Answers

U10.5 Feedforward and feedback

Following a trajectory · University · about 30 min

What this lesson is about

A ramp leaves a proportional controller permanently behind, and the model knows how to fix it.

• The arithmetic
• The better answer
• Where the model comes from
• What feedforward cannot do

Questions 6 marks in all

1. [1 mark]A proportional controller with K = 1.5 per second chases a reference moving at 10 cm/s. How far behind does it settle, in cm to 2 decimal places?

   Answer: 6.67 (accept within 0.01). In steady state K e = v, so e = 10 / 1.5 = 6.67 cm, for as long as the ramp lasts.

2. [1 mark]The reference speed is now fed forward, but the model is 10 percent out. With the same v = 10 cm/s and K = 1.5, what steady error is left, in cm to 2 decimal places?

   Answer: 0.67 (accept within 0.01). Feedback only has the model error to correct: 0.1 x 10 / 1.5 = 0.67 cm, ten times smaller.

3. [1 mark]Why can a proportional controller never catch a reference moving at constant speed?

   1. AIt only produces a command when there is an error, and the command must stay non-zero to keep moving
   2. BIts gain is always too low
   3. CThe drive's dead band stops it moving at constant speed
   4. DThe reference is sampled too slowly
   Answer: A. A non-zero command for ever needs a non-zero error for ever. Raising K shrinks it but never removes it.

4. [1 mark]This simulates a robot whose real speed is 90 percent of the model's, chasing a 10 cm/s ramp, first with gain alone and then with the reference fed forward. What does it print?

   V, K, DT = 10.0, 1.5, 0.1
   for ff in (0.0, 1.0):
       pos = 0.0
       for tick in range(300):
           ref = V * tick * DT
           err = ref - pos
           pos += 0.9 * (ff * V + K * err) * DT
       print(ff, round(err, 2))

   Answer:

   0.0 7.41
   1.0 0.74

   In steady state 0.9 (ff V + K e) = V. Gain alone gives e = 11.11 / 1.5 = 7.41 cm; with feedforward only the 1.11 cm/s shortfall is left, e = 0.74 cm.

5. [1 mark]Which statements about feedforward and feedback are right?

   Tick every answer that is true.

   1. AFeedforward has no stability cost because it is not in the loop
   2. BFeedforward alone drifts, because nothing measures the result
   3. CFeedback handles what the model does not know, such as a slope or a flat battery
   4. DWith good feedforward the feedback loop can be removed
   5. EFeedforward reduces the steady error by raising the loop gain
   Answer: A, B, C. Feedforward makes it happen and feedback fixes what it got wrong. Feedforward does not change the loop gain, and it cannot see what actually happened.

6. [1 mark]A robot running feedforward plus feedback is stuck against a chair leg. What does the feedforward term do?

   1. AIt keeps commanding cruise speed, because it never sees what actually happened
   2. BIt drops to zero, because the error has grown
   3. CIt reverses, because the model predicts a collision
   4. DIt doubles, to push past the obstacle
   Answer: A. Feedforward depends only on the reference. Anything that depends on the result has to come from the feedback path.

The task: how far behind a ramp

from bugbot import *
import math
connect()

DT = 0.1
V_REF, K = 10.0, 1.5

from bugbot import *
import math
connect()

DT = 0.1
V_MAX = 20.0
V_REF, K = 10.0, 1.5

def band(u):
    return 0.0 if abs(u) < 3.0 else max(-100.0, min(100.0, u))

def leg(seconds, feedforward, name):
    """drive a ramp reference for this long, and report the settled following error"""
    base = position()[1]
    errors = []
    for tick in range(int(seconds / DT)):
        t = tick * DT
        ref = V_REF * t
        here = position()[1] - base
        err = ref - here
        want = (V_REF if feedforward else 0.0) + K * err
        drive(band(100 * want / V_MAX), 0, 0)
        plot("ref", base + ref)
        plot("actual", position()[1])
        plot("error", err)
        if t > seconds - (4.0 if feedforward else 5.0):
            errors.append(abs(err))
        wait(DT)
    stop()
    wait(0.6)
    return sum(errors) / len(errors)

# the gain alone: the reference runs away and the error settles where the gain can just keep up
print("p only:", round(leg(9.0, False, "p"), 2))
# the same ramp with the reference speed handed straight to the motors
print("with ff:", round(leg(7.0, True, "ff"), 2))