Steady-State Error and Static Error Constants
A system can be fast and well-damped and still settle to the wrong value. The transient response tells us how the output gets there; the steady-state error tells us how close "there" is to where we asked. This chapter measures that final accuracy with three error constants and ties it all to a single structural property — the number of integrators in the loop. The result is one compact table that predicts the long-run error of any system from its type alone.
- What steady-state error is and how it differs from the transient response.
- The actuating error signal \(E(s)=R(s)/(1+G(s))\) for a unity-feedback system.
- How the final-value theorem turns \(E(s)\) into the steady-state error \(e_{ss}\).
- The three static error constants — position \(K_p\), velocity \(K_v\), and acceleration \(K_a\).
- The complete system-type error table linking type 0, 1, 2 to step, ramp, parabolic errors.
- Why each integrator eliminates one error, and the gain–accuracy trade-off it implies.
The Accuracy Problem
Chapters 7 and 8 measured how a system reaches its final value — its speed, overshoot, and settling. But reaching a final value smoothly is no use if that value is wrong. The steady-state error \(e_{ss}\) is the difference between the desired input and the actual output once all transients have died away. It is the permanent, unavoidable gap that remains forever.
How large that gap is depends on three things: the kind of input (step, ramp, or parabola), the gain of the loop, and — most importantly — the structure of the system, specifically how many integrators it contains. This chapter makes all three precise.
The Error Signal
Consider the standard unity-feedback loop. The actuating error \(E(s)\) is what the summing point produces: the difference between the reference and the fed-back output. Solving the loop expresses it in terms of the input and the open-loop transfer function \(G(s)\).
Because \(C(s)=G(s)E(s)\), substituting and solving gives this compact form. The larger \(G(s)\) is at low frequency, the smaller the error — which is why high loop gain improves accuracy.
The Final Value Theorem for Error
We rarely need the full time function \(e(t)\) — only its limit as \(t\to\infty\). The final-value theorem delivers exactly that limit directly from \(E(s)\), without any inverse transform.
Substitute the input's transform for \(R(s)\) and take the limit. One important condition: the theorem is valid only if the closed-loop system is stable. For an unstable system the output never settles, and \(e_{ss}\) is meaningless however the algebra comes out.
The Static Error Constants
For each of the three standard inputs, the limit above collapses to a single number — a static error constant — that captures the system's low-frequency gain. Each constant pairs with one input.
| Constant | Definition | Input | Steady-state error |
|---|---|---|---|
| Position \(K_p\) | \(\displaystyle\lim_{s\to 0} G(s)\) | Step \(A/s\) | \(\dfrac{A}{1+K_p}\) |
| Velocity \(K_v\) | \(\displaystyle\lim_{s\to 0} sG(s)\) | Ramp \(A/s^2\) | \(\dfrac{A}{K_v}\) |
| Acceleration \(K_a\) | \(\displaystyle\lim_{s\to 0} s^2 G(s)\) | Parabola \(A/s^3\) | \(\dfrac{A}{K_a}\) |
The names borrow from mechanics: a ramp is a constant-velocity demand, so the ramp error constant is called the velocity constant; a parabola is a constant-acceleration demand, so its constant is the acceleration constant. Note these are not rates — \(K_v\) is a gain, not a speed.
System Type and the Error Table
Recall from Chapter 6 that the type of a system is the number of poles at the origin (integrators) in \(G(s)H(s)\). Type controls the error constants completely: each integrator drives one constant to infinity, and an infinite constant means zero error. Writing \(G(s)\) in time-constant form with low-frequency gain \(K\) makes the pattern exact.
| Type | \(K_p\) | \(K_v\) | \(K_a\) | Step error | Ramp error | Parabola error |
|---|---|---|---|---|---|---|
| Type 0 | \(K\) | 0 | 0 | \(\dfrac{A}{1+K}\) | \(\infty\) | \(\infty\) |
| Type 1 | \(\infty\) | \(K\) | 0 | 0 | \(\dfrac{A}{K}\) | \(\infty\) |
| Type 2 | \(\infty\) | \(\infty\) | \(K\) | 0 | 0 | \(\dfrac{A}{K}\) |
The finite entries run down the diagonal of the table. Above the diagonal the error is zero (the system has "enough" integrators); below it, the error is infinite (too few).
Reading the Pattern
The table is not a list to memorise but a single idea to understand: each integrator buys one level of tracking. A type-0 system cannot even hold a constant output without error; add one integrator and it tracks steps perfectly but lags a ramp by a fixed amount; add a second and it tracks ramps perfectly but lags a parabola. The figure below shows the characteristic constant lag of a type-1 system following a ramp.
