Student Solutions Manual for
Design of Nonlinear Control Systems
with the Highest Derivative
in Feedback
World Scientific, 2004
ISBN 981-238-899-0
Valery D. Yurkevich
c 2007 by Valery D. Yurkevich
Copyright °
Preface
Student Solutions Manual contains complete solutions of 20 % of Exercises from the
book “Design of Nonlinear Control Systems with the Highest Derivative in Feedback”,
World Scientific, 2004, (ISBN 9812388990). The manual aims to help students understand a new methodology of output controller design for nonlinear systems in presence of
unknown external disturbances and varying parameters of the plant.
The solutions manual is accompanied by Matlab-Simulink files1 for calculations and
simulations related with Exercises. The program files provide the student a possibility
to design the discussed control systems in accordance with the assigned performance
specifications of output transients, and make a comparison of simulation results.
The distinguishing feature of the discussed throughout design methodology of dynamic output feedback controllers for nonlinear systems is that two-time-scale motions
are induced in the closed-loop system. Stability conditions imposed on the fast and slow
modes, and a sufficiently large mode separation rate, can ensure that the full-order closedloop system achieves desired properties: the trajectories of the full singularly perturbed
system approximate to the trajectories of the reduced model, where the reduced model is
identical to the reference model, by that the output transient performances are as desired,
and they are insensitive to parameter variations and external disturbances.
Robustness of the closed-loop system properties is guaranteed so far as the stability
of the fast mode and the sufficiently large mode separation rate are maintained in the
closed-loop system. Consequently, the ensuring of the fast mode stability by selection of
control law parameters is the problem requiring undivided attention and that constitutes
the core of the controller design procedure.
In general, the selection of the control law structure as well as selection of controller
parameters are not unique, inasmuch as a set of constraints has to be taken into account
such as a range of variations for plant parameters and external disturbances, required
control accuracy, requirements on load disturbance rejection as well as high frequency
measurement noise rejection. Therefore, it would be much more correctly, if the student
will take up the solution presented in the manual as an example or draft version of such
solution, and then one can make a try to extend the solution by taking into account some
additional practical limitations.
Overall, the above mentioned book, along with the Student Solutions Manual, as well
as accompanying Matlab-Simulink files are an excellent learning aid for advanced study
of real-time control system designing and ones may be used in such course as “Design of
Nonlinear Control Systems”, where prerequisites are “Linear Systems” and “Nonlinear
Systems”. Any comments about the solutions manual (including any errors noticed) can
be sent to hyurkev@mail.rui or hyurkev@ieee.orgi with the subject heading hbook i. They
will be sincerely appreciated.
Valery D. Yurkevich
1
A set of Matlab-Simulink files for the Student Solutions Manual can be downloaded from the website
http://ac.cs.nstu.ru/∼yurkev/books.html.
Student Solutions Manual for “Design of nonlinear control systems . . .”
3
Contents
Exercise 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Exercise 1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Exercise 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Exercise 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
Exercise 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Exercise 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Exercise 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
Exercise 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
Exercise 5.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .19
Exercise 6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
Exercise 6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
Exercise 7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Exercise 7.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31
Exercise 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Exercise 8.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
Exercise 9.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
Exercise 9.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
Exercise 10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .44
Exercise 10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .48
Exercise 11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .54
Exercise 11.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .55
Exercise 12.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .58
Exercise 13.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .59
Auxiliary Material (The optimal coefficients based on ITAE criterion) . . . . . . . . . . . . . . . 61
Auxiliary Material (Euler polynomials) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
Auxiliary Material (Describing functions) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
Auxiliary Material (The Laplace Transform and the Z-Transform) . . . . . . . . . . . . . . . . . . 64
Errata for the book . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
Student Solutions Manual for “Design of nonlinear control systems . . .”
4
Chapter 1
Exercise 1.2 The behavior of a dynamical system is described by the equation
x(2) + 1.5x(1) + 0.5x + µ{2x2 + [x(1) ]2 }1/2 = 0.
(1)
Determine the region of µ such that X = 0 is an exponentially stable equilibrium point
of the given system.
Solution.
Denote x1 = x, x2 = x(1) , and X = [x1 , x2 ]T . Hence, we have
Ẋ = AX + µg(X),
where
"
A=
0
1
−0.5 −1.5
#
"
,
0
2
(2x1 + x22 )1/2
g(X) =
#
.
Hence, g(X)|X=0 = 0, and so the perturbation g(X) is vanishing at the equilibrium point
of the linear nominal system Ẋ = AX. We can find that
√
kg(X)k2 = (2x21 + x22 )1/2 ≤ (2x21 + 2x22 )1/2 = 2 kXk2 .
√
Denote c5 = 2. Then, from the Lyapunov equation
P A + AT P = −Q
with Q = I, we get
"
P =
2
1
1
1
(2)
#
,
where λmin (P ) = 0.382 and λmax (P ) = 2.618. Hence, if the inequalities
0<µ<
λmin (Q)
= 0.135
2λmax (P )c5
hold, then X = 0 is an exponentially stable equilibrium point of the system (1).
The above results can be obtained by Matlab program e1 2 Lyap.m as well as the
initial value problem solution can be found by running e1 2.mdl.
Exercise 1.4 The behavior of a dynamical system is described by the equations
"
ẋ1
µẋ2
#
"
=
1 −1
2 1
#"
x1
x2
#
.
(3)
Obtain and analyze the stability of the slow-motion subsystem (SMS) and the fast-motion
subsystem (FMS).
Solution.
Student Solutions Manual for “Design of nonlinear control systems . . .”
5
From (3), we have
Ẋ = A(µ)X,
where
"
A(µ) =
1
2/µ
−1
1/µ
#
.
The characteristic polynomial of the system is given by
Ã
!
1
3
det [sI − A(µ)] = s −
+1 s+ .
µ
µ
2
By passing over in silence, we have that µ > 0 is the permissible region for parameter µ.
Hence, the system (3) is unstable for all µ from that region.
By introducing the new time scale t0 = t/µ into the system (3), we obtain
d
x1 = µ[x1 − x2 ],
dt0
d
x2 = 2x1 + x2 .
dt0
Take µ = 0. Hence, we get
d
x1 = 0, =⇒ x1 = const,
dt0
d
x2 = 2x1 + x2 .
dt0
By returning to the primary time scale t, the FMS
µ
d
x2 = 2x1 + x2
dt
is obtained, where x1 = const. The FMS is unstable.
Next, consider an equilibrium point of the FMS, that is dx2 /dt = 0. Hence, 2x1 +x2 =
0 =⇒ x2 = −2x1 . As the result, from
d
x1 = x1 − x2 ,
dt
0 = 2x1 + x2
the equation of the SMS
d
x1 = 3x1
dt
follows. The SMS is unstable too.
Finally, plot the phase portrait of the given system by means of Matlab program2
pplane.m for µ = 0.1 s.
2
Phase Plane Demo for Matlab by John Polking at Rice Univerisity contains the programs
dfield.m, dfsolve, pplane.m, and ppsolve.m. The programs can be downloaded from the website
http://calclab.math.tamu.edu/docs/math308/MATLAB-pplane/ as well as the instructions for use.
Student Solutions Manual for “Design of nonlinear control systems . . .”
6
Chapter 2
Exercise 2.1 Construct the reference model in the form of the 2nd order differential
equation given by
T n y (n) + adn−1 T n−1 y (n−1) + · · · + ad1 T y (1) + y = r
(4)
in such a way that the step response parameters of the output meet the requirements
tds ≈ 6 s, σ d ≈ 0 %. Plot by computer simulation the output response, and determine the
steady-state error from the plot for input signals of type 0 and 1.
Solution.
Take tds = 6 s and σ d = 0 %, then by3
Ã
d
θ = tan
−1
!
π
,
ln(100/σ d )
ωd =
4
,
tds
we get θd = 0 rad, ζ d = 1, and ω d = ωn = 0.6667 rad/s. By selecting the 2 roots
s1 = s2 = −0.6667, we obtain the desired characteristic polynomial given by
s2 + 1.333s + 0.4444.
Consider the desired transfer function given by
0.4444
Gdyr (s) = 2
.
s + 1.333s + 0.4444
Hence, the reference model in the form of the type 1 system
y (2) + 1.333y (1) + 0.4444y = 0.4444r
(5)
follows. Denote e(t) = r(t) − y(t).
Let r(t) be the input signal of type 0, that is, r(t) = rs 1(t), where rs = const and
rs 6= 0. Hence, r(s) = rs /s and we get
es = lim se(s)
s→0
1
= lim s[1 − Gdyr (s)] rs
s→0
s
2
s + 1.333s
= lim 2
rs = 0.
s→0 s + 1.333s + 0.4444
Let r(t) be the input signal of type 1, that is, r(t) = rv t 1(t), where rv = const and rv 6= 0.
Hence, r(s) = rv /s2 and we get
evr = es = lim se(s)
s→0
1 v
r
s→0
s2
s + 1.333
= lim 2
rv
s→0 s + 1.333s + 0.4444
1.333 v
=
r ≈ 3rv .
0.4444
= lim s[1 − Gdyr (s)]
3
z = tan−1 (x) denotes the arctangent of x, i.e., tan(z) = x.
Student Solutions Manual for “Design of nonlinear control systems . . .”
7
Run the Matlab program e2 1 Parameters.m in order to calculate the reference model
parameters. Next, run the Simulink program e2 1.mdl to get a plot for the output response
of (5). Then, from inspection of the plot, determine the steady-state error for input signals
of type 0 and 1, respectively. The simulation results are shown in Fig. 1.
Figure 1: Responses of y(t) and e(t) for input signal r(t) of type 0 and 1.
Exercise 2.3 Construct the reference model in the form of the 2nd order differential
equation as the type 1 system with the following roots of the characteristic polynomial:
s1 = −1+j, s2 = −1−j. Plot by computer simulation the output response, and determine
the steady-state error from the plot for input signals of type 0, 1, and 2.
Solution.
