System Dynamics · Module 7 of 10
First-Order System Response
One pole, one exponential. A first-order system answers a step with a smooth rise governed by a single time constant, and that one number, read from the response or the model, predicts everything.
Readiness check
This module reads the first-order response in detail. Tick only what you can do closed-notes.
- Recall the time constant τ and the 63 percent rule.
- Evaluate e−1 ≈ 0.368 and e−2 ≈ 0.135.
- Read a DC gain from a transfer function.
- Recall that a single pole gives an exponential response.
- Distinguish an initial value from a final value.
The core idea
A first-order system has the transfer function K/(τs + 1). Its step response rises exponentially toward K times the input, reaching 63 percent of the change in one time constant and effectively settling in about five.
G(s) = K/(τs + 1)y(t) = yfinal(1 − e−t/τ)63% at τ, 95% at 3τ, 99% at 5τA first-order system has one energy store and one pole, at s = −1/τ. In standard form its transfer function is G(s) = K/(τs + 1), where K is the DC gain and τ the time constant. Its response to a step input is a single rising exponential: it starts at its initial value and heads smoothly toward the final value yfinal = K times the step size, with no overshoot and no oscillation. The time constant sets the pace. After one τ the response has covered 63.2 percent of the total change, after three τ about 95 percent, and after five τ over 99 percent, effectively complete. Because the shape is always the same, knowing τ and the final value fixes the entire curve, and reading τ from measured data identifies the system.
The skills, taught in order
Five skills cover the standard form, the step response, and identifying a time constant from data.
7.1 The first-order standard form
A first-order transfer function is written G(s) = K/(τs + 1). The gain K scales the steady output; the time constant τ sets the speed and places the single pole at s = −1/τ. Putting any first-order model in this form exposes both numbers at once.
7.2 The time constant
The time constant τ is the time for the response to reach 63.2 percent of its total change. It is the reciprocal of the pole magnitude: a faster system has a smaller τ and a pole further from the origin. Everything about the timing scales with τ.
7.3 The step response
To a step input the response is y(t) = yfinal(1 − e−t/τ), starting at zero and rising smoothly to yfinal = K times the step. There is no overshoot, because a single real pole cannot produce oscillation.
| Time | Fraction of change | Remaining |
|---|---|---|
| τ | 63.2% | 36.8% |
| 2τ | 86.5% | 13.5% |
| 3τ | 95.0% | 5.0% |
| 5τ | 99.3% | 0.7% |
The universal first-order step response in units of the time constant. The same percentages hold for every first-order system.
7.4 Settling and the 63 percent rule
Practical settling is usually taken at three to five time constants, where the response is within 5 to 1 percent of its final value. The 63 percent rule runs in reverse too: the time to reach 63 percent of the change reads the time constant straight off a measured curve.
7.5 Identifying a time constant from data
Given a measured step response, find the final value, locate where the curve reaches 63 percent of the change, and read that time as τ. This simple identification turns a recorded response into a model, the experimental side of system dynamics.
Engineering connection: a sensor's time constant, identified this way, tells you the fastest signal it can faithfully follow.
Worked example 1: step response of a first-order system
A first-order system has gain K = 4 and time constant τ = 0.5 s. A step of size 2 is applied. Find the final value and the output at t = 1 s.
- ProblemFind the final value and y(1 s) for the system in Figure 1.
- Given / findK = 4, τ = 0.5 s, step size U = 2. Find yfinal and y(1 s).
- AssumptionsLinear first-order system; zero initial output; step applied at t = 0.
- Modelyfinal = K U; y(t) = yfinal(1 − e−t/τ).
- Equationsyfinal = K Uy(t) = yfinal(1 − e−t/τ)
- Solveyfinal = 4 × 2 = 8. At t = 1 s, t/τ = 1/0.5 = 2, so y = 8(1 − e−2) = 8(1 − 0.135) = 8 × 0.865 = 6.92.
- Checkt = 1 s is two time constants, where the standard response is 86.5 percent of the change: 0.865 × 8 = 6.92, confirming the result.
