What is First-order system?

A system whose transfer function is G(s) = K/(1 + Ts), where K is the steady-state gain and T the time constant. Its output for a step input is K(1 − e^(−t/T)), reaching about 63.2% of the final value at t = T, about 95% at 3T and about 99.3% at 5T, with no overshoot. Many familiar systems, such as RC circuits, thermal systems and liquid-level systems, can be approximated in this form. In the frequency response, the gain falls at −20 dB/dec above the corner angular frequency 1/T.

Japanese name: 一次遅れ系

First-order system appears in 5 practice questions across 1 topic, including Control Engineering Basics.

Field: Control Engineering Basics

Try it: questions about First-order system

A step of size 1 is applied at time 0 to a first-order system with the transfer function G(s) = 2/(1 + 0.5s). Which of the following is closest to the output at t = 1 s?

AAbout 1.26
BAbout 1.73
C2

Show answer and explanation

Answer: B. About 1.73

The step response of a first-order system K/(1 + Ts) is y(t) = K(1 − e^(−t/T)), here with K = 2 and T = 0.5 s. Since t = 1 s is twice the time constant, y = 2 × (1 − e^(−2)) ≈ 2 × 0.865 ≈ 1.73. The response reaches about 63.2% of its final value at the time constant T, about 95% at 3T and about 99.3% at 5T.

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?

A3/(1 + 0.4s)
B3/(1 + 2.5s)
C6/(1 + 0.4s)

Show answer and explanation

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.

First-order system in the explanations

The step response of a first-order system K/(1 + Ts) is y(t) = K(1 − e^(−t/T)), here with K = 2 and T = 0.5 s.
A step of size 1 is applied at time 0 to a first-order syst…
Rewriting it as G(s) = 2.5/(1 + 0.5s) shows that it is a first-order system with a steady-state gain of 2.5 and a time constant of 0.5 s.
A step of size 1 is applied to a system with the transfer f…