Answers

U5.4 The integral term

Feedback control · University · about 30 min

What this lesson is about

Removing the error that P leaves behind, and the windup that comes with it.

• In code
• Windup
• Choosing Ki

Questions 7 marks in all

1. [1 mark]A PI controller has settled exactly on target. Which term is supplying the command?

   1. AThe integral term
   2. BThe proportional term
   3. CBoth equally
   4. DNeither, the command is zero
   Answer: A. At zero error the P term is zero, so whatever push is needed to stay there comes from the integral's accumulated value.

2. [1 mark]A constant 4 cm error is held for five ticks. What does this print?

   KP, KI, DT = 1.6, 0.5, 0.1
   integral = 0.0
   for tick in range(5):
       error = 4.0
       integral += error * DT
       print(round(KP * error + KI * integral, 2))

   Answer:

   6.6
   6.8
   7.0
   7.2
   7.4

   The P term stays at 6.4 while the integral grows by 0.4 per tick, adding 0.2 each time: 6.6, 6.8, 7.0, 7.2, 7.4. A stubborn error keeps pushing harder.

3. [1 mark]Using the lesson's rule of thumb, how many seconds does the integral take to contribute as much as the proportional term when Kp = 1.6 and Ki = 0.5? Give one decimal place.

   Answer: 3.2 (accept within 0.05). Kp / Ki = 1.6 / 0.5 = 3.2 s. Make it comparable with the settling time you want.

4. [1 mark]Which of these are the standard defences against integral windup given in the lesson?

   Tick every answer that is true.

   1. AClamp the integral to a sensible range
   2. BStop integrating while the output is saturated
   3. CSubtract the clipped amount, scaled, from the integral
   4. DRaise Kp so the error shrinks faster
   5. EFilter the error before integrating it
   Answer: A, B, C. Clamping, conditional integration and back-calculation all stop the integral storing effort the actuator could not deliver. Gain or filtering changes do not.

5. [1 mark]An obstacle holds the robot at 40 cm while the target is 25 cm. Ten seconds later it is removed and the robot charges far past the target. Why?

   1. AThe integral grew the whole time and must be unwound by an error in the other direction
   2. BKp is too high
   3. CThe D term amplified the removal of the obstacle
   4. DThe sensor was biased by the obstacle
   Answer: A. That is windup: a large stored integral pins the command at full speed until an equal and opposite error cancels it.

6. [1 mark]The error is in cm, the command in percent, and time in seconds. What are the units of Ki?

   Answer: percent per centimetre per second. Ki multiplies the integral of error, in cm s, to give percent, so its units are percent per centimetre per second.

7. [1 mark]The robot sails slowly past the target, turns round, and sails past again with large, slow swings. Which gain is the likely cause?

   1. AKi too large
   2. BKd too large
   3. CKp too small
   4. DThe dead band too small
   Answer: A. Too much integral action is its own kind of oscillation: slow and large. Start with an integral that acts slower than the P term.

The task: no offset left

from bugbot import *
connect()

KP, KI = 1.5, 0.8
TARGET = 25.0
DT = 0.1
integral = 0.0

from bugbot import *
connect()

KP, KI = 1.6, 0.9
TARGET = 25.0
DT = 0.1
integral = 0.0

for tick in range(480):
    error = distance() - TARGET
    integral += error * DT
    integral = max(-60.0, min(60.0, integral))
    cmd = KP * error + KI * integral
    plot("error", error)
    plot("i term", KI * integral)
    drive(max(-60.0, min(60.0, cmd)), 0, 0)
    wait(DT)
stop()