Filters and Scaling

Four numbers define a low-pass specification:

Nothing at all is specified between \(\omega_p\) and \(\omega_s\) — that gap is the transition band, and it is where the design does its work.

Why a region and not a curve. Specifying an exact response would over-constrain the problem: no rational function passes exactly through an arbitrary curve, and demanding it forbids every realisable filter. A tolerance region asks only what the application actually needs.

The two ratios that govern everything:

The order depends only on these two numbers — never on the absolute frequencies. A 1 kHz filter and a 1 GHz filter with the same ratios need the same order, which is the whole basis of the prototype method (Problem 10).

Which parameter costs the most. Tightening each in turn from the running example:

The transition band is by far the most expensive item. Narrowing it from 2:1 to 1.5:1 raises the order from 8 to 14, because it sits inside a logarithm in the denominator. Challenge C1 takes this to its logical extreme.

Other filter types need the specification restated but nothing else changes:

Problems 14 and 15 show that all of these reduce to a low-pass problem by substitution, so only the low-pass case ever needs solving.

aThe rational-function argument. Any lumped circuit has a transfer function that is a ratio of polynomials:

Such a function is analytic — infinitely differentiable — everywhere except at its poles. A brick-wall response has a discontinuity in \(|H|\) at \(\omega_c\):

A rational function cannot be identically 1 on an interval and identically 0 on another without being constant everywhere. No finite circuit produces a brick wall.

bThe causality argument, which is stronger and more revealing. The impulse response of an ideal low-pass is

This is non-zero for \(t < 0\) — the filter would have to respond before the input arrived. No physical system can do that, and the conclusion is independent of whether the circuit is lumped, distributed, active or digital.

The Paley–Wiener condition makes this precise. A causal system's magnitude response must satisfy

If \(|H| = 0\) over any interval, \(\ln|H| = -\infty\) there and the integral diverges. A causal filter may approach zero but can never reach it over a band of frequencies. Stopband attenuation is always finite.

What is achievable instead:

Each of the four impossibilities corresponds to one number in Problem 1's specification. The specification exists because the ideal does not.

The consolation. The approximation improves without limit as the order rises:

so any finite specification can be met — the question is only what it costs in components, and Problem 4 answers that.

The definition:

The \(-3\) dB point is always at \(\omega_c\) whatever the order — which makes \(\omega_c\) a natural normalising frequency but also a trap, since a specification quoting \(A_{\max} = 1\) dB does not put \(\omega_c\) at \(\omega_p\) (Problem 4).

The flatness proof. Expand as a geometric series for small \(u = \omega/\omega_c\):

Every term between the constant and \(u^{2n}\) is absent. The first non-constant term is of order \(2n\), so

and the first surviving derivative is the \(2n\)-th. That is what "maximally flat" means — no rational function of order \(n\) can have more vanishing derivatives at the origin.

Why maximal flatness is the natural optimum. Given \(n\) poles, one has \(n\) degrees of freedom. Butterworth spends every one of them on making the response flat at DC, leaving none for shaping elsewhere. Chebyshev spends them differently — Problem 6.

The price of flatness is a leisurely transition. Comparing the loss at \(\omega = 2\omega_c\):

Each extra order adds about 6 dB per octave, and the ultimate slope is \(-20n\) dB/dec — exactly the pole count of Set 28, Problem 2.

The first-order case is familiar. With \(n = 1\):

— the \(RC\) low-pass of Set 28, Problem 12. Every simple pole is a first-order Butterworth filter, so the family is a direct generalisation of the most elementary case.

Write the two constraints. Loss in decibels is \(A = 10\log_{10}\left(1+u^{2n}\right)\), so:

Divide to eliminate \(\omega_c\):

Confirming Problem 1: only the two ratios appear, never the absolute frequencies.

Applying to the running specification:

Always round up — a fractional order is not realisable, and rounding down would miss the specification.

Placing \(\omega_c\). This is the step most often skipped. From the passband constraint:

Not 1000. The \(-3\) dB point sits above the passband edge, because 1 dB of loss is reached before 3 dB is. Taking \(\omega_c = \omega_p\) would give 3.01 dB of passband loss instead of the specified 1 dB.

