03 · Intermediate

Driven response & step response

Add forcing to the second-order dynamics without losing the natural-mode structure. Separate zero-input and zero-state response, derive the series-RLC capacitor-voltage transfer equation, solve the canonical DC-step response in over-, critical- and underdamped cases, derive peak time and percent overshoot only for the canonical underdamped all-pole step response, replace vague settling heuristics by explicit tolerance/envelope bounds, and show how nonzero initial energy changes the total waveform without changing the system poles.

180–230 minIntermediate

In this lesson

Decompose an LTI second-order response into zero-input and zero-state contributions and state the assumptions behind that decomposition.
Derive the driven series-RLC capacitor-voltage equation and normalized canonical form y''+2ζω0 y'+ω0²y=ω0²K u(t).
Find the constant-input particular solution and distinguish it from the natural transient.
Derive the zero-initial-condition step response for overdamped, critically damped and underdamped cases.
For 0<ζ<1, derive the normalized underdamped step response 1−e^(−ζω0t)[cos(ωdt)+ζ/√(1−ζ²) sin(ωdt)].
Derive tp=π/ωd and fractional overshoot Mp=exp[−πζ/√(1−ζ²)] for the canonical unit-step all-pole response, with explicit scope.
Derive an explicit underdamped error-envelope bound and compare it with common approximate settling-time rules.
Show how arbitrary nonzero initial conditions add a zero-input response while preserving the same poles.
Explain why zeros, feedthrough, nonminimum-phase behavior or a different output variable can invalidate simple overshoot formulas even when the denominator is second order.
Use final-value, initial-slope, continuity and dimensional checks to validate a step-response solution.

Starting question

The natural roots tell us how a second-order circuit can move on its own. What changes when a source is applied, and how do the familiar overshoot formulas arise without turning them into universal rules?

03.1Separate what comes from the initial state and what comes from the input

For a linear time-invariant system, the total response can be written as

y(t)=y_ZI(t)+y_ZS(t)

where yZI is the zero-input response produced by the initial stored state and yZS is the zero-state response produced by the input when the state starts at zero.

This decomposition follows from linearity. The system poles are the same in both parts because they come from the same homogeneous operator. Initial conditions change modal coefficients, not pole locations.

03.2Driven series RLC in canonical form

For capacitor voltage in a series RLC, use the same electrical references established in chapter 01:

Reference schematic — driven series RLC

The arrows and polarities below define exactly the variables used in KVL.

With those references, chapter 01 derived

LC y''+RC y'+y=v_s(t)

Divide by LC:

y''+(R/L)y'+[1/(LC)]y=[1/(LC)]v_s(t)

Using 2ζω0=R/L and ω0²=1/(LC),

y''+2ζω₀y'+ω₀²y=ω₀² v_s(t)

More generally a canonical all-pole second-order low-pass with DC gain K can be written

y''+2ζω₀y'+ω₀²y=ω₀²K u(t)

for a unit step u(t).

03.3The constant-input particular solution

For a constant input U0, try a constant particular solution yp. Its derivatives vanish:

ω₀² y_p=ω₀² K U₀

y_p=K U₀

The complete response is then

y(t)=K U₀+y_h(t)

where yh contains the same natural modes derived in chapter 02. The source changes the equilibrium point; it does not replace the natural roots.

03.4Underdamped unit-step response from zero initial state

For 0<ζ<1 and K=1, write the complete response as

y(t)=1+e^−αt[A cos(ω_dt)+B sin(ω_dt)]

Zero initial output gives y(0)=0, so A=−1. For an all-pole low-pass with no direct feedthrough, y'(0)=0. Applying that condition yields

B=−α/ω_d=−ζ/√(1−ζ²)

Therefore

y(t)=1−e^−ζω₀t[cos(ω_dt)+ζ/√(1−ζ²) sin(ω_dt)]

This formula belongs to this declared canonical transfer and initial state. It is not the generic waveform of every denominator containing a complex pole pair.

03.5Peak time comes from setting the derivative to zero

Differentiate the normalized underdamped step response. After cancellation, extrema occur when

sin(ω_dt)=0

The first nonzero extremum occurs at

t_p=π/ω_d

For the positive unit-step response this first extremum is the first peak. The result depends on the canonical numerator and zero initial state used above.