- ConclusionThe system heads to 8 and is at 6.92 after 1 s. Knowing K and τ fixes the whole curve without solving a differential equation each time.
Worked example 2: a thermometer time constant
A thermometer behaves as a first-order system with τ = 8 s. It starts at 20 °C and is plunged into a 100 °C bath. Find its reading at t = 8 s, and the times to reach 95 percent and 99 percent of the change.
- ProblemFind T(8 s) and the 95 and 99 percent times for the thermometer in Figure 2.
- Given / findτ = 8 s, T0 = 20 °C, Tfinal = 100 °C. Find T(8 s), t95, t99.
- AssumptionsFirst-order behaviour; bath temperature constant; step change at t = 0.
- ModelT(t) = Tfinal − (Tfinal − T0)e−t/τ; settling times are multiples of τ.
- EquationsT(t) = 100 − 80 e−t/τt95 ≈ 3τ, t99 ≈ 5τ
- SolveAt t = 8 s = τ: T = 100 − 80 e−1 = 100 − 80 × 0.368 = 100 − 29.4 = 70.6 °C. t95 = 3 × 8 = 24 s, t99 = 5 × 8 = 40 s.
- CheckThe 70.6 °C reading is 63.2 percent of the 80-degree rise above 20, the defining mark of one time constant. The settling multiples follow the universal table.
- ConclusionThe thermometer needs about 40 s to read the bath accurately. Its 8 s time constant is the key number for whether it can track a changing temperature.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Expecting overshoot | First-order response drawn with a peak | "How many poles does it have?" | One real pole cannot overshoot. |
| τ from the wrong point | Time constant read at the final value | "Where is the 63 percent mark?" | τ is the time to reach 63.2 percent of the change. |
| Final value error | Output settles to the wrong level | "Did I multiply gain by step size?" | yfinal = K times the step. |
| Settling too early | Declared settled at one τ | "How many τ for the tolerance?" | Use about 3τ for 5 percent, 5τ for 1 percent. |
Practice ladder
A first-order system has τ = 2 s. What fraction of a step change is reached at t = 2 s?
Show answer
At t = τ the response is 1 − e−1 = 0.632, or 63.2 percent.
A system G(s) = 5/(0.2s + 1) gets a unit step. Find the final value and the time to reach 95 percent.
Show answer
K = 5, τ = 0.2 s. Final value = 5. Time to 95 percent ≈ 3τ = 0.6 s.
A measured step response rises from 0 toward 12 and passes 7.6 at t = 3 s. Estimate the time constant.
Show answer
7.6/12 = 0.63, the 63 percent mark, so τ ≈ 3 s. The system is first order with τ = 3 s.
You record a sensor's step response and want its time constant. Describe the measurement and the single point you read.
What good work looks like
Apply a step, record the output, find the final value, and read the time at which the output reaches 63.2 percent of the total change; that time is τ. A check at 3τ (95 percent) confirms first-order behaviour.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Record or simulate a first-order step response, identify its time constant from the 63 percent point, and predict the 95 percent settling time, then verify it.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the first-order standard form.
G(s) = K/(τs + 1), with gain K and time constant τ.
2. Write the step response.
y(t) = yfinal(1 − e−t/τ).
3. What fraction is reached at one τ?
63.2 percent of the total change.
4. How many τ to settle within 1 percent?
About five time constants.
5. How do you read τ from data?
Find the time to reach 63.2 percent of the change.
Textbook mapping
This module follows Karnopp, Margolis, and Rosenberg, System Dynamics: Modeling, Simulation, and Control of Mechatronic Systems, 5th edition. Use these references to read further.
| Topic in this module | Where to read more |
|---|---|
| First-order standard form and time constant | Karnopp, Margolis & Rosenberg, Chapter 8 |
| The step response and settling | Karnopp, Margolis & Rosenberg, Chapter 8 |
| Identifying a time constant from data | Karnopp, Margolis & Rosenberg, Chapter 9 |
Chapter numbers refer to the 5th edition. The first-order response is standard, so any recent edition will align closely.