Verify both ends:

The passband is met exactly and the stopband exceeded — a consequence of rounding 7.62 up to 8. The surplus always appears in the stopband when \(\omega_c\) is placed from the passband constraint.

Confirm that \(n = 7\) fails:

Short by 3.7 dB, so 8 really is the minimum.

An alternative placement. \(\omega_c\) could instead be chosen from the stopband constraint, giving exactly 40 dB at 2000 rad/s and better than 1 dB in the passband. Any \(\omega_c\) between the two extremes satisfies the specification — the surplus from rounding up can be spent at either end, and a designer usually splits it.

Start from the magnitude-squared function. Since \(|H(j\omega)|^2 = H(s)H(-s)\) evaluated at \(s = j\omega\), substitute \(\omega^2 = -s^2\) with \(\omega_c = 1\):

The poles satisfy \(s^{2n} = -(-1)^n = (-1)^{n+1}\), which has \(2n\) roots all of modulus 1:

Every pole lies on the unit circle. That is the geometric content of maximal flatness.

Keep only the left-half-plane roots — the other \(n\) belong to \(H(-s)\), and taking them would give an unstable filter (Set 28, Problem 15). The retained poles are at

Equally spaced on the left semicircle, symmetric about the real axis, and never on the imaginary axis.

The poles, computed:

Odd orders always include a real pole at \(-1\); even orders do not. That single fact determines whether a realisation needs a first-order section.

The second-order sections, written as \(s^2+(\omega_0/Q)s+\omega_0^2\) with \(\omega_0 = 1\):

Note that \(n = 2\) gives \(Q = 0.7071\) — exactly the value at which peaking first appears (Set 28, Problem 10). A second-order Butterworth sits precisely on that boundary, which is another way of saying it is maximally flat.

Section \(Q\) rises with order. The highest-\(Q\) section is always the one whose poles lie closest to the imaginary axis:

This matters for realisation: Problem 18 shows that high-\(Q\) sections are the sensitive ones, so the order sets not just the component count but the precision required.

The idea. Butterworth spends all its freedom making the response flat at DC, which wastes accuracy where none is needed. Chebyshev instead lets the response ripple between 0 and \(A_{\max}\) across the whole passband, spending the freed-up freedom on a steeper skirt.

The Chebyshev polynomials have the key property of oscillating between \(\pm1\) inside \(|u| \le 1\) and growing explosively outside:

Compare with \(u^n\) at \(u = 2\): 2, 4, 8, 32. Chebyshev grows about \(2^{n-1}\) times faster than the Butterworth power law — and that factor is the entire advantage.

Deriving the order. In the stopband \(u > 1\), so the \(\cosh\) form applies:

The same two ratios as Butterworth, but with \(\cosh^{-1}\) replacing \(\log\) — and \(\cosh^{-1}\) grows more slowly, so the required order is smaller.

Applying to the running specification:

Three orders fewer — six fewer reactive components in an \(LC\) realisation, or two fewer op-amp sections in an active one.

Where \(\omega_p\) sits. Unlike Butterworth, the passband edge is defined by the ripple, so \(\omega_p\) needs no adjustment:

A 5.3 dB margin from rounding 4.54 up to 5. The normalising frequency is the ripple-band edge itself — simpler than Butterworth's \(\omega_c\) placement of Problem 4.

The ripple count. Inside the passband the response touches its extremes \(n\) times, alternating between 0 dB and \(A_{\max}\):

An even-order Chebyshev has loss at DC — a nuisance in a DC-coupled instrument, and a reason odd orders are often preferred.

The pole locations follow from solving \(1+\varepsilon^2C_n^2(s/j) = 0\). Defining

the left-half-plane poles are

This is an ellipse. Dividing and using \(\sin^2+\cos^2 = 1\):

Semi-minor axis \(\sinh a\) along the real axis, semi-major \(\cosh a\) along the imaginary. The Butterworth circle has been squashed horizontally — every pole pushed towards the imaginary axis, which is exactly what raises the \(Q\) values and steepens the skirt.

Computing for \(n = 5\), 1 dB ripple:

All five satisfy the ellipse equation to six decimal places. Note the poles have different \(\omega_0\) values — unlike Butterworth, where every section shares \(\omega_c\).

