567
This is often referred to as Þnite settling time or deadbeat design
because the dynamics will settle in a Þnite number of sample
periods. This estimator always drives the error to zero in time
2T or less as veriÞed below.
The estimator-error equation :
·
¸·
¸
1 − T1
x̃1 (k)
x̃(k + 1) = (Φ − LH)x̃(k) =
x̃2 (k)
T −1
For k = 0, 1, 2, 3, ...
·
x̃1 (0) − T1 x̃2 (0)
x̃(1) =
T x̃1 (0) − x̃2 (0)
· ¸
0
x̃(2) =
0
· ¸
0
x̃(3) =
0
..
.
¸
(c) Yes, since one state element is measured directly, one can estimate the other element with a Þrst-order estimator following the
development for reduced order estimators for contintuous systems in Section 7.5.2.
From the discrete state-space representation,
x2 (k + 1) = T x1 (k) + x2 (k) +
T2
u(k)
2
·
¸
1
T2
=⇒ x1 (k) =
x2 (k + 1) − x2 (k) −
u(k)
T
2
{z
}
|
known measurement and input
For K = 0,
·
¸
1
T2
x2 (1) − x2 (0) −
u(0)
T
2
·
¸
2
1
T
=
y(1) − y(0) −
u(0)
T
2
x̄2 (0) = x2 (0)
= y(0)
x̄1 (0) =
14. Single-axis Satellite Attitude Control: Satellites often require attitude control for proper orientation of antennas and sensors with respect to Earth.
568
CHAPTER 8. DIGITAL CONTROL
Figure 8.23: Satellite control schematic for Problem 14
Figure 2.6 shows a communication satellite with a three-axis attitudecontrol system. To gain insight into the three-axis problem we often consider one axis at a time. Figure 8.23 depicts this case where motion is
only allowed about an axis perpendicular to the page. The equations of
motion of the system are given by
I θ̈ = MC + MD ,
where
I = moment of inertia of the satellite about its mass center,
MC =control torque applied by the thrusters,
MD =disturbance torques,
θ =angle of the satellite axis with respect to an inertial reference with no
angular acceleration.
We normalize the equations of motion by deÞning
u=
MC
,
I
wd =
MD
,
I
and obtain
θ̈ = u + wd .
Taking the Laplace transform yields
θ(s) =
1
[u(s) + wd (s)],
s2
which with no disturbance becomes
θ(s)
1
= 2 = G1 (s).
u(s)
s
569
In the discrete case where u is applied through a ZOH, we can use the
methods described in this chapter to obtain the discrete transfer function
G1 (z) =
¸
·
θ(z)
T2
z+1
.
=
u(z)
2 (z − 1)2
(a) Sketch the root locus of this system by hand assuming proportional
control.
(b) Draw the root locus using MATLAB to verify the hand sketch.
(c) Add a lead network to your controller so that the dominant poles
correspond to ζ = 0.5 and ω n = 3π/(10T ).
(d) What is the feedback gain if T = 1 sec? If T = 2 sec.
(e) Plot the closed-loop step response and the associated control time
history for T = 1 sec.
Solution
(a) The hand sketch will show that the loci branches depart vertically
from z = 1; therefore, the system is marginally stable or unstable
for any value of gain.
(b) The Matlab version below conÞrms the situation.
G1 (z) =
(z + 1)
T 2 (z + 1)
= K0
2
2 (z − 1)
(z − 1)2
(c) To obtain the desired damping and frequency, Fig. 8.4 shows
that a root locus branch should go through the desired poles at
570
CHAPTER 8. DIGITAL CONTROL
z = 0.44 ± 0.44j. After some trial and error, you can Þnd that
this can be accomplished with the lead compensation :
D(z) = K
(z − 0.63)
(z + 0.44)
The speciÞc value of K that yields the closed-loop poles are at :
z = 0.44 ± 0.44j, −0.113
is K =
0.692
K0 .
The second-order pair give :
ωn
ζ
0.917
rad/sec
T
= 0.519
=
which is close enough to the desired speciÞcations.
locus for the compensated design is:
(d)
K
=
=
0.692
½
T2
2
1.383 for T = 1 sec
0.3458 for T = 2 sec
¾
The root
571
(e) Closed-loop step response :
Closed-Loop Step Response (T=1s)
1.5
theta
1
0.5
0
0
2
4
6
8
10
12
14
Time (sec)
Time History of Control Effort (T=1s)
1.5
1
u
0.5
0
-0.5
-1
-1.5
0
2
4
6
8
10
12
Time (sec)
15. In this problem you will show how to compute Φ by changing states so
14
573
(c) Let T =
·
a b
c d
¸
,
TA = FT =⇒ 2a = c, b = d
=⇒ Choosing c = d = 1, then a = 0.5, b = c = d = 1
·
¸
0.5 1
=⇒ T =
1 1
This T satisÞes £TAT−1 =¤F.