By selecting the 2 roots s1,2 = −1 ± j, we obtain the desired characteristic polynomial
2
s + 2s + 2. Consider the transfer function given by
Gdyr (s) =
2
.
s2 + 2s + 2
From Gdyr (s), the reference model
y (2) + 2y (1) + 2y = 2r
(6)
follows. Denote e(t) = r(t) − y(t). By the same way as in Exercise 2.1, we obtain that
esr = 0 and evr = 1. Hence, the reference model (6) is the type 1 system. Next, let us
consider the type 2 system given by
y (2) + 2y (1) + 2y = 2r(1) + 2r
(7)
= 1/2, where eacc
is the relative acceleration error
From (7), we obtain evr = 0 and eacc
r
r
due to the reference input r(t).
Run the Matlab program e2 3 Parameters.m to calculate the reference model parameters as well as to obtain the step response of the reference model (6). Next, run the
Simulink program e2 3.mdl in order to get a plot for the output response of (6), and then,
from inspection of the plot, determine the steady-state error for input signals of type 0,
1 and 2. Repeat that for (7). The simulation results are shown in Figs. 2– 3.
Chapter 3
Student Solutions Manual for “Design of nonlinear control systems . . .”
8
Figure 2: Responses of y(t) and e(t) of the system (6) for input signal r(t) of type 0, 1,
and 2.
Figure 3: Responses of y(t) and e(t) of the system (7) for input signal r(t) of type 0, 1,
and 2.
Exercise 3.1 The differential equation of a plant is given by
x(2) = x2 + |x(1) | + [1.2 − cos(t)]u,
(8)
where y(t) = x(t). The reference model for x(t) is assigned by
x(2) = −1.2x(1) − x + r.
(9)
u = k0 [F (X̂, r) − x̂(n) ]
(10)
µq x̂(q) + dq−1 µq−1 x̂(q−1) + · · · + d1 µx̂(1) + x̂ = x, X̂(0) = X̂ 0 ,
(11)
Consider the control law
with real differentiating filter
where k0 = 40, q = 2, µ = 0.1 s, d1 = 3. Determine the fast-motion subsystem (FMS)
and slow-motion subsystem (SMS) equations. Perform a numerical simulation.
Solution.
The differential equation of a plant is given by (8). Hence, n = 2 and x(n) = x(2) is
the highest derivative of the output signal. We have that the reference model for x(t) is
assigned by (9). Then, the control law with the highest derivative of the output signal in
feedback and an ideal differentiating filter is given by
u = k0 [F (x(1) , x, r) − x(2) ],
Student Solutions Manual for “Design of nonlinear control systems . . .”
9
where F (x(1) , x, r) = −1.2x(1) − x + r. As the result, we have the control law given by
u = k0 [−x(2) − 1.2x(1) − x + r].
Hence, the closed-loop system with the ideal differentiating filter is given by
x(2) = f (x(1) , x) + g(t)u,
u = k0 [F (x(1) , x, r) − x(2) ],
where f (x(1) , x) = x2 + |x(1) | and g(t) = 1.2 − cos(t). Then
x(2) = F (x(1) , x, r) +
1
[f (x(1) , x) − F (x(1) , x, r)]
1 + g(t)k0
(12)
is the SMS equation.
Next, let us consider the system given by
µ2 x̂(2) + d1 µx̂(1) + x̂ = x,
where x̂(1) and x̂(2) can be used as the estimates of x(1) and x(2) , respectively. Hence,
this system can play a role of a real differentiating filter. Then, the closed-loop system
equations with the real differentiating filter is given by
x(2) = f (x(1) , x) + g(t)k0 [F (x̂(1) , x, r) − x̂(2) ],
µ2 x̂(2) + d1 µx̂(1) + x̂ = x.
(13)
Denote x̂(1) = x̂1 and x̂(2) = x̂2 . Consider the extended system given by
x(2) = f (x(1) , x) + g(t)k0 [F (x̂1 , x, r) − x̂2 ],
µ2 x̂(2) + d1 µx̂(1) + x̂ = x,
(2)
(1)
µ2 x̂1 + d1 µx̂1 + x̂1 = x(1) ,
(2)
(1)
µ2 x̂2 + d1 µx̂2 + x̂2 = x(2) .
Substitution of the right member of the first equation into the last one yields
x(2) = f (x(1) , x) + g(t)k0 [F (x̂1 , x, r) − x̂2 ],
µ2 x̂(2) + d1 µx̂(1) + x̂ = x,
(2)
(1)
µ2 x̂1 + d1 µx̂1 + x̂1 = x(1) ,
(2)
(14)
(1)
µ2 x̂2 + d1 µx̂2 + [1 + g(t)k0 ]x̂2 = f (x(1) , x) + g(t)k0 F (x̂1 , x, r).
From (14), taking µ = 0, we get the discussed above SMS (12). The FMS of the extended
system (14) is given by
µ2 x̂(2) + d1 µx̂(1) + x̂ = x,
(2)
(1)
µ2 x̂1 + d1 µx̂1 + x̂1 = x(1) ,
(2)
(1)
µ2 x̂2 + d1 µx̂2 + [1 + g(t)k0 ]x̂2 = f (x(1) , x) + g(t)k0 F (x̂1 , x, r)
Student Solutions Manual for “Design of nonlinear control systems . . .”
10
where we assume that f = const, g = const. The behavior of x̂(2) is described by the last
differential equation, where the characteristic equation is given by µ2 s2 + d1 µs + γ = 0.
Note that γ = 1 + g(t)k0 , k0 = 40, d1 = 3, and µ = 0.1 s. Hence, γmax = 89, γmin = 9.
By taking into account (9), we obtain the degree of time-scale separation between FMS
√
and SMS given by η3 ≥ γmin /µ = 30.
Finally, run the Simulink program e3 1.mdl to perform a numerical simulation of the
closed-loop system (13). The simulation results are shown in Fig. 4.
Figure 4: Simulation results of the closed-loop system (13).
Exercise 3.2 A system is given by (8). Consider the control law in the form of (10) with
the desired dynamics given by (9) and the real differentiating filter (11), where k0 = 40
and q = 2. Determine the parameters µ and d1 of (11) such that the damping ratio
exceeds 0.5 in the FMS and the degree of time-scale separation between FMS and SMS
exceeds 10. Compare with simulation results.
Solution.
By following through solution of Exercise 3.1, we get the closed-loop system equations,
SMS and FMS equations as well, where
µ2 s2 + d1 µs + γ = 0
(15)
is the characteristic equation of the FMS for x̂(2) where
γ = 1 + g(t)k0 , g(t) = 1.2 − cos(t), k0 = 40, γmax = 89, γmin = 9.
If d21 − 4γ < 0 when γ = γmax , then from (15) we obtain
q
|d21 − 4γ|
d1
d1
s1,2 = −
±j
=⇒ ζF M S = √ =⇒
2µ
2µ
2 γ
d1
, ζFmin
= 0.5 =⇒ d1 = 9.4340.
ζFmin
= √
MS
MS
2 γmax
Next, let us find estimates for µ based on different notions for degree of time-scale separation between FMS and SMS in the closed-loop system. Let us take γ = γmin , then
µ2 s2 + d1 µs + γmin =⇒ s2 +
1 F MS
1 F MS
a1 s + 2 a0
µ
µ
Student Solutions Manual for “Design of nonlinear control systems . . .”
F MS
where a1
F MS
= d1 , a0
11
= γmin = 9, and the state matrix of the FMS is given by
"
A22 =
AF M S = µ−1 A22 ,
#
#
"
0
1
0
1
.
=
−9 −9.4340
−γmin −d1
Since d21 − 4γmin > 0, then from (15) we obtain
q
d1
s1 = −
+
2µ
d21 − 4γmin
q
d1
s2 = −
−
2µ
2µ
d21 − 4γmin
2µ
.
Hence, ωFmin
= |s1 |.
MS
From the reference model, we get
"
AS =
0
1
−1 −1.2
#
is the state matrix of the SMS, where
s2 + 1.2s + 1 = 0
is the characteristic equation of the SMS. Hence, we obtain
max = 0.6,
s1,2 = −0.6 ± j0.8 =⇒ ωSM
S
SM S
SM S
(a0
)1/2 = 1.
By solving the Lyapunov equations
PF A22 + AT22 PF = −QF ,
PS AS + ATS PS = −QS ,
where QF = I and QS = I, we obtain
"
PF =
1.0541 0.0556
0.0556 0.0589
#
"
,
PS =
1.4333
0.5
0.5
0.8333
#
.
Hence,
λmax (PF ) = 1.0572, λmin (PF ) = 0.0558,
λmax (PS ) = 1.7164, λmin (PS ) = 0.5502.
Finally, we get the following estimates for µ based on the various notions for degree of
time-scale separation between FMS and SMS in the closed-loop system, that are:
λmin (PS )
, η1 = 10 =⇒ µ = 0.052,
µλmax (PF )
ωFmin
d1
MS
η2 = max
=
max , η2 = 10 =⇒ µ = 0.1795,
ωSM S
2µωSM
S
√
F M S 1/2
γmin
(a0 )
η3 =
, η3 = 10 =⇒ µ = 0.3.
SM S 1/2 =
µ
µ(a0 )
η1 =
Student Solutions Manual for “Design of nonlinear control systems . . .”
12
Figure 5: Simulation results of the closed-loop system (13) for d1 = 9.4340 and µ = 0.052
s.
Run the Matlab program e3 2 Parameters.m to calculate d1 and the above estimates for
µ based on such criteria as η1 , η2 , and η3 . Next, run the Simulink program e3 2.mdl, to
get the step response of the closed-loop system for d1 = 9.4340 and µ = 0.052 s. The
simulation results are shown in Fig. 5.
Chapter 4
Exercise 4.2 The differential equation of a plant is
x(2) = x + x|x(1) | + {2 + sin(t)}u,
(16)
while that of the reference model is
x(2) = −3.2x(1) − x + 3.2r(1) + r.
Construct the control law in the form of
µq u(q) + dq−1 µq−1 u(q−1) + · · · + d1 µu(1) + d0 u
k0
= n {−T n x(n) − adn−1 T n−1 x(n−1) − · · · − ad1 T x(1) − x
T
+ bdρ τ ρ r(ρ) + bdρ−1 τ ρ−1 r(ρ−1) + · · · + bd1 τ r(1) + r}.
