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ESECE: Control Systems

ESECE · Work through your exam, one topic at a time.

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Control Systems

Control Systems tests modelling, stability and classical design. ESECE expects block-diagram reduction, Routh–Hurwitz, root-locus sketches, Bode plots with gain and phase margins, steady-state error, and second-order time-domain specifications.

Models and reduction

  • For negative feedback, G_closed = G/(1 + GH); watch the sign and the pickoff point.
  • The characteristic equation is the denominator of the closed-loop transfer function; the type number equals the count of free integrators in the loop.
  • Steady-state error: step 1/(1 + K_p), ramp 1/K_v, parabola 1/K_a, with K_p = lim G, K_v = lim sG, K_a = lim s²G.

Stability and response

  • Routh–Hurwitz: a zero row means roots on the jω axis - use the auxiliary polynomial and continue. Each sign change counts one right-half-plane pole.
  • Second order: given ω_n and ζ, overshoot M_p = e^(−πζ/√(1−ζ²)), peak time π/(ω_n√(1−ζ²)), and 2 % settling near 4/(ζω_n).
  • Bode: gain margin is measured at the phase crossover (−180°); phase margin at the gain crossover (0 dB). A rough link is ζ ≈ PM/100.
  • Root locus starts at open-loop poles, ends at zeros or infinity; a point lies on it when the angle condition sums to 180°.

Worked example

G(s) = 10/[s(s+2)] with unity feedback. Closed loop: s² + 2s + 10 = 0, so ω_n = 3.16 rad/s and ζ = 2/(2 × 3.16) = 0.316. Overshoot = e^(−π×0.316/√(1−0.1)) = e^(−1.047) ≈ 35 %, and settling ≈ 4/(0.316 × 3.16) = 4.0 s. For a unit ramp, type 1 gives e_ss = 1/K_v = 1/(10/2) = 0.2.

Traps

Reading gain margin at the gain crossover; abandoning a Routh array when a zero row appears; assuming an added pole always destabilises without checking the locus; computing settling with the 4 % formula while a 2 % requirement is stated; dropping the feedback sign in the closed-loop formula.