Circuit Response
every model has an edge, and the figure carries it
The floor, which bounds from below

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.
16 min read 7 figures The constants decideTwo routes to a numberEvery model has an edge

Noise out of a system is √(density × bandwidth). The density is a property of the source and the previous essay computes it. The bandwidth is a property of the response, and it is not the number almost everybody uses.

The −3 dB point is where a response has fallen by thirty per cent in voltage. It is a useful marker for a signal, and for noise it is the wrong quantity entirely: a filter does not stop at its corner, it continues to pass a diminishing amount above it, and all of that contributes.

A single pole, and the brick wall that passes the same noisecomputed 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.00.500101234frequency ÷ the −3 dB point|H|², the power that gets throughhalf the power: the −3 dB pointthe brick wall: 1.5706× the corner1.571×solved, then checked — the area integrated, not tabulated1.571× the corner, not 1×
Fig. 1 A single pole’s squared magnitude — the fraction of power that gets through at each frequency — and the rectangular filter that would pass the same total. The two shapes enclose the same area, and the rectangle is wider than the −3 dB point by a factor the figure computes rather than quotes. The slider is the number of poles.

What is being integrated, and what π/2 is

The equivalent noise bandwidth is ∫|H(f)|²df divided by |H|² in the passband: the width of the brick wall that would pass the same noise power.

For a single pole, |H|² = 1/(1 + (f/f_c)²), and the integral of that from zero to infinity is f_c × π/2. So the noise bandwidth is 1.5708 times the corner, exactly, and using the corner instead of it understates the noise power by that factor and the noise voltage by its square root, which is 21%.

The figure measures it. It sweeps the solved response of a real network — a resistor and a capacitor, assembled and inverted at every frequency on the grid — squares the magnitude, integrates trapezoidally in linear frequency on a logarithmic grid, and divides by the −3 dB point found by bisection on the same response. The answer is 1.5706, against π/2 = 1.5708.

Two decisions in that sentence are worth defending, because both were got wrong first.

The grid is logarithmic and the rule is linear. The response is logarithmic in frequency, so a linear grid would spend all its points in the last decade and none where the response is changing. The integral, though, is a linear one — ∫|H|²df — so the trapezoidal rule has to be applied in linear f on a logarithmic grid. Pairing them the other way is a silent factor of ln 10.

The stretch below the grid is not empty. A logarithmic axis cannot reach zero, so the integral starts at some small frequency and the region below it is missed. That region is not negligible: starting at f_c/1000 leaves out one hertz of a 1,571 Hz integral, which is 0.064% — and it was exactly the size of the discrepancy that led to it being found. The response there is flat by construction, since the grid starts deep in the passband, so the missing piece is a rectangle and can be added exactly.

With both corrected the measured ratio is 1.5706 and the residual is the truncation at the top of the sweep, which falls as the sweep is extended: 1.5697 at a thousand times the corner, 1.5706 at ten thousand, 1.5706 at a hundred thousand.

What the ratio does as poles are added

The single pole’s π/2 is the extreme case, and the ratio falls towards one as the skirt steepens — which is what a brick wall is, in the limit.

Measured on realised networks at a common −3 dB point: 1.5706 for one pole, 1.1107 for two, 1.0472 for three, 1.0262 for four, 1.0167 for five, 1.0115 for six.

That sequence is worth reading as a practical statement. A two-pole filter already gets most of the benefit — the excess over a brick wall falls from 57% to 11% — and beyond about three poles the noise bandwidth is within five per cent of the corner and there is very little left to gain. Filters are made steeper for reasons other than noise, and the noise argument stops paying almost immediately.

The shape of the diminishing return is worth a moment too. The excess over unity goes 0.571, 0.111, 0.047, 0.026, 0.017, 0.012 for one through six poles, and the ratios between consecutive terms settle towards about 0.7 — so each additional pole removes roughly a third of what is left. That is slow enough that the first two poles do most of the work and fast enough that nothing beyond the fourth is worth having, which is a narrower conclusion than “more poles are better” and a more useful one.

A 3-pole Butterworth, and the brick wall that passes the same noisecomputed by solving, not by drawing and integrated over 267 frequencies. The equivalent noise bandwidth is 1.0472 times the −3 dB point. A noise voltage computed with the corner frequency instead is 2.3% low. The area under the curve and the area of the rectangle are the same number.00.500101234frequency ÷ the −3 dB point|H|², the power that gets throughhalf the power: the −3 dB pointthe brick wall: 1.0472× the corner1.047×solved, then checked — the area integrated, not tabulated1.047× the corner, not 1×
Fig. 2 A three-pole Butterworth at the same corner. The squared magnitude falls much faster past the corner and the equivalent rectangle is only 4.7% wider than the −3 dB point. The area under the curve and the area of the rectangle are still the same number, which is what the ratio means.

