First-Order System Response
The first-order system is the simplest dynamic system there is — one pole, one exponential, one number that describes everything: the time constant. An RC circuit charging, a thermometer warming, a tank filling, a motor reaching speed — all of them obey the same law and the same rising curve. Master this one response completely and you have the vocabulary of speed, lag, and settling that the rest of time-domain analysis is built on.
- The standard first-order form \(1/(\tau s+1)\), the time constant \(\tau\), and the DC gain \(K\).
- The unit step response \(1-e^{-t/\tau}\) and the famous 63.2% rule.
- The first-order performance measures: delay time, rise time, and settling time.
- The impulse response and its link to the step response as a derivative.
- The ramp response and the steady-state tracking error equal to \(\tau\).
- How the single pole at \(-1/\tau\) in the \(s\)-plane sets the speed of the whole response.
The Standard First-Order Form
A first-order system has a single pole and no zeros in the basic case. Its closed-loop transfer function is written in the standard form below, where \(\tau\) is the time constant (in seconds) and \(K\) is the DC gain — the steady-state output for a unit step input.
Take \(K=1\) for the unity-gain case (the default in this chapter). The pole sits on the negative real axis at \(-1/\tau\); a smaller \(\tau\) pushes it farther from the origin and makes the system faster.
The Time Constant τ
The time constant \(\tau\) is the single most important number for a first-order system. It is the natural timescale of the exponential: every behaviour — how fast the output rises, how long it takes to settle, how far it lags a ramp — is measured in multiples of \(\tau\). Larger \(\tau\) means a more sluggish system; smaller \(\tau\) means a snappier one.
Unit Step Response
Drive the unity-gain system with a unit step, \(R(s)=1/s\). A partial-fraction split gives the signature rising exponential — the most important single result of the chapter.
At one time constant the output has reached 63.2% of its final value; at \(2\tau\) it is \(86.5\%\), at \(3\tau\) it is \(95\%\), and by \(5\tau\) it is \(99.3\%\) — effectively complete. The tangent at the origin has slope \(1/\tau\) and would reach the final value at exactly \(t=\tau\).
Performance Measures
The standard time-domain measures all reduce to multiples of \(\tau\) for a first-order system, because the response is a single clean exponential. These are the numbers a specification sheet quotes.
| Measure | Definition | First-order value |
|---|---|---|
| Time constant \(\tau\) | Time to reach 63.2% of final value | \(\tau\) |
| Delay time \(t_d\) | Time to reach 50% of final value | \(0.693\,\tau\) |
| Rise time \(t_r\) | Time from 10% to 90% of final value | \(2.2\,\tau\) |
| Settling time \(t_s\) (5%) | Time to stay within 5% of final | \(3\,\tau\) |
| Settling time \(t_s\) (2%) | Time to stay within 2% of final | \(4\,\tau\) |
Unit Impulse Response
For a unit impulse, \(R(s)=1\), so \(C(s)=G(s)\) directly. The response is a decaying exponential that starts at \(1/\tau\) and falls to zero — and it is exactly the time-derivative of the step response, as Chapter 6 promised.
Unit Ramp Response
Now drive the system with a unit ramp, \(R(s)=1/s^2\). The output eventually rises at the same rate as the input but lags permanently behind it — and that constant lag is exactly the time constant \(\tau\).
The first-order system follows a ramp with a permanent error equal to its time constant. A faster system (smaller \(\tau\)) tracks the ramp more closely — the same conclusion the type number reaches more generally in Chapter 9.
Pole Location and Speed
Everything in this chapter is encoded in one point: the pole at \(s=-1/\tau\). The single real pole sits on the negative real axis, and its distance from the origin — the quantity \(1/\tau\) — is the speed of the response. Push the pole left and the system speeds up; let it drift toward the origin and the system slows.
