Resonance, Quality Factor and Bandwidth

The impedance:

Resonance is where the reactance vanishes:

Note \(R\) does not appear. The resonant frequency of a series circuit is independent of its resistance — which is not true of the parallel case in Problem 8.

The numbers:

Behaviour either side:

The capacitor dominates below resonance because \(X_C = 1/\omega C\) grows as \(\omega\) falls — the opposite of the intuition many people start with.

At resonance the circuit forgets it contains reactance. With a 10 V source:

Exactly what a bare 10 Ω resistor would draw. Yet Problem 3 shows 316 V standing across the inductor at the same moment.

Why the cancellation is exact. The two reactances are of opposite sign because a capacitor's current leads and an inductor's lags — they are 180° apart at every frequency, not merely at \(\omega_0\). Resonance is simply where their magnitudes also match:

Where \(\omega_0\) sits geometrically. Plotting \(X_L\) and \(X_C\) against \(\omega\) on log axes gives two straight lines of slope \(+1\) and \(-1\); they cross at \(\omega_0\), and the crossing height is

This quantity is the characteristic impedance of the resonant circuit, and Problem 2 shows that \(Q\) is simply its ratio to \(R\).

aThe reactance ratio, the easiest to compute:

Three algebraically identical forms:

The third is the most useful in design, because it contains no frequency — \(Q\) is fixed by the components alone.

bThe energy definition. With \(V = 10\) V RMS the current is 1 A RMS, so its peak is \(\sqrt2\) A:

Energy dissipated per cycle:

cThe bandwidth definition, anticipating Problem 4:

All three agree to five figures. They are not three coincidences but three readings of one physical fact.

Proving the energy and reactance forms equivalent in general:

The current cancels, so \(Q\) is a property of the circuit and not of how hard it is driven — as it must be for a useful figure of merit.

Which definition to use when:

The energy form is the only one that generalises: a pendulum, a quartz crystal and a laser cavity all have a \(Q\) defined exactly this way, with no reactance or bandwidth in sight.

The current, from Problem 1:

The three element voltages:

How KVL survives this. The two large voltages are exactly antiphase:

KVL applies to phasors, not to magnitudes. Adding \(10 + 316 + 316 = 642\) V would be meaningless — and this is precisely the case where the distinction has practical consequences.

The voltages are real and measurable. A voltmeter across the inductor genuinely reads 316 V, and the insulation genuinely sees it:

A capacitor rated for 100 V would fail immediately, from a 10 V supply. Problem 19 develops this into a genuine power-system hazard.

Where the magnification comes from. Not from any energy gain — the circuit is passive. The large voltages accompany a large circulating energy that shuttles between \(L\) and \(C\) (Problem 6), and the source supplies only the small amount lost in \(R\) each cycle. The source tops up a reservoir; it does not fill it each cycle.

The corresponding parallel result. By duality (Problem 7), a parallel circuit magnifies current by \(Q\): a 1 mA source produces 31.6 mA circulating round the \(LC\) loop, again with the two branch currents antiphase and cancelling at the node.

The definition. The half-power points are where the power delivered falls to half its resonant value:

Half power in the resistor, hence \(1/\sqrt2\) of the current — the \(-3.01\) dB of Set 28, Problem 3.

The condition on impedance. Since \(I = V/|\mathbf{Z}|\) and \(|\mathbf{Z}|\) is minimum at \(R\):

The half-power points are where the net reactance equals the resistance — a clean characterisation, and the phase angle there is \(\pm45°\).

Deriving the bandwidth. At the two half-power frequencies:

Subtracting:

Problem 5 shows \(\omega_1\omega_2 = \omega_0^2 = 1/LC\), so the bracket is \(L+L = 2L\):

Evaluating:

A circuit resonating at 503 Hz responds over only 16 Hz — the selectivity that makes tuning possible.

Bandwidth depends only on \(R\) and \(L\). Changing \(C\) retunes the circuit without altering its bandwidth in rad/s:

This is why a tuned radio's bandwidth changes across the band: tuning with a variable capacitor holds BW constant in rad/s while \(\omega_0\) moves, so \(Q\) — and hence selectivity — varies across the dial.

