Bode, Nyquist, Compensators and State Space

Bode plots, gain and phase margins, Nyquist criterion, polar plots, lead and lag compensators, PID controllers, and state-space models with controllability and observability.

Frequency-response methods judge stability and robustness from G(jω). State-space methods describe a system with first-order matrix equations.

Bode plots

Magnitude in dB (20 log|G|) and phase against log ω.

FactorMagnitude slopePhase
Kflat, 20 log K0°
Pole at origin 1/s−20 dB/decade−90°
Zero at origin s+20 dB/decade+90°
First-order pole 1/(1 + s/a)0, then −20 dB/dec after ω = a0 to −90° (−45° at a)
Second-order poles−40 dB/dec after ω_n0 to −180°

Stability margins

  • Gain crossover ω_gc: |G| = 1 (0 dB). Phase margin PM = 180° + ∠G(jω_gc).
  • Phase crossover ω_pc: ∠G = −180°. Gain margin GM = 1/|G(jω_pc)| (in dB: −20 log|G|).
  • Stable (minimum-phase systems) if both margins are positive. For a second-order system, PM ≈ 100ζ degrees (for ζ < 0.7).
  • Resonant peak M_r = 1/(2ζ√(1 − ζ²)) at ω_r = ω_n√(1 − 2ζ²) (for ζ < 0.707). Bandwidth grows with ω_n.

Nyquist criterion

Plot G(jω)H(jω) for ω from −∞ to ∞. Z = N + P:

  • P = open-loop poles in the RHP; N = clockwise encirclements of (−1, 0); Z = closed-loop RHP poles.
  • For stability Z = 0, so the plot must encircle −1 counter-clockwise P times (no encirclement if P = 0).

Compensators and controllers

  • Lead compensator (zero before pole): adds phase, improves transient response and margins; acts like a PD controller (high-pass).
  • Lag compensator (pole before zero): raises low-frequency gain, reduces steady-state error; acts like PI (low-pass).
  • PID: K_p + K_i/s + K_d s. Integral action removes steady-state error (raises type by 1); derivative action adds damping.

State space

ẋ = Ax + Bu, y = Cx + Du

  • Transfer function: G(s) = C(sI − A)⁻¹B + D.
  • Eigenvalues of A = poles.
  • State transition matrix φ(t) = e^(At) = L⁻¹[(sI − A)⁻¹].
  • Controllable if [B AB … Aⁿ⁻¹B] has rank n. Observable if [C; CA; …; CAⁿ⁻¹] has rank n.

Worked example

G(s) = 10/(s(s + 1)) at ω = 3 rad/s.

  • |G| = 10/(3 × √10) = 1.054; ∠G = −90° − tan⁻¹3 = −161.6°.

Practice

GATE-styleNAT · numerical answer

A system's gain at the phase crossover frequency is 0.25. What is its gain margin in dB? (Give 2 decimal places.)

Show answer and explanationAnswer: 12.04

GM = −20 log 0.25 = 12.04 dB.

GATE-styleNAT · numerical answer

The phase of G at the gain crossover frequency is −140°. What is the phase margin in degrees?

Show answer and explanationAnswer: 40

PM = 180° + (−140°) = 40°.

GATE-styleMCQ · one correct answer

A system has A = [[0, 1], [−6, −5]]. What are its poles?

Show answer and explanationAnswer: A

det(sI − A) = s² + 5s + 6 = (s + 2)(s + 3).

GATE-styleNAT · numerical answer

An open-loop system has 1 pole in the RHP. Its Nyquist plot encircles −1 once counter-clockwise. How many closed-loop poles are in the RHP?

Show answer and explanationAnswer: 0

N = −1 (counter-clockwise), P = 1, so Z = N + P = 0: the closed loop is stable.