Control Engineering 2: Feedback Control Design: practice problems and exam questions

Step response: a first-order system G(s) = K/(τs + 1) answers a step of size U with y(t) = K·U·(1 − e^−t/τ): it reaches 63.2 % of the final value K·U after one time constant τ and 98 % after 4τ. A second-order system G(s) = K·ωₙ²/(s² + 2ζωₙs + ωₙ²) has the poles s = −ζωₙ ± jω_d with ω_d = ωₙ√(1 − ζ²). For 0 < ζ < 1 the step response overshoots: percent overshoot %OS = 100·e^{−ζπ/√(1−ζ²)}, peak time t_p = π/ω_d, and the 2 % settling time t_s ≈ 4/(ζωₙ). Compare the denominator with s² + 2ζωₙs + ωₙ² to read off ωₙ and ζ. For a mass–spring–damper ωₙ = √(k/m) and ζ = c/(2√(km)); for a series RLC circuit ωₙ = 1/√(LC) and ζ = (R/2)·√(C/L).

Frequency response: substitute s = jω in the transfer function. The magnitude in decibels is 20·log₁₀|H(jω)| and the phase is the sum of the angles of the numerator factors minus those of the denominator factors. A first-order RC or RL filter has its corner (−3 dB) frequency at ω_c = 1/(RC) or R/L, and f_c = ω_c/(2π); a low-pass filter gives |H| = 1/√(1 + (ω/ω_c)²) and φ = −atan(ω/ω_c). In the asymptotic (straight-line) Bode plot each pole bends the magnitude down by 20 dB/decade at its corner frequency, each zero bends it up by 20 dB/decade, and an integrator 1/s gives −20 dB/decade at all frequencies. Gain crossover ω_c: |L(jω)| = 1 (0 dB); phase margin PM = 180° + ∠L(jω_c). Phase crossover ω₁₈₀: ∠L(jω) = −180°; gain margin GM = −20·log₁₀|L(jω₁₈₀)| dB. The closed loop is stable when both margins are positive.