Controls and Systems · Chapter 5 of 10 · Intermediate
Steady-State Error
Will the output ever exactly reach the target, and if not, by how much will it miss? The answer depends on the input and on how many integrators the loop already has.
Readiness check
This chapter is about long-run accuracy. Tick only what you can do closed-notes.
- Apply the final value theorem.
- Take a limit of a rational function as s → 0.
- Recognise a pole at the origin (a factor of s).
- Tell a step input from a ramp input.
- Form the closed-loop error E = R − Y.
The core idea
The steady-state error is the leftover gap between input and output as time goes to infinity. It depends on the input type and on the system type, the number of integrators in the loop.
ess = lims→0 sE(s)Kp = lims→0 G, Kv = lims→0 sGstep: ess = 1/(1 + Kp)For a unity-feedback loop, the error transform is E = R/(1 + G), and the final value theorem gives the steady-state error. How it comes out depends on the system type, the number of poles G has at the origin: each integrator (a factor of 1/s) drives one class of input to zero error. A type 0 system has a finite error to a step; a type 1 system tracks a step perfectly but lags a ramp; a type 2 system tracks a ramp and lags only a parabola. The static error constants Kp, Kv, and Ka package these limits.
The skills, taught in order
Five skills define the error, the system type, the static error constants, the error table, and the role of integral action.
5.1 Steady-state error
The error is e(t) = r(t) − y(t); its steady-state value for a unity-feedback loop is ess = lims→0 sE(s) with E = R/(1 + G). It measures how well the system tracks its command once transients die out.
5.2 System type
The system type is the number of pure integrators (poles at the origin) in the open-loop transfer function G. Type 0 has none, type 1 has one factor of 1/s, type 2 has two. Each integrator removes the steady-state error for one harder class of input.
5.3 Static error constants
Three limits summarise tracking: the position constant Kp = lims→0 G for steps, the velocity constant Kv = lims→0 sG for ramps, and the acceleration constant Ka = lims→0 s²G for parabolas. A larger constant means a smaller error.
5.4 The error table
Matching input to type gives the error directly. The pattern is that a system of type n tracks inputs up to order n with zero error, has a constant error at order n, and an infinite error beyond.
| Input | Type 0 | Type 1 | Type 2 |
|---|---|---|---|
| Step | 1/(1 + Kp) | 0 | 0 |
| Ramp | ∞ | 1/Kv | 0 |
| Parabola | ∞ | ∞ | 1/Ka |
5.5 Integral action
Adding an integrator (a 1/s in the controller) raises the system type by one, driving a whole class of error to zero. This is the deep reason the integral term in a PID controller eliminates steady-state offset, as Chapter 10 shows.
Engineering connection: the demand for zero steady-state error is what motivates integral control, and it trades against stability margin.
Worked example 1: step error of a type 0 system
A unity-feedback system has open-loop G(s) = 10/[(s + 1)(s + 2)]. Find the position error constant and the steady-state error to a unit step.
- ProblemFind Kp and the step error for the system in Figure 1.
- Given / findG(s) = 10/[(s + 1)(s + 2)], unit step input. Find Kp and ess.
- AssumptionsUnity feedback; stable closed loop; the final value theorem applies.
- ModelThis is type 0 (no pole at the origin), so Kp = lims→0 G and ess = 1/(1 + Kp).
- EquationsKp = lims→0 G(s)ess = 1/(1 + Kp)
- SolveKp = 10/[(1)(2)] = 5. ess = 1/(1 + 5) = 1/6 ≈ 0.167.
- CheckA type 0 system always leaves a finite step error; raising the gain raises Kp and shrinks the error but never removes it. 0.167 means the output reaches about 83 percent of the command.
- ConclusionWithout an integrator, perfect step tracking is impossible. The position constant quantifies exactly how short the output falls.
