Standard Test Signals and System Time Response
Once a system is captured as a transfer function, the next question is the practical one: how does it actually behave in time? To answer it consistently we excite every system with the same small family of standard inputs — impulse, step, ramp, parabola — and read the response. That response splits cleanly into a transient part that fades and a steady-state part that remains, and the location of the poles decides the shape of both. This chapter sets up that whole language and is the gateway to all of time-domain analysis.
- Why analysis uses a small set of standard test signals rather than arbitrary inputs.
- The four signals — impulse, step, ramp, parabolic — their definitions and Laplace transforms.
- The integral family that links them, and what each one physically represents.
- How the total response splits into a transient part and a steady-state part.
- How pole locations in the \(s\)-plane set the shape — decaying, growing, or oscillating — of the transient.
- The meaning of the order and the type of a system, and why each matters.
Why Test Signals?
A control system in service may face any input at all — a pilot's stick motion, a fluctuating load, a noisy sensor. We cannot characterise a system against "any input," so analysis instead agrees on a handful of standard test signals that are simple to generate, simple to transform, and rich enough to expose every behaviour that matters. Because the systems we study are linear and time-invariant, the response to these few signals predicts the response to a great many real ones by superposition.
Each test signal also mimics a class of real demands. A sudden command is a step; a constant-velocity tracking demand is a ramp; a constant-acceleration demand is a parabola; a sharp jolt is an impulse. Choosing the input that resembles the worst case the system will meet is how a designer decides what to test.
The Standard Test Signals
Four signals do almost all the work. Each is defined for \(t \ge 0\) and is zero before; \(A\) is a chosen magnitude.
| Signal | \(r(t),\; t\ge 0\) | \(R(s)\) | Models |
|---|---|---|---|
| Impulse | \(A\,\delta(t)\) | \(A\) | A sudden shock or jolt |
| Step | \(A\,u(t)\) | \(\dfrac{A}{s}\) | A sudden constant command |
| Ramp | \(A\,t\) | \(\dfrac{A}{s^2}\) | A constant-velocity demand |
| Parabolic | \(\dfrac{A\,t^2}{2}\) | \(\dfrac{A}{s^3}\) | A constant-acceleration demand |
Because the impulse transforms to \(1\), the impulse response of a system is the inverse transform of its transfer function itself. Every other response can be built from it.
The Integral Family
The four signals are not independent — each is the time-integral of the one before it (and the derivative of the one after). Integrating in time divides the transform by \(s\), which is exactly why each \(R(s)\) gains one more power of \(s\) in its denominator.
The whole family is a single pattern. The generalised test signal of order \(n\) and its transform are:
Transient and Steady-State Response
Apply a test signal and the output evolves in two overlapping parts. The transient response is the portion that dies away with time; the steady-state response is what remains once the transient has vanished. The total response is their sum:
For a stable system the transient always decays to zero, so the steady-state part is what the output settles to. The transient governs how the system gets there — its speed, overshoot, and oscillation.
Poles Shape the Transient
Where do the transient terms come from? From the poles of the response \(C(s)\). A partial-fraction expansion turns each pole into one time-domain term, and the pole's location in the \(s\)-plane fixes that term's character. A real pole gives a pure exponential; a complex-conjugate pair gives a decaying (or growing) sinusoid.
| Pole location | Time-domain term | Behaviour |
|---|---|---|
| Real, \(s=-\sigma\) (left half) | \(e^{-\sigma t}\) | Decaying exponential |
| Real, \(s=+\sigma\) (right half) | \(e^{+\sigma t}\) | Growing — unstable |
| Complex pair, \(s=-\sigma\pm j\omega\) | \(e^{-\sigma t}\sin(\omega t+\phi)\) | Damped oscillation |
| Imaginary pair, \(s=\pm j\omega\) | \(\sin(\omega t+\phi)\) | Sustained oscillation |
Order of a System
The order of a system is the highest power of \(s\) in the denominator of its (closed-loop) transfer function — equivalently, the number of poles. Order counts the independent energy-storage elements in the physical system and sets how many transient terms the response can contain.
The next two chapters dissect the two simplest and most important cases in detail: the first-order system (one pole, one exponential) in Chapter 7, and the second-order system (a complex pair, a damped oscillation) in Chapter 8. Almost every real response is read as one of these two, or a combination of them.
Type of a System
Distinct from order is the type of a system: the number of poles at the origin (\(s=0\), i.e. pure integrators) in the open-loop transfer function \(G(s)H(s)\). Where order describes the transient richness, type describes the system's ability to track inputs without steady-state error.
