Resonance, bandwidth & quality factor
Derive resonance from the actual transfer quantity rather than treating it as a universal property of the denominator. For the ideal series RLC, derive Z(jω), current resonance at ω0=1/√(LC), resistor-output band-pass response, exact half-power frequencies and Δω=R/L, Q=ω0L/R=1/(ω0RC)=1/(2ζ), voltage magnification across L/C, and the energy meaning of Q. Then contrast this with capacitor-voltage peaking, whose peak exists only for ζ<1/√2 and occurs at a different frequency.
- Derive the series-RLC impedance Z=R+j(ωL−1/ωC) from element impedances and the declared circuit references.
- Show that current magnitude is maximized at ω0=1/√(LC) because the net reactance vanishes there.
- Derive the resistor-voltage transfer H_R=R/Z and express it as a normalized band-pass response in Ω=ω/ω0 and Q.
- Derive the exact half-power frequencies of the ideal series-RLC current/resistor-power response and prove Δω=ω_H−ω_L=R/L.
- Prove ω_Lω_H=ω0², so ω0 is the geometric mean of the half-power edge frequencies.
- Derive Q=ω0/Δω=ω0L/R=1/(ω0RC)=1/(2ζ) for the ideal series RLC.
- Derive the resonance voltage magnification |V_L|=|V_C|=Q|V_s| at current resonance and explain how the two large reactive voltages cancel in KVL.
- Derive the energy definition Q=ω0 W_max/P_loss for the resonant series RLC and relate it to energy lost per cycle.
- Derive the capacitor-output magnitude peak condition ω_peak=ω0√(1−2ζ²), showing that a peak exists only for ζ<1/√2.
- Explain why current resonance, natural damped frequency ωd, and an output-specific magnitude peak are distinct concepts and why bandwidth/Q formulas must state the response and criterion.
A series RLC can ring in time, but under sinusoidal forcing it can also exhibit a sharp frequency-selective response. Which quantity is actually resonating, where does the peak occur, and when is the familiar relation Q=ω0/BW exact?
We answer those questions from the declared circuit and output variable rather than from the word “resonance.”
04.1Start from the electrical circuit and its impedance
Reference schematic — driven series RLC
The arrows and polarities below define exactly the variables used in KVL.
For sinusoidal steady state, the same series current flows through R, L and C. With the references shown,
Series impedances add:
The current phasor is
This equation already contains the current resonance. No separate resonance formula is required.
04.2Current resonance occurs when the net reactance vanishes
For fixed source amplitude and R>0, |I| is largest when the denominator is smallest. The squared reactance is nonnegative, so its minimum is zero:
At this frequency,
The source sees a purely resistive impedance and current is in phase with source voltage. For the ideal series RLC, the undamped natural angular frequency from the homogeneous equation and the forced-current resonance frequency coincide at ω0. That coincidence does not make every other response peak occur at ω0.
04.3The resistor output is an exact band-pass response
Because VR=RI,
Introduce
Since 1/(ωC)=ω0L/Ω and ωL=ω0LΩ,
Therefore
It tends to zero for both Ω→0 and Ω→∞ and reaches unity at Ω=1: a true band-pass response with peak at the current resonance.
04.4Derive the exact half-power frequencies
Average resistor power is proportional to |I|². Half of the maximum resistor power therefore corresponds to
Using the impedance form, this requires
The lower edge satisfies 1/(ωC)−ωL=R:
The upper edge satisfies ωL−1/(ωC)=R:
Figure 4.1 — Half-power bandwidth of the resistor output
The one question is: which edges are defined by |H_R|=1/√2 of its maximum?
For the ideal series RLC observed across R, the maximum occurs at ω0. The two half-power edges bracket that resonance exactly and their separation is R/L.
04.5Bandwidth and geometric-mean identities are exact here
Subtract the two exact edge expressions:
Multiply them:
Thus for this ideal series-RLC current/resistor-power response, the resonant frequency is exactly the geometric mean of the two half-power edges, not their arithmetic midpoint.
04.6Quality factor from bandwidth and damping
Using the exact bandwidth above, define the series-resonator quality factor
Since ω0²=1/(LC), equivalent forms are
For the series-RLC natural equation, ζ=R/(2Lω0), therefore
These equalities are exact for the declared ideal series-RLC model and its conventional second-order Q. They should not be copied to an arbitrary resonator without deriving the corresponding loss and bandwidth model.
04.7Reactive voltage magnification at current resonance
At ω0, the current magnitude is |Vs|/R. Hence
The inductor and capacitor voltages are opposite in phase, so their phasors cancel:
Thus each reactive element can carry a voltage much larger than the source when Q≫1 even though the source sees only the resistor drop. This is a physical energy-exchange effect, not a violation of KVL.
04.8Energy meaning of Q
At steady-state current resonance, let Ipk be the peak current. The maximum magnetic energy is
Average resistor loss is
Therefore
Since one cycle lasts 2π/ω0, the energy dissipated per cycle is ΔW=P_loss·2π/ω0, so
Large Q means little energy is lost per cycle compared with the peak energy stored. The source continuously replenishes that loss in steady state.
04.9Not every output peaks at ω0
For capacitor voltage in the same series RLC,
Its squared denominator magnitude is
A nonzero magnitude peak requires dD/dΩ=0:
Therefore
and such a peak exists only for ζ<1/√2. By contrast, the natural transient oscillates at
So ωpeak,C, ωd and the series-current resonance ω0 are three different quantities. “The resonance frequency” is incomplete unless the response variable and criterion are declared.
04.10Mastery exercises
- For R=20 Ω, L=10 mH and C=1 μF, calculate ω0, Q and Δω.
- For the same circuit, calculate ωL and ωH from the exact formulas and verify both Δω and ωLωH.
- At current resonance with source amplitude 2 V, calculate |I|, |VL| and |VC|.
- Derive H_R(jΩ)=1/[1+jQ(Ω−1/Ω)].
- Show from the half-power condition that Δω=R/L exactly.
- For ζ=0.2, compare ωd/ω0 and the capacitor-output ωpeak/ω0.
- For ζ=0.8, does the capacitor-output magnitude have a nonzero resonance peak?
- Explain why Q=ω0/BW must not be quoted without stating which response and bandwidth criterion are being used.
Solutions
1. ω0=10,000 rad/s; Q=ω0L/R=5; Δω=R/L=2000 rad/s.
2. √(R²+4L/C)=√40400≈200.998; ωL≈9049.88 rad/s and ωH≈11049.88 rad/s. Difference=2000 rad/s and product≈10⁸ rad²/s²=ω0².
3. |I|=2/20=0.1 A; |VL|=|VC|=Q|Vs|=10 V.
4. Divide Z by R and substitute ωL/R=QΩ and 1/(ωCR)=Q/Ω.
5. The exact edge equations differ only by ±R in the numerator; subtraction gives R/L.
6. ωd/ω0=√0.96≈0.9799; ωpeak,C/ω0=√0.92≈0.9592.
7. No. 1−2ζ²<0, so no positive stationary peak exists away from DC.
8. Different outputs can have different peaks and edge locations; Q/bandwidth identities are tied to a declared resonant response and criterion.
Chapter 05 will close DY-02 by putting roots, step response, resonance, Q, bandwidth and energy decay on one map and showing which quantities can—and cannot—be inferred from each other.