TRACKING THE REFERENCE PATH
CHAPTER 7
REGULATING THE REFERENCE PATH
(CONTINUOUSLY-ACTING ACTUATORS)
This Chapter considers the regulation problem restricted to the case in which the
control force is produced by a continuously-acting actuator. The control force is a
continuous function of time. We’ll consider in this chapter control forces that are
linearly proportional to displacements, integrals of displacements, velocities, and
linear combinations of them. In each case, we’ll see how the dynamic performance
is affected by the control.
1. Displacement Feedback
Consider an undamped single degree of freedom system. In the absence of a
control force, the equation governing the motion and the uncontrolled system
response are
v
mxu  kxu  0, xu  x0 cos( 0 t )  0 sin(  0 t )
(7 – 1)
0
In the presence of displacement feedback, we get
(7 – 2)
mx  kx  0, f   gx, x  x0 cos( t ) 
v0

sin( t )
where g is called the displacement feedback control gain. In Eq. (7 – 1) and Eq. (7
– 2), we have
k
(k  g )
0 

m
m
in which  is called the closed-loop frequency of the system (See Fig. 7 – 1).
Fig. 7 - 1
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Since displacements are continuous functions of time, so too are displacement
feedback control forces. Therefore, displacement feedback can be relatively easy
to produce in devices that can generate forces that are continuous functions of
time.
Now let’s see how displacement feedback changes system performance. First,
consider peak-overshoot. From Eq. (7 – 1), the peak-overshoot of the uncontrolled
system is
v
POu  [ x   0
 0
(7 – 3)
2
0
2
 1/ 2
 ]

and from Eq. (7 – 2), the peak-overshoot of the controlled system is
1/ 2
v
 2

x0  ( 0 ) 2 
2

v 
 
(7 – 4)
PO  [ x02   0  ]1 / 2  
POu
v



2
2
0
 0
 x0  ( ) 
0


In Eq. (7 – 4) we find that peak-overshoot can not be reduced significantly if
v
x0  0 . Indeed, if a system is initially at rest when it’s initially displaced, its
0
peak-overshoot will be equal to its initial displacement regardless of the stiffness
in the system. On the other hand, if x0  0, Eq. (7 – 4) reduces to
0
POu

This case also arises when the system is intermittently subjected to impulsive
forces. From Eq. (7 – 5), the desired closed-loop frequency is (See Fig. 7 – 2)
(7 – 5)
PO 
(7 – 6)

POu
0
PO
Fig. 7 – 2
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Next, let’s examine the effect of displacement feedback on settling time. From Eq.
(7 – 1) and (7 – 2) displacement feedback clearly has no effect of settling time.
Finally, turning to steady-state error, let’s subject the system to a unit step function
in order to create a steady-state error. The steady-state response in the absence of
displacement feedback and the steady-state response in the presence of
displacement feedback are
(7  7 )
1
1
1  PO
xu  , x 
 
k
k  g k  POu



2
Notice that displacement feedback can reduce steady-state error but not eliminate
it without applying a very large control force.
The displacement feedback control force is realized using Eq. (7 – 2) in which the
control gain is
 PO  2 
2
2
  1
(7  8)
g  m(   0 )  k 
 POu 



In summary, displacement feedback is predominantly used to control peakovershoot, but can also reduce steady-state error. It has no effect on settling time.
The displacement feedback control force depends on the static properties of the
uncontrolled system, i.e., its stiffness.
2. Velocity Feedback
Consider an undamped single degree of freedom system subjected to a velocity
feedback control force. The uncontrolled response was given in Eq. (7 – 1). The
equation governing the controlled response is
(7 – 9)
mx  kx  f , f  hx
Velocity feedback is realized using forcing functions that imitate velocities.
Velocities are piece-wise continuous functions. The functions are continuous
except during instants when jump discontinuities occur. Jump discontinuities occur
when the system is subjected to instantaneous impulses caused by impact loads. It
follows that velocity feedback forces can be achieved in devices that can generate
continuous functions of time but they’ll have difficulty when jump discontinuities
occur.
Dividing Eq. (1.105) by m, we get
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(7  10)
x  2x   02 x  0,  
h
2m
Equation (7 – 10) describes the closed-loop system, and admits the homogeneous
solution
(7 – 11)
v  x 0
x  e t [ x0 cos( t )  0
sin( t )],   [ 02   2 ]1 / 2

