MAE 524 course notes – Spring 2002, Copyrighted L. Silverberg
- - 56
H. Nonlinear Feedback (Regulation)
 Nonlinear feedback represents a feedback method that
controls motion using forces that are nonlinearly related
to the state variables.
 Nonlinear feedback is necessary in actuators that
produce discontinuous forcing functions. Examples of
actuators that, by design, produce discontinuous forcing
functions are solenoids, pneumatic and hydraulic
devices, and braking systems.
 These actuators frequently operate in a bang-off-bang
manner. The bangs refer to constant levels of force in
one direction or another. At one extreme, the actuator is
always on – banging in one direction or another. This is
called bang-bang control. At the other extreme, the
actuator on-time is very small – banging over short
bursts of time. This is called impulse control. It turns
out that time-optimal control is of the bang-bang type,
and that fuel-optimal control is of the impulse type.
NONLINEAR FEEDBACK (REGULATION)
Lecture notes prepared by L.Silverberg and J.Morton
MAE 524 course notes – Spring 2002, Copyrighted L. Silverberg
- - 57
H.1 Bang-bang feedback
 Consider an undamped single degree of freedom system
subject to bang-bang feedback. Assume that the bangbang feedback is produced by dry friction with a friction
coefficient denoted by  . The equation governing the
motion of the system and the bang-bang force are
(1.141)
 mg, x  0
mx  kx  f , f  
 mg, x  0
 Notice that the bang-bang force is a dry friction force of
constant value in a direction that opposes the velocity.
Dividing Eq. (1.141) by m,
2
(1.142) x   0 x   g sgn( x )
where sgn( x ) is the sign function, valued at 1 or –1
depending on the sign of x .
 The particular solution and the homogeneous solution
over the i-th time interval are
(1.143)
 g sgn( x )
xp 
, x h  B cos( 0 (t  t Li ))  C sin(  0 (t  t Li )),
2
0
t Li  t  tUi , x(t Li )  x(t L (i 1) ), x (t Li )  x (tUi )  0
NONLINEAR FEEDBACK (REGULATION)
Lecture notes prepared by L.Silverberg and J.Morton
MAE 524 course notes – Spring 2002, Copyrighted L. Silverberg
- - 58
 From Eq. (1.143) the complete response over the i-th
time interval is
(1.144)
g sgn( x (t Li ))
g sgn( x (t Li ))
x

(
x
(
t
)

) cos( 0 t ),
Li
2
2
0
0
t Li  t  tUi
 As shown in the figure, the bang-bang response decays
linearly at the linear rate
2 g


(1.145)
0

From Eq. (1.145), the response decays at a rate that is
inversely proportional to the system’s natural frequency
of oscillation.
H.2 Impulse feedback
 Let’s examine the effect of a unit impulse on the
response of the system. First, notice when the unit
impulse is applied that the associated fuel cost is
C F   f (t ) dt    (t )dt  1. Indeed, a unit amount of
fuel is consumed.
NONLINEAR FEEDBACK (REGULATION)
Lecture notes prepared by L.Silverberg and J.Morton
MAE 524 course notes – Spring 2002, Copyrighted L. Silverberg
- - 59
 Secondly, notice that the reduction of energy caused by
the unit impulse depends on the velocity of the system at
the time of the impulse. Indeed,
mv 2 m(v  v) 2
1
in which v  is the change
E 

2
2
m
in velocity caused by the unit impulse.
 Now, consider the undamped mass-spring system subject
to impulses applied at local values of the maximum
velocity (every time x = 0). The equation governing the
motion of the system and the impulse force are
(1.146)
n
mx  kx  f , f   I r  (t  t r ), I r   I sgn( x ), (r  1,2,, n)
r 1
where I r is the magnitude of the r-th impulse. When
the impulses are set to be uniform, the impulse
response decays linearly, as shown. The linear decay
rate of the displacement is
I
(1.147)  
m
Notice that the linear decay rate associated with uniform
impulses is independent of the system’s natural frequency.
Thus, the time required to bring the system to rest is
independent of the system’s natural frequency.
NONLINEAR FEEDBACK (REGULATION)
Lecture notes prepared by L.Silverberg and J.Morton