(17)
where q = 3. Determine the FMS and SMS equations from the closed-loop system equations.
Solution.
From (16), we have n = 2 and x(2) is the highest derivative of the output signal, where
x(2) = f (x(1) , x) + g(t)u
and f (x(1) , x) = x + x|x(1) | and g(t) = 2 + sin(t).
The reference model is given by x(2) = F (x(1) , x, r(1) , r), where
F (x(1) , x, r) = −3.2x(1) − x + 3.2r(1) + r.
Take q = 3 and consider the control law given by
µ3 u(3) + d2 µ2 u(2) + d1 µu(1) + d0 u = k0 {F (x(1) , x, r(1) , r) − x(2) },
(18)
Student Solutions Manual for “Design of nonlinear control systems . . .”
13
that is
µ3 u(3) + d2 µ2 u(2) + d1 µu(1) + d0 u = k0 {−x(2) − 3.2x(1) − x + 3.2r(1) + r}.
Then, the closed-loop system equations are given by
x(2) = f (x(1) , x) + g(t)u,
µ3 u(3) + d2 µ2 u(2) + d1 µu(1) + d0 u = k0 {F (x(1) , x, r(1) , r) − x(2) }.
Denote x1 = x, x2 = x(1) , u1 = u, u2 = µu(1) , and u3 = µ2 u(2) . From the above closed-loop
system equations, we obtain
d
x1
dt
d
x2
dt
d
µ u1
dt
d
µ u2
dt
d
µ u3
dt
= x2 ,
= f (x1 , x2 ) + g(t)u1 ,
= u2 ,
= u3 ,
(
= −d0 u1 − d1 u2 − d2 u3 + k0
)
d
F (x2 , x1 , r , r) − x2 .
dt
(1)
Substitution of the right member of the second equation into the last one yields the
closed-loop system equations in the following form:
d
x1
dt
d
x2
dt
d
µ u1
dt
d
µ u2
dt
d
µ u3
dt
= x2 ,
= f (x1 , x2 ) + g(t)u1 ,
= u2 ,
= u3 ,
(19)
= −{d0 + k0 g(t)}u1 − d1 u2 − d2 u3
n
o
+k0 F (x2 , x1 , r(1) , r) − f (x1 , x2 ) .
In order to find the FMS equations, let us introduce the new fast time scale t0 = t/µ into
the closed-loop system equations given by (19). We obtain
d
x1 = µx2 ,
dt0
d
x2 = µ{f (x1 , x2 ) + g(t)u1 },
dt0
d
u1 = u2 ,
dt0
Student Solutions Manual for “Design of nonlinear control systems . . .”
14
d
u2 = u3 ,
dt0
d
u3 = −{d0 + k0 g(t)}u1 − d1 u2 − d2 u3
dt0
n
o
+k0 F (x2 , x1 , r(1) , r) − f (x1 , x2 ) .
If µ → 0, then we get the FMS equations in the new time scale t0 , that is
d
x1
dt0
d
x2
dt0
d
u1
dt0
d
u2
dt0
d
u3
dt0
= 0,
= 0,
= u2 ,
= u3 ,
= −{d0 + k0 g(t)}u1 − d1 u2 − d2 u3
n
o
+k0 F (x2 , x1 , r(1) , r) − f (x1 , x2 ) .
Then, returning to the primary time scale t = µt0 , we obtain the following FMS equations:
x1
d
µ u1
dt
d
µ u2
dt
d
µ u3
dt
= const,
x2 = const,
= u2 ,
= u3 ,
(20)
= −{d0 + k0 g(t)}u1 − d1 u2 − d2 u3
n
o
+k0 F (x2 , x1 , r(1) , r) − f (x1 , x2 ) .
These equations may be rewritten as
µ3 u(3) + d2 µ2 u(2) + d1 µu(1) + {d0 + k0 g(t)}u
= k0 {F (x2 , x1 , r(1) , r) − f (x1 , x2 )},
(21)
where x1 = const, x2 = const, and g(t) = const during the transients in the FMS (21).
Next, by letting µ → 0 in (19), we find the SMS equations in the following form:
ẋ1 = x2 ,
ẋ2 = F (x2 , x1 , r(1) , r)
d0
+
{f (x1 , x2 ) − F (x2 , x1 , r(1) , r)}.
d0 + k0 g(t)
(22)
Student Solutions Manual for “Design of nonlinear control systems . . .”
15
At the same time, we can find the above SMS by some another way. Suppose the
FMS (20) is stable. Taking µ → 0 in (21) we get u(t) = us (t), where us (t) is a steady
state (more precisely, quasi-steady state) of the FMS (20) and
us =
k0
{F (x2 , x1 , r(1) , r) − f (x1 , x2 )}.
d0 + k0 g(t)
Substitution of us into (18) yields the SMS equation given by
x(2) = F (x(1) , x, r(1) , r)
d0
+
{f (x, x(1) ) − F (x(1) , x, r(1) , r)},
d0 + k0 g(t)
which is the same as (22).
Chapter 5
Exercise 5.1 The differential equation of a plant model is given by
x(2) = x + x|x(1) | + {1.5 + sin(t)}u.
(23)
Assume that the specified region is given by the inequalities |x(t)| ≤ 2, |x(1) (t)| ≤ 10, and
|r(t)| ≤ 1, where t ∈ [0, ∞). The reference model for x(t) is chosen as x(2) = −2x(1) −x+r.
Determine the parameters of control law to meet the requirements: εF = 0.05, εr = 0.02,
ζF M S ≥ 0.5, η3 ≥ 20, q = 2. Compare simulation results with the assignment. Note that
η3 is the degree of time-scale separation between stable fast and slow motions defined by
F MS
(a
)1/m
η3 = 0 SM S 1/n .
µ(a0 )
Solution.
Consider the system given by (23). Then n = 2 and x(2) = f (x(1) , x) + g(t)u, where
(1)
f (x , x) = x + x|x(1) | and g(t) = 1.5 + sin(t).
We have n = 2 and x(2) is the highest derivative of the output signal. The reference
model is given by x(2) = F (x(1) , x, r), where F (x(1) , x, r) = −2x(1) − x + r. As far as
the requirement on the high frequency sensor noise attenuation is not specified, then take
q = n = 2. Therefore, consider the control law given by
µ2 u(2) + d1 µu(1) + d0 u = k0 {F (x(1) , x, r) − x(2) },
(24)
µ2 u(2) + d1 µu(1) + d0 u = k0 {−x(2) − 2x(1) − x + r}.
(25)
that is
Consider the closed-loop system equations given by
x(2) = f (x(1) , x) + g(t)u,
µ2 u(2) + d1 µu(1) + d0 u = k0 {F (x(1) , x, r) − x(2) }.
(26)
(27)
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16
From the above closed-loop system equations, we get the FMS given by
µ2 u(2) +d1 µu(1) +[d0 + k0 g]u = k0 {F (x(1) , x, r)−f (x(1) , x)},
(28)
where F = const, f = const, and g = const during the transients in (28), as well as the
SMS given by
x(2) = F (x(1) , x, r) +
d0
{f (x(1) , x) − F (x(1) , x, r)}.
d0 + k0 g(t)
(29)
We have that the region of x, x(1) , r is specified by the inequalities |x(t)| ≤ 2, |x(1) (t)| ≤ 10,
|r(t)| ≤ 1. Hence, we obtain
fmax = |x + x|x(1) ||max = 2 + 2 · 10 = 22,
Fmax = | − 2x(1) − x + r|max = 2 · 10 + 2 + 1 = 23,
gmin = 0.5, gmax = 2.5,
F
emax = εF Fmax = 0.05 · 23 = 1.15.
We have g(t) > 0 ∀ t. Hence, take k0 > 0. We have εr = 0.02 6= 0. Hence, take d0 = 1.
From the requirement |eF (us )| ≤ eFmax = 1.15, we obtain

|k0 | ≥
max |F (X, R) − f (X, w)|
d0  ΩX,R,w
gmin


− 1
eFmax
·

¸
1 23 + 22
− 1 ≈ 90.
=
0.5
1.15
Consider a steady state of the SMS, that is x(2) = x(1) = 0. Hence, we obtain
d0
{f (x(1) , x) − F (x(1) , x, r)} =⇒
d0 + k0 g
(2)
(1)
x
− x + r}
|{z} = |−2x {z
x(2) = F (x(1) , x, r) +
=0
=es
d0
+
{x + x|x(1) | −{−2x(1){z− x + r}}} =⇒
d0 + k0 g | {z } |
s
=e
=xs =r−es
es = −
d0
r.
k0 g − d0
From the requirement
¯
¯
¯
¯
d0
d0 rmax
¯
|e | ≤ ¯−
r¯¯ ≤
≤ esmax = εr rmax = 0.02,
¯ k0 g − d0 ¯
k0 gmin − d0
s
we obtain
"
#
·
¸
1
d0 rmax
1·1
1
|k0 | ≥
− d0
=
−1
= 98.
s
emax
gmin
0.02
0.5
Student Solutions Manual for “Design of nonlinear control systems . . .”
17
Let us take k0 = 100.
From the SMS and reference model equations, it follows that
s2 + 2s + 1 = 0
SM S
is the characteristic equation of the SMS, where a0
= 1, and
µ2 s2 + d1 µs + d0 + k0 g = 0
is the characteristic equation of the FMS. Hence, we obtain
q
√
F M S 1/2
d0 + k0 gmin
(a0 )
d0 + k0 g
=
≥
≥ η3min = 20
η3 =
SM S
SM S
SM S
µ(a0 )1/2
µ(a0 )1/2
µ(a0 )1/2
q
√
d0 + k0 gmin
1 + 100 · 0.5
=⇒ µ ≤
=
≈ 0.3571 s.
SM S 1/2
min
20 · 1
η3 (a0 )
From the characteristic equation of the FMS, we get
q
F MS
s1,2
d21 − 4(d0 + k0 g)
d1
±j
= α ± jβ,
=−
2µ
2µ
where we assume that d21 − 4(d0 + k0 g) < 0 when g = gmax . Hence, we can find
q
ζF M S = cos(θF M S ) = |α|/ α2 + β 2
d1
= √
≥ ζFmin
= 0.5 =⇒
MS
2 d0 + k0 g
q
d0 + k0 gmax
d1 ≥ 2ζFmin
MS
√
= 2 · 0.5 1 + 100 · 2.5 ≈ 15.84.
