Circuit Response
every model has an edge, and the figure carries it
Concept

Equivalent noise bandwidth — where it appears

The width of the brick-wall filter that would pass the same noise power as a given response at its own reference gain. For a single pole it is π/2 times the half-power point, which is where a floor computed from the −3 dB frequency is two decibels light.

Named by 15 essays across 3 fields — each of them below, with the objects they name alongside it.

The noise of a 1.6 kΩ resistor through a 10.0 kHz filter. computed by solving, not by drawing. A seeded white sequence of 5.06 nV/√Hz marched through the network gives 619.3 nV across six seeds, spread 1.79%. Integrating the same density against the solved |H(f)|² gives 620.6 nV — -0.21% apart, well inside the spread. The noise bandwidth is 15.03 kHz against a −3 dB point of 10.00 kHz.

The floor a resistor sets

A kilohm at room temperature produces 4.00 nanovolts per root hertz, and it does so because it is warm rather than because of anything about how it was made. That is the first boundary in this collection that bounds a model from below — gain does not help, because gain amplifies it too — and it is the only field here whose figures are samples.

noise · Johnson noise 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.

noise · Noise bandwidth Five resistances, five corner frequencies, and one total on the capacitor. computed by solving, not by drawing. The noise density at a 1 pF capacitor charged through 100 Ω, 1 kΩ, 10 kΩ, 100 kΩ and 1 MΩ, at 290 K. The densities are 100 times apart and the noise bandwidths 1.0e+4 times apart, in opposite directions, so the area under every curve is the same: 63.2762 µV against 63.2762 µV, and √(kT/C) is 63.2762 µV. The resistance has cancelled out of the answer, and the reason is that ½C⟨v²⟩ is the ½kT a degree of freedom in contact with a bath holds — which no arrangement of resistors can change. The claim is about the whole frequency axis and nothing less: inside a 15.9 MHz band the same five networks give 5.05 µV to 63.07 µV, a factor of 12.5.

The total that has no resistor in it

A larger resistor is noisier and makes a narrower filter, and the two dependences are exactly reciprocal: the density goes as the square root of the resistance and the noise bandwidth as its inverse. Five decades of resistance charging one picofarad therefore give five decades of corner frequency, two and a half decades of density, and one total — 63.2762 microvolts at every one of them, which is the square root of kT over C and contains no resistance at all.

noise · kT/C noise The switches set the floor below 60.2 MHz and the amplifier sets it above. computed by solving, not by drawing. The two contributions to a switched-capacitor stage's noise floor, against the clock frequency. √(kT/C) is 63.3 µV on a 1 pF hold capacitor and does not move with the clock at all. The amplifier's own 4 nV/√Hz is white and is sampled, so all of it folds into the band: settling to 12 bits in half a clock period needs 8.32 time constants, the closed loop's noise bandwidth is then 8.32 clocks over two, and the number of folds is 8.32 — the same number, exactly. The marched route is a seeded white sequence put through the settling exponential and sampled once per clock, and it agrees with the closed form to 0.05% of variance. The two floors are equal at 60.2 MHz, above which a larger capacitor buys nothing.

The amplifier inside the sample

kT/C is exactly independent of the clock, of the capacitor ratio and of the switch resistance — two essays measured that and found it identically true rather than nearly so. The amplifier in the same loop behaves in the opposite way in every respect: its noise is white, it is sampled, and the number of times it folds into the band is exactly the number of time constants the settling needs. So the switches set the floor below sixty megahertz and the amplifier sets it above, and asking for two more bits of settling costs fifteen per cent more noise before anything else has changed.

noise · kT/C noise The quietest capacitor is 18× the fastest one, and the margin prefers neither. computed by solving, not by drawing. The total noise at the load of a capacitively loaded stage against its compensation capacitor, with the settling time on the same axis at ten microseconds to the microvolt. Three independent sources are put in the netlist and solved separately — the amplifier's own 4 nV/√Hz at its input, and √(4kTR) in series with each of the two feedback resistors — and added in power. The noise falls monotonically with the capacitor, from 50.7 µV at 1 pF to 12.1 µV at 220 pF. The peak in the noise gain falls with every larger capacitor and is gone entirely from 12 pF upward, where the uncompensated stage's peaks at 3.85 times its own low-frequency value. What the capacitor costs is settling: the fastest is 12 pF at 0.74 µs — the same capacitor that flattens the noise gain, because one handover decides both — and the quietest takes 20.3 µs, at a margin above 40° everywhere in that range.

