Robotics Fundamentals 1, Chapter 5: Challenge 2 solution (filtering and averaging)

Formulas: mean = sum / N          moving average of N: mean of the last N          median of N: the middle value of the last N sorted
          exponential filter: y(k) = y(k-1) + alpha ( x(k) - y(k-1) )          sigma(mean) = sigma / sqrt(N)
          EMA step response after n samples: 1 - (1 - alpha)^n          time constant about dt (1 - alpha) / alpha          moving average of N lags (N - 1) / 2 samples

Readings (cm), true value 150.0:  150.2, 149.6, 150.5, 149.9, 168.0, 150.1, 149.7, 150.4, 150.0, 149.8

(a) Sum = 1518.2, so the mean = 151.82 cm (the outlier pulls it 1.8 cm off).
    Sorted: 149.6, 149.7, 149.8, 149.9, 150.0, 150.1, 150.2, 150.4, 150.5, 168.0.  The middle two are 150.0 and 150.1, so the median = 150.05 cm.
    Mean without the outlier: (1518.2 - 168.0) / 9 = 1350.2 / 9 = 150.02 cm.

(b)  k    raw      MA3       Med3
     3   150.5    150.10    150.2        (mean of 150.2, 149.6, 150.5;   median of the same three)
     4   149.9    150.00    149.9
     5   168.0    156.13    150.5        <-- the outlier enters the window
     6   150.1    156.00    150.1
     7   149.7    155.93    150.1
     8   150.4    150.07    150.1        <-- the outlier has left the window
     9   150.0    150.03    150.0
    10   149.8    150.07    150.0
    MA3 is wrong for three readings (k = 5, 6, 7), by up to +6.13 cm (156.13 against about 150.0): the outlier is averaged in until it leaves the window.
    Med3 is never more than 0.5 cm off, no more than the ordinary reading-to-reading noise: a single wild value is never the middle one of three.

(c) EMA, alpha = 0.4, y(1) = 150.2:
        y(2) = 150.2 + 0.4 (149.6 - 150.2) = 149.96         y(3) = 149.96 + 0.4 (150.5 - 149.96) = 150.18        y(4) = 150.18 + 0.4 (149.9 - 150.18) = 150.07 (150.066)
        y(5) = 150.066 + 0.4 (168.0 - 150.066) = 157.24     y(6) = 157.24 + 0.4 (150.1 - 157.24) = 154.38        y(7) = 154.38 + 0.4 (149.7 - 154.38) = 152.51
        y(8) = 151.67      y(9) = 151.00      y(10) = 150.52
    Errors from the outlier onward (true 150.0):  k = 5 to 10:  +7.24, +4.38, +2.51, +1.67, +1.00, +0.52 cm.  Each step keeps 60 % of the previous error (1 - alpha), so it fades but does not vanish at once.

(d) sigma(mean) = 0.4 / sqrt(N) is at most 0.1 when sqrt(N) is at least 4, so N = 16 independent readings.
    At 25 readings a second, 16 readings take 16 / 25 = 0.64 s.  A robot at 0.3 m/s moves 0.3 x 0.64 = 0.192 m = 19.2 cm in that time.
    For 0.05 cm:  sqrt(N) = 0.4 / 0.05 = 8, N = 64, taking 64 / 25 = 2.56 s (0.77 m of travel at 0.3 m/s).  Halving the noise costs four times as many readings.

(e) Time constant about dt (1 - alpha) / alpha = 0.04 x 0.6 / 0.4 = 0.06 s.
    Fraction of a step covered after n samples = 1 - 0.6^n:   n = 1: 0.400    2: 0.640    3: 0.784    4: 0.870    5: 0.922.
    It reaches 63 % after 1 / (-ln 0.6) = 1.96 samples = 0.078 s (the 0.06 s is the rule-of-thumb value, close for alpha this large).

(f) A moving average of 3 lags (3 - 1) / 2 = 1 sample = 0.04 s.   At 0.3 m/s that is 1.2 cm of travel: small, which is why a short average or median is fine even in a control loop,
    but N = 25 would lag 12 samples = 0.48 s, 14 cm of travel, which a fast robot cannot afford.