Note : Let T = v1 v2 ,
TFT−1
£
¤−1 £
A =⇒ v1 v2
F v1
½
¾
Fv1 = (−1)v1
=⇒
Fv2 = (−2)v2
=
v2
¤
=
·
−1 0
0 −2
Thus, the columns v1 and v2 are the eigenvectors of F.
(d)
eFT
= TeAT T−1
·
¸ · −T
0.5 1
e
=
1 1
0
·
−T
−2T
−e + 2e
=
−2e−T + 2e−2T
¸·
−2 2
2 −1
e−2T
¸
−T
−2T
e −e
2e−T − e−2T
0
¸
16. It is possible to suspend a mass of magnetic material by means of an
electromagnet whose current is controlled by the position of the mass
(Woodson and Melcher, 1968). The schematic of a possible setup is shown
in Fig. 8.24, and a photo of a working system at Stanford University is
shown in Fig. 2.35. The equations of motion are
mẍ = −mg + f (x, I),
where the force on the ball due to the electromagnet is given by f (x, I).
At equilibrium the magnet force balances the gravity force. Suppose we
let I0 represent the current at equilibrium. If we write I = I0 + i, expand
f about x = 0 and I = I0 , and neglect higher-order terms, we obtain the
linearized equation
mẍ = k1 x + k2 i.
(1)
Reasonable values for the constants in Eq. (1) are m = 0.02 kg, k1 = 20
N/m, and k2 = 0.4 N/A.
(a) Compute the transfer function from I to x, and draw the (continuous)
root locus for the simple feedback i = −Kx.
¸
574
CHAPTER 8. DIGITAL CONTROL
Figure 8.24: Schematic of magnetic levitation device for Problems16 and 17
(b) Assume the input is passed through a ZOH, and let the sampling
period be 0.02 sec. Compute the transfer function of the equivalent
discrete-time plant.
(c) Design a digital control for the magnetic levitation device so that the
closed-loop system meets the following speciÞcations: tr ≤ 0.1 sec,
ts ≤ 0.4 sec, and overshoot ≤ 20%.
(d) Plot a root locus with respect to k1 for your design, and discuss the
possibility of using your closed-loop system to balance balls of various
masses.
(e) Plot the step response of your design to an initial disturbance displacement on the ball, and show both x and the control current i.
If the sensor can measure x only over a range of ±1/4cm and the
ampliÞer can only provide a current of 1 A, what is the maximum
displacement possible for control, neglecting the nonlinear terms in
f (x, I)?
Solution:
(a)
G(s) =
=
X(s)
k2 /m
= 2
I(s)
s − k1 /m
20
s2 − 1000
(2)
575
(b) T = 0.02 sec,
G(z) = (1 − z
−1
)Z
= 0.004135
½
G(s)
s
¾
z+1
(z − 0.5313)(z − 1.8822)
(c) The speciÞcations imply that :
tr
ts
Mp
1.8
= 18 rad/sec
0.1
4.6
≤ 0.4 sec =⇒ σ ≥
= 11.5 rad/sec
0.4
=⇒ r = |z| ≤ e−11.5×0.02 = 0.7945 (←− z = esT )
≤ 20% =⇒ ζ ≥ 0.48
≤
0.1 sec =⇒ ω n ≥
Thus, the closed-loop poles must be pulled into the unit circle
near r = 0.8 and ζ = 0.5. Using the template of Fig. 8.4, we
experiment with lead compensation and select,
D(z) = 116
The closed-loop poles are :
z − 0.5313
z − 0.093
z = 0.75 ± 0.39j, 0.53
Performance :
tr
ts
Mp
= 0.072
= 0.397
= 18.3%
which meet all the speciÞcations. The root locus is below.
576
CHAPTER 8. DIGITAL CONTROL
The step response shows Mp
Closed-Loop Step Response
8
7
6
x [m]
5
4
3
2
1
0
0
0.1
0.2
0.3
0.4
Time (sec)
0.5
0.6
0.7
0.8
577
Time History of Control Effort
200
100
i [A]
0
-100
-200
-300
-400
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Time (sec)
(d) As can be seen from Eq. (2), the loop gain and the open loop pole
locations depend on the mass of the ball. Changing the mass
will therefore affect the dynamic characteristics of the system
and may render it unstable. A root locus of the closed-loop
poles versus k1 shows how the locus changes as a function of the
mass:
The closed-loop system becomes unstable for k1 ≥ 24. Since a
small increase in k1 makes the system unstable and a decrease
0.8
578
CHAPTER 8. DIGITAL CONTROL
in m has the same effect on the system, it is difficult to balance
balls of smaller masses.