What the second path costs at the floor

The arrangement that repaired a capacitively loaded amplifier was suspected of paying for itself in noise, because that is how compensations usually pay. It does not: it has no peak in its noise gain at all, and the total at the load falls from 54.9 microvolts to 28.9 as the capacitor is added. What it costs is settling, and the capacitor that is quietest is eighteen times the capacitor that settles fastest — a trade the phase margin says nothing about, because the margin is comfortable at both.

feedback · Capacitive load Which resistor the noise of a Chebyshev 5 actually comes from. computed by solving, not by drawing, one solve per resistor. Each bar is that resistor's share of the noise power at the output, found by splitting its node, putting a source of √(4kTR) in series with it and re-solving the whole network — so what is drawn is not how much noise each resistor makes but how much of it arrives. The largest contributor is F2R at 64.1 per cent, the smallest F0R at 9.0, and the shares add to 1.000000000000 because noise powers add. A resistor's share of the noise is not its share of the resistance: the largest departure between the two is 3.9 percentage points.

The resistor the noise comes from

Two identical 1.59 kΩ resistors in one third-order filter contribute 60.0 and 40.0 per cent of its output noise, because a resistor's noise is filtered by everything after it and by nothing before it. Solved one resistor at a time, the rule everybody carries — the resistance in the noise bandwidth — comes out 1.551 times the truth on a seventh-order Chebyshev and 0.791 times on a sixth-order Bessel. It is wrong in both directions on the same axis, so no factor repairs it.

noise · Johnson noise 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.

noise · Noise bandwidth 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.

noise · Noise bandwidth 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.

noise · Noise bandwidth 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.

noise · Noise bandwidth The order is worth 0.68 bits and a factor of 29 in clock. computed by solving, not by drawing. A butterworth design at a 20 kHz passband, at each order from two to ten and at three impedance levels, with the sample rate it demands for eighty decibels and the resolution its own resistors' noise allows. The clock demand falls steeply — 50.50×, 11.27×, 5.50×, 3.65× and on down to 1.76×, a factor of 28.8 — and the bar widths carry it. The noise rises: 2.00 µV at order two against 3.22 µV at ten, which is 0.68 of a bit. So the order is cheap in the currency everybody worries about and expensive in nothing. What is expensive is the impedance level: every resistance goes as one over the capacitance and a noise voltage as the square root of a resistance, so a decade of capacitance is 1.66 bits — the same at every order, to 0.000 of a bit — and the three curves are ten, one and a tenth of a nanofarad a section. The ceiling on resolution is set by a choice nobody prices, and the order, which everybody argues about, moves it by two thirds of a bit.

Two thirds of a bit for a factor of twenty-eight

An anti-alias filter costed in clock rate gets steeply cheaper with order — fifty times Nyquist at the second and 1.76 at the tenth — and the obvious objection is noise, since the filter is in the signal path and every section adds resistors. Measured, the objection barely holds: from order two to order ten the clock demand falls by a factor of 28.7 and the resolution the filter's own noise allows falls by 0.69 of a bit. What does cost resolution is the impedance level, at five thirds of a bit per decade of capacitance, and nobody argues about that at all.

digital · Anti-alias 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.

noise · Noise bandwidth 1 pA in 10.0 kHz: 312 carriers a window, and a Gaussian of the same variance that looks right and is not in its tails. computed by solving, not by drawing. A direct current of 1 pA counted in windows of 1/(2B) for a noise bandwidth of 10.0 kHz: the mean count is N = I/(2qB) = 312.1, and the count is Poisson, drawn as bars, against a Gaussian of the same mean and variance, drawn as a curve. The variance is N exactly, so 2qI is right; the signal-to-noise ratio is √N = 17.67 and the skewness 1/√N = 0.0566. The count exceeds its mean by five standard deviations with probability 7.84 × 10⁻⁷, against the Gaussian's 2.87 × 10⁻⁷. A seeded arrival train counted in 20,000 windows gives a mean of 312.200 and a variance of 314.461.

