Concept

Complementary sensitivity — where it appears

The fraction of a change in a feedback network that reaches a closed-loop gain, T/(1 + T) for a return ratio T. It and the sensitivity to the amplifier add to one at every frequency, so a loop can hand its accuracy to the resistors but cannot hide the gain from both.

Named by 2 essays across one field — each of them below, with the objects they name alongside it.

$The gain's sensitivity to the device and to the divider add to one at every frequency, and at crossover each is exactly one half (34.9° of margin). computed by solving, not by drawing, three times at each of 241 frequencies: nominal, with the amplifier's gain raised by one part in 10⁵, and with the feedback fraction raised by one part in 10⁵. The fraction of the device's change reaching the closed-loop magnitude is 9.97 × 10⁻⁴ at 10.0 Hz and 1.60 at its peak near 7.94 kHz; the fraction of the divider's change is 0.99899 at 10.0 Hz and falls to −0.60 at 7.94 kHz. They add to one within 1.16 × 10⁻⁵ everywhere, and at the loop's crossover, 5.73 kHz, they are 0.5000 and 0.5000: the real parts of 1/(1 + T) and T/(1 + T), which are one half at |T| = 1 for any phase. The cut loop, a second route, agrees to 9.55 × 10⁻⁶.$

Half to the amplifier, half to the resistors

A feedback amplifier is supposed to hand its accuracy to the resistors, and at low frequency it does — 99.9 per cent of a change in the divider reaches the gain and 0.1 per cent of a change in the device. The two fractions are measured here by nudging each and solving again, and they add to one at every frequency, to five parts in a hundred thousand. At the loop's crossover each is exactly one half, whatever the margin: the gain's magnitude answers the amplifier and the resistors equally. As complex numbers both are 1/(2 sin(φ/2)) there, above one for any margin under sixty degrees, and around the crossover lies a band where the gain is more sensitive to both than it would be with no feedback at all.

feedback · Desensitivity

On a linear frequency axis the loop's desensitisation is an area, and the area it gives back above crossover is the same: 6.74 and 6.74 kilohertz-nepers at 34.9° of margin. computed by solving, not by drawing. The logarithm of the device sensitivity |1/(1 + T)|, from the cut loop at 3601 frequencies between 0.01 Hz and 10 GHz, drawn in decibels against linear frequency to 60 kHz. It is −60.0 dB at low frequency, crosses zero at 4.47 kHz (the crossover is 5.73 kHz), peaks at +5.70 dB at 6.87 kHz, and returns towards zero as the loop gain goes. The area below zero is 6.74 kHz·Np (58.6 dB·kHz); the area above, integrated to 10 GHz, is 6.74, of which 6.11 lies inside the drawing. The net is −1.48 × 10⁻⁸ Hz·Np. Bode's sensitivity integral says it must be zero for a loop that falls two poles or more and is stable when cut.

The area that feedback only moves

A feedback amplifier divides its device's variation by the loop gain, sixty decibels of it at low frequency, and pays for it with a sensitivity peak near crossover. The payment is exact. Integrated over linear frequency, the logarithm of the sensitivity has as much area above zero as below — 6.74 kilohertz-nepers each side at 34.9° of margin, net zero to eight parts in a billion — at every margin and every gain, because the loop falls by more than one pole. Margin decides only the shape of the repayment: a 14-decibel spike at 11.7°, a one-decibel hump spread past half a megahertz at 80.4°. Only a loop that falls as a single pole keeps any of it, (π/2) times its crossover, and no amplifier is one.

feedback · Desensitivity

Named alongside it

The objects these essays reach for when they reach for this one.

Crossover frequency Desensitivity Phase margin Component sensitivity Integration Loop gain Return ratio

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