(e) The response to an initial x displacement is shown :
Closed-Loop Initial Response
1
0.5
x [m]
0
-0.5
-1
-1.5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Time (sec)
Time History of Control Effort
100
i [A]
50
0
-50
-100
0
0.1
0.2
0.3
0.4
Time (sec)
0.5
0.6
0.7
0.8
579
The assumption here is that an allowable transient must stay in
the range of the sensor and not require more than the limit of
the current. From I(z) = D(z)(0 − X(z)), we have a difference
equation :
i(k) − 0.093i(k − 1) = −116{x(k) − 0.5313x(k − 1)}
For k = 0, i(0) = −116x(0). We see that if x(0) = 1 then
i(0) = −116. Note that i(0) = −D(∞)x(0).
Thus, if i is to be kept below 1A then x(0) must be kept below
1/116 = 0.00862 m = 0.862 cm = 0.339 inch, which is greater
than the sensor range. The current control can handle any displacement in the range of ±0.25 inch.
17. In Problem 16 we described an experiment in magnetic levitation described
by Eq. (1) which reduces to
ẍ = 1000x + 20i.
Let the sampling time be 0.01 sec.
(a) Use pole placement to design a controller for the magnetic levitator
so that the closed-loop system meets the following speciÞcations: settling time, ts ≤ 0.25 sec, and overshoot to an initial offset in x that
is less than 20%.
(b) Plot the step response of x, x̃, and i to an initial displacement in x.
(c) Plot the root locus for changes in the plant gain, and mark the pole
locations of your design.
(d) Introduce a command reference input r (as discussed in Section 7.8)
that does not excite the estimate of x. Measure or compute the
frequency response from r to the system error r − x and give the
highest frequency for which the error amplitude is less than 20% of
the command amplitude.
Solution
(Note: part (d) of this problem is a stretch for the students. The text
doesn’t cover the relevant discrete development and the student would be
helped by referring to Digital Control of Dynamic Systems, by Franklin,
Powell, and Workman.)
(a) State-space representation of the plant :
¸ ·
¸
·
x
x1
=
, y = x, u = i
x=
xú
x2
d
dt
·
x1
x2
¸
·
0
1
1000 0
= Fx + Gu
=
¸·
x1
x2
¸
+
·
0
20
¸
u
606
CHAPTER 8. DIGITAL CONTROL
Step Response (T=0.1)
1.2
1
0.8
y
______ continuous
0.6
- - - tustin and MPZ
0.4
0.2
0
0
1
2
3
4
5
time (sec)
6
7
8
9
10
7
8
9
10
Step Response (T=0.01)
1.5
y
1
______ Continuous
0.5
- - - Tustin and MPZ
0
0
1
2
3
4
5
time (sec)
6
22. Repeat Problem 5.29 by constructing discrete root loci and performing
the designs directly in the z-plane. Assume that the output y is sampled,
607
the input u is passed through a ZOH as it enters the plant, and the sample
rate is 15 Hz.
Solution
(a) The most effective discrete design method is to start with some idea
what the continuous design looks like, then adjust that as necessary
with the discrete model of the plant and compensation. We refer to
the solution for Probelm 5.29 for the starting point. It shows that
the specs can be met with a lead compensation,
D1 (s) = K
(s + 1)
(s + 60)
and a lag compensation,
D2 (s) =
(s + 0.4)
.
(s + 0.032)
Although it is stated in the solution to Problem 5.29 that a gain,
K = 240 will satisfy the constraints, in fact, a gain of about K = 270
is actually required to meet the rise time constraint of tr ≤ 0.4 sec.
So we will assume here that our reference continuous design is
D1 (s) = 270
(s + 1) (s + 0.4)
(s + 60) (s + 0.032)
1
0.4
It yields a rise time, tr ∼
)( 0.32
)=
= 0.38, Mp ∼
= 15%, and Kv = (270)( 60
56. So all specs are met. For interest, use of the damp function shows
that ω n = 6.4 rad/sec for the dominant roots, and those roots have
a damping ratio, ζ ∼
= 0.7. For the discrete case with T = 15 Hz,
we should expect some degradation in performance, especially the
damping, because the sample rate is approximately 15×ω n .