The family that comes out below one

The three families this collection builds — Butterworth, Chebyshev and Bessel — are all normalised to the same measured −3 dB point, so their noise bandwidths are directly comparable. They are not the same, and one of them is surprising.

Three families at order 3, and how much noise each lets throughcomputed by solving, not by drawing. All three are normalised to the same −3 dB point, so the bars compare the noise each admits for the same nominal corner: butterworth 1.047, chebyshev 0.999, bessel 1.074. Chebyshev's is the lowest and at order five it drops below one — its response is already falling inside the band while its −3 dB point sits beyond the ripple, so it passes less noise than a brick wall at its own corner would.-40-2001001k10kfrequency (hertz)gain (decibels)the −3 dB point they sharenoise bandwidth ÷ the −3 dB pointButterworth1.0472×Chebyshev0.9992×Bessel1.0736×a brick wall at the cornersolved, then checked — three integrals, one cornerChebyshev at 0.999× the corner
Fig. 3 The three families at order three, drawn together, with their noise bandwidths as bars beneath. The marked line on the bars is where a brick wall at the shared corner frequency would sit, and Chebyshev is below it.

At order three: Butterworth 1.0472, Bessel 1.0736, Chebyshev 0.9992. At order five: 1.0167, 1.0385, 0.9637. At order six: 1.0115, 1.0386, 1.0687.

Two things in that table need explaining and both are real.

Chebyshev, at odd order, is below one. Its noise bandwidth is less than its own −3 dB point, which sounds impossible for a filter whose response extends above the corner. It is not: the response is already falling inside the band, because an odd-order Chebyshev starts at unity, dips into its ripple band, and comes back — so the area lost inside the corner exceeds the area gained outside it. The −3 dB point of a Chebyshev sits beyond the ripple band edge rather than at it, and that gap is where the deficit comes from.

Even orders behave differently. At order six Chebyshev is back above one, at 1.0687. An even-order Chebyshev starts at the bottom of its ripple band and comes up, so its passband shape relative to its corner is different and the sign of the effect reverses. The figure’s assertion is therefore made for odd orders only, and the restriction is the interesting part rather than a convenience: the claim is about the shape, so it is asserted where the shape holds.

Where the odd-order deficit comes from, precisely

The Chebyshev result is odd enough to deserve more than a sentence, and it is entirely explicable once the two frequencies involved are separated.

A Chebyshev filter is defined by a ripple band: its response oscillates between one and 1/√(1 + ε²) up to a frequency, and falls monotonically after it. That frequency — the ripple band edge — is where the definition ends, and it is not the −3 dB point. For a 0.5 dB ripple the band edge is where the response is 0.5 dB down, and the −3 dB point is further out, by an amount that grows with the order.

So when three families are normalised to a common −3 dB point, as they are here and in the filters field, the Chebyshev’s ripple band is pushed inside that point. Its response is already well below unity for a substantial part of the interval from zero to the corner, and the noise it fails to pass there exceeds the noise it does pass beyond the corner.

The deficit therefore grows with order, because the gap between the ripple band edge and the −3 dB point grows: 0.9992 at order three, 0.9637 at order five. And it reverses for even orders, because an even-order Chebyshev’s response starts at the bottom of the ripple band rather than the top — the same normalisation then places its band differently and the arithmetic goes the other way.

None of that makes Chebyshev a quiet filter in any absolute sense. It makes the ratio to its own −3 dB point less than one, which is a statement about a comparison and not about a component. Two filters compared at their ripple band edges would rank differently again, and the reason the comparison here uses the −3 dB point is that it is the number a datasheet gives.

Why any of this matters

The consequence is one sentence and it is the reason this essay exists rather than a paragraph in the previous one.

Two filters with the same corner frequency admit different amounts of noise, and the difference is up to fifty per cent in power. A designer who chooses a filter for its skirt, and then computes the noise from its corner frequency, has made an error whose size depends on a property of the filter that the corner frequency does not report.

Put concretely: the same source through a one-pole filter at 10 kHz and through a five-pole Chebyshev at 10 kHz produces noise voltages in the ratio √(1.5706/0.9637) = 1.28. Twenty-eight per cent, from a choice made for entirely unrelated reasons.

