Steady-state error and error constants

Why it matters

A CNC axis that stops 20 µm short of its target, or a conveyor that lags behind its speed command, fails on accuracy however good its transient response looks. Steady-state error tells you how closely a stable loop follows a command after the transients have died out, and the error constants let you predict it — and size a gain or add an integrator — without simulating anything.

Key ideas

Definition. The actuating error is e(t) = r(t) − c(t) for unity feedback. The steady-state error is ess = lim(t→∞) e(t). It is only meaningful if the closed loop is stable; always check stability (Routh) first — the final value theorem gives a finite but meaningless number for an unstable loop.

Final value theorem. lim(t→∞) e(t) = lim(s→0) s·E(s), valid when all poles of s·E(s) are in the open left half-plane.

System type. Write the open-loop transfer function as G(s)H(s) = K·(1 + T_a·s)… / (s^N·(1 + T₁·s)…). The number N of open-loop poles at the origin (pure integrators) is the type of the system. Type is not the same as order. Each integrator lets the loop hold a non-zero output with zero error, which is why type decides which inputs are tracked exactly.

Standard test inputs (magnitudes A):

• Step r = A → R(s) = A/s (position command).
• Ramp r = A·t → R(s) = A/s² (constant velocity).
• Parabola r = A·t²/2 → R(s) = A/s³ (constant acceleration).

Error constants (unity feedback):

• Position constant Kp = lim(s→0) G(s)
• Velocity constant Kv = lim(s→0) s·G(s) (unit s⁻¹)
• Acceleration constant Ka = lim(s→0) s²·G(s) (unit s⁻²)

The type table (unit inputs):

Type	Step	Ramp	Parabola
0	1/(1 + Kp)	∞	∞
1	0	1/Kv	∞
2	0	0	1/Ka

A finite, non-zero error appears on the diagonal; raising K reduces it but cannot remove it. Raising the type (adding an integrator, e.g. PI control) removes it but tends to reduce stability margins — the classic accuracy–stability trade-off.

Superposition. For a combined input such as r(t) = a + b·t, add the errors for each part. If any part gives ∞, the total is ∞.

Non-unity feedback. With H(s) ≠ 1 the error constants above do not apply directly. Either convert to an equivalent unity-feedback system G_eq = G/(1 + GH − G) or compute E(s) = R(s) − C(s) from the closed-loop transfer function. With a constant sensor gain H = k_s, the output settles at the command divided by k_s, so define the error accordingly.

Disturbances. A step load torque D entering between the controller and the plant gives a steady-state output error that falls with controller gain and becomes zero only if there is an integrator before the disturbance entry point (in the controller).

Formulas

E(s) = R(s) / (1 + G(s)) — unity feedback.

ess = lim(s→0) s·R(s) / (1 + G(s)) — final value theorem.

Kp = lim(s→0) G(s), Kv = lim(s→0) s·G(s), Ka = lim(s→0) s²·G(s)

ess(step A) = A/(1 + Kp); ess(ramp A·t) = A/Kv; ess(parabola A·t²/2) = A/Ka

• A in the input's units (e.g. rad, rad/s, rad/s²); ess in output units (e.g. rad).

G_eq(s) = G(s) / (1 + G(s)H(s) − G(s)) — equivalent unity-feedback forward path for a non-unity H.

Worked examples

Example 1 (standard). A unity-feedback speed loop has G(s) = 10/(s + 2). Find the steady-state error for a unit step and for a unit ramp.

1. Closed loop: 10/(s + 12), pole at −12, stable.
2. Type 0 (no pole at the origin). Kp = G(0) = 10/2 = 5.
3. Step: ess = 1/(1 + Kp) = 1/6 = 0.167 (16.7 % of the command).
4. Ramp: Kv = lim s·G(s) = 0, so ess = ∞ — a type-0 loop cannot follow a ramp.

Example 2 (GATE level, gain sizing). A position servo with unity feedback has G(s) = K(s + 2)/(s(s + 4)(s + 5)) (output in rad). (a) Find the minimum K for a steady-state error of at most 0.05 rad to a unit ramp. (b) With that K, find ess for r(t) = 2 + 3t rad.

1. Type 1, so step error is zero and ramp error is finite.
2. Kv = lim(s→0) s·G(s) = K × 2/(4 × 5) = K/10 s⁻¹.
3. Requirement: 1/Kv ≤ 0.05 → Kv ≥ 20 → K ≥ 200.
4. Stability check: characteristic equation s(s + 4)(s + 5) + K(s + 2) = s³ + 9s² + (20 + K)s + 2K = 0. Routh condition 9(20 + K) > 2K holds for every K > 0, so K = 200 is stable (poles −1.94 and −3.53 ± j13.9).
5. Input 2 + 3t: step part gives 0 (type 1); ramp part gives 3/Kv = 3/20.
6. ess = 0.15 rad.

Common mistakes

• Applying the error formulas to an unstable closed loop. Check stability first.
• Counting the order instead of the type: 1/(s(s + 1)(s + 2)) is third-order but type 1.
• Forgetting that G(s) must be in time-constant form (or evaluated as a limit) to read K: Kp for 10/(s + 2) is 5, not 10.
• Using Kv or Ka formulas directly with non-unity feedback.
• Scaling: for a ramp of slope A the error is A/Kv, not 1/Kv.
• Thinking a derivative term changes steady-state error to a step — it does not; integral action does.

For GATE ME

Expect: identifying the system type and the finite error for a given input; computing Kp, Kv or Ka; finding the minimum gain for a specified ramp error (often combined with a Routh stability limit, so the answer is a range of K); error for combined inputs such as a + bt; and conceptual questions on the effect of adding an integrator. Practise always stating stability before quoting an error.

Quick check

1. A type-1 unity-feedback system has Kv = 10 s⁻¹. What is the steady-state error to a unit ramp?
2. What is the steady-state error of a type-0 system with Kp = 5 to a unit step?
3. What is the type of G(s) = 20/(s²(s + 3))?
4. A type-1 system receives a unit parabolic input. What is ess?
5. Why must stability be checked before using the final value theorem?

Answers: 1. 0.1. 2. 1/6 ≈ 0.167. 3. Type 2. 4. Infinite. 5. The theorem only holds when s·E(s) has all poles in the left half-plane; for an unstable loop the error grows without bound.