Comparing the section \(Q\) values:

Chebyshev's highest section \(Q\) is 3.4 times Butterworth's. That is the hidden cost of the three saved orders: Problem 18 shows high-\(Q\) sections are disproportionately sensitive to component tolerance, and Challenge C1 finds a design needing \(Q = 35\).

Why more ripple means higher \(Q\). Increasing \(A_{\max}\) increases \(\varepsilon\), hence decreases \(a\) and \(\sinh a\):

and poles near the axis mean high \(Q\) (Set 29, Problem 14). The ripple, the steepness and the sensitivity are three views of one geometric fact.

The limiting cases:

Butterworth is the zero-ripple limit of Chebyshev, so the two families are not rivals but ends of one continuum.

The four families differ in where they permit ripple:

Each concession buys order. Permitting ripple in one band saves three orders; permitting it in both saves four.

Why elliptic is the most economical. It places zeros in the stopband — finite frequencies of infinite attenuation — rather than sending all its zeros to infinity:

An elliptic filter's stopband does not keep falling — it levels off at \(A_{\min}\). That is fine if only \(A_{\min}\) was ever required, and a serious problem if something far above the stopband edge must also be rejected.

What each family costs:

The choice is made by the application, not the mathematics:

The last is worth noting: an elliptic filter's finite zeros can be positioned deliberately, so a known interferer can be nulled exactly rather than merely attenuated.

A caution on the order comparison. Order is not the same as cost. A 4th-order elliptic needs finite zeros, which in an \(LC\) realisation means resonant traps — more components per order — and in an active realisation means additional feedback paths. An 8th-order Butterworth may be cheaper to build than a 4th-order elliptic and will certainly be easier to tune.

What Bessel optimises. Set 28, Problem 17 established that constant group delay means no waveform distortion. A Bessel filter is maximally flat in delay rather than in magnitude:

The same optimisation principle applied to a different quantity — which is why the two families look so different despite sharing a design philosophy.

The comparison at \(n = 5\), normalised, with delay in units of \(1/\omega_c\):

Bessel's delay varies by under 0.3% across the whole passband; Chebyshev's more than doubles. The sharper the magnitude cut-off, the worse the delay — a rule that holds across every family.

The cost. Bessel's magnitude roll-off is very gentle:

A 5th-order Bessel is barely better than a 2nd-order Butterworth in the transition band. Delay flatness is expensive in selectivity — the reverse of Chebyshev's bargain.

Why it matters for pulses. A rectangular pulse through each filter:

The ringing comes from the high-\(Q\) sections of Problem 7 — \(Q = 5.56\) rings for \(5.56/\pi = 1.8\) cycles (Set 29, Problem 18), and that is visible on every edge.

When each is right:

A useful test: does the application look at the waveform, or measure a spectrum? If the waveform matters, delay matters, and Bessel is the family to start from.

The hybrid solution. Where both are needed, use a sharp filter followed by an all-pass delay equaliser (Set 28, Problem 16) that flattens the delay without touching the magnitude. This costs extra sections but decouples the two requirements — the only way to have both.

The prototype is a filter designed with two arbitrary normalisations:

Both are conventions with no physical meaning — chosen because they make the tabulated element values simplest.

Why this loses nothing. Problem 1 established that the order depends only on ratios. Two further facts complete the argument:

Three transformations reach every filter from one prototype. That is why textbooks tabulate a handful of prototype values rather than an infinite catalogue of designs.

The tabulated values. For a doubly-terminated Butterworth \(LC\) ladder:

A closed form, so no table is strictly necessary. The values alternate between shunt capacitors and series inductors, and are symmetric — a consequence of the source and load being equal.

Reading the ladder. The \(g\) values are used in either of two dual arrangements:

Both give identical responses — Set 18's duality again. The choice is made on practical grounds: fewer inductors is usually better, so capacitor-input is preferred for odd orders.

A worked check. For \(n = 3\) the prototype is \(C_1 = 1\) F, \(L_2 = 2\) H, \(C_3 = 1\) F between 1 Ω terminations. Its transfer function is

exactly the 3rd-order Butterworth polynomial of Problem 5 ✓. The ladder is the polynomial, realised in components.

