Circuit Response
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Series

Noise bandwidth — the series

6 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. A single pole, and the brick wall that passes the same noise. computed by solving, not by drawing and integrated over 267 frequencies. The equivalent noise bandwidth is 1.5706 times the −3 dB point, and π/2 is 1.5708. A noise voltage computed with the corner frequency instead is 20.2% low. The area under the curve and the area of the rectangle are the same number.

    The bandwidth noise sees

    A single pole passes π/2 times as much noise power as a brick wall at its own corner frequency, so a noise voltage computed with the −3 dB point is twenty-one per cent low. Measured by integrating the solved response rather than taken from the table it usually comes from, the ratio is 1.5706 and π/2 is 1.5708. A five-pole Chebyshev's is 0.964 — less than one.

    part 2 · noise
  2. The noise bandwidth of four families at 8 orders, and the one that has none. computed by solving, not by drawing. Every one of the 32 entries is an integral of the realised network's own squared magnitude, divided by that network's own measured −3 dB point. At order one the four families are the same filter and return 1.5706, which is π/2 — the calibration the rest of the table is quoted against. Only Butterworth then does what the ratio is usually said to do: it falls at every order, to 1.0065 at 8. Bessel is least at order 5 (1.0385) and rises to 1.0441; Chebyshev alternates with parity, 0.9637 at five against 1.0686 at six; and an even-order elliptic has no noise bandwidth at all, because its stopband comes back up to a constant — its magnitude at the top of the range moves by 0.00 decades per decade of frequency, so the integral grows with whatever limit it is stopped at.

    The ratio that does not walk to one

    A single pole passes π/2 times as much noise as a brick wall at its corner, and every account of it says the ratio falls towards one as the skirt steepens. Over thirty-two realised filters only Butterworth does that. Bessel is least at order five, 1.0385, and rises again; Chebyshev alternates with parity and the two branches separate only above 0.1968 decibels of ripple; and an even-order elliptic has no noise bandwidth at all, its integral returning 380 or 38,005 depending on where it was stopped.

    part 3 · noise
  3. Three averagers passing the same noise, and three different half-power points. Integrated by eight-point quadrature on every lobe, with the tail past the last lobe in closed form. A mean over 20 ms has the response sin(πfT)/(πfT), and the area under its square is 25.000000 Hz against the 25 Hz of 1/(2T) — the brick wall drawn shaded. A mean over the window passes half its power at 22.147 Hz, so the noise bandwidth is 1.12880 times that frequency; A one-pole averager passes half its power at 15.915 Hz, so the noise bandwidth is 1.57080 times that frequency; Two means in cascade pass half their power at 23.919 Hz, so the noise bandwidth is 1.04521 times that frequency. Every curve drawn encloses the same area; they differ in where they spend it.

    The filter an average is

    A mean taken over a window is a filter, and the area under its squared response is exactly one over twice the window — 25 hertz of noise bandwidth for twenty milliseconds, passing half its power at 22.15. Built to the same noise, a one-pole averager passes half its power at 15.92 hertz and takes 2.33 times as long to settle to one per cent, and two means in cascade pass half at 23.92 and take 1.25 times as long. Between its nulls a mean rejects the mains no better than the one-pole does, and one per cent off a null it rejects it by forty decibels however many cycles the window holds.

    part 4 · noise
  4. The bandwidth a windowed density is divided by, against the one bin an analyser assumes. Computed from the weights, from the coefficients in closed form, and out of the bins by Parseval — three routes agreeing to a part in 10¹². A record of N samples weighted by a window and transformed gives bins whose noise bandwidth is N·Σw²/(Σw)² in units of the bin spacing: 1 exactly for an unwindowed record, 1.5000 for Hann — three halves, exactly, because (a₀² + ½a₁²)/a₀² is (0.25 + 0.125)/0.25 with no record length in it — 1.3628 for Hamming, 1.7268 for Blackman and 2.0044 for Blackman–Harris. A density divided by one bin instead is wrong by the figure beside each bar. The bandwidth is not ordered with the sidelobe: Hamming's first sidelobe is 11.2 dB below Hann's and it costs less bandwidth as well, so the received sequence of windows fails at its second entry. What Hann buys is the far skirt instead, and that is a different measurement.

