Library / Control engineering / CTL-02
Control engineering
First-order step response
Output of a first-order system after a unit step, starting from rest.
- Ref
- CTL-02
- Rev
- v1
- Field
- Control engineering
- Result
- y(t)
y(t) is the output of a first-order system K/(τs + 1) after a unit step at t = 0, starting from rest. K is the final value (the DC gain) and τ is the time the response takes to cover 63.2 % of the way there. For a step of size R multiply the result by R.
Output of a first-order system after a unit step, starting from rest. It reaches 63.2 % of its final value after one time constant and about 98 % after four.
Calculator
- Fraction of the final value reached86.47 %
- Time to reach 63.2 %, τ0.5 s
- 2 % settling time, 4τ2 s
- Initial slope, K/τ4 units/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.
- At t = 1 s, which is 2 time constants, the output has covered 86.47 % of its final value. After 4τ = 2 s it has covered 98.2 %, which is why 4τ is quoted as the settling time.
- The initial slope is K/τ = 4 per s: a tangent drawn at t = 0 would reach the final value at t = τ. That gives a quick way to read τ from a recorded step.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| K DC gain | 0.5 to 5 | 0.4323 | 4.323 | 3.891 | +10% | +10 % |
| t Time since the step | 0 to 5 s | 0 | 2 | 2 | +10% | +2.84 % |
| τ Time constant | 0.1 to 2 s | 0.7869 | 2 | 1.213 | +10% | −3.12 % |
Sweep of Time since the step
| Time since the step (s) | Output at time t |
|---|---|
| 0 | 0 |
| 0.625 | 1.427 |
| 1.25 | 1.836 |
| 1.875 | 1.953 |
| 2.5 | 1.987 |
| 3.125 | 1.996 |
| 3.75 | 1.999 |
| 4.375 | 2 |
| 5 | 2 |
Sweep one variable
Variables and how each one contributes
Output at time t
What the equation gives you. Everything else feeds it.
DC gain
The final value reached for a unit step: the DC gain of the system. Directly proportional: the whole curve scales with K at every instant. Raising it by 10 % (to 2.2) moves the result by +10 %, from 1.729 to 1.902.
Time constant
The time constant, in s: the time to reach 63.2 % of the final value. A larger τ slows the response, so at a fixed time the output is lower. Raising it by 10 % (to 0.55) moves the result by −3.12 %, from 1.729 to 1.675.
Time since the step
Time since the step was applied, in s (same unit as τ). The output rises with time but more slowly as it approaches K, so the same extra time gains less each time. Raising it by 10 % (to 1.1) moves the result by +2.84 %, from 1.729 to 1.778.
Worked example with the values above
- Number of time constants: t/τ = 1 / 0.5 = 2.
- Remaining fraction: e−2 = 0.1353, so the response has covered 1 − 0.1353 = 0.8647 of its way.
- y = K × 0.8647 = 2 × 0.8647 = 1.729.
Assumptions and limits
- A linear first-order system starting from rest, with a step applied at t = 0.
- Constant K and τ, with no time delay, saturation or nonlinearity.
Common mistakes
- Taking the 63 % point as the final value. The response only approaches K and needs about 4τ to settle within 2 %.
- Mixing time units, for example τ in minutes and t in seconds.
Statistics and uncertainty
Reading τ from the time the curve reaches 63.2 % is robust, because the curve is still steep there: noise of 2 % of the final value shifts that time by only about 5 % of τ (0.02/0.368). Near 98 % of the final value the curve is nearly flat, and the same noise would shift the time reading by a full τ, so the settling time is a poor place to measure it.
Where it is used
- Fitting a time constant to a thermal, motor-speed or tank-level step test.
- Estimating when a sensor reading has settled after a change.
Maths you need
- Exponential decay: e^(−x) falls to 37 % at x = 1
- Step input: a sudden change that is then held constant