Natural response, roots & damping regimes
Solve the source-free second-order equation from its characteristic roots instead of memorizing waveform labels. Derive s₁,₂=−α±√(α²−ω₀²), classify over-, critical- and under-damping from the discriminant or damping ratio ζ, derive the correct time-domain form and constants from physical initial conditions, distinguish the undamped natural frequency ω₀ from the damped natural frequency ωd, derive the series-RLC critical resistance, and show why increasing resistance beyond critical damping can actually slow the dominant non-oscillatory mode.
- Derive the two characteristic roots s₁,₂=−α±√(α²−ω₀²) of the canonical source-free second-order equation.
- Classify overdamped, critically damped and underdamped responses from α versus ω₀ or ζ versus 1.
- Derive the two-real-exponential form for distinct real roots and solve its constants from x(0) and ẋ(0).
- Derive the repeated-root response (A+Bt)e^(−αt) at critical damping rather than treating it as a memorized exception.
- Derive the underdamped form e^(−αt)[A cos(ωdt)+B sin(ωdt)] with ωd=√(ω₀²−α²).
- Distinguish ω₀, ωd and the later frequency-response resonance frequency; do not treat them as synonyms.
- For a series RLC, derive Rcrit=2√(L/C) and connect α, ζ and resistance.
- Use physical initial conditions vC(0+) and iL(0+) to determine response constants consistently.
- Explain the energy interpretation of oscillation as exchange between electric and magnetic storage while resistance removes total energy.
- Show from the large-R root asymptotes that excessive resistance creates a slow mode approximately −1/(RC), so 'more damping means faster' is false outside the appropriate regime.
Chapter 01 produced the natural characteristic equation
How can the same circuit family decay as two exponentials, return without oscillation at one boundary case, or ring while decaying? The answer is entirely in the two roots.
02.1Solve the characteristic quadratic first
The sign of the discriminant controls the root geometry:
Figure 2.1 — The discriminant moves the two roots
The one question is: what happens to the two roots as α crosses ω₀?
The classification is not a memorized vocabulary table. It follows directly from the sign of α²−ω₀².
With ζ=α/ω₀ the same boundaries are ζ>1, ζ=1 and 0<ζ<1.
02.2Overdamped: two real decays
For distinct real roots s₁ and s₂, linearity gives
If x(0)=x₀ and ẋ(0)=u₀, then
so
The root closest to the imaginary axis decays most slowly and dominates at long time. “Overdamped” does not mean one time constant; it means two non-oscillatory natural modes.
02.3Critical damping: derive the t e−αt term
At α=ω₀ the two roots coalesce at s=−α. Two copies of e−αt would not be linearly independent, so the second independent solution is t e−αt:
Apply the initial conditions:
Critical damping is the exact boundary between distinct-real and complex-conjugate roots. It is often useful as the fastest boundary case without oscillatory roots, but a blanket statement such as “critical damping always gives the shortest settling time” needs an output, initial condition and settling criterion before it becomes a precise optimization claim.
02.4Underdamped: derive the damped natural frequency
For α<ω₀ define
The roots become
Combining the conjugate exponentials into a real response gives
From x(0)=x₀, A=x₀. Differentiating at zero gives
The oscillation period of the natural response is Td=2π/ωd. The envelope decays with e−αt.
02.5Series RLC: derive the critical resistance
For a series RLC, use the same current and voltage references as in chapter 01:
Reference schematic — source-free series RLC
The source is suppressed; the i, vR, vL and vC references remain exactly those used in the equations.
Critical damping requires α=ω₀:
Thus
for this specific series-RLC form. This compact ratio should not be transferred unchanged to other second-order topologies without rederiving their coefficients.
02.6One L and C, three damping cases
Take pedagogical values L=10 mH and C=1 μF. Then
Underdamped, R=100 Ω:
For vC(0)=1 V and i(0)=0, u₀=dvC/dt=0:
Critical, R=200 Ω:
Overdamped, R=300 Ω:
All three responses satisfy the same initial voltage and zero initial current. Only R changes the root geometry.
02.7Why very large resistance can become slow
For large α compared with ω₀, expand the square root:
The two roots are approximately
As R becomes very large, the slow root moves toward zero because −1/(RC) becomes less negative. The system becomes strongly non-oscillatory but its long tail becomes slower.
02.8Energy interpretation of ringing
Chapter 01 established
When the response is underdamped, electric energy in C and magnetic energy in L exchange repeatedly, producing sign changes and oscillation in state variables. R removes part of the total energy during each exchange.
The state coordinates have an e−αt envelope, so an energy scale based on squared amplitude falls roughly on an e−2αt envelope. The exact instantaneous total energy need not be a pure exponential because the partition between C and L changes during the cycle; the exact statement is dw/dt=−Ri².
02.9Root locations are model statements
The formulas above assume a linear, time-invariant, lumped second-order model with two independent states. Real inductors add winding resistance, core nonlinearity/loss and parasitic capacitance; real capacitors add ESR, leakage and dielectric effects; wiring can add further states.
If measured decay cannot be fit by the predicted two-root structure over the relevant range, the appropriate response is to improve the model—not to force a damping label onto the data.
02.10Mastery exercises
- Derive the roots of s²+2αs+ω₀²=0.
- For ζ=0.25, express ωd/ω₀.
- For L=20 mH and C=5 μF, calculate Rcrit.
- For an underdamped state with x0=2, u0=0, α=100 s⁻¹ and ωd=400 rad/s, find A and B.
- For critical damping with x0=3 and u0=−50 s⁻¹ in the same units, α=20 s⁻¹, find A and B.
- For α=3000 s⁻¹ and ω₀=2000 rad/s, calculate both roots and identify the dominant one.
- Show the large-R series-RLC root approximations −1/(RC) and −R/L.
- Explain why ωd and the later resonance-peak frequency should not be assumed identical.
Solutions
1. Quadratic formula gives −α±√(α²−ω₀²).
2. ωd/ω₀=√(1−0.25²)=0.968246.
3. Rcrit=2√(0.02/5×10⁻⁶)=126.49 Ω.
4. A=2; B=(0+100×2)/400=0.5.
5. A=3; B=u0+αx0=−50+60=10.
6. √(9−4)×10⁶=2236.07; roots ≈−763.93 and −5236.07 s⁻¹. The −763.93 root dominates long time.
7. Use √(α²−ω₀²)≈α−ω₀²/(2α), with α=R/(2L), ω₀²=1/(LC).
8. ωd describes the source-free transient roots; resonance is a driven frequency-response property whose peak location also depends on the transfer numerator and damping.
Chapter 03 will add forcing and step inputs, separating zero-input and zero-state response and deriving overshoot/settling properties only where their assumptions are explicit.