Start from the half-power condition \(|X| = R\) of Problem 4:

Solving the quadratic and keeping the positive roots:

The geometric mean. Multiplying the two roots:

Exactly, for every \(Q\). The resonant frequency is the geometric mean of the half-power points, never the arithmetic mean — which is why a resonance curve is symmetric on a logarithmic frequency axis, as Set 28, Problem 11 also found.

Evaluating for the running example (\(Q = 31.62\)):

The separation is exactly \(\text{BW}\) regardless of \(Q\); only the placement is shifted. The arithmetic mean exceeds \(\omega_0\) by 0.0125%, so the resonance sits slightly below the midpoint of its own half-power band.

Why the approximation is so good here. The correction term is

It improves as \(1/Q^2\), so the usual rule "\(Q > 10\) and the symmetric approximation is fine" is well founded.

A useful corollary. Since \(\omega_1\omega_2 = \omega_0^2\), knowing any two of \(\omega_1\), \(\omega_2\), \(\omega_0\) gives the third at once — handy when a measured curve gives the half-power points but not the peak, which is common when the peak is noisy or the sweep is coarse.

The instantaneous stored energies. With \(i = I_m\cos\omega_0t\), the capacitor voltage lags the current by 90°:

At resonance \(\omega_0^2 = 1/LC\), so the capacitor's coefficient becomes

The two coefficients are identical, so

The energy does not pulsate — it merely moves, from the magnetic field to the electric field and back, twice per cycle.

The numbers. With \(I = 1\) A RMS, \(I_m = \sqrt2\) A:

Check via the capacitor, whose peak voltage is \(316.2\sqrt2 = 447.2\) V:

Compare with what the source supplies. Per cycle:

The circuit holds five times more energy than it loses in a whole cycle. That reservoir is the physical origin of the voltage magnification of Problem 3 — 316 V is what 0.1 J stored in a 1 µF capacitor looks like.

Why this only happens at resonance. Away from \(\omega_0\) the two coefficients differ, and the total oscillates at \(2\omega\):

The source must then supply and reabsorb energy each half-cycle — which is exactly the reactive power \(Q\) of Set 23, and it vanishes at resonance because the exchange becomes entirely internal.

The general principle. Set 23's Problem 7 established \(Q_{\text{reactive}} = 2\omega \times \text{average stored energy}\). At resonance the inductor's and capacitor's contributions are equal and opposite:

Resonance is precisely the condition of zero net reactive power — the definition used in power engineering, and identical to \(X_L = X_C\).

Work with admittance, the natural quantity for parallel elements:

Resonance is where the susceptance vanishes:

The same expression as the series case — for an ideal parallel circuit. Problem 8 shows what happens when the coil has resistance.

The quality factor, in three equivalent forms:

Note the inversion. In series, \(Q = \omega_0L/R\) — resistance in the denominator. In parallel, resistance is in the numerator. A high-\(Q\) series circuit needs small \(R\); a high-\(Q\) parallel circuit needs large \(R\). Both mean "little loss".

The bandwidth:

The dual of \(R/L\) under \(R \to 1/R\), \(L \to C\) ✓.

Current magnification. With a 1 mA source at resonance:

31.6 mA circulating from a 1 mA source — the dual of Problem 3's voltage magnification. The two branch currents cancel at the node (KCL, phasors again), so the source sees only the 1 mA through \(R\).

The full duality:

The last row is the practical distinction: put a series circuit in the signal path to pass one frequency; put a parallel circuit across it to block one. A parallel tank in series with a load is the standard band-pass arrangement, because its high impedance at \(\omega_0\) becomes a high output voltage.

The impedance:

Unity power factor requires the imaginary part of \(\mathbf{Z}\) to vanish. Rationalising, the imaginary part of the numerator is

Below \(\omega_0 = 3162.28\) by 0.050%. Unlike the series case, resistance now shifts the resonant frequency.

