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Control engineering

Proportional feedback on a first-order plant

Closed-loop time constant when a first-order plant b/(s + a) is controlled by a pure proportional gain with negative unity feedback.

Ref
CTL-01
Rev
v1
Field
Control engineering
Result
τcl [s]

The plant is b/(s + a), so on its own it has a time constant 1/a. Proportional feedback with gain Kc moves its pole from −a to −(a + Kcb), which makes the loop faster by the factor (a + Kcb)/a. The closed-loop DC gain is Kcb/(a + Kcb), which is always below 1, so with this controller the output never quite reaches a constant reference.

Closed-loop time constant when a first-order plant b/(s + a) is controlled by a pure proportional gain with negative unity feedback. A larger gain speeds the response up, but it leaves a steady-state error.

Calculator

Solve for
s
1/s
0.25
1/s
0.25
0.220
  • Closed-loop pole, −(a + Kcb)−5 1/s
  • Closed-loop DC gain, Kcb / (a + Kcb)0.6
  • Steady-state error to a unit step, 1 − DC gain0.4
  • Open-loop time constant, 1/a0.5 s

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Closed-loop time constant (s) current value 0.2 s
KcProportional controller gainKey driver
0.04550.455
bPlant gain numerator b (plant is b/(s + a))
0.05880.385
aPlant pole a
0.1250.313
  • At the defaults the loop has a time constant of 0.2 s, against 0.5 s for the plant alone: feedback made it 2.5 times faster. The price is that the output reaches only 0.6 of a unit step, a steady error of 40 %.
  • Doubling the gain to 6 shortens the time constant to 0.125 s and cuts the steady error to 25 %, but it never reaches zero. Removing the error entirely needs integral action.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (s)Local stepChange in result
Kc Proportional controller gain0.2 to 20 0.045450.45450.4091+10%−5.66 %
b Plant gain numerator b (plant is b/(s + a))0.2 to 5 1/s0.058820.38460.3258+10%−5.66 %
a Plant pole a0.2 to 5 1/s0.1250.31250.1875+10%−3.85 %

Sweep of Proportional controller gain

Proportional controller gainClosed-loop time constant (s)
0.20.4545
2.6750.2139
5.150.1399
7.6250.1039
10.10.08264
12.570.06861
15.050.05865
17.520.05122
200.04545

Sweep one variable

Closed-loop time constant (s)Hover, touch, or use the arrow keys on the chart
0.10.20.30.4 5101520 Proportional controller gain now 0.2

Variables and how each one contributes

τcl

Closed-loop time constant

Result · s

What the equation gives you. Everything else feeds it.

b

Plant gain numerator b (plant is b/(s + a))

1/s · typical 0.2 to 5

The numerator of the plant b/(s + a), in 1/s. The plant's own DC gain is b/a. Raising b strengthens the loop in the same way as raising Kc, because only the product Kcb enters the closed-loop pole. Raising it by 10 % (to 1.1) moves the result by −5.66 %, from 0.2 to 0.1887 s.

a

Plant pole a

1/s · typical 0.2 to 5

The pole of the plant, in 1/s: left alone the plant settles as e−at. A larger a means a faster plant. A larger a makes the loop faster, but the effect is smaller than the change in a itself because feedback already contributes part of the pole. Raising it by 10 % (to 2.2) moves the result by −3.85 %, from 0.2 to 0.1923 s.

Kc

Proportional controller gain

· typical 0.2 to 20

The proportional controller gain: the control signal is Kc times the error. It is dimensionless when the plant gain b/a is dimensionless. A higher gain pushes the pole further left and shortens the time constant, but the benefit fades as Kcb grows larger than a. Raising it by 10 % (to 3.3) moves the result by −5.66 %, from 0.2 to 0.1887 s.

Worked example with the values above

  1. Closed-loop pole: a + Kcb = 2 + 3 × 1 = 5 1/s, so the pole is at s = −5.
  2. Time constant: τcl = 1 / 5 = 0.2 s.
  3. Final output for a unit step: Kcb / (a + Kcb) = 3 / 5 = 0.6. The time constant is 0.2 s.

Assumptions and limits

  • A linear first-order plant with no time delay, no actuator limit and an ideal sensor.
  • Negative unity feedback with a purely proportional controller (no integral or derivative action).

Common mistakes

  • Treating the loop as accurate because it is fast. With proportional action alone the output stops short of the reference by 1/(1 + Kcb/a) of the step.
  • Raising Kc without limit. Real actuators saturate, sensors add noise and unmodelled lag makes the loop unstable at high gain, so the first-order picture stops holding.

Statistics and uncertainty

The time constant is 1/(a + Kcb), so a relative error in the sum maps into τcl with the opposite sign. At the defaults the plant pole supplies 40 % of the sum and feedback supplies the rest. The higher the gain, the less the response depends on uncertainty in a, and the more it depends on Kcb, the quantity a designer controls.

Where it is used

  • Estimating how fast a speed, temperature or level loop responds after a gain change.
  • Showing why integral action is added: it removes the finite steady-state error this formula leaves.

Maths you need

  • Poles and time constants: a pole at −σ gives a decay e^(−σt), time constant 1/σ
  • Negative feedback: closed loop T = L / (1 + L)

Related formulas

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