Why Fine Frequency Costs Slow Time

Ada re-derives this chapter’s own numbers step by step, at full precision

ADA · CALCULATION AUDIT

Why Fine Frequency Costs Slow Time

The lab runs a phone microphone at a 48 kHz sample rate and lets you pick the FFT size. A 2048-point FFT resolves the spectrum into 23.4 Hz bins over a 42.7 ms window; double it to 4096 and the bins sharpen to 11.7 Hz. This audit carries the window duration alongside the bin width to answer why fine frequency has to cost slow time.

Companion to the chapter Lab: Mobile PWA and Audio Sensing — every number here comes from that chapter.

Ada: The Practitioner layer makes two numbers from one 48 kHz context: a 2048-point FFT gives 23.4 Hz bins, and doubling to 4096 gives 11.7 Hz bins. Both are right, but they hide the price. Let me carry the window duration alongside the bin width.

• Bin width, 2048-point FFT: 48000 / 2048 = 23.4375 Hz (the chapter’s 23.4 Hz).
• Window duration, same FFT: 2048 / 48000 = 0.0426667 s = 42.7 ms (the chapter’s window figure).
• Their product: 23.4375 Hz x 0.0426667 s = 1.000000.
• Double the FFT to 4096: bin width 48000 / 4096 = 11.71875 Hz (11.7 Hz) and window 4096 / 48000 = 85.3 ms; product 11.71875 x 0.0853333 = 1.000000 again.

The product is pinned to 1 because bin width is sampleRate / fftSize and window is fftSize / sampleRate – the sampleRate and fftSize cancel, leaving df x T = 1 for every setting. So the two sliders are really one budget: each halving of the bin width you buy in frequency is paid for by an exact doubling of the analysis window in time. An audio lab that wants to separate a 100 Hz hum from a 110 Hz hum must accept a slower spectrum, and one chasing a fast knock must accept coarser bins – the trade is not a tuning preference but a fixed one Hz-second the physics will not discount.

Every number above is taken from the chapter’s own material and re-derived step by step.