Answers

U1.4 Time, rate and latency

The robot as a system · University · about 25 min

What this lesson is about

How often the loop runs, how late the answer arrives, and what both do to a controller.

• Three different times
• The same controller at two rates
• Why lateness turns into instability
• Measuring your own loop

Questions 6 marks in all

1. [1 mark]This drive takes about a quarter of a second to get most of the way to a new speed. Which of the three times in the lesson is that?

   1. AThe time constant
   2. BThe period
   3. CThe latency
   4. DThe deadman timeout
   Answer: A. The time constant is how long the machine itself takes to react. The period is how often your loop runs, and latency is how old a measurement is when you act on it.

2. [1 mark]The drive's time constant is about 0.25 s. Which loop rate does the lesson's rule of thumb favour?

   1. A10 Hz
   2. B4 Hz, one pass per time constant
   3. C2 Hz
   4. D1 Hz
   Answer: A. Sample well faster than the thing you are controlling. At 10 Hz the loop runs several times within one time constant; at 2 Hz the error grows for half a second before anything is done about it.

3. [1 mark]Why can too much delay make a feedback loop unstable?

   1. AThe loop pushes against an error that has already moved on, and with enough delay it pushes the wrong way, so negative feedback becomes positive
   2. BLate measurements carry more noise than fresh ones
   3. CA slow loop always saturates the motors
   4. DDelay adds a constant bias to the measured heading
   Answer: A. Feedback only works if it pushes against the current error. Delay turns negative feedback into positive feedback, which is the whole story of instability in one sentence.

4. [1 mark]A loop oscillates because its information is late. Which of these are fixes from the lesson?

   Tick every answer that is true.

   1. ARun the loop faster
   2. BTurn the gain down
   3. CPredict where the error is going rather than where it was
   4. DTurn the gain up, so the correction is finished before the information goes stale
   Answer: A, B, C. Fresher information, gentler pushes, or pushing against a prediction. More gain makes a late push do more harm, not less.

5. [1 mark]A loop asks for wait(0.05), and each pass also spends 0.012 s in distance() and 0.018 s in scan(). What does this print?

   work = 0.012 + 0.018
   requested = 0.05
   period = requested + work
   print("period:", round(period, 3), "s")
   print("rate:", round(1 / period, 1), "Hz")

   Answer:

   period: 0.08 s
   rate: 12.5 Hz

   The pass costs 0.05 + 0.03 = 0.08 s, so the loop runs at 1 / 0.08 = 12.5 Hz, not the 20 Hz the wait suggests.

6. [1 mark]The heading-hold loop works out its error with wrapped(). With clockwise-positive heading, what does this print?

   def wrapped(a):
       return (a + 180) % 360 - 180
   
   print(wrapped(0 - 350), wrapped(0 - 10), wrapped(190))

   Answer:

   10 -10 -170

   A robot at heading 350 is 10 degrees anticlockwise of 0, so the error is +10 (turn clockwise), not -350. At heading 10 it is -10, and 190 wraps to -170.

The task: hold the heading

from bugbot import *
connect()

def wrapped(a):
    return (a + 180) % 360 - 180

# drive, correcting as you go, and keep correcting once you are there
forward(70, distance=130)
stop()

from bugbot import *
connect()

def wrapped(a):
    return (a + 180) % 360 - 180

while position()[1] < 130:
    error = wrapped(0 - heading())
    drive(70, 0, error * 3)
    wait(0.1)
stop()
for i in range(12):
    error = wrapped(0 - heading())
    drive(0, 0, error * 3)
    wait(0.1)
stop()