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TUM exam-prep problem class

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It contains university-style exam work, ACE-specific notation, and a complete solution walkthrough. Switch modes to include it in navigation and progress.

Return to PT1 step responses: the meaning of K and T

Module 2 · Exam problem class · 35 min problem class

Exam workshop: PT1 closed-loop analysis

Solve a complete university-style PT1 feedback problem before revealing the written and video solutions.

  • Derive a closed-loop transfer function without skipping algebraic steps.
  • Read closed-loop gain, time constant, and steady-state error from the result.
  • Translate a steady-state accuracy requirement into a controller-gain bound.
Unofficial TUM ACE alignment: ACE §2 and §5: PT1-Glied, Standardregelkreis, stationärer Fehler

Original exam-style problem · 20 points

Recommended time: 25 min

A plant is modeled by

G(s)=25s+1.G(s)=\frac{2}{5s+1}.

It is controlled by a proportional controller C(s)=KPC(s)=K_P in a standard negative-feedback loop with unity measurement. The reference is a unit step and the initial conditions are zero.

  1. 4 points: Derive the closed-loop reference transfer function Gyw(s)G_{yw}(s).
  2. 6 points: For KP=3K_P=3, determine the closed-loop static gain and time constant.
  3. 4 points: Determine the steady-state tracking error for the unit step.
  4. 6 points: Find the smallest KPK_P that limits the steady-state error to 5%.

Submit the essential results

Concept check

Which expression has the correct closed-loop structure?

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s

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Reveal guided video and written solution

Guided solution video storyboard ready · recording next

The walkthrough will emphasize setup, normalization, and efficient exam notation.

Written solution

The standard negative-feedback formula gives

Gyw(s)=C(s)G(s)1+C(s)G(s)=2KP5s+1+2KP.G_{yw}(s)=\frac{C(s)G(s)}{1+C(s)G(s)} =\frac{2K_P}{5s+1+2K_P}.

For KP=3K_P=3,

Gyw(s)=65s+7=6/7(5/7)s+1.G_{yw}(s)=\frac{6}{5s+7} =\frac{6/7}{(5/7)s+1}.

Therefore Kcl=6/7K_\mathrm{cl}=6/7 and Tcl=5/7 sT_\mathrm{cl}=5/7\,\mathrm{s}. For a unit step, y∞=6/7y_\infty=6/7, hence e∞=1/7e_\infty=1/7.

In general, e∞=1/(1+2KP)e_\infty=1/(1+2K_P). Requiring e∞≤0.05e_\infty\leq0.05 gives

11+2KP≤0.05⟹KP≥9.5.\frac{1}{1+2K_P}\leq0.05 \quad\Longrightarrow\quad K_P\geq9.5.

This algebraic requirement alone does not check actuator limits, noise amplification, or robustness.

Retrieve it later

Derive the PT1 closed-loop gain, time constant, and step error from memory without looking at the solution.

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