Robotics Fundamentals 1, Chapter 7: Challenge 1 solution (the motor model, open loop, bang-bang and proportional control)

Formulas: K = 1 / k ;  tau(m) = J R / k^2 ;  d = R T / k ;  step response fraction = 1 - e^(-t / tau(m))
          open loop: u = r / K + d ;  speed = K ( u - d )
          P control: e(ss) = ( r + K d ) / ( 1 + K Kp ) ;  final speed = r - e(ss)
          bang-bang: t = tau ln( (final - start) / (final - end) ) ;  final = K ( U - d )

(a) k = 0.04, R = 1.5, J = 2.4e-4.
    K = 1 / 0.04 = 25 rad/s per volt.
    tau(m) = J R / k^2 = 2.4e-4 x 1.5 / 0.0016 = 3.6e-4 / 0.0016 = 0.225 s.
    63.2 % after one time constant:   0.225 s.
    95.0 % after three:               3 x 0.225 = 0.675 s.
    98.2 % after four:                4 x 0.225 = 0.900 s.

(b) d = 1.2 V.  d = R T / k, so T = d k / R = 1.2 x 0.04 / 1.5 = 0.032 N m.
    Open loop voltage for 75 rad/s:   u = r / K + d = 75 / 25 + 1.2 = 3.0 + 1.2 = 4.2 V.
    Battery sag (every voltage 10 % lower):   u = 0.9 x 4.2 = 3.78 V,  speed = 25 x ( 3.78 - 1.2 ) = 25 x 2.58 = 64.5 rad/s.
        Error = 75 - 64.5 = 10.5 rad/s = 14 % of the target.
    Load rises to d = 2.0 V (voltage still 4.2 V):   speed = 25 x ( 4.2 - 2.0 ) = 25 x 2.2 = 55 rad/s.   Error = 20 rad/s = 26.7 % of the target.
    Open loop trusts the model, so every change in battery or load goes straight into the speed.

(c) Kp = 0.16, so K Kp = 25 x 0.16 = 4 and 1 + K Kp = 5.
    With the 1.2 V load:   e(ss) = ( 75 + 25 x 1.2 ) / 5 = ( 75 + 30 ) / 5 = 21 rad/s.    Final speed = 75 - 21 = 54 rad/s.
    With no load:          e(ss) = 75 / 5 = 15 rad/s.                                    Final speed = 60 rad/s.
    For e(ss) = 3 rad/s with the load:   ( 75 + 30 ) / ( 1 + 25 Kp ) = 3,  so  1 + 25 Kp = 35,  25 Kp = 34,  Kp = 1.36 V per rad/s.
    Why it is not practical:  at the start the error is 75 rad/s, so the controller asks for 1.36 x 75 = 102 V from a 6 V supply: the output saturates for a long time, and the formula
        only describes the steady state.  The loop gain K Kp = 34 is also far above what any real loop with sensor delay and a finite update rate can stand without ringing or going
        unstable (sections 7 and 8).  The right answer is the integral term, which removes the offset with a modest Kp.

(d) Supply 6 V, load 1.2 V, target 75 rad/s, hysteresis +-4:  switch on below 71, off above 79.
    With the supply on:   the speed heads for K ( 6 - 1.2 ) = 25 x 4.8 = 120 rad/s.
    With the supply off:  the speed heads for K ( 0 - 1.2 ) = -30 rad/s (a limit of the formula, never reached; only the part of the curve between 71 and 79 is used).
    Climb from 71 to 79:   t(on)  = 0.225 x ln( (120 - 71) / (120 - 79) ) = 0.225 x ln( 49 / 41 ) = 0.225 x 0.17824 = 0.0401 s.
    Fall from 79 to 71:    t(off) = 0.225 x ln( (79 + 30) / (71 + 30) ) = 0.225 x ln( 109 / 101 ) = 0.225 x 0.07623 = 0.0172 s.
    Period = 0.0401 + 0.0172 = 0.0573 s,   switching frequency = 1 / 0.0573 = 17.5 Hz.
    With +-8 (on below 67, off above 83):   t(on) = 0.225 x ln( 53 / 37 ) = 0.0809 s;   t(off) = 0.225 x ln( 113 / 97 ) = 0.0344 s;   period 0.1152 s;   frequency 8.7 Hz.
    Doubling the band halves the switching rate (17.5 Hz to 8.7 Hz) and doubles the ripple (+-4 to +-8 rad/s): hysteresis trades ripple against switching.