Three filter families at order 5, all with the same half-power pointAt three times the corner the Chebyshev is -64.0 dB down, the Butterworth -47.7 dB and the Bessel -28.3 dB. The inset is the passband at forty times the vertical magnification, which is the only place the Chebyshev's half-decibel of ripple is visible at all.-90-60-3001001k10kfrequency (hertz)gain (decibels)ButterworthChebyshevBesselhalf power1.00 kHzthe passband, magnified-1-0.500000.2000.4000.6000.8001solved, then checked — three networks, 133 frequencies eachall normalised to a measured −3 dB at 1.00 kHz
Fig. 4 The three families at one order and one corner, from the filters field. That figure is about what each family costs in ripple, skirt and delay; this page adds a fourth column to the same comparison — what each costs in noise — and it does not rank the families the same way.

The trade this adds to

What a steep skirt costs measures the price of a sharper cutoff in delay variation and in ringing: at order five, Chebyshev is 16 dB further down at three times the corner than Butterworth and its passband deviation is the same, and what it pays is 49% of delay variation against Bessel’s 0.06%.

Noise is a fourth axis on that comparison and it does not agree with the other three. Chebyshev, which is the worst of the three for delay, is the best of the three for noise at every odd order. Bessel, which is the best for delay, is the worst for noise at every order tested.

There is no ordering that makes one family best, which is the honest state of the subject and the reason all three exist. What this page contributes is that the noise column is computable and usually is not computed — the table it would come from is in some textbooks and not in most, and the integration takes a few lines given a solved response.

What each family costs, at order 5Measured on the solved networks. The Chebyshev is 36 dB further down at three times the corner than the Bessel, and pays for it in delay: its group delay varies 49.0% across the passband against the Bessel's 0.06%.passband deviationdecibels, peak to trough below 0.8 f_cButterworth0.443 dBChebyshev0.500 dBBessel1.882 dBattenuation at three times the cornerdecibels downButterworth47.7 dBChebyshev64.0 dBBessel28.3 dBgroup-delay variation across the passbandper cent, slowest against fastestButterworth48.0%Chebyshev49.0%Bessel0.1%solved, then checked — nine measurements, three networksevery number here moves with the order
Fig. 5 The trade-off between the families as the filters field measures it: ripple, skirt, delay variation and ringing, at one order. Noise bandwidth is the column that is missing from it, and adding it changes the ranking rather than reinforcing it.

What the measurement rests on

The whole page is one integral repeated, so it is worth being explicit about what makes it trustworthy.

The response being integrated is a solved network rather than a polynomial. That matters here for a specific reason: the first version used the polynomial recovered by sampling the determinant, and for the higher-order families that recovery quietly returned a different filter — a fifth-order cascade’s roots are spread far enough that the trim discards a genuine coefficient, and the −3 dB point came back at 657 Hz for a filter normalised to 1 kHz. Nothing about the resulting numbers looked wrong; they were simply about a filter nobody had asked for.

Solving the network directly at each frequency is slower and cannot fail that way. Every point on every curve here is an assembled matrix, inverted, with current law rebuilt from the element relations and the energy counted twice before the answer is returned — the same check every other figure on this site runs, applied a few thousand times per figure.

The noise of a 1.6 kΩ resistor through a 10.0 kHz filtercomputed 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.±619.3 nV, the root mean square1200 of 192001 stepssampled, this seed621.2 nVsampled, 6 seeds619.3 nV ± 1.79%integrated from |H|²620.6 nV…ignoring the hold626.3 nVnoise bandwidth15.03 kHzthe −3 dB point10.00 kHzsolved, then checked — a sample against an integralagreeing to 0.21%
Fig. 6 The two-route comparison from the previous essay, which is where the noise bandwidth computed here is used. The sampled route knows nothing about π/2; it generates a sequence and marches it through the network. That the two agree to a fraction of the sampling spread is the check that the integral on this page is the right integral.

A last observation about π/2

There is something worth noticing about the single-pole result that is easy to walk past.

π/2 is not an approximation, a fitted constant or a convention. It is ∫₀^∞ dx/(1 + x²), which is arctan evaluated at infinity, and it comes out of the shape of a one-pole response and nothing else. Any single-pole system — a resistor and a capacitor, a thermal time constant, a mechanical damper — passes exactly π/2 times its corner frequency’s worth of white noise power. The constant belongs to the shape rather than to electronics.

