The maths

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On the switch panel

The lines etched in bare board between the toggles, as they read:

Vr = 10^(dB/20)    Rs = 2.25×Z (1 − 1/Vr)    Rp = 2.25×Z / (Vr − 1)    L1 = 0.1125×Z mH
Rp in 0    Rs in out    Rs/Rp = (Vr − 1)²/Vr    P = 60 × 10^(−dB/10)
L1 VAC 1B    Rs3 VAC 2B    Rs4 VAC 2B 3½    Rs2 2B 1B    Rp1 1B 0

Netlists read part, node, node; node 0 is ground, VAC the amp input.

Notation

Z: nominal speaker load, Ω. A: a stage's attenuation, dB.
$V_{r} = 10^{A / 20} > 1$ : the voltage ratio.
$R_{0} = k Z$ , $k = \frac{9}{4}$ : the design resistance (the panel's 2.25×Z).
A source is written as its Thévenin pair (E, R): open-circuit voltage, source resistance.

A switched stage

Claim. Feed a stage from (E, $R_{0}$ ), with $R_{p}$ from in to 0 and $R_{s}$ from in to out. The Thévenin pair seen at out is ( $E / V_{r}$ , $R_{0}$ ) if and only if

R_{p} = \frac{R_{0}}{V_{r} - 1}, R_{s} = R_{0} (1 - \frac{1}{V_{r}})

Proof. No current flows in $R_{s}$ with out open, so the open-circuit voltage is the divider at in, and the output resistance is $R_{s}$ plus the source and shunt in parallel:

\frac{E R_{p}}{R_{0} + R_{p}} = \frac{E}{V_{r}}, R_{s} + \frac{R_{0} R_{p}}{R_{0} + R_{p}} = R_{0}

The first solves to $R_{p} = R_{0} / (V_{r} - 1)$ ; then the parallel term is $R_{0} / V_{r}$ and the second gives $R_{s}$ . ∎

Consequences.

A stage maps $(E, R_{0}) \mapsto (E / V_{r}, R_{0})$ . Any subset of stages, in any order, maps $E \mapsto E / \prod V_{r, i}$ : the dB add, and the source resistance after the ladder is $R_{0}$ in every setting.
The speaker, $Z$ on the last stage, therefore gets $E Z / ((Z + R_{0}) \prod V_{r, i})$ . Between two settings the speaker voltage changes by exactly $10^{- Δ A / 20}$ , whatever Z, provided the network ahead of the ladder has source resistance $R_{0}$ .
The ratio depends on A alone; $R_{0}$ only sets the scale:

\frac{R_{s}}{R_{p}} = \frac{{(V_{r} - 1)}^{2}}{V_{r}}, R_{s} R_{p} = \frac{{R_{0}}^{2}}{V_{r}}

With k = 9/4, the panel's stages, as multiples of Z:

Stage	Rs / Z	Rp / Z
1¾ dB	0.411	10.1
3½ dB	0.746	4.53
7 dB	1.24	1.82
14 dB	1.80	0.561

Work one

Watts

An ideal attenuator of A dB passes the fraction $V_{r}^{- 2}$ of the power:

P = P_{in} \cdot 10^{- A / 10}

The panel takes $P_{in}$ = 60 W: 27 W at 3½ dB, 12 W at 7 dB, 0.028 W at 33¼ dB.

The first stage

From the panel's netlist: $L_{1}$ VAC–1B, $R_{p1}$ 1B–0, $R_{s2}$ 2B–1B, $R_{s3}$ VAC–2B, output at 2B; $R_{s4}$ VAC–2B in the 3½ dB position only. Let V be its voltage ratio. Take the two ends of the band.

ω → 0. The coil is a short, VAC ≡ 1B: $R_{p1}$ shunts the input and $R_{s2} ∥ R_{s3}$ is in series. A shunt-first pad, so by the claim

R_{p1} = \frac{R_{0}}{V - 1}, R_{s2} ∥ R_{s3} = R_{0} (1 - \frac{1}{V})

ω → ∞. The coil is open: $R_{s3}$ is in series, then $R_{s2} + R_{p1}$ to ground. A series-first pad. Fed from $R_{0}$ , asking again for (E/V, $R_{0}$ ):

\frac{R_{0} + R_{s3} + R_{sh}}{R_{sh}} = V, (R_{0} + R_{s3}) ∥ R_{sh} = R_{0}

with $R_{sh} = R_{s2} + R_{p1}$ . The second gives $R_{0} + R_{s3} = V R_{0}$ , so

R_{s3} = R_{0} (V - 1), R_{s2} + R_{p1} = \frac{R_{0} V}{V - 1}

Four equations, three unknowns. They are consistent, with the unique solution

R_{p1} = \frac{R_{0}}{V - 1}, R_{s2} = R_{0}, R_{s3} = R_{0} (V - 1)

Check: $R_{0} ∥ R_{0} (V - 1) = R_{0} (1 - 1 / V)$ and $R_{0} + R_{0} / (V - 1) = R_{0} V / (V - 1)$ . Note $R_{p1} R_{s3} = {R_{0}}^{2}$ and $R_{s2} = R_{0}$ : the network is dual about $R_{0}$ . So the first stage gives A dB and leaves $R_{0}$ at both ends of the band; in between, the coil makes the impedance the amp sees rise with frequency, as a speaker's does.

The coil's scale: $L_{1} = 0.1125 Z$ mH $= R_{0} \times 50$ µH, so $L_{1} / R_{0} = 50$ µs, a characteristic frequency of 1/(2π · 50 µs) ≈ 3.18 kHz. The transfer between the limits is solved numerically below.

The lower position. $R_{s4}$ in parallel with $R_{s3}$ lowers the high-frequency loss. It has no closed form here: the designer finds it by bisection on $log R_{s4}$ so that the first stage alone gives the lower dB at the check frequency.

Work one

A whole design

At each frequency the circuit is solved by nodal analysis. The amp is a Norton source, $I = E / R_{src}$ in parallel with $R_{src}$ . Each part is an admittance: 1/R, $- j / (ω L)$ , $j ω C$ . The node voltages solve

Y (ω) v = i

by Gaussian elimination with partial pivoting (about a dozen nodes). The speaker is Z, or a model: $R_{e} + j ω L_{e}$ in series with a parallel RLC peaking at $f_{s}$ to $Z_{max}$ with sharpness $Q_{ms}$ .

Attenuation: $A (ω) = 20 {log}_{10} | v_{direct} / v_{spk} |$ , against the same speaker on the amp directly.
The amp sees $v_{VAC} / i_{VAC}$ ; the speaker sees the Thévenin impedance of the network with the speaker removed.
Watts: E is set so the amp would deliver its rated P into Z, $E^{2} = P {(R_{src} + Z)}^{2} / Z$ ; each resistor dissipates $| Δ v |^{2} / R$ , each coil carries $| Δ v | / (ω L)$ . The designer takes the maximum over every setting and 25 log-spaced frequencies from 20 Hz to the top of the full-power band.

Sweep a whole design on the Sweeps page.

What the model leaves out

Parts are ideal: no resistor tolerance, no coil resistance, no heat.
The amp is a source with a fixed output impedance. Real amps change with level, feedback and the presence and resonance controls.
The speaker model is a sketch of a speaker, not your speaker.