Steady state and system type
Each integrator in the loop makes it possible to follow, without error, an input one degree harder. A motor position control follows a step exactly because the motor already integrates.
01The error once everything has settled
The transient response tells how the system gets there; the steady state, where it ends up. In a loop with unity feedback and loop gain \( G(s) \), the error between the reference and the output is \( E(s) = R(s)/(1 + G(s)) \), and its final value is obtained with the final value theorem of the Laplace transform, if the closed loop is stable:
02System type and the error constants
The type of a system is the number of poles at the origin of its loop gain, that is, of integrators. Each integrator makes it possible to follow, without error, an input one degree harder:
| Type | Step | Ramp | Parabola |
|---|---|---|---|
| 0 | \( 1/(1 + K_p) \) | Infinite | Infinite |
| 1 | 0 | \( 1/K_v \) | Infinite |
| 2 | 0 | 0 | \( 1/K_a \) |
A motor position control, which already integrates the velocity, is type 1: it follows a fixed position without error and a ramp with a constant error. The oven with proportional control in the introduction is type 0: that is why it left an error. Adding an integrator in the controller raises the type by one, at the cost of phase and stability.
Three loops with unity feedback: type 0, \( K/((s+1)(0.2s+1)) \); type 1, \( K/(s(0.2s+1)) \); type 2, \( K(s+0.5)/(s^2(0.2s+1)) \). They are simulated for 20 s with Runge-Kutta; the gray is the reference and the blue, the output.
03Classification of control systems
Besides the type, control systems are classified by what is asked of them. A regulator keeps the output at a fixed value despite disturbances, like an oven or a power supply; a servomechanism makes the output follow a changing reference, like the axis of a robot. A distinction is also made between linear and nonlinear, continuous and discrete, single-input single-output and multivariable, and fixed-parameter and adaptive systems. The same question, how much error remains, is also asked in the face of a constant disturbance, and the answer depends on where the integrators are with respect to it.
04In the lab
Close a proportional speed loop around a motor and measure the steady-state error for several gains. Compare with \( 1/(1+K_p) \).
With a position loop, apply a ramp reference and measure the tracking error. Verify that it goes down as the gain goes up.
Add integral action to the speed loop and check that the position error disappears.
05Common mistakes
- Applying the final value theorem to an unstable system. The result makes no sense; stability has to be checked first.
- Counting the plant's integrators without looking at the controller. The type is that of the complete loop gain.
- Raising the gain only to lower the error. The error goes down, but the overshoot and the risk of instability grow.
- Confusing error with respect to the reference with error with respect to disturbances. They depend on where each signal enters.
06Self-assessment
A type 0 system with Kp = 19, what error does it have for a unit step?
\( 1/20 = 0.05 \).
What error does a type 1 system with Kv = 50 s⁻¹ have for a ramp of 2 units per second?
\( 2/50 = 0.04 \).
What type is \( G(s) = 10/(s(s+2)(s+5)) \) and what is Kv?
Type 1; \( K_v = 10/10 = 1 \) s⁻¹.
Why does a motor position control follow a step without error?
Because the motor integrates the velocity: the loop has a pole at the origin and is type 1.
07Further reading
- Katsuhiko Ogata. Modern Control Engineering. 5th ed., Prentice Hall, 2010 (Spanish edition: “Ingeniería de control moderna”, Pearson). Steady-state errors and the error constants.
- Richard C. Dorf and Robert H. Bishop. Modern Control Systems. 10th ed., Pearson, 2005 (Spanish edition: “Sistemas de control moderno”). Steady-state error and the sensitivity of feedback systems.
- Norman S. Nise. Control Systems Engineering. 8th ed., Wiley, 2019. Steady-state error with respect to references and disturbances.