The workflow this enables:

Only the first two steps involve the specification; everything after is mechanical. That separation is what makes filter design a procedure rather than an art.

The requirement. A response identical in shape but stretched along the frequency axis by \(k_f\) means

Whatever the prototype does at 1 rad/s, the new filter does at \(k_f\) rad/s — and at every other point in proportion.

What each element must become. The substitution must leave every impedance unchanged at the corresponding frequency:

Both reactive elements are divided by \(k_f\); resistors are untouched. That asymmetry is the whole content of the rule — resistance has no frequency dependence to scale.

A worked case. Take the \(n = 3\) prototype (\(C_1 = 1\) F, \(L_2 = 2\) H, \(C_3 = 1\) F) to \(\omega_c = 10\,000\) rad/s:

The cut-off is now correct, but the impedance level is absurd — 100 µF driven from 1 Ω. Problem 12 fixes that independently.

Verify the shape is preserved. Every resonance in the network scales together:

and every section \(Q\) is unchanged:

The \(k_f\) cancels in the ratio. Frequencies move; damping does not — which is exactly what "same shape" means.

Poles move radially. In the \(s\)-plane the substitution \(s \to s/k_f\) multiplies every pole by \(k_f\):

Butterworth's unit circle becomes a circle of radius \(k_f\); Chebyshev's ellipse scales likewise. The pattern is unaltered, and so is every angle — hence every \(Q\).

The requirement. Multiply every impedance in the network by \(k_m\), at every frequency:

Element by element:

Resistance and inductance multiply; capacitance divides. The inversion for \(C\) is because capacitive impedance goes as \(1/C\).

Why the voltage transfer function is unchanged. Any voltage ratio in a network is a ratio of impedances:

The \(k_m\) cancels identically. Every voltage in the network is the same; every current is divided by \(k_m\); every power is divided by \(k_m\). This is Set 27's referring operation, applied uniformly instead of across a transformer.

Why one would want to. Impedance level is a free choice, and it is used to make the components buildable:

There is always a tension: raising \(k_m\) shrinks capacitors and grows inductors, and vice versa. The optimum is where both are reasonable.

Continuing the worked case. Applying \(k_m = 600\) to the frequency-scaled network of Problem 11:

Now entirely practical: standard capacitor values and a buildable inductor.

The two scalings are independent and commute, so they can be combined into one step:

Note \(k_m\) appears in the numerator for \(L\) and the denominator for \(C\), while \(k_f\) is in the denominator for both. Getting either wrong is caught immediately by checking \(1/\sqrt{L'C'}\).

1Prototype values from Problem 10:

2The scaling factors:

3Apply the combined rules:

4Check one — the characteristic frequency:

Not \(\omega_c\) itself — this is the resonance of one pair of elements, and for the 3rd-order prototype \(1/\sqrt{L_2C_1} = 1/\sqrt2 = 0.7071\) in normalised terms. Scaled, that is \(0.7071 \times 10^4\) ✓ — the ratio is preserved, which is the real test.

5Check two — the impedance level:

In the prototype, \(\sqrt{L_2/C_1} = \sqrt2 = 1.414\ \Omega\). Multiplying by \(k_m = 600\) gives \(848.5\) ✓. Frequency and impedance checks together confirm both factors, since one tests \(1/\sqrt{LC}\) and the other \(\sqrt{L/C}\).

6Check three — the response:

A practical note on the terminations. A doubly-terminated ladder has 6 dB of insertion loss at DC, because the source and load form a divider. That is not a defect — it is the condition for the ladder to realise the Butterworth polynomial exactly, and it buys the lowest sensitivity of any realisation (Problem 18).

Choosing \(k_m\) in practice. Had 600 Ω not been dictated, the choice would be a compromise:

The substitution:

This maps \(\omega = 0 \leftrightarrow \omega = \infty\) and leaves \(\omega = \omega_c\) fixed — it turns the frequency axis inside out about the cut-off. What the prototype does at low frequency, the transformed filter does at high.

What happens to each element:

Every \(L\) becomes a \(C\) and every \(C\) becomes an \(L\), in the same position. Series stays series, shunt stays shunt — only the element type changes. Resistors are untouched.

