Steady-state errors and error constants

Why it matters

A level controller that settles 2 cm below set point, or a positioning table that lags a moving target by 0.5 mm, has a steady-state error. Specifications often state the allowed error directly ("offset less than 1 %", "following error under 0.05 rad at 1 rad/s"), and the error constants tell you, in one line, how much gain or how many integrators the loop needs to meet them.

Key ideas

Steady-state error is what is left of the error e(t) = r(t) − c(t) as t → ∞, after the transients have died out. It only makes sense if the closed loop is stable — always check stability first; the formulas below give nonsense for an unstable loop.

System type = the number of open-loop poles at the origin (pure integrators) in G(s)H(s). Write G(s)H(s) = K·(1 + T_a s)(1 + T_b s)… / [s^N·(1 + T_1 s)(1 + T_2 s)…]; N is the type and K is the Bode (time-constant-form) gain.

Why the type matters: an integrator keeps producing output as long as its input is non-zero, so a loop with one integrator can hold a constant output with zero error. Each extra integrator lets the loop follow one higher-order polynomial input with zero error.

Error constants (unity feedback):

• Position constant Kp — governs the error to a step.
• Velocity constant Kv — governs the error to a ramp.
• Acceleration constant Ka — governs the error to a parabola.

Error table (unity feedback; inputs: unit step u(t), unit ramp t·u(t), unit parabola (t²/2)·u(t))

• Type 0: step 1/(1 + Kp), ramp ∞, parabola ∞.
• Type 1: step 0, ramp 1/Kv, parabola ∞.
• Type 2: step 0, ramp 0, parabola 1/Ka. For a type-N system, the error constant of its own order equals the Bode gain K; lower-order constants are infinite and higher-order ones are zero.

Inputs of other sizes: errors scale with amplitude. For r(t) = A·u(t), e_ss = A/(1 + Kp); for r(t) = A·t, e_ss = A/Kv; for r(t) = A·t²/2, e_ss = A/Ka. For a sum of inputs, add the errors (superposition).

Non-unity feedback: if H(s) is not 1, either compute E(s) = R(s) − C(s) from the closed-loop transfer function directly, or convert to an equivalent unity-feedback loop with G_eq = G/(1 + GH − G).

The trade-off: raising gain or adding integrators improves steady-state accuracy but adds phase lag and reduces stability margins. Lag compensators and PI controllers are the usual way to raise low-frequency gain without upsetting the transient response.

Formulas

• e_ss = lim(s→0) s·E(s) = lim(s→0) s·R(s) / (1 + G(s)H(s)) — final value theorem, unity feedback (H = 1).
• Kp = lim(s→0) G(s)H(s)
• Kv = lim(s→0) s·G(s)H(s)
• Ka = lim(s→0) s²·G(s)H(s)
• e_ss(step A) = A/(1 + Kp), e_ss(ramp A·t) = A/Kv, e_ss(parabola A·t²/2) = A/Ka

Units: Kp is dimensionless; Kv has units of s⁻¹; Ka has units of s⁻². The error carries the unit of the input (for a ramp of A rad/s, the error is in rad). The final value theorem is valid only when s·E(s) has all its poles in the left half-plane.

Worked examples

Example 1 — type-0 temperature loop Given: unity feedback, G(s) = 10/[(s + 1)(s + 2)].

1. No pole at the origin, so the system is type 0.
2. Kp = lim(s→0) G(s) = 10/(1 × 2) = 5.
3. Unit step: e_ss = 1/(1 + 5) = 0.167, i.e. 16.7 % offset.
4. Unit ramp: Kv = lim(s→0) s·G(s) = 0, so e_ss = ∞.

Result: e_ss(step) = 0.167, e_ss(ramp) = ∞. To remove the offset, an integrator (PI action) is needed; raising gain alone only shrinks it.

Example 2 — designing gain for a ramp specification (GATE level) Given: unity feedback, G(s) = K(s + 2)/[s(s + 4)(s + 5)]. (a) Find K so that the error to a unit ramp is 0.05. (b) Confirm stability. (c) With this K, find e_ss for r(t) = 5 + 2t.

1. Type 1. Kv = lim(s→0) s·G(s) = K × 2/(4 × 5) = K/10.
2. Need 1/Kv = 0.05, so Kv = 20 s⁻¹ and K = 200.
3. Characteristic equation: s(s + 4)(s + 5) + K(s + 2) = s³ + 9s² + (20 + K)s + 2K = 0 → s³ + 9s² + 220s + 400 = 0.
4. Routh first column: 1, 9, (9 × 220 − 400)/9 = 175.6, 400 — no sign change, so the loop is stable.
5. Input 5 + 2t: the step part gives zero error (type 1); the ramp part gives 2/Kv = 2/20 = 0.1.

Result: K = 200, loop stable, and e_ss = 0.1 (in the unit of the input) for r(t) = 5 + 2t.

Common mistakes

• Applying the error formulas without checking that the closed loop is stable.
• Using 1/Kp instead of 1/(1 + Kp) for the step error.
• Reading Kp from the pole-zero form without setting s = 0 properly: for 10/[(s + 1)(s + 2)], Kp = 5, not 10.
• Forgetting that a unit parabola is t²/2; for r(t) = t² the error is 2/Ka.
• Applying unity-feedback formulas directly to a non-unity-feedback loop.
• Thinking a lead compensator is the main tool for steady-state error — it mainly improves transient response; lag or PI action fixes steady-state error.

For GATE IN

Expect questions that ask for the type of a system, an error constant, the steady-state error for a step, ramp or parabola (sometimes with a combined input), or the gain needed to meet an error specification — often combined with a stability check. Practise reading the type from block diagrams with inner loops, and using the final value theorem directly when the feedback is not unity.

Quick check

1. G(s)H(s) = 20/[s(s + 10)]. What is Kv?
2. For the system in Q1, what is the error to a ramp r(t) = 4t?
3. A type-2 system has Ka = 8 s⁻². What is the error to r(t) = t²?
4. Why can't a type-0 system follow a ramp?

Answers: 1. 2 s⁻¹; 2. 4/2 = 2; 3. r = 2·(t²/2), so e_ss = 2/8 = 0.25; 4. Its Kv = 0, so the error grows without bound.