Take µ = 0.3 s and d1 = 16.
Control law implementation. The discussed control law (25) can be rewritten in the
form given by
d1 (1) d0
u + 2u
µ
µ
d
k0 a
k0
k0
k0
= − 2 x(2) − 2 1 x(1) − 2 2 x + 2 2 r =⇒
µ
µT
µT
µT
(2)
(1)
(2)
(1)
u + a1 u + a0 u = b2 x + b1 x + b0 x + c0 r
u(2) +
where
a1 =
b2 = −
k0
,
µ2
d1
,
µ
a0 =
d0
,
µ2
k0 ad1
k0
, b0 = − 2 2 ,
2
µT
µT
k0
c0 = 2 2 .
µT
b1 = −
(30)
Student Solutions Manual for “Design of nonlinear control systems . . .”
18
Then, in order to find the block diagram of the discussed control law, from (30), we get
u(2) − b2 x(2) + a1 u(1) − b1 x(1)
= −a0 u + b0 x + c0 r =⇒
|
{z
}
=u̇2
u(1) − b2 x(1) + a1 u − b1 x = u2 =⇒
u(1) − b2 x(1) = u
− a1{zu + b1 x} =⇒
|2
=u̇1
u = u1 + b2 x.
Hence, we obtain the equations of the controller given by
u̇1 = u2 − a1 u + b1 x,
u̇2 = −a0 u + b0 x + c0 r.
u = u1 + b2 x.
(31)
From (31), we obtain the block diagram of the controller as shown in Fig. 6.
Figure 6: Block diagram of (30) represented in the form (31).
In conclusion, run the Matlab program e5 1 Parameters.m to calculate the controller
parameters. Next, run the Simulink program e5 1.mdl, to get the step response of the
closed-loop system.4 It can be verified that the simulation results confirm the analytical
calculations. The simulation results are shown in Fig. 7.
Note that the control law (25) may be expressed in terms of transfer functions as
u(s) =
k0 (s2 + 2s + 1)
k0
r(s)
−
x(s).
µ2 s2 + d1 µs + d0
µ2 s2 + d1 µs + d0
(32)
Take d0 = 0, then from (32) the conventional PID controller with low-pass filtering
½
u(s) =
¾
1
1
k −2x(s) + [r(s) − x(s)] − sx(s)
τLP F s + 1
s
results, where the low-pass filter (LPF) is given by 1/(τLP F s + 1) and
k=
k0
,
µd1
τLP F =
µ
.
d1
4
Throughout the simulation the following solver options are used: variable-step, ode113(Adams),
relative tolerance equals 1e-6.
Student Solutions Manual for “Design of nonlinear control systems . . .”
19
Figure 7: Simulation results of the closed-loop system given by (23) and (31) for k0 = 100,
d1 = 15, d0 = 1 and µ = 0.3 s.
Exercise 5.10 The differential equation of a plant model is given by
x(2) = 2x(1) + x + 2u,
(33)
where y(t) = x(t). Determine the parameters of the control law such that εr = 0, tds ≈ 1
s, σ d ≈ 10 %, ζF M S ≥ 0.3, and η3 ≥ 10. The additional requirement
|Guns (jω)| ≤ εuns (ω),
ns
∀ ω ≥ ωmin
(34)
ns
= 103 rad/s. Compare simulation
should be provided such that εuns (ω) = 103 and ωmin
results with the assignment.
Solution.
Reference model. From (33), we have x(2) = f (x, x(1) )+gu, where f (x, x(1) ) = 2x(1) +x
and g = gmin = gmax = 2. We have n = 2 and x(2) is the highest derivative of the output
signal. Hence, consider the reference model given by
x(2) = F (x(1) , x, r).
Take tds = 1 s, σ d = 10 %, then by
Ã
d
θ = tan
−1
!
π
,
ln(100/σ d )
ωd =
4
,
tds
we get θd = 0.9383 rad, ζ d = 0.5912, ω d = 4 rad/s and ωn = 6.7664 rad/s. By selecting
the 2 roots s1,2 = −4 ± j5.4575, where Re(s1,2 ) = −ω d = −ωn cos(θd ) and |Im(s1,2 )| =
ωn sin(θd ), we obtain the desired characteristic polynomial s2 + 8s + 45.78. Hence, the
desired transfer function is given by
Gdxr (s) =
s2
45.78
+ 8s + 45.78
and, from the above, the reference model in the form of the type 1 system
x(2) = −8x(1) − 45.78x + 45.78r
follows.
(35)
Student Solutions Manual for “Design of nonlinear control systems . . .”
20
Control law of the 2-nd order. At the beginning, let us take q = 2. Therefore, the
control law will be constructed in the form (24), where the reference model is given by
(35). Hence, the control law is
µ2 u(2) + d1 µu(1) + d0 u
= k0 {−x(2) − 8x(1) − 45.78x + 45.78r}.
(36)
The closed-loop system equations are given by (26)–(27). Hence, the FMS and SMS
equations are given by (28) and (29), respectively.
Selection of control law parameters, when q = 2. The control law parameters k0 , d0 , d1 , µ
can be selected by following through solution of Exercise 5.1, if d0 = 1.
Let us consider a simplified version for the gain k0 selection. In order to provide the
requirement εr = 0, take d0 = 0. Then the gain k0 can be selected such that k0 gmin = 10.
Hence, we get k0 = 5.
From (28), (29), and (35), we have that
s2 + 8s + 45.78 = 0
SM S
is the characteristic equation of the SMS, where a0
= 45.78, and
µ2 s2 + d1 µs + d0 + k0 g = 0
is the characteristic equation of the FMS. Hence, we obtain
q
√
F MS
d0 + k0 gmin
(a0 )1/2
d0 + k0 g
η3 =
=
≥
≥ η3min = 10
SM S
SM S
SM S
µ(a0 )1/2
µ(a0 )1/2
µ(a0 )1/2
q
√
d0 + k0 gmin
0+5·2
=⇒ µ ≤
=
≈ 0.04673 s.
SM S 1/2
min
10 · 6.766
η3 (a0 )
From the characteristic equation of the FMS, we get
q
F MS
s1,2
d21 − 4(d0 + k0 g)
d1
=−
±j
= α ± jβ,
2µ
2µ
where we assume that the FMS is underdamped-stable, that is
d21 − 4(d0 + k0 g) < 0
when g = gmax . Hence, we can find
q
ζF M S = cos(θF M S ) = |α|/ α2 + β 2
d1
= √
≥ ζFmin
= 0.3 =⇒
MS
2 d 0 + k0 g
q
√
min
d1 ≥ 2ζF M S d0 + k0 gmax = 2 · 0.3 0 + 5 · 2 ≈ 1.8974.
Student Solutions Manual for “Design of nonlinear control systems . . .”
21
High-frequency sensor noise attenuation, when q = 2. Let us replace x(t) by y(t) =
x(t) + ns (t). Then, from the above, we can obtain that
Ad (s)
DF M S (s)
(1/45.78)s2 + (8/45.78)s + 1
5 · 45.78
=
· 2
d0 + k0 g [µ /(d0 + k0 g)]s2 + [d1 µ/(d0 + k0 g)]s + 1
Guns (s) = kuns
is the input sensitivity function with respect to noise for high frequencies, where the requirement on high-frequency sensor noise attenuation is given by the following inequality:
ns
∀ ω ≥ ωmin
= 103 rad/s.
(37)
ns
Take µ = 0.04673 s and d1 = 1.8974, then |Guns (jωmin
)| ≈ 2208 (where 20 lg 2208 ≈
66.88 dB), or, for the sake of simplicity, we can find the limit given by
|Guns (jω)| ≤ εuns = 103 ,
lim |Guns (jω)| =
ω→∞
|k0 |
≈ 2289,
µq
where 20 lg 2289 ≈ 67.19 dB. Hence, the requirement (37) on high-frequency sensor noise
attenuation doesn’t hold. The Bode plots of Guns (jω) and Guf (jω) are shown in Fig. 8.
Figure 8: The Bode plots of Guns (jω) and Guf (jω).
Note, the same conclusion can be obtained by inspection the Bode amplitude plot of
Guf (jω), where
1
,
DF M S (s)
k0
kuf =
,
d 0 + k0 g
µ2
d1 µ
DF M S (s) =
s2 +
s + 1.
d0 + k0 g
d0 + k0 g
Guf (s) = kuf
Student Solutions Manual for “Design of nonlinear control systems . . .”
22
Hence, we get
ns
ns
Luf (ωmin
) = 20 lg |Guf (jωmin
)| ≈ −53 dB
and
HF A
ns
ns
Lmax (ωmin
= −60 dB,
) = 20 lg εns − 20[n + ϑ] lg ωmin
where ϑ = 0. Hence, the requirement for high-frequency sensor noise attenuation
HF A
ns
ns
Luf (ωmin
) ≤ Lmax (ωmin
)
doesn’t hold.
Run the Matlab program e5 10 A Parameters.m to calculate the reference model
parameters, Bode plots of Guns (jω) and Guf (jω), as well as parameters of the controller
given by (36), where q = 2. Next, run the Simulink program e5 10 a.mdl, to get a step
response as well as ramp response (by using Switch 1) of the closed-loop system. The
simulation results are shown in Fig. 9.
Figure 9: Simulation results of the closed-loop system (33), (36) for k0 = 5, d0 = 0,
d1 = 1.8974, and µ = 0.04673 s, where y(t) = x(t).
Note that, by Switch 2, the type of the reference model can be changed from 1 to 2
in the program e5 10 a.mdl.
By Switch 3, add the hign frequiency sensor noise ns (t) to the output y(t) = x(t) +
ns (t), where ns (t) = 10−3 sin(103 t). The simulation results are shown in Fig. 10.
Figure 10: Simulation results of the closed-loop system (33), (36) for k0 = 5, d0 = 0,
d1 = 1.8974, and µ = 0.04673 s in the presence of the noise ns (t), where y(t) = x(t)+ns (t).
