Closed loop speed control design conditions

Block diagram

 r (set speed)
   -->(+)--e--> [Controller] --u--> [Engine / throttle]
       ^ -        Kp                     |
       |                                 v   d (slope)
       |                              (+)<--
       |                                 |
       |                        [Car: 1/(m s + b)]
       |                                 |
       +----[Speed sensor]<--------------+--> v

Model and controller

m·dv/dt = -b·v + u + d, controller u = Kp·(r - v).

Sampling with period T: v[k+1] = a·v[k] + c·(u[k] + d[k]), a = 1 - b·T/m, c = T/m.

Design steps

1. Substitute the controller: v[k+1] = a·v[k] + c·Kp·(r - v[k]) + c·d[k].
2. Collect terms: v[k+1] = (a - c·Kp)·v[k] + c·Kp·r + c·d[k].
3. The closed-loop pole is p = a - c·Kp = 1 - (b + Kp)·T/m.

Condition for no unbounded output

The output stays bounded if |p| < 1, i.e. 0 < (b + Kp)·T/m < 2, so Kp < 2m/T - b.

Condition for no oscillation

The response is non-oscillatory if the pole is non-negative: 0 ≤ p < 1, which gives (b + Kp)·T/m ≤ 1, so Kp ≤ m/T - b.

So choose 0 < Kp ≤ m/T - b. For a larger gain (m/T - b < Kp < 2m/T - b) the speed is bounded but oscillates; beyond that it grows without bound.

Steady state

v_ss = (Kp·r + d)/(b + Kp). The error r - v_ss = (b·r - d)/(b + Kp) falls as Kp rises, and the effect of the slope is d/(b + Kp), better than d/b of open loop.

Final selection

Pick Kp as large as allowed by the no-oscillation limit to get small error, then add an integral term if zero steady-state error is needed. Verify by simulation.