where we can now see in Eq. (7 – 11) that  represents the closed-loop damping
rate of the system.
Now, let’s see how velocity feedback changes system performance. First, consider
peak-overshoot. When the damping rate is considerably smaller than the natural
frequency of the system, the peak-overshoot of the closed-loop system can be
approximated by Eq. (7 – 3)
v  x0 2 1 / 2
PO  [ x02  ( 0
) ]  POu ,    0
(7 – 12)

Indeed, velocity feedback does not significantly reduce peak-overshoot when the
closed-loop damping rate is small compared to the system’s natural frequency.
However, as the damping rate increases, the peak-overshoot decreases (See Fig. 7
– 3).
Fig. 7 - 3
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Next consider the effect of velocity feedback on settling time. From Eq. (7 – 9)
and Eq. (7 – 10) velocity feedback causes the system motion to dampen at the rate
 depending on the control gain h that is applied. On the other hand, velocity
feedback has no effect on steady-state error, since steady-state error is a static
phenomenon.
From Eq. (7 – 10), the velocity feedback gain used to achieve this performance is
h  2m  2
(7  13)
ln( 10)
m
Ts
In summary, velocity feedback is predominantly used to control settling time, but
also can reduce peak-overshoot. It has no effect on steady-state error. The velocity
feedback control force depends on the dynamic properties of the uncontrolled
system, i.e., its inertia.
3 State Feedback
Let’s now consider a damped single degree of freedom system subject to a
combination of displacement and velocity feedback, i.e., state feedback. State
feedback is capable of controlling both peak-overshoot and settling time. The
equation governing the motion of the system is
(7 – 14)
mx  cx  kx  f
where f denotes the state feedback control force. In the absence of control, Eq. (7 –
14) reduces to
c
k
xu  2 0 x u   02 xu  0,  0  ,  0 
(7 – 15)
m
m
The uncontrolled under-damped system response is
(7 – 16)
xu  e t [ x0 cos(t ) 
v0  x0

sin( t )],  2   02   2
Now, apply the state feedback control force
(7 – 17)
f   gx  hx
and substitute Eq. (7 – 17) into Eq. (7 – 14) to get
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x  2x  ( 2   2 ) x  0,
(7 – 18)
ch
h
k  g ch
g
2
2

 0 
, 2 

  0   
2m
2m
m
m
 2m 
2
where  is the closed-loop frequency of the system, and  is its closed-loop
frequency of oscillation. From Eq. (7 – 18), the state feedback control gains used
to get this performance are
g  m( 2   2   02 ), h  2m(   0 )
(7 – 19)
The displacement and velocity feedback gains in Eq. (7 – 19) can be rewritten as
   2 
 POu  2 
g  m  m    1  m 2  k 
  1,
PO
  0 





