a part in a million, from $k = -30$ to $k = 0.9$. At $k = -15$ the effective ratio is 1,600 and the residue is 0.00125, a sixteenth of the earthed network’s 0.0196. By the earlier essay’s arithmetic that should lift the ceiling from five to twenty.

It does not, and the panel beside the curve says so before any loop is closed. The ceiling at first order is the slope of the network’s transfer through its null divided by twice its residue, $s/2\rho$. At $k = 0$ that is 4.902; at $k = -3$, 4.975; at $k = -15$, 4.994. The residue fell sixteenfold and the slope fell with it, almost exactly in proportion, so their ratio moved by two per cent — towards 5.000, $\sqrt{100}/2$, and no further.

Why the slope must fall #

The reason is one line, and it is the most useful thing in this essay.

A selective loop of gain $A$ around a network $T$ has the closed-loop response $A/(1 + AT)$. As $A$ grows without limit this tends to $1/T$, and the poles of $1/T$ are the zeros of $T$: the loop’s ceiling is the quality factor of the network’s own transmission zeros. For the earthed bridged-T the zeros are the roots of $1 + 2sCR_2 + s^2C^2R_1R_2$, whose quality factor is $\sqrt{R_1R_2}/2R_2 = \sqrt n/2$ exactly. That is where “about five” comes from.

Now return the shunt resistor to $k$ times the output. At a transmission zero the output is zero, so the returned node is at $k \times 0 = 0$ — at ground, exactly as if it had never been lifted. The network at its zeros is the earthed network, and its zeros are the earthed network’s zeros. A trim that acts through the output cannot move the transmission zeros, and so it cannot move the ceiling.

What it moves is everything else: the poles of $T$, and with them the residue and the slope at the null, which are properties of the whole transfer rather than of its zeros alone. The residue can be pushed towards zero, but the ceiling was never the residue’s to set.

The loops confirm it. Located at gains from thirty to a million, the four curves have the same shape, displaced along the gain axis, and at a gain of a million all four read 5.00 — the zeros’ $\sqrt n/2$ to three digits. The first-order estimate $s/2\rho$ approached 5.000 as the residue fell because the estimate is exact only in the limit of a small residue; the loop’s actual ceiling was 5.000 all along.

What the trim costs instead #

The trim changes one thing that matters, and it is not an improvement. Near the ceiling the loop’s Q is $Q_\infty\,A\rho/(1 + A\rho)$, so the gain needed to reach a given fraction of the ceiling is inversely proportional to the residue. Nine-tenths is reached at $A\rho = 9$.

The earthed network reaches nine-tenths of its ceiling at a gain of 449. Returned to −31 times its output, with its residue thirty-two times smaller, it needs 14,400 for the same Q. So the inverting return takes a bridged-T that reached its ceiling with a modest amplifier and makes it need a much better one — and delivers exactly the same selectivity when it gets there. The residue was the network’s gain requirement, not its Q limit.

A low ratio, where the first-order estimate misleads #

At a ratio of a hundred the first-order ceiling $s/2\rho$ is within two per cent of the true one, and the distinction above is a refinement. At low ratios it is not.

With $n = 10$ the earthed network’s $s/2\rho$ is 1.32 while its zeros have a Q of 1.58, and the inverting return appears to raise the ceiling from 1.32 to 1.56. It has not raised anything: the loop at unlimited gain was always going to reach 1.58, and the first-order estimate simply becomes accurate as the residue it divides by becomes small. A design that read the estimate as the ceiling, and bought an inverting buffer to lift it, would have bought an amplifier requirement and nothing else.

A non-inverting return goes the other way on the estimate. At $k = 0.9$ the effective ratio is ten, the residue is 0.167 and the first-order ceiling is 4.17; the zeros are still at $\sqrt{100}/2$, and the loop at unlimited gain still reaches five. What falls is the gain-limited Q at any practical gain, since the residue is eight times larger and $A\rho$ saturates eight times sooner in the other direction.

The sign, which the trim cannot change #

The earlier essay’s hope was that a trim might carry the residue close to zero from the positive side, where the parts’ tolerances would make its sign uncertain. The closed form says why that point never arrives: $2/(2 + n(1 - k))$ is positive for every $k$ below one and approaches zero only as $k \to -\infty$. At $k = 1$ the returned node follows the output exactly, the shunt resistor carries no current, and the residue is one — the network has become transparent at its null. Above $k = 1 + 2/n$ the residue is negative, but it arrives there through infinity, not through zero.

So a trim through the output cannot make a bridged-T into a twin-T. The residue is a ratio in disguise, and ratios are positive. The twin-T’s zero residue comes from a cancellation between two paths — the sign the null leaves behind is the arithmetic of it — and nothing in the bridged-T’s single path can imitate a cancellation.

The passive candidate the earlier essay named does no better. A resistance in series with the dual’s shunt capacitor is not a trim through the output; it changes the network’s zeros directly, and it changes them the wrong way. A solve of the dual with a hundred ohms in series with its shunt capacitor raises the residue from 0.0196 to 0.029 and moves the null down by one per cent, and a thousand ohms raises it to 0.107. The residue moves away from zero, not towards it; what the resistance does to the zeros’ own Q, and so to the ceiling, is a property of the new zeros that this solve did not compute. The one thing it cannot be is the trim the earlier essay was looking for.

A high ratio, where nothing changes #

At the other end the trim is simply invisible.

With $n = 1000$ the earthed first-order ceiling is 15.78 against the zeros’ 15.81, and the inverting return closes the last two-tenths of a per cent. A bridged-T with a high ratio is already at its ceiling’s first-order estimate, and there is no headroom for a trim to appear to buy.