Caveats and Non-Unity Feedback
The static error constants are powerful but narrow. They apply only to the three standard inputs, only to unity-feedback systems, and only when the closed loop is stable. Step outside those bounds and the constants must be used with care.
| Situation | What to do |
|---|---|
| Combined input (e.g. \(r=A+Bt\)) | Superpose: add the errors from each component separately. |
| Non-standard input | Return to \(e_{ss}=\lim_{s\to0}sR(s)/(1+G)\) directly; the constants may not apply. |
| Non-unity feedback \(H(s)\) | Use \(E(s)=R(s)/(1+G(s)H(s))\); the constants use \(G(s)H(s)\) in place of \(G(s)\). |
| Unstable closed loop | \(e_{ss}\) is undefined — check stability first. |
For non-unity feedback, note a subtlety: the actuating error \(E(s)=R-B\) is what the constants measure, which is not the same as the true tracking error \(R-C\) unless \(H=1\). When the genuine output error matters, form \(R(s)-C(s)\) explicitly and apply the final-value theorem to that.
Worked Examples
Problem. For \(G(s)=\dfrac{10}{(s+1)(s+2)}\) (unity feedback), find the steady-state error to a unit step.
Solution. This is type 0. Compute the position constant:
A ramp or parabola would give infinite error, since \(K_v=K_a=0\).
Problem. For \(G(s)=\dfrac{20}{s(s+5)}\), find the errors to a unit step and a unit ramp.
Solution. One pole at the origin → type 1, so the step error is zero. For the ramp:
Problem. For \(G(s)=\dfrac{100}{s^2(s+10)}\), find the error to a unit parabola.
Solution. Two poles at the origin → type 2; step and ramp errors are zero. For the parabola:
Problem. A type-1 system has \(K_v=5\). Find the steady-state error to \(r(t)=2 + 3t\).
Solution. Treat the step and ramp parts separately and add. The step part gives zero error (type 1); the ramp part gives \(3/K_v\):
Problem. For \(G(s)=\dfrac{K}{s(s+4)}\), choose \(K\) so the unit-ramp error is at most 0.1.
Solution. The system is type 1, so \(K_v=\lim sG = K/4\). Require the ramp error to meet the bound:
Accuracy demands large gain — but the larger \(K\) is, the lower the damping of this loop, the very trade-off flagged in Section 9-6.
Problem. Without computing constants, state which standard inputs \(G(s)H(s)=\dfrac{50(s+2)}{s(s+1)(s+8)}\) tracks with finite or zero error.
Solution. One pole at the origin → type 1. By the diagonal rule: step error zero, ramp error finite, parabola error infinite.
Chapter Summary
Unity feedback: \(E(s)=R(s)/(1+G(s))\); higher gain → smaller error.
\(e_{ss}=\lim_{s\to0}sR(s)/(1+G(s))\) — valid only if the loop is stable.
\(K_p=\lim G\), \(K_v=\lim sG\), \(K_a=\lim s^2G\), one per input.
Step \(A/(1{+}K_p)\), ramp \(A/K_v\), parabola \(A/K_a\).
Each integrator zeros one error; finite errors lie on the table's diagonal.
Gain improves accuracy but harms damping — the case for compensators.
Problems
For each system, identify the type first, then compute only the relevant error constant. Assume unity feedback and a stable closed loop unless told otherwise. Difficulty rises down the list.
- State the type of \(G(s)=\dfrac{8}{(s+2)(s+4)}\) and find its position error constant.
- Find the unit-step error of the system in Problem 1.
- For \(G(s)=\dfrac{15}{s(s+3)}\), find \(K_v\) and the unit-ramp error.
- For \(G(s)=\dfrac{40}{s^2(s+5)}\), find \(K_a\) and the unit-parabola error.
- A type-1 system has \(K_v=10\). Find its steady-state error to \(r(t)=4t\).
- A type-0 system has \(K_p=9\). Find its error to a step of magnitude 5.
- For \(G(s)=\dfrac{K}{s(s+6)}\), choose \(K\) so the unit-ramp error is at most 0.05.
- A type-1 system tracks \(r(t)=1+2t\). Given \(K_v=8\), find the total steady-state error.
- State, without full computation, which standard inputs \(G(s)=\dfrac{20(s+1)}{s^2(s+4)}\) tracks with zero error.
- For non-unity feedback \(G(s)=\dfrac{10}{s+2}\), \(H(s)=2\), find the actuating-error constant \(K_p\) and the step error.
- Explain why raising the loop gain reduces steady-state error but can degrade the transient response.
- A system must track a unit ramp with error \(\le 0.02\) and overshoot \(\le 10\%\). Explain why these two specs can conflict and what kind of compensation might satisfy both.