03.6Derive the fractional overshoot

At tp, cos(π)=−1 and sin(π)=0, so

y(t_p)=1+e^−απ/ωd

The fractional overshoot relative to the final step change is

M_p=e^−απ/ωd

Using α=ζω0 and ωd=ω0√(1−ζ²),

M_p=exp[−πζ/√(1−ζ²)]

Percent overshoot is 100Mp%.

03.7Settling time: use a tolerance and an envelope

The normalized error is

e(t)=1−y(t)=e^−αt[cos(ωdt)+α/ωd·sin(ωdt)]

The bracket is a sinusoid with amplitude

√[1+(α/ωd)²]=ω0/ωd=1/√(1−ζ²)

Therefore a sufficient envelope bound is

|e(t)|≤e^−αt/√(1−ζ²)

To guarantee |e(t)|≤ε from that bound, it is sufficient that

t≥[ln(1/(ε√(1−ζ²)))]/α

Rules such as ts≈4/(ζω0) for a nominal 2% criterion are convenient approximations, not exact definitions. The exact settling time of an oscillatory trace depends on the declared band and its last crossing.

03.8Critical and overdamped step responses

For critical damping with zero initial state and unit DC gain, the step response is

y(t)=1−(1+ω₀t)e^−ω₀t

It follows directly from the repeated-root homogeneous form and the initial conditions.

For distinct real roots s1 and s2, write

y(t)=1+A e^s₁t+B e^s₂t

with A+B=−1 and s1A+s2B=0 for zero initial output and zero initial slope. Thus

A=s₂/(s₁−s₂)

B=−s₁/(s₁−s₂)

No separate memorized waveform family is needed; it is the same “particular equilibrium + natural modes” construction.

03.9Nonzero initial energy changes coefficients, not poles

Suppose a source step is applied while C or L already stores energy. The total response is

y(t)=y_ZS(t)+y_ZI(t)

The zero-state part is the step response above; the zero-input part is the source-free response from chapter 02 using the actual vC(0+) and iL(0+).

The resulting first peak need not obey the zero-state overshoot formula because the initial-state modes can add or subtract at the same time. This is why initial conditions must be declared before comparing a measured transient with textbook overshoot curves.

03.10Mastery synthesis

For ζ=0.5 and ω0=1000 rad/s, compute ωd, tp and Mp.
Derive the critical unit-step response from y=1+(A+Bt)e^(−ω0t).
Show that the final value of the series-RLC capacitor voltage equals the DC step value.
For ζ=0.2, compute the sufficient 2% envelope-bound settling time in units of 1/ω0.
Why can a right-half-plane zero invalidate the simple Mp formula?
Explain why nonzero initial inductor current changes peak height without moving the poles.
For overdamped roots s1=−100 and s2=−500 s⁻¹, find A and B for a zero-state unit step.
List four independent checks on a derived step-response waveform.

Solutions

1. ωd=866.025 rad/s; tp=π/ωd≈3.628 ms; Mp=exp[−π·0.5/√0.75]≈0.1630, or 16.3%.

2. y(0)=0 gives A=−1; y'(0)=0 gives B=−ω0, so y=1−(1+ω0t)e^(−ω0t).

3. At DC the ideal capacitor carries no current, so resistor and inductor drops vanish and vC→Vs.

4. α=0.2ω0. Bound: t≥ln[1/(0.02√0.96)]/(0.2ω0)≈19.66/ω0.

5. A zero changes the numerator and therefore the transient shape/initial derivative; poles alone no longer determine peak behavior.

6. Initial state changes modal coefficients while the homogeneous characteristic equation—and thus the poles—stays the same.

7. A=s2/(s1−s2)=−1.25; B=−s1/(s1−s2)=0.25.

8. Initial value/continuity, initial slope, final value/equilibrium, substitution into the ODE; dimensions and limiting cases provide additional checks.

Chapter 04 will move from time-domain forcing to sinusoidal forcing and derive resonance, bandwidth and quality factor from the actual RLC transfer functions rather than from slogans.