What the three ratios together say is the design rule. The ceiling is chosen with the ratio, as $\sqrt n/2$, and nothing done through the output changes it. A higher selectivity from a bridged-T means a higher ratio — the dual that is the same function found that a high ratio belongs in resistors — and past a Q of fifteen or twenty, which needs ratios of a thousand or more, the twin-T and its unlimited Q come back into consideration, together with its sensitivity. That is the answer to the earlier essay’s question about whether a high-Q bridged-T is ever a good idea: not by trimming one.

The same argument wherever a loop is closed around a network #

The argument about the zeros is not about the bridged-T. It holds for any network closed inside a loop of large gain, and it is worth stating generally because it explains a family of trims that look promising and do nothing.

A loop of gain $A$ around a network $T$ tends, as $A$ grows, to $1/T$: its poles migrate from wherever they start towards the zeros of $T$. Where the behaviour is written down is the reading of a response from its poles; in a high-gain loop the poles that matter were written down by the feedback network’s zeros. Anything that changes the network only where its output is non-zero — a buffer fed from the output, a load on the output, a second path from the output back to an internal node — leaves those zeros alone, and so leaves the loop’s high-gain limit alone. It changes how much gain the loop needs to reach that limit and how the loop behaves on the way.

The corollary is what to look for when the limit itself has to move: something that changes the network at its zeros. In a filter designed with finite transmission zeros, the zeros that buy an order places them on purpose with the network’s own elements; in a selective amplifier, the zeros belong to the feedback network’s passive parts, and only a change to those parts, or to a node that is not silent at the zeros, can move them.

What a selective amplifier’s designer takes #

Choose the ceiling with the ratio. The loop can never be more selective than $\sqrt n/2$, so a Q of five needs a ratio of a hundred and a Q of ten needs four hundred. Nothing added later raises it, and a design review that finds the Q short should look at the two resistors before it looks at the amplifier.

Choose the gain with the residue. Reaching nine-tenths of the ceiling needs $A\rho = 9$, and the earthed residue is $2/(2 + n)$, so the gain needed is about $4.5n$ — 449 for a ratio of a hundred. An amplifier with that much gain at the null frequency reaches the ceiling; one with ten times more reaches it and gains nothing further, which is the steadiness the earlier essay valued: past the knee the Q is the network’s, not the amplifier’s.

Do not trim the residue. An inverting return lowers it and raises the gain the loop needs in the same proportion, which moves the Q further from the ceiling at any practical gain. A non-inverting return raises the residue and lowers the gain needed, which sounds like an economy and is one: at $k = 0.5$ the loop reaches nine-tenths of the same ceiling at a gain of about 234 instead of 449, the figure’s own measurement of that point lying on the same line $A = 9/\rho$ as the others. That is a real use for the trim, the opposite of the one it was proposed for — a way to reach the ceiling with a weaker amplifier, at the price of a Q that depends on the buffer’s gain until the ceiling is reached. The Q the amplifier decides is the regime in which that dependence is the whole story.

How the numbers were found #

The residues are solved, by a nodal solve of the bridged-T with its shunt resistor returned through a voltage-controlled source to $k$ times the output, at the null frequency, and required to equal $2/(2 + n(1 - k))$ to a part in a million at fifteen return gains from −30 to 0.9. The slope is the central difference of the transfer’s imaginary part across the null, and the first-order ceiling is that slope over twice the residue; it is required never to exceed $\sqrt n/2$ and to come within two per cent of it at $k = -30$.

The loops are located at gains from thirty to a million — stability by winding, peak by golden section, edges by bisection — and at a million every return gain is required to reach $\sqrt n/2$ to eight per cent; they reach it to three digits. The gain for nine-tenths of the ceiling is found by bisection on the located Q, and the product of that gain and the residue is required to be nine to within a tenth at every return gain.

An ideal buffer, an ideal amplifier and exact ratios #

The return buffer is ideal. A real inverting buffer has its own bandwidth and output resistance, which put a pole into the network’s denominator; at the frequencies here — a null at 1.6 kHz — an ordinary amplifier is ideal enough, and the argument about the zeros holds for any buffer, since at a zero it outputs nothing whatever it is.

The loop’s amplifier has no pole of its own. The peak the amplifier moves finds that a real stage’s lag moves the peak off the null, and since the trim makes the ceiling need more gain, it makes that lag matter more.

And the ratios are exact. With toleranced parts the zeros move and the ceiling with them, by half the ratio’s error, which is the spread the dual that is the same function measured.

Still open: the zeros moved on purpose, the loop with a ceiling, and the buffer’s own pole #

A trim that moves the zeros. Anything that acts at the junction rather than through the output — a current injected in proportion to the input, say — changes the network at its zeros and can change the zeros’ Q. Whether such a trim can raise the ceiling without making the residue’s sign uncertain, and what it costs in sensitivity, is the question the output-referred trim cannot answer.

The loop that needs a ceiling. A gain-controlled oscillator or an automatic gain control wants a loop that cannot be pushed past a Q, whatever its amplifier does. The bridged-T’s ceiling is exactly that, set by passive parts; marching such a loop with its amplifier saturating would say whether the ceiling survives the nonlinearity.

The buffer’s pole. A return buffer with a pole moves the network’s poles further and leaves its zeros alone, so it should cost gain requirement and nothing else — until its phase at the null is large enough to shift the residue’s phase as well, when the residue stops being real and the loop’s peak moves off the null. Where that begins is a number for the buffer’s bandwidth.