Formulas

Lorentzian lineshape. A resonance of centre frequency f₀ and quality factor Q has the normalised response

• L(f) = 1 / [ 1 + (2Q (f − f0) / f0)² ]
• Full width at half maximum: Δf = f0 / Q, i.e. the half-power points sit at f0(1 ± 1/2Q).
• More generally, the band where L ≥ α is f0(1 ± k) with k = √((1−α)/α) / (2Q). For α = ½ this reduces to k = 1/(2Q).

Why the spacing is geometric. The linewidth is proportional to the centre frequency, so a resonance at 1 THz is ten times wider than one at 100 GHz. Uniform spacing would therefore be wasteful at the top of the band and leave gaps at the bottom. Demanding that the upper band edge of resonance n coincide with the lower band edge of resonance n+1,

• fn(1 + k) = fn+1(1 − k)  ⇒  fn+1 / fn = r = (1 + k)/(1 − k), a constant ratio.
• For α = ½: r = (2Q + 1)/(2Q − 1).
• The centres therefore form a geometric progression — equally spaced on a logarithmic frequency axis.

Number of resonances. Covering [fmin, fmax] requires

• N = ⌈ ln(fmax/fmin) / ln r ⌉
• Large-Q limit (α = ½): ln r ≈ 1/Q, so N ≈ Q · ln(fmax/fmin). One octave needs about 0.693 Q resonances, one decade about 2.303 Q.

Placement. Rounding N up leaves a little slack, which is spread evenly in log-frequency. Two limits fix the end resonances: the first centre may sit no higher than A = fmin/(1−k) (otherwise its lower band edge would leave a gap below f_min), and the last centre no lower than B = fmax/(1+k). Pinning both and distributing the rest geometrically gives

• f1 = A = fmin/(1 − k),   fN = B = fmax/(1 + k)
• ρ = (B/A)1/(N−1), which is guaranteed ≤ r because N ≥ ln(span)/ln r
• fn = f1 · ρ(n−1),   n = 1 … N
• The band of the first resonance then starts exactly at f_min and that of the last ends exactly at f_max, so the envelope touches α at the two range edges and stays above it everywhere in between.
• If a single resonance already suffices (span ≤ r, so N = 1) it is placed at √(A·B), centring the available slack.

Note that the naive choice ρ = span1/N with centres at the geometric midpoints of equal log-bins does not work: the upper edge requires √ρ ≤ 1+k, which is a stricter condition than ρ ≤ r, and the top of the range ends up slightly uncovered.

Comparison with uniform spacing. If instead you spaced the centres uniformly, you would have to use the narrowest linewidth (the one at f_min) everywhere, needing Nuniform = ⌈ Q (fmax/fmin − 1) ⌉ resonances. For a decade that is 9 Q instead of 2.3 Q — a factor of about 4 wasted.

Note: the criterion here is that the envelope (the maximum over all resonances) never falls below α. If instead you coherently sum overlapping resonances the combined response is slightly higher in the gaps, so this placement is the conservative choice.