The impedance at \(\omega_r\). Substituting the resonance condition into the real part gives, after simplification,

Called the dynamic resistance. Note the inversion: less coil resistance gives more impedance at resonance — a lossless coil across a capacitor would present an infinite impedance.

Three candidate frequencies, all distinct:

The impedance is not maximum where the power factor is unity. The peak of \(|\mathbf{Z}|\) sits essentially at \(\omega_0\), while the unity-pf point is 1.58 rad/s lower. Textbooks that use the two interchangeably are relying on \(Q\) being large.

How the discrepancy scales with \(Q\):

At \(Q \le 1\) the square root turns imaginary — a sufficiently lossy coil across a capacitor never reaches unity power factor at any frequency. The circuit remains inductive throughout.

The practical rule. For \(Q > 10\), treat \(\omega_r = \omega_0\) and \(Z_r = L/CR\); below that, use the exact expressions. Since most tuned circuits are built with \(Q\) of 50 to 200, the approximation is nearly always safe — but knowing why it is safe is what stops it being applied to a \(Q = 2\) circuit.

Equate the admittances. For the series form:

Matching real and imaginary parts against \(1/R_p + 1/jX_p\):

using \(Q = X_s/R_s\) for the element itself.

Applying at \(\omega_0 = 3162.3\), where \(X_s = 316.23\ \Omega\) and \(Q = 31.623\):

Compare with the exact result of Problem 8:

They differ by 0.1%, and the reason is instructive: the two are evaluated at slightly different frequencies. \(L/CR\) is the impedance at the unity-pf frequency \(\omega_r\); \(R_s(1+Q^2)\) is the parallel equivalent at \(\omega_0\). Neither is wrong.

The high-\(Q\) simplification. When \(Q \gg 1\):

The reactance barely changes, and the resistance is multiplied by \(Q^2\). Here \(Q^2R_s = 10\,000\ \Omega\) — recovering \(L/CR\) exactly, since \(Q^2R = (L/CR^2)R = L/CR\).

Why the conversion is worth having:

The conversion is frequency-dependent. \(Q = X_s/R_s\) changes with \(\omega\), so \(R_p\) and \(X_p\) are valid only near the frequency where they were computed. This is an equivalence at a point, not an identity — the same qualification as the \(\Delta\)–Y transformation's frequency dependence when reactances are involved.

A coil's \(Q\). Every winding has resistance, so a real inductor is \(L\) in series with \(R_s\):

Typical values: 50–200 for an air-cored RF coil, 10–100 for a ferrite-cored one, over 10 000 for a superconducting cavity.

A capacitor's dissipation factor. Losses in the dielectric and the leads appear as a small series resistance (the ESR):

The angle \(\delta\) is the departure from a perfect 90° between voltage and current. Good film capacitors reach \(D = 10^{-4}\), so \(Q_C = 10\,000\); electrolytics may be \(D = 0.1\), so \(Q_C = 10\).

How the two combine. Convert both to parallel form (Problem 9); the loss conductances then simply add:

The same form as resistors in parallel — the total is always below the smaller of the two.

Worked cases:

Improving the better component is nearly pointless. In the last row, doubling \(Q_L\) from 200 to 400 raises the total only from 18.2 to 19.0 — the capacitor is the bottleneck, and no amount of coil quality will help.

Why the coil is usually the culprit. Three loss mechanisms, all growing with frequency:

Because \(Q_L = \omega L/R_s\) and \(R_s\) grows as \(\sqrt f\) from skin effect alone, \(Q_L\) rises only as \(\sqrt f\) rather than as \(f\) — and eventually falls once core and proximity losses take over. Every coil has a frequency of peak \(Q\).

The design consequence. Set 27's transformer needed a low-loss core for efficiency; a tuned circuit needs it for selectivity. The two requirements point the same way, which is why the same grain-oriented and ferrite materials serve both.

A load appears in parallel with the tank, so the two resistances combine:

Tabulating, with \(\omega_0L = 316.23\ \Omega\):

A load equal to the dynamic resistance halves \(Q\) and doubles the bandwidth. A load ten times smaller destroys the resonance almost entirely.

