Why Fine Frequency Costs Slow Time
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.
See the relationship before changing it
The figure reads from left to right. The blue card is fft size. The middle card applies the page rule. The green card is bin width. Walk the arrows once: set the input, apply the rule, then read the result with its unit.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline is 2048 samples.
- 2
Name the relationship. bin width = 48,000 samples/s / FFT size
- 3
Substitute with units. 48,000 / 2,048 = 23.44 Hz
- 4
Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.
Predict, then change fft size
Try Predict the direction of bin width = 48,000 samples/s / FFT size. Test another fft size, then compare bin width.
Observe Finer frequency bins require a longer time window at the same sample rate. Reset fft size to 2048 and compare bin width.
Explain Finer frequency bins require a longer time window at the same sample rate.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
Ready: use the stated baseline inputs, then compare each displayed result.
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 window4096 / 48000 = 85.3 ms; product11.71875 x 0.0853333 = 1.000000again.
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.