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Module 6 · Exam problem class · 40 min problem class

Exam workshop: PI design by pole cancellation

Design a PI controller from a target closed-loop time constant and examine the assumptions behind ideal pole cancellation.

  • Choose the PI reset time for nominal pole cancellation.
  • Derive the resulting closed-loop first-order model.
  • Select proportional gain from a required settling speed and discuss robustness.
Unofficial TUM ACE alignment: ACE §6: PI-Regler, Polkompensation und Führungsverhalten

Original exam-style problem · 18 points

Recommended time: 30 min

Consider the plant and PI controller

G(s)=14s+1,C(s)=KR(1+1TNs).G(s)=\frac{1}{4s+1}, \qquad C(s)=K_R\left(1+\frac{1}{T_Ns}\right).

They form a standard unity-feedback loop. The desired closed-loop reference response is first order with time constant Tcl=0.8 sT_\mathrm{cl}=0.8\,\mathrm{s}.

  1. 4 points: Choose TNT_N so the controller zero cancels the nominal plant pole.
  2. 6 points: Derive the closed-loop transfer function after cancellation.
  3. 4 points: Choose KRK_R to obtain the specified time constant.
  4. 4 points: Explain one practical reason not to trust exact pole cancellation blindly.

Submit the design values

s

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Concept check

Why is exact pole cancellation fragile in a real plant?

Reveal guided video and written solution

Guided solution video storyboard ready · recording next

The walkthrough will separate the nominal design calculation from its robustness assumptions.

Written solution

Rewriting the controller gives

C(s)=KRTNs+1TNs.C(s)=K_R\frac{T_Ns+1}{T_Ns}.

The nominal plant pole is at −1/4-1/4, so choose TN=4 sT_N=4\,\mathrm{s}. Then

C(s)G(s)=KR4s+14s14s+1=KR4s.C(s)G(s)=K_R\frac{4s+1}{4s}\frac{1}{4s+1}=\frac{K_R}{4s}.

Closing the loop yields

Gyw(s)=KR4s+KR=1(4/KR)s+1.G_{yw}(s)=\frac{K_R}{4s+K_R} =\frac{1}{(4/K_R)s+1}.

Thus Tcl=4/KRT_\mathrm{cl}=4/K_R. Setting this equal to 0.8 s0.8\,\mathrm{s} gives KR=5K_R=5. The ideal loop has unit static gain and therefore zero steady-state step error.

Exact cancellation is a nominal-model argument. If the real time constant is not exactly 4 s4\,\mathrm{s}, a residual pole–zero pair remains. Saturation and integral windup also remain outside this ideal linear calculation.

Retrieve it later

Reconstruct the pole-cancellation PI design and name two reasons the nominal result can fail in practice.

Answer from memory tomorrow, then return to check your reasoning.