Control law of the 3-rd order. In order to provide the requirement for high-frequency
sensor noise attenuation given by (37), let us take q = 3 and consider the control law
given by
µ3 u(3) + d2 µ2 u(2) + d1 µu(1) + d0 u = k0 {F (y (1) , y, r) − y (2) },
Student Solutions Manual for “Design of nonlinear control systems . . .”
23
where y(t) = x(t) + ns (t) and the reference model is the same as (35). Hence, the control
law can be rewritten as
µ3 u(3) +d2 µ2 u(2) +d1 µu(1) +d0 u = k0 {−y (2) −ād1 y (1) −ād0 y+ād0 r},
(38)
where
ād1 = 8,
ād0 = 45.78.
Note that the control law (38) may be expressed in terms of transfer functions as
u(s) =
k0 (s2 + ād1 s + ād0 )
k0 ād0
r(s)
−
x(s).
µ3 s3 + d2 µ2 s2 + d1 µs + d0
µ3 s3 + d2 µ2 s2 + d1 µs + d0
(39)
Take d0 = 0, then from (39) the conventional PID controller with low-pass filtering
(
)
ād
u(s) = 2 2
k −ād1 x(s) + 0 [r(s) − x(s)] − sx(s)
τlpf s + aLP F τLP F s + 1
s
1
2
results, where the low-pass filter is given by 1/(τLP
s2 + aLP F τLP F s + 1) and
F
k=
k0
,
µd1
µ
τLP F = √ ,
d1
aLP F τLP F =
µd2
.
d1
The FMS characteristic polynomial in the closed-loop system is given by
µ3 s3 + d2 µ2 s2 + d1 µs + d0 + k0 g.
Let us consider the selection of the control law parameters based on Bode amplitude plot
of the closed-loop FMS given by
Luf (ω) = 20 lg |Guf (jω)|,
where
k0
,
DF M S (s)
d0 + k0 g
µ3
d2 µ2 2
d1 µ
3
DF M S (s) =
s +
s +
s + 1.
d0 + k0 g
d0 + k0 g
d 0 + k0 g
Guf (s) = kuf
1
,
kuf =
(40)
By the same way as was shown above, take d0 = 0 and k0 = 5. Hence, kuf = 0.5. Then,
let us perform Gduf (s) in the corner frequency factored form given by
Gduf (s) = kuf
[T12 s2
1
.
+ 2ζ1 T1 s + 1][T2 s + 1]
Then, the roots of quadratic factor
T12 s2 + 2ζ1 T1 s + 1
(41)
Student Solutions Manual for “Design of nonlinear control systems . . .”
24
are the dominant poles of Gduf (s), where the damping ratio ζ1 is selected such that
ζ1 = ζFmin
= 0.3.
MS
Take tds ≈ tds,SM S and tds,F M S = tds /η, where η = 10. Then, we can obtain
4
ω1 ≈
tds,F M S
=
4η
ζ1 tds
4 · 10
1
= 0.0075 s.
≈ 133 =⇒ T1 =
ζ1 · 1
ω1
=
Let us calculate a lower bound for T2 from the condition
HF A
ns
ns
Lduf (ωmin
) = Lmax (ωmin
),
where
HF A
ns
ns
) = 20 lg εns − 20[n + ϑ] lg ωmin
Lmax (ωmin
= 20 lg 103 − 20[2 + 0] lg 103 = −60 dB.
ns
)n+ϑ = 10−3 . Hence, we get
Denote L = εns /(ωmin
ns
|Gduf (jωmin
, T2min )| = L =⇒
kuf
= L =⇒
ns
ns
2 ns
+ 1]|
]][jT2min ωmin
|[1 − T1 [ωmin ]2 + j2ζ1 T1 [ωmin
1/2

[kuf /L]2
T2min = ns 
− 1
ns
ns
)2
]2 )2 + (2ζ1 T1 ωmin
ωmin (1 − [T1 ωmin
1
"
=⇒
#1/2
1
[0.5/10−3 ]2
T2min = 3
−1
10 (1−[0.0075·103 ]2 )2 +(2·0.3·0.0075·103 )2
≈ 0.009 s.
The time constant T2 should be selected such that the inequalities T2min ≤ T2 ≤ T1 hold.
We see, there is apparent contradiction. Therefore, let us replace the degree of time-scale
separation between fast and slow modes η = 10 by η = 8 and redesign the parameter
T1 again. We get T1 = 0.0094 s. Accordingly, by the same way as above, we obtain
T2min = 0.0057 s. Hence, the condition T2min ≤ T2 ≤ T1 holds and then, we can take
T2 = T2min = 0.0057 s.
As a result of the above, we obtain
DFd M S (s) = [T12 s2 + 2ζ1 T1 s + 1][T2 s + 1]
= 4.9699 · 10−7 s3 + 1.197 · 10−4 s2 + 0.0113s + 1.
From (40), and by taking into account that d0 = 0 as well as the requirement
DF M S (s) = DFd M S (s),
Student Solutions Manual for “Design of nonlinear control systems . . .”
25
we obtain
µ = {dqd k0 g}1/3 ≈ 0.0171,
d1d [k0 g](2)/3
≈ 6.6097,
[d3d ]1/3
d d [k0 g](1)/3
d2 = 2 d 2/3 ≈ 4.1101.
[d3 ]
d1 =
(42)
Finally, the control law (38) can be rewritten in the form given by
d2 (2) d1 (1) d0
u + 2u + 3u
µ
µ
µ
d
k0
k0 ā
k0
k0
= − 3 y (2) − 3 1 y (1) − 3 2 y + 3 2 r =⇒
µ
µT
µT
µT
(3)
(2)
(1)
(2)
u +a2 u +a1 u +a0 u = b2 y +b1 y (1) +b0 y+c0 r.
u(3) +
(43)
From (43), we can obtain the equations of the controller given by
u̇1
u̇2
u̇3
u
=
=
=
=
a2 =
d2
,
µ
u2 − a2 u1 + b2 y,
u3 − a1 u1 + b1 y,
−a0 u1 + b0 y + c0 r,
u1 ,
(44)
where
b2 = −
k0
,
µ3
b1 = −
a1 =
k0 ād1
,
µ3
d1
,
µ2
a0 =
b0 = −
d0
,
µ3
k0 ā0
,
µ3
c0 =
k0 ā0
.
µ3
The Bode plots of Guns (jω) and Guf (jω) are shown in Fig. 11, where the parameters
µ, d1 , d2 are given by (42) with d0 = 0 and k0 = 5.
In conclusion, run the Matlab program e5 10 B Parameters.m to calculate the reference model parameters, Bode plots of Guns (jω), and Guf (jω), as well as the parameters
of the controller given by (44), where q = 3. Next, run the Simulink program e5 10 b.mdl,
to get a step response as well as ramp response (by using Switch 1) of the closed-loop
system. By Switch 3, add the hign frequiency sensor noise ns (t) to the output, that is
y(t) = x(t) + ns (t), where ns (t) = 10−3 sin(103 t). The simulation results are shown in
Fig. 12. It can be verified that the simulation results confirm the analytical calculations.
Note that in the program e5 10 b.mdl, by Switch 2, the type of the reference model
can be changed from 1 to 2. Then, instead of (43), we have the controller given by
u(3) + a2 u(2) + a1 u(1) + a0 u
= b2 y (2) + b1 y (1) + b0 y + c1 r(1) + c0 r,
(45)
Student Solutions Manual for “Design of nonlinear control systems . . .”
26
Figure 11: The Bode plots of Guns (jω) and Guf (jω).
Figure 12: Simulation results of the closed-loop system (33), (44) for k0 = 5, d0 = 0,
µ = 0.0171 s, d1 = 6.6097, d2 = 4.1101 in the presence of the noise ns (t), where y(t) =
x(t) + ns (t).
where c1 = −b1 and c0 = −b0 . Hence, instead of (44), we get
u̇1
u̇2
u̇3
u
=
=
=
=
u2 − a2 u1 + b2 y,
u3 − a1 u1 + b1 y + c1 r,
−a0 u1 + b0 y + c0 r,
u1 .
(46)
From (46), we can obtain the block diagram of the controller as shown in Fig. 13.
Chapter 6
Exercise 6.1 The plant model is given by
x(2) (t) = 0.5x(t)x(1) (t) + 0.1x(1) (t)
+0.5x(t) + 0.5 sin(0.5t) + gu(t − τ ),
(47)
Student Solutions Manual for “Design of nonlinear control systems . . .”
27
Figure 13: Block diagram of (45) represented in the form (46).
where g = 1 and the reference model x(2) = F (x(1) , x, r) is assigned by
x(2) = T −2 {−a1 T x(1) − x + r}.
(48)
The control law has the form
µ2 u(2) + d1 µu(1) + d0 u = k0 {F (x(1) , x, r) − x(2) },
(49)
where T = 1 s, a1 = 2, k0 = 10, µ = 0.1 s, d0 = 0, d1 = 4. Determine the region of
stability for τ of the FMS. Compare with simulation results of the closed-loop system.
Solution.
The closed-loop system equations are given by
x(2) = f (x(1) , x, t) + gu(t − τ ),
µ2 u(2) + d1 µu(1) + d0 u = k0 {F (x(1) , x, r) − x(2) },
where f (x(1) , x, t) = 0.5x(t)x(1) (t) + 0.1x(1) (t) + 0.5x(t) + 0.5 sin(0.5t), F (x(1) , x, r) =
T −2 {−ad1 T x(1) − x + r}, g = 1, T = 1 s, ad1 = 2, k0 = 10, µ = 0.1 s, d0 = 0, and d1 = 4.
Hence, from the above equations, we get the FMS given by
µ2 u(2) +d1 µu(1) +d0 u+k0 gu(t−τ ) = k0 {F (x(1) , x, r)−f (x(1) , x)},
(50)
where F = const and f = const during the transients in (50). The block diagram
representation of the FMS (50) is shown in Fig. 14, where
D(µs) = µ2 s2 + d1 µs + d0 .
(51)
Figure 14: Block diagram of the FMS (50) with delay τ , where F = const, f = const.