ln( 10)
h  2m
c
Ts
2
(7 – 20)
2
0
The displacement feedback control gain given in Eq. (7 – 20) has been separated
into two terms; the first term depends on the system’s inertia and the second term
depends on the system’s stiffness. Notice that the first term is responsible for
damping the system and that the second term is responsible for stiffening the
system. Next, look at the velocity feedback gain. It controls the damping. Notice
that it depends on the system’s inertia and the system’s damping.
Frequency-invariant State Feedback
In some circumstances, the interest lies in controlling settling time and not in
controlling peak-overshoot and steady-state error. In this case, set the closed-loop
frequency equal to the system’s natural frequency. Substitute 0 into Eq. (7 –
18) and (7 – 19) to get the frequency-invariant state feedback control force
(7 – 21)
f  m 2 x  (2m  c) x
Notice that the control force depends on the system’s inertia and its damping, and
is independent of its stiffness. Also, note that by setting the closed-loop frequency
equal to the natural frequency, wasted power associated with stiffening the system
was avoided.
In summary, state feedback is predominantly used to control peak-overshoot and
settling time. The damping components of the control force depend on the
system’s dynamic properties, i.e. its inertia and damping. The stiffening
component of the control force depends on the system’s static parameter, i.e., its
stiffness.
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4. Integral Feedback
The previous three sections examined the effect of displacement feedback and
velocity feedback on the performance of single degree of freedom systems. This
section examines the effect of integral feedback on performance. We’ll see that
integral feedback controls steady-state error in addition to having another
unexpected property. In order to examine steady-state error, an under-damped
system is subjected to a constant unit force, producing a steady-state error, in
addition to being subjected to an integral feedback control force. The equation
governing the motion of the under-damped system subject to a unit force and
integral feedback are
(7 – 22)
t
mx  cx  kx  1  f , f  i  xdt
0
Integral feedback is realized using forcing functions that imitate integrals of
displacements. These functions are continuous. Hence, they can be achieved easily
in devices that can generate continuous functions of time. However, it’s important
that the integral not grow unbounded. The unbounded growth of the integral is
called integral wind-up.
Differentiating Eq. (7 – 22) with respect to time and dividing by m, yields
(7 – 23)
x  2 0 x   02 x 
i
x0
m
Eq. (7 – 24) is a linear third-order homogeneous differential equation. Try
solutions of the form
(7 – 24)
x  e st
Substituting Eq. (7 – 24) into (7 – 23) and dividing by est yields a cubic. The cubic
can be expressed in the factored form
(7 – 25)
0  s 3  2 0 s 2   02 s 
i
 ( s  s1 )( s  s 2 )( s  s3 )
m
where s1, s2, and s3 are the roots of the equation. By the principle of linear
superposition, a linear combination of the three solutions provides a general
solution of Eq. (7 – 23). If the real part of each of the roots is negative or zero, the
solution is stable. Moreover, if the solution is stable, it follows by the homogeneity
of Eq. (7 – 23) that the steady-state solution is zero. Indeed, integral feedback of a
stable closed-loop system drives the steady-state error to zero. This is the result
that we are looking for, but the question remains whether the solution is stable.
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Since the coefficients of the cubic equation are real, it follows that either all three
of its roots are real or that one of its roots is real and the other two are a complex
conjugate pair. First assume that one root is real and that the other two are a
complex conjugate pair. The roots are then of the general form
(7 – 26)
s1  1 , s 2   2  j , s3  s 2   2  j
where 1 is the closed-loop steady-state decay rate, 2 is the closed-loop vibration
decay rate, and  is the closed-loop frequency of oscillation. Substituting Eq. (7 –
26) into (7 – 25) and equating coefficients of powers of s, we get expressions for
the closed-loop decay rates and the closed-loop frequency of oscillation in terms of
the parameters in Eq. (7 – 25). From the equation resulting from equating powers
in s2, we get
(7 – 27)
2 0  1  2 2
Equation (7 – 27) reveals that the natural damping rate is equal to a positive
combination of the closed-loop damping rates. In the absence of natural damping,
that is when 0 = 0, either 1or 2 < 0. Indeed, the closed-loop system is
unstable in the absence of natural damping.
Next, consider the second possibility that all three roots of Eq. (7 – 25) are real,
i.e., that
(7 – 28)
s1  1 , s 2   2 , s3   3
Substituting Eq. (7 – 28) into (7 – 25) and equating powers in s2, we get
(7 – 29)
2 0  1   2   3
Again, the natural damping rate is equal to a sum of the closed-loop decay rates. In
the absence of natural damping, either 12 < 0, or 3 Clearly, this
possibility also leads to an unstable system. Hence, integral feedback of an
undamped single degree of freedom system creates an unstable system, regardless
of the value of the integral control gain.
We have just found that integral feedback destabilizes undamped systems. It can’t
be used by itself when the system is undamped. It needs to be used together with
velocity feedback to offset the “negative damping” produced by the integral