The count under the density

Shot noise's density, 2qI, is exact at every current, because it is the variance of a count of independent carriers, and that count is Poisson at any size. Reading that variance as a Gaussian floor is not exact. A picoampere in a 10 kHz noise bandwidth is 312 carriers a window, and it passes five standard deviations above its mean 2.74 times as often as a Gaussian of the same variance says. At 10 femtoamperes it is 359 times as often, and 4.4 per cent of the windows are empty. A single pole reads a third more skewness than a count does, (4/3)/√N against 1/√N. The density is right and the shape is wrong, and the shape is what a threshold uses.

noise · Shot noise One, two, three and four means of twenty milliseconds in cascade: the response, and the band each holds forty decibels down. computed by solving, not by drawing: |sin(πfT)/(πfT)|ⁿ with T = 20 ms, held against running means on samples at three frequencies. Each mean nulls 50 Hz, so n of them null it to order n, and their first sidelobe sits at −13.26 dB a mean. Forty decibels down, one mean holds 49.50–50.51 Hz, two 45.40–55.69, three 40.71–69.49 — its sidelobe at −39.78 dB just above the line — and four everything above 36.90 Hz: the null has become a stopband.

The mean that makes a stopband

A cascade of n means of one mains period holds, at a depth D, exactly the band a single mean holds at D/n — to a part in a million, because the cascade's response is the single mean's raised to the nth power. That one identity answers whether the third mean pays. Forty decibels down, two means hold 45.4 to 55.7 Hz, three 40.7 to 69.5, and four hold everything above 36.9 Hz: once D/n is shallower than one mean's first sidelobe, 13.26 dB, the null stops being a null and becomes a stopband. The price is a mains period of settling per mean and a noise bandwidth that exceeds a single mean of the same length by √(3n/π) — a factor that never stops growing.

noise · Noise bandwidth A gain of ten round an amplifier with a second pole: its half-power bandwidth and its noise bandwidth, each over what the gain–bandwidth product predicts. computed by solving, not by drawing: the stage as a netlist round a two-pole amplifier of 1 MHz, the second pole placed at r times GBW/G. The half-power bandwidth agrees with the second-order closed form √r·g(√r/2) to two parts in a thousand at every placement: 1.000 at r = 0.5, 1.272 at r = 1, 1.414 at r = 2, 1.231 at r = 5, 1.110 at r = 10, 1.010 at r = 100 times GBW/G. It is 1.414 times the product's prediction at r = 2 and within ten per cent only beyond r = 10. The noise bandwidth is (π/2)·GBW/G to half a per cent at every r: the second pole moves the one and leaves the other where it was.

The bandwidth the product still gets right

Gain times bandwidth is a constant for an amplifier with one pole, and real amplifiers have a second. Put it at r times the closed-loop bandwidth the product predicts and the half-power bandwidth becomes √r·g(√r/2) times GBW/G — 41.4 per cent wide at r = 2, within ten per cent only beyond r = 10 — so a follower round a 1 MHz part with its second pole at 2.5 MHz runs 40 per cent wider than the rule says. The noise bandwidth does not move at all. A second-order loop's noise bandwidth is set by one coefficient the second pole cannot touch, and it stays (π/2)·GBW/G at every r and every gain, to half a per cent on the netlist. A third pole is what breaks it.

feedback · Gain-bandwidth

Named alongside it

The objects these essays reach for when they reach for this one.

Spectral densityBrick wall filterIntegrationSettling timeVerificationAveragingDesign tradeoffFilter familiesJohnson noiseSeeded generatorModel rangeAliasing

All concepts