Integral Windup in a Saturated Speed Loop
A DC-motor speed loop uses a PI controller followed by a four-quadrant converter. The speed reference is a step from 0 to 1000 rpm. The controller gains are Kp = 0.8 V/rpm and Ki = 2.0 V/(rpm·s). The converter output saturates at ±10 V, and the motor plus load has a first-order speed response with a steady-state gain of 100 rpm/V and a time constant of 0.5 s. At 1000 rpm the steady-state armature voltage required is exactly 10 V. The measured response overshoots to about 1350 rpm and takes more than 3 s to settle. When the reference is later stepped down to 800 rpm, the speed first falls to about 700 rpm before recovering. The programmer has not implemented any anti-windup. The team wants to know whether retuning the PI gains alone will remove the overshoot, and how to handle a future reference of 1050 rpm.
1. Explain why the integral term produces a large overshoot even though the plant is first-order, and determine the steady-state error that would remain if the integral term were switched off.
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During the initial rise, the speed error is positive, so the PI controller asks for more than 10 V. The converter cannot deliver more than 10 V, but the integrator continues to accumulate the positive error because it does not know the actuator is saturated. By the time the speed crosses 1000 rpm, the integrator state may correspond to a controller output well above 10 V. The error then becomes negative, but the output remains saturated until the integrator unwinds. That delayed reversal is the overshoot and the long settling time.
With only the proportional term, steady state requires u = Kp e and speed = 100u. For 1000 rpm, u = 10 V, so e = 10/0.8 = 12.5 rpm. The integral term removes that steady-state error when the operating point is feasible, but it also causes windup when the actuator saturates.
2. Describe a back-calculation anti-windup scheme for this loop, give a reasonable tracking time constant, and state what it changes in the observed response.
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Back-calculation feeds the difference between the saturated and unsaturated controller outputs back into the integrator. One common form is: integrator input = Ki e + (1/Tt)(u_sat − u), where u is the raw PI output and u_sat is the limited output. When the controller is saturated high, u_sat − u is negative, so the integrator is driven down instead of continuing to accumulate positive error.
A reasonable starting value is Tt = Ti = Kp/Ki = 0.8/2.0 = 0.4 s. The scheme leaves the PI law unchanged when the actuator is not saturated, so it does not slow the small-signal response. It reduces the 1350 rpm overshoot and the 700 rpm undershoot after the reference step-down, because the integrator no longer has to discharge a large stored value. It does not change the fundamental saturation limit: the drive still cannot exceed 1000 rpm at 10 V.
3. The next production recipe requires 1050 rpm. Using the saturation limit, determine the maximum steady speed and explain why anti-windup alone cannot solve this. Suggest one hardware or control change.
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The maximum converter output is 10 V. With a plant gain of 100 rpm/V, the maximum steady speed is 100 × 10 = 1000 rpm. At a 1050 rpm reference, the required steady voltage is 10.5 V, which is outside the actuator range. The integrator will saturate at the limit and the speed will remain at about 1000 rpm, leaving a steady-state error of about 50 rpm.
Anti-windup prevents the integrator from accumulating a huge value during the transient, so the recovery is cleaner, but it cannot create the missing 0.5 V. The operating point is infeasible. A hardware or control change is required: increase the converter voltage rating, change the gear ratio or motor so that 1050 rpm needs less than 10 V, or use field weakening if the motor and load permit it.