Worked Examples
Problem. A system is \(\dfrac{C}{R}=\dfrac{1}{0.5s+1}\). Write \(c(t)\) for a unit step and find the time to reach 63.2% and to settle within 2%.
Solution. Here \(\tau = 0.5\,\text{s}\):
Problem. For \(G(s)=\dfrac{5}{2s+1}\), find the unit step response and its final value.
Solution. Read off \(K=5\) and \(\tau=2\,\text{s}\). The shape is the same exponential, scaled by \(K\):
The final-value theorem confirms it: \(\lim_{s\to0} s\cdot G(s)\cdot\frac{1}{s} = G(0) = 5\).
Problem. A thermometer behaves as a first-order system with \(\tau = 4\,\text{s}\). Estimate its delay time and rise time.
Solution. Apply the multiples from Section 7-4:
It takes roughly 8.8 s for the reading to swing from 10% to 90% of a temperature change.
Problem. Find the unit impulse response of \(G(s)=\dfrac{1}{0.25s+1}\).
Solution. With \(\tau=0.25\,\text{s}\), the impulse response starts at \(1/\tau=4\):
Problem. A unity-gain first-order system with \(\tau=0.2\,\text{s}\) tracks a unit ramp. Find the steady-state error.
Solution. From Section 7-6, the steady-state ramp error equals the time constant:
The output runs parallel to the ramp but a fixed 0.2 units behind it forever.
Problem. In a step test, the output reaches 95% of its final value after 6 s. Estimate the time constant and the pole.
Solution. The 95% point occurs at \(t=3\tau\), so:
This is how a first-order model is fitted from a measured step response — read a known percentage point, divide by its multiple of \(\tau\).
Chapter Summary
\(\dfrac{K}{\tau s+1}\): one pole at \(-1/\tau\), time constant \(\tau\), DC gain \(K\).
\(c(t)=1-e^{-t/\tau}\); 63.2% at \(\tau\), 95% at \(3\tau\), 99.3% at \(5\tau\).
\(t_d=0.693\tau\), \(t_r=2.2\tau\), \(t_s=3\tau\) (5%) or \(4\tau\) (2%).
\(\dfrac{1}{\tau}e^{-t/\tau}\) — the derivative of the step response.
\(t-\tau+\tau e^{-t/\tau}\); steady-state tracking error \(e_{ss}=\tau\).
Distance \(1/\tau\) from the origin sets the speed; real pole means no overshoot.
Problems
Use the standard form and the \(\tau\)-multiples of Section 7-4 throughout. Difficulty rises down the list.
- Write the pole location and time constant of \(G(s)=\dfrac{1}{3s+1}\).
- For \(\tau=0.5\,\text{s}\), find the unit step response \(c(t)\) and the value at \(t=\tau\).
- A first-order system has \(\tau=2\,\text{s}\). Find its delay time, rise time, and 2% settling time.
- For \(G(s)=\dfrac{4}{s+2}\), identify \(K\) and \(\tau\), then write the unit step response and its final value.
- Find the unit impulse response of \(G(s)=\dfrac{1}{0.1s+1}\) and state its initial value.
- Show that the impulse response of a first-order system is the derivative of its step response.
- A unity-gain system with \(\tau=0.4\,\text{s}\) tracks a unit ramp. Write \(c(t)\) and find \(e_{ss}\).
- A step test reaches 86.5% of its final value after 5 s. Find \(\tau\) and the pole.
- Two first-order systems have \(\tau_1=1\,\text{s}\) and \(\tau_2=0.25\,\text{s}\). Compare their settling times and pole locations.
- A first-order system has DC gain 10 and reaches 63.2% of its final value at \(t=0.8\,\text{s}\). Write its transfer function.
- Explain, using the pole location, why a first-order system can never overshoot its final value.
- An RC low-pass filter has \(R=10\,\text{k}\Omega\) and \(C=10\,\mu\text{F}\). Find \(\tau\), the pole, and the 2% settling time of its step response.