Worked example 2: ramp error of a type 1 system
A unity-feedback system has open-loop G(s) = 20/[s(s + 5)]. Find the velocity error constant and the steady-state errors to a unit step and a unit ramp.
- ProblemFind Kv and the step and ramp errors for the system in Figure 2.
- Given / findG(s) = 20/[s(s + 5)]. Find Kv, the step error, and the ramp error.
- AssumptionsUnity feedback; stable closed loop; one pole at the origin makes this type 1.
- ModelType 1 gives zero step error; for the ramp use Kv = lims→0 sG and ess = 1/Kv.
- EquationsKv = lims→0 sG(s)ess,ramp = 1/Kv
- SolveKv = lims→0 s · 20/[s(s + 5)] = 20/5 = 4. Step error = 0; ramp error = 1/4 = 0.25.
- CheckThe integrator removes the step error entirely, exactly as the error table predicts for type 1. The finite ramp error means the output keeps pace in slope but trails by a constant 0.25.
- ConclusionAdding an integrator buys perfect step tracking at the cost of only a constant ramp lag, which more integral action could remove in turn.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Wrong error constant for the input | Kp used for a ramp | "Is the input a step, ramp, or parabola?" | Match Kp to steps, Kv to ramps, Ka to parabolas. |
| Miscounting system type | Predicting finite step error for type 1 | "How many poles at the origin?" | Count factors of 1/s in the open-loop G. |
| Final value theorem on unstable loop | A finite error for a diverging system | "Is the closed loop stable?" | Steady-state error is meaningful only if the loop is stable. |
| Expecting integral action for free | Adding 1/s without a stability check | "Did the extra pole hurt stability?" | Integral action raises type but can erode the margin; verify stability. |
Practice ladder
A type 0 system has G = 8/[(s + 2)(s + 4)]. Find Kp and the unit-step error.
Show answer
Kp = 8/[(2)(4)] = 1. ess = 1/(1 + 1) = 0.5. A low Kp gives a large error.
Raise the gain of the Worked Example 1 system so G = 50/[(s + 1)(s + 2)]. What is the new step error?
Show answer
Kp = 50/2 = 25. ess = 1/(1 + 25) = 0.038. Five times the gain cut the error roughly fivefold, but it is still nonzero.
A type 1 system has G = 100/[s(s + 4)(s + 10)]. Find Kv and the ramp error.
Show answer
Kv = lims→0 s · 100/[s(s+4)(s+10)] = 100/(4·10) = 2.5. Ramp error = 1/2.5 = 0.4. Step error is 0 because it is type 1.
Find a system that must track a moving target (a tracking antenna, a CNC axis). Argue what system type it needs and why integral action is or is not required.
What good work looks like
A link from the input (ramp-like motion) to the required type (at least 1 for zero step error, 2 for zero ramp error), with integral action justified by the tracking spec.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Take a real tracking system, determine its system type, compute the relevant error constant, and state the steady-state error for its actual input, then say whether integral action is needed.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. What is steady-state error?
The leftover gap ess = lims→0 sE(s) between input and output as t → ∞.
2. What is system type?
The number of pure integrators (poles at the origin) in the open-loop G.
3. Give the step error for a type 0 system.
1/(1 + Kp), with Kp = lims→0 G.
4. What error does a type 1 system have to a ramp?
A constant error 1/Kv, with Kv = lims→0 sG; its step error is zero.
5. What does adding an integrator do?
Raises the system type by one, driving a class of error to zero, at some cost to stability.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Ogata, Modern Control Engineering, Chapter 5 (Steady-State Error Analysis) |
| Cross-reference | Nise, Ch. 7 · Dorf and Bishop, Ch. 5 |
| Core topics | 5.1 Steady-state error · 5.2 System type · 5.3 Static error constants · 5.4 Error table · 5.5 Integral action |
| Engineering connection | Zero-error demands motivate integral control and trade against margin. |
| Read next | Chapter 6: Stability and the Routh-Hurwitz Criterion. |