The reciprocal rule. Defining an "external \(Q\)" from the load alone:

Check with \(R_L = 10\ \text{k}\Omega\): \(Q_{\text{ext}} = 31.62\), so \(1/Q_L = 1/31.62+1/31.62\) and \(Q_L = 15.81\) ✓. The same reciprocal structure as Problem 10's component losses — because loading is just another loss mechanism.

The designer's dilemma. Coupling a tank to anything useful degrades it:

Selectivity and power transfer are in direct conflict, and the resolution is the same as Set 27's: use a transformer or a tap to present the load as a larger equivalent resistance.

The tapped-coil solution. Connecting the load across a fraction \(n\) of the coil turns transforms it by \(1/n^2\):

To restore \(Q_L = 25\) with the 10 kΩ load, we need \(R_{\text{total}} = 25(316.23) = 7906\ \Omega\), hence \(R_L' = 37\,750\ \Omega\):

Tapping at about half the turns. The load still receives power; it simply no longer sees the full tank voltage — Set 27's \(a^2\) doing exactly the job it did for the loudspeaker.

Where each regime is wanted:

The normalised response. For a series \(RLC\), writing the detuning as

Everything about the shape is contained in the product \(Q\delta\) — the basis of Problem 13's universal curve.

Bandwidth at any attenuation. Setting the response to \(1/r\):

The shape factor measures how nearly rectangular the response is:

An ideal filter would have SF = 1. A single tuned circuit is 1000 — about as far from rectangular as a response can be while still being called selective.

Why radio needs better. AM broadcast channels are spaced 9 or 10 kHz apart, and a receiver must pass its own channel while rejecting the neighbour by at least 40 dB:

A single circuit sharp enough to reject the adjacent channel would be far too narrow to pass the wanted one's sidebands. The requirement is not more \(Q\) but a different shape — steeper skirts with a flat top.

The three solutions, all of which appear later:

Two critically coupled circuits reduce the shape factor to about 30; a six-pole Chebyshev filter reaches 2. That progression is the whole subject of filter design.

The underlying reason. A single resonance is a two-pole response, so its ultimate roll-off is \(-40\) dB/dec (Set 28, Problem 9) — and 40 dB per decade is simply not steep when the channels are 0.4% apart in frequency. Steeper skirts require more poles, and there is no way round it.

Approximate the detuning. Writing \(\omega = \omega_0+\Delta\omega\):

For \(\Delta\omega \ll \omega_0\) the numerator's second factor is about \(2\omega_0\) and the denominator about \(\omega_0^2\):

The universal form. Substituting into Problem 12's response:

This is exactly the simple-pole table of Set 28, Problem 5. Near resonance, a two-pole band-pass behaves like a one-pole low-pass in the variable \(x\) — which is why one curve serves every resonant circuit ever built.

Why this is worth having. The variable \(x\) contains \(Q\), \(\omega_0\) and \(\Delta\omega\) in one number:

A 500 Hz circuit with \(Q = 30\) and a 100 MHz circuit with \(Q = 3000\) have identical curves in \(x\). Before computers this was plotted once and reused for every design.

The accuracy, checked against the exact \(\delta\) for the running example:

Under 1% within the half-power band and under 2% within a full bandwidth. The approximation is excellent exactly where the response matters and degrades only far out on the skirts, where the response is small anyway.

The asymmetry it conceals. The approximation makes the curve symmetric in \(\Delta\omega\), whereas the true response is symmetric in \(\log\omega\) (Problem 5). The two agree near \(\omega_0\) and diverge on the skirts — the exact curve falls off faster below resonance than above it.

The characteristic equation, from the series impedance:

Comparing coefficients:

This is the bridge to Set 28. Everything said there about \(\zeta\) translates directly.

The poles:

What each coordinate means:

The pole's radius is the resonant frequency and its distance from the axis is half the bandwidth. A high-\(Q\) circuit has poles hugging the imaginary axis; pushing them onto it would give zero bandwidth and perpetual oscillation.