Student Solutions Manual for “Design of nonlinear control systems . . .”
28
By the Nyquist stability criterion, the FMS (50) is marginally stable if ωc 6= 0 exists
such that the condition
k0 ge−jτm ωc
= −1 + j0,
(52)
D(jµωc )
holds, where ωc is the crossover frequency and τm is an upper bound for delay τ . The
value τm determines the region of stability for τ of the FMS.
From (52) and by taking into account the condition d0 = 0, we get
¯
¯
¯
k0 ge−jτm ωc ¯¯
¯
¯
¯ = 1 =⇒
¯ jµωc (jµωc + d1 ) ¯
µ4 ωc4 + d21 µ2 ωc2 − k02 g 2 = 0 =⇒
y = ωc2 > 0, µ4 y 2 + d21 µ2 y − k02 g 2 = 0 =⇒
y=
=
−42 +
−d21 +
√
(53)
q
d41 + 4k02 g 2
2µ2
44 + 4 · 102 · 12
√
≈
481
=⇒
ω
=
y ≈ 22 rad/s.
c
2 · 0.12
From (52), we have5
τm = [π/2 − tan−1 (µωc /d1 )]/ωc .
(54)
Hence, the region of stability for τ of the FMS is defined as
0 < τ < τm = 0.049 s.
Finally, the discussed control law can be rewritten in the form given by
d1 (1) d0
u + 2u
µ
µ
d
k0
k0 a
k0
k0
= − 2 x(2) − 2 1 x(1) − 2 2 x + 2 2 r =⇒
µ
µT
µT
µT
(2)
(1)
(2)
(1)
u + a1 u + a0 u = b2 x + b1 x + b0 x + c0 r,
u(2) +
where
a1 =
d0
k0
k0 ad
k0
k0
d1
, a 0 = 2 , b 2 = − 2 , b 1 = − 2 1 , b 0 = − 2 2 , c0 = 2 2 .
µ
µ
µ
µT
µT
µT
Run the Matlab program e6 1 Parameters.m to calculate the region of stability for τ of
the FMS and the controller parameters. Next, run the Simulink program e6 1.mdl, to get
the step response of the closed-loop system.
Exercise 6.4 Determine the phase margin and gain margin of the FMS based on the
input data of Exercise 6.1 for the time delay τ = 0.3τm , where τ = τm corresponds to the
marginally stable FMS.
5
y = tan−1 (x) denotes the arctangent of x, i.e., tan(y) = x.
Student Solutions Manual for “Design of nonlinear control systems . . .”
29
Solution.
By following through the solution of Exercise 6.1, the block diagram representation
of the FMS (50) is shown in Fig. 14, where the phase margin (PM) of the FMS (50) is
given by
P M = π − ArgD(jµωc ) − τ ωc .
(55)
From (51) and d0 = 0, we get
ArgD(jµωc ) =
π
+ tan−1 (µωc /d1 ).
2
(56)
Hence, we obtain
π
− tan−1 (µωc /d1 ) − τ ωc .
2
In order to find the gain margin (GM) of the FMS, consider the equation
PM =
O
Arg[GF M S (jωπ )] = −π.
(57)
(58)
From (58), we get
π
− tan−1 (µωπ /d1 ) − τ ωπ = 0,
(59)
2
where τ, τm , ωπ and ωc are calculated by joint numerical resolution of (53), (54), and (59).
Finally, we can find that
lπ
¯
¯
¯
¯
k0 ge−jτ ωπ
¯
¯
= |GF M S (jωπ )| = ¯
¯
¯ jµωπ (jµωπ + d1 ) ¯
¯
¯
¯
¯
k0 g
¯
¯
¯
= ¯
¯ jµωπ (jµωπ + d1 ) ¯
O
k0 g
= q
(µωπ )4 + (d1 µωπ )2
Hence, the gain margin (GM) of the FMS is given by GM = 1/lπ .
By running the Matlab program e6 4 Parameters.m, we obtain
ωc ≈ 22 rad/s, ωπ ≈ 47.7 rad/s, τm ≈ 0.049 s,
τ = 0.3τm ≈ 0.0146 s, P M = 0.7486 rad, GM ≈ 2.9685.
Next, run the Simulink program e6 4.mdl, to get the step response of the closed-loop
system where τ = 0.0146 s.
Chapter 7
Exercise 7.1 The differential equations of a plant model are given by
ẋ1
ẋ2
ẋ3
y
=
=
=
=
x1 + x2 ,
x1 + x2 + x3 + u,
2x1 − x2 + 2x3 + a u,
x1 .
(60)
Student Solutions Manual for “Design of nonlinear control systems . . .”
30
Verify the invertibility and internal stability of the given system (60), where (a) a = 1,
and (b) a = 3. Find the degenerated system.
Solution.
From (60), we obtain
ẏ = x1 + x2 =⇒ ÿ = ẋ1 + ẋ2 =⇒ ÿ = 2x1 + 2x2 + x3 + u.
Hence, the system (60) is invertible and the relative degree is given by α = 2. Let us
introduce a state-space transformation defined by
y 1 = y = x1 ,
y2 = ẏ = x1 + x2 ,
z = x3
(61)
x1 = y1 ,
x2 = y2 − y1 ,
x3 = z.
(62)
From (61), we have
By the change of variables (62), we get the normal form of (60) given by
ẏ1
ẏ2
ż
y
=
=
=
=
y2 ,
2y2 + z + u,
y1 + y2 + 2z + a u,
y1 .
(63)
Let the desired stable output behavior is defined by ÿ = F (ẏ, y, r), which can be rewritten
as ÿ1 = F (ẏ1 , y1 , r) or ẏ1 = y2 , ẏ2 = F (y2 , y1 , r). In order to find the equations of the
internal subsystems, take F (y2 , y1 , r) = 2y2 +z +u. Hence, we obtain the inverse dynamics
solution given by
uid = F (y2 , y1 , r) − 2y2 − z.
Substitution of u = uid into (63) yields
ẏ1
ẏ2
ż
y
=
=
=
=
y2 ,
F (y2 , y1 , r),
(2 − a)z + y1 + (1 − 2a)y2 + aF (y2 , y1 , r),
y1 ,
(64)
where ż = (2−a)z +y1 +(1−2a)y2 +aF (y2 , y1 , r) is the equation of the internal subsystem
and y2 , y1 , r are treated as bounded external disturbances of the internal subsystem. If
a = 1, then 2 − a > 0 and the unique equilibrium point of the internal subsystem is
unstable. If a = 3, then 2 − a < 0. Hence, the solutions of the internal subsystem are
bounded when variables y2 , y1 , r are bounded, that is the bounded-input-bounded-state
Student Solutions Manual for “Design of nonlinear control systems . . .”
31
(BIBS) stability of the internal subsystem. Next, from (64), by taking y1 = r = const
(hence, y2 = 0 and F (y2 , y1 , r) = 0), we find the degenerated system given by
ż = (2 − a)z + r,
where the unique equilibrium point of the degenerated subsystem is exponentially stable
when 2 − a < 0.
Exercise 7.10 Verify the invertibility and internal stability of the system
ẋ1 = x21 − x32 + u,
ẋ2 = |x2 | − u,
y = x1 ,
(65)
Assume that the inequalities |x1 (t)| ≤ 1.5, |x2 (t)| ≤ 1.5, |r(t)| ≤ 1 hold for all t ∈ [0, ∞).
Find the control law such that εr = 0, tds ≈ 3 s, σ d ≈ 0%. Run a computer simulation
of the closed-loop system with zero initial conditions. Compare simulation results of the
output response with the assignment for r(t) = 1, ∀ t > 0.
Solution.
Invertibility and internal stability. By the change of variables y = x1 and z = x2 , we
get
ẏ = y 2 − z 3 + u,
ż = |z| − u.
(66)
Hence, the system (66) is invertible and the relative degree is given by α = 1.
Let the desired stable output behavior is defined by ẏ = F (y, r). Take F (y, r) =
y 2 − z 3 + u. Hence, we obtain the inverse dynamics solution given by
uid = F (y, r) − y 2 + z 3 .
Substitution of u = uid into (66) yields
ẏ = F (y, r),
ż = f (z) + φ(y, r),
(67)
where φ(y, r) = y 2 − F (y, r) and f (z) = |z| − z 3 .
Denote c = φ(y, r) where c is an arbitrary real number. Take ż = 0 in the internal
subsystem given by
ż = f (z) + c.
(68)
Hence, from f (z) + c = 0, depending on value c, it can be up to three equilibrium points
of the internal subsystem. It can be easily verified, all solutions of the internal subsystem
(68) are bounded.
Control law. We have α = 1. Hence, y (1) is the highest derivative of the output signal.
As far as the requirement on the high frequency sensor noise attenuation is not specified,
then, for simplicity, take q = α = 1 and consider the control law given by
µu(1) + d0 u = k0 [F (y, r) − y (1) ],
Student Solutions Manual for “Design of nonlinear control systems . . .”
32
where the reference model is
y (1) = F (y, r).
Take tds = 3 s, σ d = 0 %, then we get θd = 0 rad, ζ d = 1, a = ω d = ωn = 1.3333 rad/s.
By selecting the root s1 = −a, we obtain the desired characteristic polynomial s + a.
Consider the desired transfer function given by
Gdyr (s) =
a
.
s+a
(69)
Hence, we get the reference model given by
y (1) = −ay + ar.
Finally, the control law is
µu(1) + d0 u = k0 [−y (1) − ay + ar].
(70)
Closed-loop system. The closed-loop system equations are given by
ẏ = y 2 − z 3 + u,
ż = |z| − u,
µu̇ + d0 u = k0 [F (y, r) − ẏ].
(71)
Denote f (y, z) = y 2 − z 3 . From (71), by following through solution of Exercise 4.2, we
obtain the FMS equation, that is
µu̇ + [d0 + k0 ]u = k0 [F (y, r) − f (y, z)],
where F = const, f = const during the transients in (72), and
µs + d0 + k0 g = 0
is the characteristic equation of the FMS, where g = 1.
