⚙ Why Time Response Matters in Real Systems
- Elevator floor positioning — settle within 2 cm without oscillation → ζ ≈ 0.7
- Missile guidance — fast rise time, minimal overshoot to avoid target offset
- Servo motor in CNC machine — settles in <10 ms, 0% overshoot (critically damped)
- Autopilot altitude hold — steady-state error must be 0 for step command (Type 1 needed)
- Suspension system — underdamped for comfort, overdamped for stability tradeoff
- Blood pressure regulation — body uses Type 1-like integral action to eliminate DC error
1. First-Order System
A first-order system has a single energy storage element (one pole). Standard form:
where τ = time constant (seconds). The pole is at s = −1/τ.
Step Response of First-Order System
For unit step input R(s) = 1/s:
| Time | Output Value | Significance |
|---|---|---|
| t = τ | 63.2% | Definition of time constant |
| t = 2τ | 86.5% | — |
| t = 3τ | 95.0% | 5% settling time |
| t = 4τ | 98.2% | 2% settling time (ts) |
| t → ∞ | 100% | Final/steady-state value |
2. Second-Order System — Standard Form
ωₙ = natural frequency (rad/s) | ζ = damping ratio (dimensionless)
Characteristic equation: s² + 2ζωₙs + ωₙ² = 0 → poles at s = −ζωₙ ± jωₙ√(1−ζ²)
Damped natural frequency: ωd = ωₙ√(1−ζ²)
Damping Cases
| Damping Ratio | Type | Poles Location | Step Response |
|---|---|---|---|
| ζ = 0 | Undamped | On jω axis: ±jωₙ | Sustained sinusoidal oscillation |
| 0 < ζ < 1 | Underdamped | Complex: −ζωₙ ± jωd | Oscillatory, decaying, overshoots |
| ζ = 1 | Critically damped | Repeated real: −ωₙ (×2) | Fastest response with no overshoot |
| ζ > 1 | Overdamped | Two distinct real: −ζωₙ ± ωₙ√(ζ²−1) | Slow approach, no oscillation |
3. Transient Response Specifications
These specs quantify how quickly and smoothly a system responds to a step input. GATE frequently asks you to compute them given ζ and ωₙ.
β = cos⁻¹(ζ)
ωd = ωₙ√(1−ζ²)
c(t) = 1 − (e^(−ζωₙt) / √(1−ζ²)) · sin(ωdt + φ) where φ = cos⁻¹(ζ)
Key Relationships
| Parameter | Increases When | Practical Trade-off |
|---|---|---|
| %Overshoot | ζ decreases | Lower ζ → faster but more overshoot |
| Settling time tₛ | ζωₙ decreases | Need large ζωₙ (= σ, real part of poles) |
| Rise time tᵣ | ωₙ decreases | Larger bandwidth → faster rise |
| Peak time tₚ | ωd decreases | Overshoot occurs earlier with larger ωd |
⚙ Lab: Measuring Transient Specs
- Connect 2nd order RLC circuit: R=200Ω, L=10mH, C=1μF → ωₙ=10000 rad/s, ζ=1
- Apply square wave input at 100 Hz on oscilloscope — observe step response each half-cycle
- Reduce R to 100Ω: ζ drops to 0.5, observe overshoot appear (≈16.3%)
- Increase R to 1kΩ: ζ=5 (overdamped) — output slowly climbs, no oscillation
- Measure tₚ from rising edge to first peak — verify tₚ = π/ωd
4. Steady-State Error
Steady-state error e_ss is the difference between desired and actual output as t → ∞. For a unity feedback system:
This is the Final Value Theorem applied to the error signal E(s) = R(s)/(1+G(s)).
System Type and Error Constants
System type = number of open-loop poles at origin (s = 0).
| System Type | Open-loop G(s) form | Position Kp | Velocity Kv | Acceleration Ka |
|---|---|---|---|---|
| Type 0 | K(s+z₁)…/[(s+p₁)…] | K (finite) | 0 | 0 |
| Type 1 | K(s+z₁)…/[s(s+p₁)…] | ∞ | K (finite) | 0 |
| Type 2 | K(s+z₁)…/[s²(s+p₁)…] | ∞ | ∞ | K (finite) |
Steady-State Error Table
| Input | R(s) | Type 0 | Type 1 | Type 2 |
|---|---|---|---|---|
| Unit Step | 1/s | 1/(1+Kp) | 0 | 0 |
| Unit Ramp | 1/s² | ∞ | 1/Kv | 0 |
| Unit Parabola | 1/s³ | ∞ | ∞ | 1/Ka |
5. Determining System Type Quickly
Method: Count Poles at Origin
Express G(s) in factored form and count s-factors in denominator that aren't cancelled by zeros.