feedback.
5. PID Feedback
Let’s now consider the most general form of linear feedback control considered
thus far. Let’s combine the feedback of displacements, velocities and integrals.
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The displacement feedback is also called proportional feedback and the velocity
feedback is also called derivative feedback. Therefore, the feedback of
displacements, integrals, and velocities is commonly known as PID feedback for
the P in proportional, the I in integral, and the D in derivative.
Let’s consider an under-damped system subjected to a constant force and PID
feedback, written
(7 – 30)
t
mx  cx  kx  1  f , f   gx  hx  i  xdt
0
Dividing Eq. (7 – 30) by m, we get
(7 – 31)
x  2x   02 x  
g
h
i t
x  x   xdt
m
m
m 0
Next, let’s try a solution in the form of Eq. (7 – 24). Substituting Eq. (7 – 24) into
(7 – 31) and dividing by s, we get the cubic equation
h 2
g
i
) s  ( 02  ) s   0
m
m
m
Equation (7 – 32) can be rewritten in a factored form assuming an under-damped
response having roots in the form of Eq. (7 – 26). We get
(7 – 32)
(7 – 33)
s 3  (2 0 
( s  s1 )( s  s 2 )( s  s3 ) 
s 3  (1  2 2 ) s 2  (2 1 2   22   2 ) s   1 ( 22   2 )
Equating powers of s,
s 2 : 2 0 
(7 – 34)
s :  02 
1:
h
  1  2 2
m
g
 2 1 2   22   2
m
i
  1 ( 22   2 )
m
Equations (7 – 34) are three linear algebraic equations expressed in terms of the
unknown control gains g, h, and i. The PID control gains become
(7 – 35)
g  m(21 2   22   2  02 )
h  m(1  2 2  2 0 )
i  m1 ( 22   2 )
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Equation (7 – 35) can be used to produce control forces that yield any desirable
dynamic performance – expressed in terms of the closed-loop steady-state decay
rate 1, the closed-loop vibration decay rate2, and the closed-loop frequency of
oscillation.
Uniform Damping and Uniform Stiffening
The cost associated with a control force increases with the closed-loop decay rate.
Furthermore, the settling time of a closed-loop system depends only on the
smallest of the two closed-loop decay rates (this will be shown shortly). It follows
that it is wasteful to expend effort to select a steady-state decay rate 1 and a
vibration decay rate 2 that are different from one another. Instead, cost efficient
feedback control forces uniformly dampen the motion.
Before proceeding with uniform damping and uniform stiffening, let’s examine
these remarks more closely. To this end, recall that settling time is associated with
an exponential envelope of the response. For example, let x (t )  A1e 1t  A2 e 2t
represent the response of a fictitious system. It is said that x(t) is exponentially
stable at the rate  only if x(t )  Ae t for some  and for some A. The settling
time is associated with , and not explicitly associated with either  or
However, it can be shown that  = min{, } for this x(t). The larger of the
decay rates  and is responsible for increasing the control force without
increasing and hence without decreasing the settling time. Since cost increases
with  and it would be wasteful to select  and different form one another
(See fig. 7 – 4).
Fig. 7 - 4
Let’s now select the dynamic performance parameters as follows:
(7 – 36)
1   2   ,    0
where is the prescribed uniform damping rate and  is the prescribed uniform
stiffening factor. Substituting Eq. (7 – 36) into Eq. (7 – 35) yields the uniform
damping and uniform stiffening PID control gains
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g  3 2 m  ( 2  1)k
h  3m  c
(7 – 37)
i   2 k
In Eq. (7 – 37) the uniform damping components of g and h depend on the
system’s dynamic parameters m and c, and that the stiffening components of g and
i depend on the static parameter k.
6. Time Delays
In systems in which the control forces are determined from sensor measurements
using digital electronics, a significant time delay can exist between the control
force and the sensor measurements. Consider the delayed state feedback of a
damped single degree of freedom system, represented by
(7 – 38)
mx  cx  kx  f , f (t )   gx(t  T )  hx (t  T )
where T denotes the time delay. The delayed displacement and the delayed
velocity can be approximated using a Taylor series expansion, by
(7 – 39)
x(t  T )  x(t )  xT  h.o.t.
x (t  T )  x (t )  xT  h.o.t.
Substituting Eq. (7 – 39) into Eq. (7 – 38), and neglecting higher-order terms,
yields the closed-loop system
(7 – 40)
(m  hT ) x  (c  h  gT ) x  (k  g ) x  0
Equation (7 – 40) reveals the effects of time delay on closed-loop response. From
Eq. (7 – 40) the time delay in the velocity feedback component of the force
produces a reduced effective mass me = m – hT. The time delay in the
displacement feedback component of the force produces a reduced effective
damping ce = c + h – gT. The time delay has no effect on the closed-loop stiffness.
Notice in the case of an undamped system (c = 0) being controlled by
displacement feedback (h = 0), that the time delay causes the system to become
unstable, regardless of the size of the time delay.
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7. Summary
The material covered in this chapter described feedback control of single degree of
freedom systems in terms of relationships between different types of parameters –
physical parameters, dynamic performance parameters (open-loop and closedloop), and control parameters. The parameters are listed below.
Physical Parameters
Mass
Damping
Stiffness
m
c
k
Performance Parameters
Open-loop (natural)