A type-0 system has no pole at the origin, a type-1 system one, a type-2 system two. Each added integrator lets the system follow one rung higher on the test-signal ladder — step, then ramp, then parabola — with zero steady-state error.
| Open-loop \(G(s)H(s)\) | Type | Order |
|---|---|---|
| \(\dfrac{K}{(s+2)(s+5)}\) | 0 | 2 |
| \(\dfrac{K}{s(s+4)}\) | 1 | 2 |
| \(\dfrac{K(s+1)}{s^2(s+6)}\) | 2 | 3 |
Worked Examples
Problem. An input is \(r(t) = 2\,u(t) + 3t\). Find \(R(s)\).
Solution. Transform each test signal and add (linearity):
Problem. An open-loop transfer function is \(G(s)H(s) = \dfrac{20}{s(s+2)(s+5)}\). State the order and the type.
Solution. The denominator has degree three, so the order is 3. There is exactly one pole at the origin, so the type is 1.
Problem. A first-order system \(\dfrac{C}{R} = \dfrac{1}{\tau s + 1}\) is driven by a unit step. Identify the transient and steady-state parts of \(c(t)\).
Solution. With \(R(s)=1/s\), expand \(C(s)\) in partial fractions:
The steady-state value is \(1\); the transient \(-e^{-t/\tau}\) decays to zero, faster for smaller \(\tau\).
Problem. A response \(C(s)\) has poles at \(s=-4\) and \(s=-1\pm j3\). Without full computation, state the form of the transient.
Solution. The real pole gives an exponential; the complex pair gives a damped sinusoid:
All poles lie in the left half-plane, so the transient decays — the system is stable. The \(e^{-4t}\) term vanishes quickly; the slower \(e^{-t}\) envelope dominates the tail.
Problem. Find the impulse response of \(G(s) = \dfrac{5}{s+5}\).
Solution. For an impulse \(R(s)=1\), so \(C(s)=G(s)\) and the response is the inverse transform of \(G(s)\):
Integrating this in time gives the step response \(1 - e^{-5t}\) — confirming that the step response is the integral of the impulse response.
Problem. A pure integrator \(G(s)=\dfrac{1}{s}\) receives a unit ramp \(r(t)=t\). Find \(c(t)\).
Solution. With \(R(s)=1/s^2\):
The integrator turns a ramp into a parabola — the integral-family ladder of Section 6-3, seen as a system response.
Chapter Summary
Impulse, step, ramp, parabola: transforms \(1,\;A/s,\;A/s^2,\;A/s^3\).
Each integrates to the next; \(A t^n/n! \leftrightarrow A/s^{n+1}\).
\(c(t)=c_{\text{tr}}(t)+c_{\text{ss}}(t)\); the transient decays for a stable system.
Real poles → exponentials, complex pairs → damped sinusoids; left half = stable.
Degree of the closed-loop denominator = number of poles = transient richness.
Number of open-loop poles at the origin; governs steady-state tracking.
Problems
Use the transform pairs of Section 6-2 and the pole-to-transient map of Section 6-5 throughout. Difficulty rises down the list.
- Write the Laplace transforms of a unit impulse, a unit step, a unit ramp, and a unit parabola.
- Find \(R(s)\) for \(r(t) = 4 + 2t\).
- Find \(R(s)\) for \(r(t) = 3t^2\) (note the factor of \(2!\)).
- State the order and type of \(G(s)H(s) = \dfrac{K}{(s+1)(s+3)}\).
- State the order and type of \(G(s)H(s) = \dfrac{K(s+2)}{s^2(s+4)}\).
- A first-order system \(\dfrac{1}{2s+1}\) is given a unit step. Find \(c(t)\) and name its transient and steady-state parts.
- A response has poles at \(s=-2\) and \(s=-5\). Write the general form of the transient.
- A response has poles at \(s=-1\pm j4\). Write the general form of the transient and state whether it is stable.
- Find the impulse response of \(G(s)=\dfrac{10}{s+2}\), then obtain the step response by integration.
- A pure integrator \(1/s\) is driven by a unit step. Find \(c(t)\) and identify which test signal it equals.
- Explain why a system with a pole in the right half-plane has no steady-state response in the usual sense.
- Two systems have the same order but different type. Sketch in words how their step responses can settle to the same value yet differ in steady-state error, pointing ahead to Chapter 9.