Applying to Problem 13's design (\(\omega_c = 10^4\) rad/s, 600 Ω). Transform the prototype first, then scale:

Checking the first: \(L_1' = k_mL/k_f = 600(1)/10^4 = 60\) mH ✓. And the second: \(C_2' = 0.5/(600\times10^4) = 83.33\) nF ✓.

The response, mirrored:

Exactly exchanged about \(\omega_c\), on a logarithmic axis. The transformation is a reflection in \(\log\omega\), which is why the pairs of frequencies satisfy \(\omega_a\omega_b = \omega_c^2\).

What happens to the poles. The substitution maps a pole at \(s_k\) to one at \(\omega_c/s_k\):

The Butterworth circle maps to itself, and the angles are conjugated. Additionally, \(n\) zeros appear at the origin — which is what makes the response rise at \(+20n\) dB/dec below cut-off, and why the order does not change.

A caution on the order formula. For a high-pass specification, \(\omega_s < \omega_p\), so the ratio must be inverted before use:

Otherwise the logarithm is negative and the order comes out negative — an error that at least announces itself clearly.

The substitution:

where \(\omega_0\) is the centre frequency and \(B\) the bandwidth. Evaluating on the imaginary axis:

This is exactly Set 29's detuning variable \(\delta\) scaled by \(\omega_0/B = Q\). The band-pass transformation and the universal resonance curve are the same idea.

Why the order doubles. The substitution is quadratic in \(s\), so each prototype pole becomes two:

The last row explains the doubling physically: the prototype's single stopband maps to two stopbands, one on each side, and each needs its own poles.

What each element becomes. A series inductor sees

a series \(LC\) resonator, and a shunt capacitor similarly becomes a parallel \(LC\) resonator:

Every new pair resonates at \(\omega_0\) — that is the check to apply after every band-pass transformation.

Applying to the 3rd-order prototype at \(\omega_0 = 10^4\), \(B = 2000\) rad/s (so \(Q = 5\)), with \(k_m = 600\):

Three prototype elements became six, confirming the doubling — a 6th-order band-pass from a 3rd-order low-pass.

The narrow-band difficulty. As \(Q = \omega_0/B\) rises, the element values spread apart:

At \(Q = 50\) the spread would be 5000:1, and no set of practical components covers it. Narrow-band filters cannot be built by transforming a low-pass prototype — coupled-resonator methods (Set 29, Problem 17) are used instead.

Analyse the ladder — 1 Ω source, shunt \(C_1 = 1\), series \(L_2 = 2\), shunt \(C_3 = 1\), 1 Ω load. Working back from the load:

Carrying through the voltage division from source to load gives

Exactly the 3rd-order Butterworth polynomial of Problem 5. The factor of \(\frac12\) is the 6 dB insertion loss of the two equal terminations — a constant, so it does not affect the shape.

Verify the response:

Why doubly terminated. Three properties follow from having equal resistive terminations:

The sensitivity argument is worth spelling out, because it is elegant. In the passband a doubly-terminated ladder is designed for maximum power transfer, so the transmitted power is at a stationary point:

A first-order change in any element produces only a second-order change in the passband response. That is why \(LC\) ladders tolerate 5% components where an active realisation of the same filter would need 1% — Problem 18 quantifies the contrast.

The limitations that push designers towards active realisations:

At radio frequencies none of these bites hard and \(LC\) ladders remain standard. Below about 100 kHz they become impractical, and Problem 17's active sections take over.

The topology. Two \(RC\) sections feeding a non-inverting amplifier of gain \(K\), with the first capacitor returned to the output rather than to ground. That feedback path is what creates the complex poles — a passive \(RC\) network alone can only have real poles.