Natural undamped frequency
Natural damped frequency
Natural damping rate



Closed-loop
Closed-loop frequency
Closed-loop vibration damping rate
Closed-loop steady-state damping rate



Control Parameters
Control gains
Maximum force
Power
Fuel
g, h, i
Cm
Cp
Cf
The usefulness of the material covered in this chapter is both quantitative and
qualitative. The designer can develop approximate values of the parameters listed
above as part of a control system design process for a system. The designer can
also use the material covered in this section qualitatively in the selection of sensors
and actuator types and in the selection of the type of feedback to use.
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PROBLEM STATEMENTS
The problems in this chapter consider the systems described in Chapter 1.
Problem 7 – 1: State Feedback
The system is acted on by linear state (proportional-derivative) feedback control.
Design the feedback controller to reduce the peak-overshoot by a factor of 2 and to
dampen 90% of the motion in 6 oscillations. There is no applied load.
(a)
(b)
Determine the controller’s control gains.
Determine the response of the uncontrolled system and the control
system by the Euler method. Plot both over about 6 oscillations.
Measured from equilibrium, let (0) =  and d (0)/dt = 0.1 rad/s.
Problem 7 – 2: PID Feedback
The system is acted on by PID (proportional-integral-derivative) feedback control.
The feedback controller is designed to reduce peak-overshoot by a factor of 2, and
to dampen 90% of the motion, including the errors associated with a “bias,” in 4
natural periods. A natural period is the period of the uncontrolled system. A bias is
a constant load acting on the system that causes its equilibrium position, if
uncontrolled, to be non-zero. You’ll need to assume, in addition to the controller,
that the bias load acts on the system. Select a bias load that increases the
equilibrium position of the uncontrolled system by 20○. Design the feedback
controller to reduce the peak-overshoot by a factor of 2 and to dampen 90% of the
motion in 6 oscillations.
(a)
(b)
(c)
Determine and graph the bias load.
Find the control gains g, h, and i.
Determine the response by the Euler method. Plot the uncontrolled
and controlled responses over about 6 oscillations. Initially the
system is at rest at equilibrium.
Note: In order to solve this problem, you’ll need to introduce the state variables
t
x1  0  ( s)ds, x2   , x3   .
The corresponding initial conditions are then
0
x1 (0)  0  ( s)ds  0, x2 (0)   (0),
x3 (0)  (0).
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH
TRACKING THE REFERENCE PATH
Problem 7 – 3: The Effect of Time Delay
Like in Problem 7 – 1, the system is acted on by linear state (proportionalderivative) feedback control and the feedback controller is designed to reduce the
peak-overshoot by a factor of 2 and to dampen 90% of the motion in 6 oscillations.
But, now we assume that there is a time delay of a fraction f of the system’s natural
period. Again, there is no other applied load.
Use the control gains found in Problem 7 – 1. But now put in a time delay.
(a) Find the largest time delay for which the system remains stable.
Hint: The time delay can be introduced in the computer program by setting it
equal to a multiple p of the step size T, i.e., let Td = pT. In order to retrieve delayed
controls you will need to store the values of the controls over the period of the time
delay (p values). The initial p values of the control are zero.
CONTROL OF DYNAMICAL SYSTEMS: AN INTRODUCTORY APPROACH