Why the peak is so tall. The frequency response is \(|H(j\omega)|\) evaluated along the imaginary axis. As \(\omega\) passes the pole's height, the distance from the evaluation point to the pole shrinks to just 50 — the pole's tiny offset:

The nearer the pole to the axis, the sharper and taller the peak — which is the geometric picture behind \(M_r = 1/2\zeta = Q\).

Three descriptions, one object:

The time constant of the ringing envelope is \(1/50 = 20\) ms, and its reciprocal is half the bandwidth in rad/s. Bandwidth and decay rate are literally the same number — Problem 18 develops this.

Where \(\omega_d\) differs from \(\omega_0\). Here by 0.0125%, entirely negligible. At \(Q = 1\) (\(\zeta = 0.5\)) the difference is 13.4%, and at \(Q = 0.5\) the poles become real and there is no oscillation at all — Set 19's critical damping, reappearing as \(Q = 0.5\).

The required \(Q\):

Find \(L\) from \(Q\) and \(R\). The source resistance is the circuit's resistance, so \(R = 50\ \Omega\):

Find \(C\) from the resonance condition:

Check both specifications:

The characteristic impedance \(\sqrt{L/C} = 5000\ \Omega\), which is \(Q\) times the resistance — as Problem 1 anticipated.

The component voltages. With a 1 V source at resonance:

A 1 V input places 100 V across a 31.8 pF capacitor. Component ratings must be checked at \(Q\) times the input, not at the input — Problem 3's warning made concrete.

Is the design realisable? Three checks:

The second is the trap. The design assumed \(R = 50\ \Omega\) is the only resistance, but the coil adds its own. If \(Q_L = 100\) the coil contributes another 50 Ω, halving the circuit \(Q\) to 50 and doubling the bandwidth to 20 kHz — the specification is missed by a factor of two.

The corrected design. To achieve a loaded \(Q\) of 100 with a coil of \(Q_L = 200\), use Problem 11's reciprocal rule:

so the external resistance must be halved to 25 Ω — achieved with a matching network or a tap, since the source is fixed at 50 Ω.

Invert the resonance condition:

The two extremes:

The square law is the awkwardness. A modest 3:1 frequency range demands a nearly 9:1 capacitance range — and mechanically variable capacitors are limited to about 10:1 before they become impractically large.

Why stray capacitance makes it worse. Wiring and valve or transistor capacitance add perhaps 20 pF in parallel, which cannot be tuned out:

The stray capacitance nearly doubles the required tuning ratio, and puts it beyond what a single gang can achieve. This is a real and historically important design constraint.

The bandwidth across the band. With a coil of \(Q_L = 100\), the circuit \(Q\) is roughly constant, so:

The bandwidth varies threefold across the dial — too narrow at the low end (audio sidebands lost) and too wide at the high end (adjacent-channel breakthrough). A single tuned circuit cannot serve the whole band well.

The historical solution: the superheterodyne. Rather than filtering at the signal frequency, mix everything down to a fixed intermediate frequency and filter there:

Almost every receiver built since 1930 works this way, and Problem 12's shape-factor requirement is met by cascading several fixed IF transformers — Problem 17.

A design check on the numbers. At 1000 kHz the tank's characteristic impedance is

so a coil \(Q\) of 100 corresponds to a series resistance of about 15 Ω and a dynamic resistance of about 151 kΩ — comfortably high, which is why the following stage must present a very light load or use a tap (Problem 11).

The transfer function, from Set 26's coupled mesh equations with both loops tuned:

The \((\omega M)^2\) in the denominator is the reflected impedance of Set 26, Problem 13 — and it is what creates the interesting behaviour.

At resonance both \(\mathbf{Z}_{11}\) and \(\mathbf{Z}_{22}\) reduce to \(R\):

Maximise over \(M\):

The general form covers unequal circuits. Critical coupling maximises the transfer at resonance, and the maximum value is \(1/2R\) — half what an ideal transformer would give, because the reflected resistance exactly matches the source.

Three regimes, verified numerically:

The peak height stops rising at critical coupling. Beyond it, extra coupling does not increase the maximum transfer — it merely splits the single peak into two of the same height. This is the key result.