This collection has a habit of finding relations that look exact and are not, so it is worth marking one that is. Resonance and its bandwidth finds a half-power width of exactly f₀/Q — and finds, in the same figure, that the band is not centred on the resonance, which every textbook picture draws as though it were. The pattern there was an exact relation sitting beside a false one.

Here the exact relation sits beside a widespread substitution: the −3 dB point used where the noise bandwidth belongs. The substitution is not an approximation to π/2; it is a different number, twenty-one per cent away, and the reason it survives is that it is close enough to look like a rounding. An error of a factor of two would have been caught decades ago. An error of twenty-one per cent gets absorbed into a margin.

Measuring it — integrating the response rather than reading the corner — takes about ten lines given a solved network, and this page’s whole contribution is that nothing about it needs to be quoted.

A single-pole low-pass with its corner at 995 HzSolved at 209 frequencies. The straight-line sketch, drawn faintly, is 3.01 dB wrong at the corner and within a tenth of a decibel only below 152 Hz. The phase is already −5.7° a decade before the corner and −84° a decade after it.-60-40-200gain (decibels)the sketch: flat, then −20 dB/decade−3.01 dB-90-450101001k10k100kfrequency (hertz)phase (degrees)sketch within 0.1 dB below 152 Hzsolved, then checked — checked against a chain-matrix productthe sketch is 3.01 dB wrong at 995 Hz
Fig. 7 The single-pole response and the straight-line sketch drawn over it, from the frequency field. That figure measures a different error in the same picture: the asymptotic sketch is 3.01 dB wrong at the corner. Between the two pages, the same one-pole response has now been shown to be badly described by its asymptotes at the corner and badly summarised by its corner for noise — two independent ways in which the simplest response in the subject resists being reduced to a number.

What the integral says about measurement time

There is a use for the noise bandwidth that is not about filters at all, and it is worth ending on because it is where the quantity is most often needed and least often recognised.

Any measurement that averages is a filter. Averaging N samples taken at intervals of T is a low-pass filter whose noise bandwidth is 1/(2NT) — the reciprocal of twice the total measurement time — and that expression contains no reference to any corner frequency because a boxcar average does not have one in any useful sense.

So the noise on an averaged measurement is √(density/(2t)) where t is how long it took, and the resolution improves as the square root of the time. Resolving a microvolt against a ten nanovolt-per-root-hertz floor at a signal-to-noise ratio of ten needs a noise bandwidth of 100 Hz, which is five milliseconds of averaging. Ten times the resolution needs a hundred times the time — half a second — and a hundred times the resolution needs fifty seconds.

That relation is the third essay’s subject and it is the practical form of everything on this page. A bandwidth is not only a property of a filter someone chose; it is a property of how long a measurement was allowed to take, and the two are the same quantity computed the same way.

Where this goes next

±619.3 nV, the root mean square1200 of 192001 stepssampled, this seed621.2 nVsampled, 6 seeds619.3 nV ± 1.79%integrated from |H|²620.6 nV…ignoring the hold626.3 nVnoise bandwidth15.03 kHzthe −3 dB point10.00 kHzsolved, then checked — a sample against an integralagreeing to 0.21% 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. -90-60-3001001k10kfrequency (hertz)gain (decibels)ButterworthChebyshevBesselhalf power1.00 kHzthe passband, magnified-1-0.500000.2000.4000.6000.8001solved, then checked — three networks, 133 frequencies eachall normalised to a measured −3 dB at 1.00 kHz Three families, one corner Butterworth is flat, Chebyshev is steep, Bessel has good delay. None of those is a number, so the table they appear in cannot answer the question anybody has. Here each family's poles are computed from its definition, built as an actual network, and then measured — starting with the step every comparison skips. passband deviationdecibels, peak to trough below 0.8 f_cButterworth0.443 dBChebyshev0.500 dBBessel1.882 dBattenuation at three times the cornerdecibels downButterworth47.7 dBChebyshev64.0 dBBessel28.3 dBgroup-delay variation across the passbandper cent, slowest against fastestButterworth48.0%Chebyshev49.0%Bessel0.1%solved, then checked — nine measurements, three networksevery number here moves with the order What a steep skirt costs A filter's order buys attenuation at a known rate — twenty decibels per decade per pole, and no arrangement of components changes it. What varies between families is how quickly the slope is reached, and the currency it is paid for in is delay: the steepest of the three distorts delay eight hundred times more than the gentlest.