Answers

U4.2 Averaging

Noise and filtering · University · about 25 min

What this lesson is about

The square root of n, why it stops paying, and what it costs in time.

• The square root of n
• The cost is time
• Independence, again

Questions 6 marks in all

1. [1 mark]A depth sensor has sigma = 4 cm. How many independent readings must be averaged to get an estimate with sigma = 1 cm?

   Answer: 16. sigma / sqrt(n) = 1 needs sqrt(n) = 4, so n = 16.

2. [1 mark]What does this print?

   SIGMA = 6.0
   for n in (1, 4, 16, 64):
       print(n, SIGMA / n ** 0.5, round(n * 0.1, 1))

   Answer:

   1 6.0 0.1
   4 3.0 0.4
   16 1.5 1.6
   64 0.75 6.4

   Each fourfold increase in n halves the noise, 6 to 3 to 1.5 to 0.75 cm, while the time taken at 0.1 s per reading grows fourfold. That is the diminishing return.

3. [1 mark]With the noise set to sigma = 8 cm and a new reading every 0.1 s, how many seconds of standing still does an estimate good to 1 cm take?

   Answer: 6.4 (accept within 0.05). 8 / sqrt(n) = 1 needs n = 64 readings, and 64 x 0.1 s = 6.4 s.

4. [1 mark]Averaging 100 readings has already been done. What does going to 400 readings buy?

   1. AThe noise halves again, at four times the time
   2. BThe noise falls to a quarter
   3. CThe noise halves and the bias halves
   4. DNothing, because 100 readings is the limit
   Answer: A. sqrt(400 / 100) = 2, so the noise halves while the time goes from 10 s to 40 s. Bias is untouched however many readings are taken.

5. [1 mark]Averaging 16 readings should divide the noise by 4. In which of these cases will it not?

   Tick every answer that is true.

   1. AThe 16 readings are taken in a tight loop faster than the sensor updates
   2. BA slow draught nudges every reading in the window the same way
   3. CThe 16 readings are taken 0.1 s apart with the robot still
   4. DThe sensor's sigma is 8 cm instead of 4 cm
   Answer: A, B. Root n only holds for independent readings. Repeated samples and noise that is correlated across the window both break independence; a larger sigma is still divided by 4.

6. [1 mark]A robot drives at 20 cm/s while averaging 60 readings taken 0.1 s apart. How many centimetres does it travel during the average?

   Answer: 120. 60 x 0.1 s = 6 s, and 6 s x 20 cm/s = 120 cm. The average describes over a metre of driving, which is why the rule is average while still, filter while moving.

The task: average it down

from bugbot import *
connect()

# how many readings put two sigma at 1 cm, with sigma = 4?

from bugbot import *
connect()

n = 80
readings = []
for i in range(n):
    readings.append(distance())
    wait(0.1)
print("readings:", n)
print("distance:", round(sum(readings) / n, 2))