Control Engineering Basics

When a step of size 2 was applied to a first-order system, the output finally settled at 6 and reached about 63% of its final value 0.4 s after the input. Which of the following is the transfer function of this system?

  1. A3/(1 + 0.4s)
  2. B3/(1 + 2.5s)
  3. C6/(1 + 0.4s)
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Answer: A. 3/(1 + 0.4s)

The steady-state gain is the final output divided by the input size: 6 ÷ 2 = 3. The time a first-order step response takes to reach about 63% (1 − e^(−1)) of its final value is the time constant, so T = 0.4 s. The transfer function is therefore G(s) = 3/(1 + 0.4s). In this way, a first-order model can be built easily from the final value and the rise time of a step response.

  • A. Correct. A steady-state gain of 3 and a time constant of 0.4 s.
  • B. This puts 2.5, the reciprocal of the time constant, as the coefficient of s. The coefficient of s is the time constant itself.
  • C. This uses the final output 6 directly as the gain. It must be divided by the input size 2.
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  1. Which of the following is the range of the damping ratio ζ for which the step response of the second-order system G(s) = ωn²/(s² + 2ζωn s + ωn²) overshoots the final value and approaches it while oscillating?
  2. Which of the following is the correct pair of natural angular frequency ωn and damping ratio ζ for the second-order system with the transfer function G(s) = 25/(s² + 4s + 25)?
  3. A step is applied to a second-order system (standard form without zeros) with a damping ratio ζ = 0.5. Which of the following is closest to the overshoot (the maximum excursion beyond the final value, as a fraction of the final value)?
  4. Which of the following is the settling time, until the step response of the system G(s) = 25/(s² + 4s + 25) stays within ±2% of its final value, estimated with the approximation ts ≈ 4/(ζωn)?
  5. A second-order system without zeros has its poles at s = −σ ± jωd. Which of the following correctly describes how the step response changes when only ωd is increased while σ is kept the same?

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