    The bandwidth a bin is not

    A density read off a transform is divided by a bandwidth, and the bandwidth belongs to the window rather than to the bin spacing — exactly three halves of a bin for a Hann window, so a density divided by one bin is 1.761 decibels high in power. Measured three ways that share only the weights. And the received sequence of windows fails at its second entry: Hamming's first sidelobe is 11.2 decibels below Hann's and it costs less bandwidth as well, with Hann collecting its debt at the fourth sidelobe and every one after it.

    part 5 · noise
  5. Where "the noise is the density over twice the window" is true for a mean of 600 samples. The exact sampled autocorrelation of a one-pole front end, summed in closed form and checked against seeded samples through the same filter. The solid curve is the reading's noise power divided by what the rule gives. It crosses one at 0.1355 of half the sample rate and nowhere else, so the rule is right at one front-end bandwidth rather than over a range: it is within one per cent from 0.0712 to 0.205 of half the sample rate, a window a factor of 2.88 wide. Above it the front end admits noise from past half the sample rate and the sampler folds it in, the ratio tending to (x/2)·coth(x/2) with x = 1/(τ·fₛ), which has no N in it — so the sample rate must be 9.7 times the front end's noise bandwidth for one per cent, not twice it. Below it the front end has already removed noise the rule charges for, and the faint curve — the √N law's own error, which is a claim about the samples rather than about a density — shows why: consecutive samples are correlated by ρ = e^(−1/(τ·fₛ)) and a mean of 600 of them is a mean of fewer things.

    The window the square-root law has

    Averaging N readings divides the noise power by N only if the readings are independent, and samples of a filtered noise are independent only over a band. Measured on the exact sampled autocorrelation, the rule that a reading's noise is the density over twice the window is right at one front-end bandwidth and wrong on both sides — a factor of three wide at six hundred samples, asking the sample rate to exceed the front end's noise bandwidth by about ten rather than by two. Above it the folded noise arrives at full amplitude; below it the front end has removed noise the rule is still charging for.

    part 6 · noise
  6. What widens a null: the order of the zero, not the placement of two. Integrated by quadrature on the union of both windows' nulls, at a fixed total length of 40 ms so that every arrangement settles in the same time. One mean of 40 ms holds -40 dB over 2.002 per cent of 50 Hz and passes 12.50 Hz of noise. Two means of 20 ms in cascade — a second-order null in the same place — hold it over 20.59 per cent — a factor of 10.3, which is 10^(−dB/40) because a first-order null rises linearly through the target and a second-order one quadratically — for 1.333 times the noise bandwidth. Splitting the two lengths puts two first-order nulls either side instead, and the widest trough is at a split of 10.12 per cent — 17.98 and 22.02 ms — holding 29.93 per cent for 16.53 Hz — which is exactly √2 times the equal pair's band, at every depth, for slightly LESS noise. It is also a cliff: a split of 10.118 per cent — 0.01 per cent further — leaves no band at 50 Hz at all, because the peak between the two nulls has climbed through the target.

    The order of the null, not the number of them

    A mean rejects the mains exactly and rejects it one per cent off frequency by forty decibels, whatever the window length. Two means in cascade of the same total length settle in the same time, cost a third more noise bandwidth and hold that forty decibels over ten times the band — the factor being ten raised to minus the depth in decibels over forty, because a first-order null rises linearly through a target and a second-order one quadratically. Splitting the two lengths to straddle the frequency buys a further factor of exactly root two, for no noise at all, and puts the design two parts in ten thousand from a cliff.

    part 7 · noise

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