The discrete transfer function for the plant described by G(s) and
preceeded by a ZOH is:
½
¾
G(s)
G(z) = (1 − z −1 )Z
s
½
¾
z−1
10
=
Z
z
s(s + 1)(s + 10)
This is most easily determined via Matlab,
sysC = tf([10],[1 11 10 0]);
T=1/15;
sysD = c2d(sysC,T,’zoh’);
which produces:
608
CHAPTER 8. DIGITAL CONTROL
G(z) = 0.00041424
(z + 3.136)(z + 0.2211)
(z − 1)(z − 0.9355)(z − 0.5134
The essential elements of the compensation are that the lead provides
velocity feedback with a TD = 1 and the lag provides some high
frequency gain. The discrete equivalent of the proportional plus
lead would be (see Eq. 8.42):
D1 (z) = K(1 +
TD
(1 + TD /T )z − TD /T
(1 − z −1 )) = K
T
z
which for T = 1/15 = 0.0667 and TD = 1 reduces to
D1 (z) = 270
16z − 15z −1
z − 0.9375
= 4320
.
z
z
The lag equivalent is best introduced by use of one of the approximation techniques, such as the matched pole-zero:
D2 (z) =
z − e−0.4T
z − .9737
=
−0.032T
z−e
z − .9979
as its whole function is to raise the gain at very low frequencies for
error reduction. Examining the resulting discrete root locus and
picking roots with rlocfind to yield the required damping shows that
the gain, K = 60. While the use of damp indicates that the frequency
and damping are acceptable, the time response shows an overshoot
of about 20% and the rise time is slightly below spec. We therefore
need to increase the lead (move the lead zero closer to z = +1) to
decrease the overshoot and increase gain to speed up the rise time.
Several iterations on these two quantities indicates that moving the
lead zero from z = 0.9375 to z = 0.96 and increasing the gain from
K = 60 to K = 65 meets both specs. The velocity coefficient is
found from and is also satisÞed.
Kv =
lim (z − 1)D(z)G(z)
z→1
Tz
The time response of the Þnal design below shows that all specs are
met.
609
Step Response
1.2
1
y
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
time (sec)
1.2
1.4
1.6
1.8
2
23. Design a digital controller for the antenna servo system shown in Figs. 3.63
and 3.64 and described in Problem 3.30. The design should provide a step
response with an overshoot of less than 10% and a rise time of less than
80 sec.
(a) What should the sample rate be?
(b) Use emulation design with the matched pole-zero method.
(c) Use discrete design and the z-plane root locus.
Solution
(a) The equation of motion is :
J θ̈ + Bθ = Tc
where
J = 600000 kg.m2 , B = 20000 N.m.sec
If we deÞne :
a=
B
1
Tc
=
, U=
J
30
B
after Laplace transform, we obtain :
G(s) =
θ(s)
1
=
u(s)
s(30s + 1)
609
Step Response
1.2
1
y
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
time (sec)
1.2
1.4
1.6
1.8
2
23. Design a digital controller for the antenna servo system shown in Figs. 3.63
and 3.64 and described in Problem 3.30. The design should provide a step
response with an overshoot of less than 10% and a rise time of less than
80 sec.
(a) What should the sample rate be?
(b) Use emulation design with the matched pole-zero method.
(c) Use discrete design and the z-plane root locus.
Solution
(a) The equation of motion is :
J θ̈ + Bθ = Tc
where
J = 600000 kg.m2 , B = 20000 N.m.sec
If we deÞne :
a=
B
1
Tc
=
, U=
J
30
B
after Laplace transform, we obtain :
G(s) =
θ(s)
1
=
u(s)
s(30s + 1)
610
CHAPTER 8. DIGITAL CONTROL
From the speciÞcations,
Mp
tr
µ
¶
ζ
1−
100 =⇒ ζ > 0.54
0.6
1.8
∼ ω BW > 0.0225
< 80 sec =⇒ tr ∼
< 80 =⇒ ω n =
=
ωn
< 10% =⇒ MP ∼
=
Note that ω pn ∼
= 1/30 < ω n .
If designing by emulation, a sample rate of 20 times the bandwidth
is recommended. If using discrete design, the sample rate can be
lowered somewhat to perhaps as slow as 10 times the bandwidth.
However, to reject random disturbances, best results are obtained by
sampling at 20 times the closed-loop bandwidth or faster. Thus, for
both design methods, we choose :
T
ωs
= 10 sec
= 0.628 rad/sec, which is > 20ω n = 0.45 rad/sec
(b) Continuous design :
Use a proportional controller :
u(s) = D(s)(θr (s) − θ(s)) = K(θr (s) − θ(s))
Root locus :
Choose K = 0.210.
