MEGR 3171 — First- & Second-Order System Response

Learning Objectives

Characterize first-order system step response: time constant, rise time, steady-state value.
Identify second-order system parameters from the standard form: ω_n, ζ, and damped natural frequency ω_d.
Calculate rise time, peak time, settling time, and percent overshoot from ω_n and ζ.
Apply the Routh-Hurwitz criterion to determine stability from the characteristic equation.
Calculate steady-state error for step, ramp, and parabolic inputs based on system type.

1. First-Order Systems

First-Order Step Response

G(s) = K / (τs + 1)

Step response: y(t) = K·A·[1 - e^(-t/τ)]  for A = step magnitude

Time constant τ: time to reach 63.2% of final value
  At t = τ:   y = 0.632 · K · A
  At t = 2τ:  y = 0.865 · K · A
  At t = 3τ:  y = 0.950 · K · A
  At t = 5τ:  y = 0.993 · K · A  (considered settled)

Rise time (10% to 90%): t_r ≈ 2.2τ

2. Second-Order Systems: Standard Form

Standard Second-Order Transfer Function

G(s) = ω_n² / (s² + 2ζω_n·s + ω_n²)

ω_n = undamped natural frequency (rad/s)
ζ   = damping ratio (dimensionless)

Damped natural frequency: ω_d = ω_n · √(1 - ζ²)  [for ζ < 1]

Damping Regimes

Underdamped (0 < ζ < 1)

Oscillatory step response. Complex conjugate poles. Systems with ζ ≈ 0.7 give fast response with small overshoot.

Critically Damped (ζ = 1)

Fastest non-oscillatory response. Real repeated poles at s = −ω_n.

Overdamped (ζ > 1)

Slow, non-oscillatory response. Two distinct real poles. Sluggish but no overshoot.

Undamped (ζ = 0)

Purely oscillatory (sinusoidal). No energy dissipation. Unstable in most practical systems.

Underdamped Step Response Metrics

Peak time:     t_p = π / ω_d

% Overshoot:   %OS = e^(-πζ/√(1-ζ²)) × 100%

Settling time: t_s ≈ 4 / (ζω_n)   (2% criterion)
               t_s ≈ 3 / (ζω_n)   (5% criterion)

Rise time (0 to 100%, underdamped): t_r ≈ (1.8) / ω_n  (approx.)

3. Routh-Hurwitz Stability Criterion

A closed-loop system is stable if and only if all roots of the characteristic polynomial have negative real parts (are in the left half s-plane). The Routh-Hurwitz test determines this from the polynomial coefficients without computing roots explicitly.

Routh Array Construction

Characteristic polynomial: a_n s^n + a_(n-1) s^(n-1) + ... + a_1 s + a_0

Row 1: a_n,  a_(n-2),  a_(n-4), ...
Row 2: a_(n-1), a_(n-3), a_(n-5), ...
Row 3 onward: computed from cross-products

Stability condition: All elements in the FIRST COLUMN must have the SAME SIGN.
Number of sign changes = number of RHP poles (unstable poles).

Necessary Condition (Quick Check) For a polynomial to possibly be stable, all coefficients must be present (none zero) and all must have the same sign. If any coefficient is missing or has the wrong sign, the system is definitely unstable. But satisfying this does NOT guarantee stability (sufficient condition requires the full Routh array).

4. Steady-State Error and System Type

Error Constants and Steady-State Errors

System Type = number of open-loop poles at s = 0 (integrators)

Type 0: K_p = lim_{s→0} G(s)    e_ss(step) = 1/(1+K_p)
Type 1: K_v = lim_{s→0} s·G(s)   e_ss(ramp) = 1/K_v
Type 2: K_a = lim_{s→0} s²·G(s)  e_ss(parabola) = 1/K_a

Higher system type: zero error for lower-order inputs
but may be harder to stabilize.

Practice Problems

Problem 1 — Second-Order Metrics A second-order system has ω_n = 10 rad/s and ζ = 0.5. Calculate: (a) damped natural frequency, (b) percent overshoot, (c) settling time (2% criterion).

(a) ω_d = ω_n √(1-ζ²) = 10 × √(1-0.25) = 10 × √0.75 = 10 × 0.866 = 8.66 rad/s

(b) %OS = e^(-π×0.5/√(1-0.25)) × 100 = e^(-1.5708/0.866) × 100 = e^(-1.8138) × 100 = 16.3%

ω_d = 8.66 rad/s, %OS = 16.3%, t_s = 0.8 s

Problem 2 — Routh-Hurwitz Is the system with characteristic equation s³ + 4s² + 5s + 2 stable?

Row 1: 1, 5 Row 2: 4, 2

Row 3, col 1: (4×5 - 1×2)/4 = (20-2)/4 = 18/4 = 4.5

Row 4, col 1: (4.5×2 - 4×0)/4.5 = 9/4.5 = 2

First column: 1, 4, 4.5, 2 — all positive, no sign changes.

Stable. All first-column elements positive, zero sign changes, zero RHP poles.