Suppose the FMS (72) is stable. Then, in order to provide the requirement εr = 0,
we take d0 = 0. Next, taking µ → 0 in (72) we get u(t) = us (t), where us (t) is a steady
state (more precisely, quasi-steady state) of the FMS (72) and
us = uid = F (y, r) − f (y, z).
Substitution of us into the equation of the plant model (67) yields the SMS.
Selection of control law parameters. Let us consider a simplified version for the gain
k0 selection. We have d0 = 0, then the gain k0 can be selected such that k0 gmin = 10,
where gmin = gmax = g = 1. Hence, we get k0 = 10.
From (69), we have that the natural frequency of the reference model is given by
d
ωn = 1.3333 rad/s. Denote
max = ω d = 1.3333 rad/s,
ωSM
n
S
ωFmin
=
MS
d0 + k0 gmin
rad/s.
µ
Student Solutions Manual for “Design of nonlinear control systems . . .”
33
Then, without the taking into account the rate of dynamics of the internal subsystem, let
us consider the ratio
ωFmin
MS
η2 = max
(72)
ωSM S
as a criterion for the degree of time-scale separation between fast and slow motions. Take
η2min = 20. Hence, we obtain
η2 =
d0 + k0 gmin
min = 20 =⇒
max ≥ η2
µωSM
S
d0 + k0 gmin
µ ≤ µmax =
η min ω max
2
SM S
0 + 10 · 1
=
≈ 0.375 s.
20 · 1.3333
As a result, take µ = 0.375 s.
Control law implementation. Finally, the control law (70) can be rewritten in the
form given by
d0
k0
k0
k0
u = − y (1) −
y+
r =⇒
µ
µ
µT
µT
u(1) + a0 u = b1 y (1) + b0 y + c0 r =⇒ u(1) − b1 y (1) = −a0 u + b0 y + c0 r,
u(1) +
|
{z
}
(73)
=u̇1
where T = 1/a. From (73), we obtain the equations of the controller given by
u̇1 = −a0 u + b0 y + c0 r,
u = u1 + b1 y,
(74)
where
a0 =
d0
,
µ
b1 = −
k0
,
µ
b0 = −
k0
,
µT
c0 =
k0
.
µT
From (74), we obtain the block diagram of the controller as shown in Fig. 15. Run the
Figure 15: Block diagram of (73) represented in the form (74).
Matlab program e7 10 Parameters.m to calculate the reference model parameter T , as
Student Solutions Manual for “Design of nonlinear control systems . . .”
34
well as the parameters k0 , µ of the controller given by (70). Next, run the Simulink
program e7 10.mdl, to get a step response of the closed-loop system with zero initial
conditions.
Chapter 8
Exercise 8.1 Verify the invertibility and internal stability of the system given by
ẋ1 = x1 + 3x2 + u1 + u2 ,
ẋ2 = x1 + x2 + u1 − 2u2 ,
y1 = x1 + x2 , y2 = −x1 + 2x2 .
(75)
Assume that the inequalities |xj (t)| ≤ 2 ∀ j and |r(t)| ≤ 1 hold for all t ∈ [0, ∞). Find
the control law of the form
(qi )
(q −1)
µqi i ũi
(1)
+ di,qi −1 µqi i −1 ũi i + · · · + di,1 µũi + di,0 ũi
= ki eFi , Ũi (0) = Ũi0 , i = 1, . . . , p,
u = K0 ũ, u = {u1 , u2 , . . . , up }T , ũ = {ũ1 , ũ2 , . . . , ũp }T ,
where
µi > 0,
(1)
ki > 0,
(qi −1) T
Ũi = {ũi , ũi , . . . , ũi
} ,
(76)
qi ≥ αi .
Provide the following requirements: εr1 = 0, εr2 = 0, tds1 ≈ 1 s, σ1d ≈ 0%, tds2 ≈ 3 s,
σ2d ≈ 0%. Compare simulation results for the step response of the closed-loop control
system with the assignment.
Solution.
Invertibility and internal stability. From (75), by following through solution of Exercise 7.1, we get
ẏ1 = ẋ1 + ẋ2 =⇒ ẏ1 = 2x1 + 4x1 + 2u1 − u2 ,
ẏ2 = −ẋ1 + 2ẋ2 =⇒ ẏ2 = x1 − x2 + u1 − 5u2 ,
where
"
∗
det G = det
2 −1
1 −5
#
= −9 6= 0.
Hence, the system (75) is invertible and the relative degrees are given by α1 = α2 = 1.
We have that α1 + α2 = n = 2. Then, the internal subsystem does not exist.
Reference model. By following through solution of Exercise 7.10, take tds1 = 1 s,
σ1d = 0 %, then we get θ1d = 0 rad, ζ1d = 1, a1 = ω1d = ω1,n = 4 rad/s. By selecting the root
s1 = −a1 , we obtain the desired characteristic polynomial s + a1 . The desired transfer
function Gdy1r1 (s) = y1 (s)/r1 (s) is given by
Gdy1r1 (s) =
a1
.
s + a1
Hence, we get the reference model for y1 given by
(1)
(1)
y1 = −a1 y1 + a1 r1 =⇒ y1 =
1
[r1 − y1 ],
T1
(77)
Student Solutions Manual for “Design of nonlinear control systems . . .”
35
where T1 = 1/a1 . By the same way as above, take tds2 = 3 s, σ2d = 0 %, then
(1)
(1)
y2 = −a2 y2 + a2 r2 =⇒ y2 =
1
[r2 − y2 ],
T2
(78)
where a2 = 1.3333 and T2 = 1/a2 . Hence, the reference model (77)–(78) has be constructed as
(1)
(1)
y1 = F (y1 , r1 ), y2 = F (y2 , r2 ).
(79)
Control law . Take q1 = q2 = 1, then the control law is
µ1 ũ1 + d1,0 ũ1 = k1 [−y1 − a1 y1 + a1 r1 ],
(1)
(1)
(80)
(1)
µ2 ũ2
(1)
k2 [−y2
(81)
(82)
+ d2,0 ũ2 =
− a2 y2 + a2 r2 ],
u1 = k11 ũ1 + k12 ũ2 ,
u2 = k21 ũ1 + k22 ũ2 .
Take
"
∗ −1
K0 = [G ]
=
k11 k12
k21 k22
#
"
=
0.5556 −0.1111
0.1111 −0.2222
#
.
Hence, the controller of the discussed 2I2O system consists of 2 separate linear controllers
generating the auxiliary controls ũ1 , ũ2 and accompanied by the matching matrix K0
where the linear controllers are described by (80) and (81), as well as the matrix K0 is
implemented by (82).
Fast-motion subsystem and slow-motion subsystem. From the closed-loop system
equations given by (75) and (80)–(82), by following through solution of Exercise 7.10
again, we obtain the FMS equations, that are
(1)
µ1 ũ1 + [d1,0 + k1 ]ũ1 = k1 [F (y1 , r1 ) − f1 (x1 , x2 )],
(83)
(1)
µ2 ũ2
(84)
+ [d2,0 + k2 ]ũ2 = k2 [F (y2 , r2 ) − f2 (x1 , x2 )],
where y1 = const, y2 = const, x1 = const, x2 = const, f1 = const, and f2 = const during
the transients in (83)–(84). Due to K0 = [G∗ ]−1 , the characteristic polynomial of the
FMS is factorized, this is
(µ1 s + d10 + k1 )(µ2 s + d20 + k2 ).
(85)
In order to provide the requirements εr1 = 0 and εr2 = 0, take d10 = 0, d20 = 0.
Suppose the FMS (83)–(84) is stable. Then, it easy to show, the SMS equations are the
same as the reference model given by (79) as µ1 → 0 and µ2 → 0.
Selection of control law parameters. Let us take k1 = 10 and k2 = 10. The natural
d
= 4 rad/s. Denote
frequency of the reference model for y1 is given by ω1,n
max = ω d = 4 rad/s,
ω1,
1,n
SM S
min =
ω1,
F MS
d1,0 + k1
rad/s.
µ1
Student Solutions Manual for “Design of nonlinear control systems . . .”
Then, let us consider the ratio
min
ω1,
F MS
max
ω1,
SM S
η2 =
36
(86)
as a criterion for the degree of time-scale separation between fast and slow motions in the
first input-output channal. Take η2min = 20. Hence, we obtain
η2 =
d1,0 + k1
max
µ1 ω1,
SM S
≥ η2min = 20 =⇒
d1,0 + k1
min
η
ω max
µ1 ≤ µ1,max =
2
1,SM S
0 + 10
=
= 0.125 s.
20 · 4
As a result, take µ1 = 0.125 s. By the same way, we get
µ2 ≤ µ2,max =
d2,0 + k2
min
η
ω max
2
2,SM S
0 + 10
=
≈ 0.375 s.
20 · 1.3333
As a result, take µ2 = 0.375 s.
Much more conservative selection of µ1 , µ2 is to take µi = µ ∀ i, where
µ ≤ µmax =
mini {di,0 + ki }
η min maxi {ω max }
2
i,SM S
0 + 10
=
= 0.125 s.
20 · 4
Control law implementation. Finally, the separate controllers given by (80) and (81)
can be rewritten as
(1)
ũi +
di,0
ki (1)
ki
ki
ũi = − yi −
yi +
ri =⇒
µi
µi
µi T i
µi T i
(1)
(1)
ũi + ai,0 ũi = bi,1 yi + bi,0 yi + ci,0 ri ,
(87)
where i = 1, 2. From (87), we obtain
(1)
ũi,1 = −ai,0 ũi + bi,0 yi + ci,0 ri ,
ũi = ũi,1 + bi,1 yi , i = 1, 2
(88)
where
ai,0 =
di,0
,
µi
bi,1 = −
ki
,
µi
bi,0 = −
ki
,
µ i Ti
ci,0 =
ki
.
µ i Ti
From (88), the block diagram follows, which is similar to the block diagram as shown in
Fig. 15 (see p. 33).
Student Solutions Manual for “Design of nonlinear control systems . . .”
37
Run the Matlab program e8 1 Parameters.m to calculate the reference model parameters Ti , as well as the parameters ki , µi , and kij of the controller given by (80)–(82)
for η2 = 20. Next, run the Simulink program e8 1.mdl, to get a step response of the
closed-loop system with zero initial conditions. Make calculations and simulations for the
degree of time-scale separation between fast and slow motions assigned as η2 = 5 and
η2 = 10. Compare the simulation results.