The transfer function (unity-gain case, \(K = 1\)):

Comparing with the standard form \(s^2+(\omega_0/Q)s+\omega_0^2\):

aSetting \(Q\) by the gain. With \(R_1 = R_2 = R\) and \(C_1 = C_2 = C\) and an amplifier of gain \(K\):

At \(K = 3\) the poles reach the imaginary axis and the circuit oscillates. Every equal-component Sallen–Key section is an oscillator that has been detuned by a controlled amount — which is exactly why Problem 18 finds it so sensitive.

bSetting \(Q\) by component ratios. With unity gain (\(K = 1\), a simple follower) and \(R_1 = R_2 = R\):

The capacitor ratio goes as \(4Q^2\), so \(Q = 5\) needs a 100:1 spread and \(Q = 10\) needs 400:1. The unity-gain form trades gain sensitivity for component spread — a different problem, not a lesser one.

Which to choose:

Above \(Q \approx 10\) neither serves, and a different topology — state-variable or biquad — is used, at the cost of three op-amps per section instead of one.

Why active realisation matters. A Sallen–Key section contains no inductor, which removes every one of Problem 16's four objections at once. That is why every filter below about 100 kHz — audio, instrumentation, anti-aliasing — is built this way.

The definition — a normalised derivative, giving the percentage change in \(y\) per percentage change in \(x\):

A sensitivity of 1 means a 1% component error gives a 1% parameter error. Anything much above 1 is a warning.

Compute for the Sallen–Key. With \(Q = 1/(3-K)\):

and substituting \(K = 3-1/Q\):

Sensitivity grows linearly with \(Q\). This is not a defect of the topology but a consequence of operating near oscillation.

What it costs:

At \(Q = 10\), a 1% gain error — easily caused by resistor tolerance in the feedback divider — produces a 29% error in \(Q\). The section is unusable above \(Q \approx 5\) with ordinary components.

Contrast with the passive ladder. Problem 16 showed a doubly-terminated ladder has zero first-order passband sensitivity:

This is why ladder-simulating active filters — leapfrog and gyrator topologies — exist. They reproduce the ladder's internal structure with op-amps and inherit its low sensitivity, at the price of many more amplifiers.

The sensitivity of \(\omega_0\) is far better behaved:

Only \(\pm\frac12\), independent of \(Q\). Centre frequency is easy; \(Q\) is hard — which is why a filter that has drifted usually shows the right cut-off with the wrong shape.

The practical rules that follow:

The sections, from Problem 7:

The overall response has only 1 dB of ripple, but section A alone peaks nearly 15 dB. The sections' peaks are cancelled by the others' roll-off — and that cancellation happens only at the output, not inside.

Why the order matters at all. Mathematically the product is commutative, so the overall transfer function is identical whatever the sequence. Physically it is not:

aThe dynamic-range argument. Suppose a strong out-of-band signal near \(\omega_0 = 0.994\) arrives:

Once a stage clips, filtering afterwards cannot undo it — clipping is non-linear and creates new frequencies inside the passband. This argues for placing the highest-\(Q\) section last.

bThe noise argument, which points the other way. Noise generated in the first section is amplified by every section after it:

Putting the high-gain section first means its noise is not subsequently boosted — arguing for A first. The two considerations conflict.

The resolution depends on which constraint binds:

Ascending \(Q\) is the standard default, because clipping is a catastrophic non-linear failure while excess noise is a graceful degradation.

A refinement: gain distribution. Each section's DC gain can be adjusted freely provided the product is correct. Choosing them so that every section's peak output is equal maximises the overall dynamic range:

This is scaling for maximum dynamic range, and it is done routinely in digital filter implementations for exactly the same reason.

The complete procedure, in six steps:

Only step 2 requires judgement. Steps 3 to 5 are arithmetic, and step 6 is where the judgement is validated.

The running specification, all four ways:

Every row meets the same specification. They differ only in what they sacrifice — which is the entire content of filter design.

The three conservation laws that no design escapes:

The second is the one most often overlooked: a 4th-order elliptic with 1% components may perform worse than an 8th-order Butterworth with 5% ones.

What has been assumed throughout. Every result in Sets 28 to 30 rests on a single unexamined premise: that \(H(j\omega)\) describes the circuit completely. But it describes only the steady state:

Every one of these connects the frequency domain to the time domain, and none of them can be derived from \(H(j\omega)\) alone — because \(s\) has been used only as shorthand for \(j\omega\).

What comes next:

Set 31 is the pivot. Letting \(s\) be genuinely complex converts every observation in the table above from a noticed coincidence into a theorem.