Why the peak splits. Above critical coupling the two resonators interact strongly enough that the combined system has two normal modes:

The same phenomenon as two coupled pendulums swinging in phase and in antiphase at slightly different rates.

Why overcoupling is deliberately used. A slightly overcoupled pair gives a flat-topped, steep-sided response — exactly the shape Problem 12 said a receiver needs:

Two critically coupled circuits reduce the shape factor from 1000 to roughly 30 — a thirtyfold improvement in selectivity from a single extra resonator.

The transient, from Problem 14's poles at \(-\zeta\omega_0 \pm j\omega_d\):

The envelope time constant:

using \(\zeta = 1/2Q\). For the running example, \(\tau = 2(31.62)/3162.3 = 20\) ms.

Count the cycles. The period is \(T = 2\pi/\omega_0\), so

And to fall to 1%:

The link to bandwidth. The envelope decay rate is

The ringing time constant is fixed entirely by the bandwidth, with no reference to the centre frequency. Here \(\tau = 2/100 = 20\) ms ✓ — a 1 MHz circuit and a 500 Hz circuit with the same bandwidth ring for the same time, though wildly different numbers of cycles.

The trade-off made explicit:

Selectivity and speed are the same trade-off seen twice. A narrow filter must ring for a long time — this is not a design flaw but a theorem, and it is the circuit form of the uncertainty relation between bandwidth and duration.

Where it matters:

Consistency with Set 28. Problem 28.19 gave \(t_rf_c = 0.35\) for a low-pass — bandwidth and rise time reciprocally linked. Here \(\tau = 2/\text{BW}\) for a band-pass. Both say the same thing: a circuit cannot respond faster than its bandwidth permits, whatever the shape of its response.

The circuit. Seen from the load busbar, the supply is inductive (the transformer and line reactance) and the capacitor bank is capacitive. They are in parallel — so this is a parallel resonance, presenting a high impedance to any harmonic current injected by the load.

The reactances at 50 Hz:

The resonant harmonic. At harmonic \(h\), \(X_L\) scales up by \(h\) and \(X_C\) down by \(h\):

The 5th is the worst possible answer. Six-pulse rectifiers — variable-speed drives, UPS units, DC supplies — inject strong 5th-harmonic current, typically 20% of fundamental.

What happens. The harmonic current source sees a very high impedance, so it develops a large harmonic voltage, which drives large circulating currents between the supply and the bank:

The screening calculation. The resonant harmonic depends only on the ratio of fault level to bank size:

This one-line check should precede every capacitor installation, and frequently is not done.

The remedy: a detuning reactor. Add a small inductor in series with each capacitor, typically 6% of \(X_C\):

The bank is now series-resonant at the 4.08th harmonic, so above that frequency it looks inductive rather than capacitive — and an inductive bank in parallel with an inductive supply cannot resonate at all. Every harmonic of concern (5th, 7th, 11th, 13th) lies safely above 4.08.

The series-resonance hazard too. Problem 3's voltage magnification is equally dangerous:

An accidental series resonance — a long cable's capacitance with a transformer's inductance — can put kilovolts across equipment while the incoming supply meter reads a perfectly normal 240 V.

Two numbers describe every resonant circuit. \(\omega_0\) says where, \(Q\) says how sharply — and \(Q\) appears in six guises:

All six give 31.62 for the running example. They are not analogies but identities.

The running example in one place:

The recurring caution. Three problems found that a familiar term hides an ambiguity or a shortfall:

The unavoidable trade-off. Problem 18 showed it is a theorem rather than a design weakness:

Narrow means slow. No amount of cleverness escapes it, because it is the same statement as Set 28's \(t_rf_c = 0.35\) and, ultimately, the same as the Fourier relation between the width of a pulse and the width of its spectrum — which Set 34 will state properly.

What Set 30 must add. Problem 12 identified the gap precisely: a receiver needs a shape factor near 2, and one resonance gives 1000. Problem 17 showed that coupling two resonators improves it to about 30 by placing two poles deliberately rather than accepting one pair.