Kv = lim(s→0) s·G(s) = lim(s→0) 10s(s+2)/[s(s+1)(s+5)] = 10×2/(1×5) = 4
ess for ramp = 1/Kv = 0.25
Kp = lim(s→0) G(s) = 5/(1×2) = 2.5
ess for step = 1/(1+Kp) = 1/(1+2.5) = 1/3.5 ≈ 0.286
6. Impulse and Ramp Responses
Impulse Response
For a second-order underdamped system, impulse response = inverse LT of T(s):
Ramp Response Steady-State
For a Type 1 system with ramp input r(t) = t, the output c(t) eventually tracks with a constant lag:
The steady-state error is the constant lag 1/Kv — higher Kv means the output tracks the ramp more closely.
GATE Previous Year Questions
A second order system has ωₙ = 4 rad/s and ζ = 0.5. The percentage overshoot for a unit step input is approximately:
%OS = exp(−πζ/√(1−ζ²)) × 100 = exp(−π×0.5/√(1−0.25)) × 100
= exp(−π×0.5/√0.75) × 100 = exp(−π×0.5/0.866) × 100
= exp(−1.814) × 100 = 0.163 × 100 = 16.3%
Note: ωₙ does not affect %OS — only ζ does.
For G(s) = 10/[s(s+2)], unity feedback, the velocity error constant Kv = ?
Kv = lim(s→0) s·G(s) = lim(s→0) s·10/[s(s+2)] = lim(s→0) 10/(s+2) = 10/2 = 5
System is Type 1 (one pole at s=0), so step error = 0, ramp error = 1/Kv = 0.2.
A unity feedback system has G(s) = K/[(s+1)(s+2)]. For zero steady-state error to a unit step input, K must be:
G(s) has no pole at s=0 → system is Type 0. Kp = lim(s→0) G(s) = K/(1×2) = K/2.
ess = 1/(1+Kp) = 1/(1+K/2). For ess=0: K→∞. No finite K gives zero step error for Type 0 system.
To get zero step error, add an integrator (make it Type 1) — i.e., use integral control.
A second order system has settling time ts = 4 s (2% criterion) and peak overshoot = 16.3%. Find ωₙ.
%OS = 16.3% → ζ = 0.5 (from exp(−πζ/√(1−ζ²))×100 = 16.3).
ts = 4/(ζωₙ) = 4 → ζωₙ = 1 → ωₙ = 1/ζ = 1/0.5 = 2 rad/s
For a Type 2 system, the steady-state error for a unit ramp input is:
For a Type 2 system: Kv = lim(s→0) s·G(s) = ∞ (because G has s² in denominator).
ess (ramp) = 1/Kv = 1/∞ = 0. Type 2 systems track both step and ramp with zero steady-state error, but may have non-zero error for parabolic inputs (1/Ka).
A first-order system T(s) = 1/(1+10s) is subjected to a unit step. The time at which output = 0.5 is approximately:
c(t) = 1 − e^(−t/τ) = 0.5 → e^(−t/10) = 0.5 → −t/10 = ln(0.5) = −0.693
t = 10 × 0.693 = 6.93 s. Note: τ=10 s (from 1+10s → τ=10).
The damping ratio of a second-order system is 0.707. Which statement is correct about the step response?
ζ = 0.707 = 1/√2. Since 0 < ζ < 1, the system is underdamped (some overshoot).
%OS = exp(−π×0.707/√(1−0.5)) × 100 = exp(−π×0.707/0.707) × 100 = exp(−π) × 100 = 4.3%
This is the ITAE-optimal damping ratio — minimizes integrated time-weighted absolute error.
For a unity feedback system with G(s) = 4/[s(s+4)], the natural frequency ωₙ and damping ratio ζ are:
Closed-loop T(s) = G/(1+G) = 4/[s(s+4)+4] = 4/[s²+4s+4]
Compare with ωₙ²/(s²+2ζωₙs+ωₙ²): ωₙ² = 4 → ωₙ = 2
2ζωₙ = 4 → ζ = 4/(2×2) = 1 (critically damped — no overshoot)
A second-order system has poles at s = −2 ± j2. For unit step input, the peak time tₚ = ?
Poles at −2±j2: real part σ = −2, imaginary part ωd = 2.
tₚ = π/ωd = π/2 ≈ 1.57 s
Also: ωₙ = √(σ²+ωd²) = √(4+4) = 2√2, ζ = σ/ωₙ = 2/(2√2) = 1/√2 = 0.707
G(s) = 10(s+2)/[s²(s+1)]. The system type and steady-state error for unit ramp input are:
G(s) = 10(s+2)/[s²(s+1)] has TWO poles at s=0 → Type 2.
For Type 2: Kv = lim(s→0) s·G(s) = ∞ → ess = 1/Kv = 1/∞ = 0.
Also: Ka = lim(s→0) s²·G(s) = 10×2/1 = 20. For parabolic: ess = 1/Ka = 0.05.