The closed-loop pole location in s-plane :
s = −0.0167 ± 0.0205j
The corresponding natural frequency and damping :
ω n = 0.0265, ζ = 0.6299
611
Digitized the continuous controller with matched pole-zero method :
D(z) = 0.0210
Tc (z) = Bu(z) = 420(θr (z) − θ(z))
Performance :
Mp
tr
= 0.119
= 67.3 sec
(c) With u(k) applied through a ZOH, the transfer function for an equivalent discrete-time system is :
G(z) = K
z+b
(z − 1)(z − eaT )
where
K
aT − 1 + e−aT
1 − e−aT − aT e−aT
, b=
a
aT − 1 + e−aT
z + 0.8949
=⇒ G(z) = 1.4959
(z − 1)(z − 0.7165)
=
Use a proportional control of the form :
D(z) = K
Root locus :
612
CHAPTER 8. DIGITAL CONTROL
The speciÞcation can be achieved with the proportional control. However, we try to achieve the same closed-loop poles as the emulation
design (part (b)) for comparison. These closed-loop pole locations
are denoted by ” + ” in the root locus.
Use a PD control of the form :
D(z) = K
z−α
z
Root locus :
Choose K = 0.0294, α = 0.3.
The resulting z-plane roots :
z = 0.8280 ± 0.1725j, 0.0165
This corresponds to the s-plane roots :
s = −0.0167±0.0205j (the design point of emulation design), −0.4104
which satisfy the speciÞcation :
ωn
ζ
= 0.0265, 0.4104
= 0.6321, 1.000
The control law :
z − 0.3
z
z − 0.3
Tc (z) = Bu(z) = 588
z
D(z) = 0.0294
613
Performance :
= 0.079
= 71.3 sec
Mp
tr
Step response :
Step Response
1.2
emulation design
1
discrete design
x (m)
0.8
0.6
0.4
0.2
0
0
50
100
150
200
250
300
time (sec)
Time History of Control Effort
0.04
0.03
discrete design
u (Tc/B)
0.02
emulation design
0.01
0
-0.01
0
50
100
150
time (sec)
24. The system
G(s) =
1
(s + 0.1)(s + 3)
200
250
300
614
CHAPTER 8. DIGITAL CONTROL
is to be controlled with a digital controller having a sampling period of
T = 0.1 sec. Using a z-plane root locus, design compensation that will
respond to a step with a rise time tr ≤ 1 sec and an overshoot Mp ≤ 5%.
What can be done to reduce the steady-state error?
Solution
(a) Continuous plant :
G(s) =
1
, Type 0 system
(s + 0.1)(s + 3)
Discrete model of G(s) preceeded by a ZOH (T = 0.1 sec) :
G(z) = 0.0045
z + 0.9019
(z − 0.7408)(z − 0.99)
SpeciÞcations :
tr
Mp
≤ 1 sec −→ ω n ≥ 1.8 rad/sec
≤ 5% −→ ζ ≥ 0.7
Discrete design : A simple proportional feedback, D(z) = K = 4.0,
will bring the closed-loop poles to :
z = 0.8564 ± 0.1278j
which are inside the specs region.
ω n = 2.07 rad/sec, ζ = 0.70
Root locus :
615
Step response :
Closed-Loop Step Response
1
0.9
0.8
0.7
y
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
8
6
7
8
Time (sec)
Time History of Control Effort
4
u
3
2
1
0
0
1
2
3
4
5
Time (sec)
The step response shows that :
tr
Mp
∼
= 1.02 sec
∼
= 4.7%
However, since the system is type 0, steady-state error exists and is
7% in this case. An integral control of the form,
D(z) =
K Tz
TI z − 1
616
CHAPTER 8. DIGITAL CONTROL
can be added to the proportional control to reduce the steady-state
error, but this typically occurs at the cost of reduced stability.
25. The transfer function for pure derivative control is
D(z) = KTD
z−1
,
Tz
where the pole at z = 0 adds some destabilizing phase lag. Can this phase
lag be removed by using derivative control of the form
D(z) = KTD
(z − 1)
?
T
Support your answer with the difference equation that would be required,
and discuss the requirements to implement it.
Solution: (8-25)
(a) No, we cannot use derivative control of the form :
D(z) = KTD
z−1
T
to remove the phase lag. The difference equation corresponding to
D(z) = Kp TD
is
u(k) = Kp TD
z−1
U (z)
=
T
E(z)
e(k + 1) − e(k)
T
This is not a causal system since it needs the future error signal
to compute the current control. In real time applications, it is not
possible to implement a non-causal system.