Exercise 8.2 Verify the invertibility and internal stability of the system given by
ẋ1 = x21 + x1 x2 + 0.5u1 − [1 + 0.2 sin(t)]u2 ,
ẋ2 = x1 + sin(x2 ) − u1 − 2[1 + 0.5 sin(2t)]u2 ,
y 1 = x1 − x2 , y 2 = x1 + x2 .
(89)
Assume that the inequalities |xj (t)| ≤ 2 ∀ j and |r(t)| ≤ 1 hold for all t ∈ [0, ∞). Find
the control law of the form (76) such that εr1 = 0, εr2 = 0, tds1 ≈ 3 s, σ1d ≈ 0%, tds2 ≈ 3
s, σ2d ≈ 0%. Compare simulation results for the step response of the closed-loop control
system with the assignment.
Solution.
Invertibility and internal stability. From (89), we get
ẏ1 = ẋ1 − ẋ2 =⇒
ẏ1 = − x1 + x1 x2 − sin(x2 ) + 1.5u1 + [1 − 0.2 sin(t) + sin(2t)]u2 ,
ẏ2 = ẋ1 + ẋ2 =⇒
2
ẏ2 = x1 + x1 + x1 x2 + sin(x2 ) − 0.5u1 − [3 + 0.25 sin(t) + sin(2t)]u2 ,
x21
where
"
∗
G =
g11 g12
g21 g22
#
and g11 = 1.5, g21 = −0.5, g12 = 1 − 0.2 sin(t) + sin(2t), g22 = −3 − 0.25 sin(t) − sin(2t).
By Matlab program e8 2 Parameters.m, we can find that
det G∗ (t) ∈ [−5.4, −2.6],
∀ t ∈ [0, ∞).
Hence, the invertibility condition det G∗ (t) 6= 0 holds ∀ t ∈ [0, ∞) and the relative degrees
of the system (89) are given by α1 = α2 = 1. We have that α1 + α2 = n = 2. Then, the
internal subsystem does not exist.
Reference model. By following through solution of Exercise 8.1, take tds1 = 3 s, σ1d =
0 %, tds2 = 3 s, σ2d = 0 %, then we get the reference model for y1 and y2 given by (77)–(78),
max = ω d rad/s.
where a1 = a2 = ω d = 1.3333 rad/s. Denote ωSM
S
Control law . Take q1 = q2 = 1, then the control law can be constructed in the form
(80)–(82). Denote ḡ12 and ḡ22 as the average values of g12 and g22 , respectively, where
ḡ12 = 1 and ḡ22 = −3. Denote
"
∗
Ḡ =
g11 ḡ12
g21 ḡ22
#
"
=
1.5
1
−0.5 −3
#
.
Student Solutions Manual for “Design of nonlinear control systems . . .”
Take, for instance,
∗ −1
K0 = [Ḡ ]
1
=
8
"
6
2
−1 −3
38
#
.
Fast-motion subsystem. From the closed-loop system equations given by (89) and
(80)–(82), we obtain the FMS equation, that is
µũ(1) + {D0 + K1 G∗ K0 }ũ = K1 {F − H ∗ },
(90)
where µ = diag{µ1 , µ2 }, K1 = diag{k1 , k2 }, D0 = diag{d10 , d20 }, ũ = [ũ1 , ũ2 ]T , and
F = const, H ∗ = const during the transients in the FMS (90). Assume that G∗ (t) is the
matrix with frozen parameters during the transients in (90).
If K0 = [G∗ (t)]−1 , then the characteristic polynomial of the FMS is factorized as
shown by (85). As far as K0 6= [G∗ (t)]−1 , then the characteristic equation of the FMS
can not be factorized on the two separate equations. Take, for simplicity, d10 = d20 = 0,
µ = µ1 = µ2 and k = k1 = k2 = 10, and denote µ̃ = µ/k, then the characteristic equation
of the FMS (90) is given by
µ̃2 s2 + a1 µ̃s + a2 ,
where
a1 = 1 − 0.125g12 − 0.375g22 ,
3
30
1
a2 = g12 g22 − g22 − g12 ,
32
64
8
and
g12 ∈ [−0.144, 2.144],
g22 ∈ [−1.82, −4.182],
∀ t ∈ [0, ∞).
Let µ̃ > 0, then, it can be verified, that
(−1)µ−1 K1 G∗ K0
is Hurwitz matrix. For example, by Matlab program e8 2 Parameters.m, we get that6
max Re λi (−G∗ K0 ) ≤ −0.6628
t∈[0,∞)
for all i = 1, 2. Take ω̄Fmin
= 0.6628 and η2min = 10, then µ can be selected such that the
MS
inequality
k ω̄Fmin
MS
µ ≤ µmax =
η min ω max
2
SM S
10 · 0.6628
=
≈ 0.4971 s.
10 · 1.3333
holds. As a result, take µ1 = µ2 = 0.4971 s. Note, the control law implementation of
(80)–(82) was discussed in Exercise 8.1.
6
Re λi (A) is the real part of the eigenvalue λi of A.
Student Solutions Manual for “Design of nonlinear control systems . . .”
39
Run the Matlab program e8 2 Parameters.m to calculate the reference model parameters of the controller given by (80)–(82) for η2 = 10. Next, run the Simulink program
e8 2.mdl, to get a step response of the closed-loop system with zero initial conditions.
Make calculations and simulations for the degree of time-scale separation between fast
and slow motions assigned as η2 = 5 and η2 = 20. Compare the simulation results.
Chapter 9
Exercise 9.1 Stabilize the internal subsystem by selective exclusion of redundant control variables in the system given by
ẋ1 = x1 + x2 + u1 + u2 ,
ẋ2 = x1 + 3x2 − 2u1 ,
y = x1 .
(91)
Solution.
Assume that x1 , x2 are measurable state variables. We have y = x1 and denote
z = x2 . Hence, from (91), we get
ẏ = y + z + u1 + u2 ,
ż = y + 3z − 2u1 .
The reference model can be constructed in the form
ẏ = F (y, r) =⇒ ẏ = −ay + ar.
(92)
Exclusion of control variable u1 . Take u1 = 0 ∀ t. From (92), we get
ẏ = y + z + u2 ,
ż = y + 3z.
Let assume that the control law is given by
µu̇2 + d0 u2 = k0 {F (y, r) − ẏ} =⇒
µu̇2 + d0 u2 = k0 {−ẏ − ay + ar}.
(93)
Hence, the closed-loop system equations are given by
ẏ = y + z + u2 ,
ż = y + 3z,
µu̇2 + d0 u2 = k0 {−ẏ − ay + ar}.
(94)
The closed-loop system equations (94) can be rewritten as
ẏ = y + z + u2 ,
ż = y + 3z,
µu̇2 + [d0 + k0 ]u2 = k0 {−ay + ar − y − z}.
(95)
Student Solutions Manual for “Design of nonlinear control systems . . .”
40
By following through solutions of Exercises 4.2 and 4.3, from (95), we obtain the FMS
equation, that is
µu̇2 + [d0 + k0 ]u2 = k0 {−ay + ar − y − z},
(96)
where y, r, u2 are frozen variables during the transients in (96).
Suppose the FMS (96) is stable and take, for simplicity, d0 = 0. Hence, taking µ → 0
in (96) we get u2 (t) = us2 (t), where us2 (t) is a steady state (more precisely, quasi-steady
state) of the FMS (96) and
us2 = −ay + ar − y − z.
Substitution of us2 into (93) yields the SMS given by
ẏ = −ay + ar,
ż = 3z + y,
(97)
(98)
where the internal subsystem (98) is unstable.
Exclusion of control variable u2 . Take u2 = 0 ∀ t. From (92), we get
ẏ = y + z + u1 ,
ż = y + 3z − 2u1 .
(99)
(100)
Let assume that the control law is given by
µu̇1 + d0 u1 = k0 {−ẏ − ay + ar}.
(101)
Hence, the closed-loop system equations are given by
ẏ = y + z + u1 ,
ż = y + 3z − 2u1 ,
µu̇1 + d0 u1 = k0 {−ẏ − ay + ar}.
(102)
The closed-loop system equations (102) can be rewritten as
ẏ = y + z + u1 ,
ż = y + 3z − 2u1 ,
µu̇1 + [d0 + k0 ]u1 = k0 {−ay + ar − y − z}.
(103)
From (103), we obtain the FMS equation, that is
µu̇1 + [d0 + k0 ]u1 = k0 {−ay + ar − y − z},
(104)
where y, r, u2 are frozen variables during the transients in (104).
Suppose the FMS (104) is stable and take, for simplicity, d0 = 0. Hence, taking
µ → 0 in (104) we get u1 (t) = us1 (t), where us1 (t) is a steady state of the FMS (104) and
us1 = −ay + ar − y − z. Substitution of us1 into (99), (100) yields the SMS given by
ẏ = −ay + ar,
ż = 5z + 3y + 2a(y − r),
(105)
(106)
Student Solutions Manual for “Design of nonlinear control systems . . .”
41
where the internal subsystem (106) is unstable again. Hence, the method of selective
exclusion of redundant control variables does not allow to obtain the stable internal subsystem.
Internal dynamics stabilization by redundant control u2 . Let us consider the system
(92) with the control law given by (101). Hence, the closed-loop system equations are
given by
ẏ = y + z + u1 + u2 ,
ż = y + 3z − 2u1 .
µu̇1 + d0 u1 = k0 {−ẏ − ay + ar},
(107)
where u2 can be utilized for internal dynamics stabilization. From (107) we obtain
ẏ = y + z + u1 + u2 ,
ż = y + 3z − 2u1 ,
µu̇1 + [d0 + k0 ]u1 = k1 {−ay + ar − y − u2 }.
(108)
From (108), we obtain the FMS equation, that is
µu̇1 + [d0 + k1 ]u1 = k1 {−ay + ar − y − u2 },
(109)
where y, r, u2 are frozen variables during the transients in (109).
Suppose the FMS (109) is stable and take, for simplicity, d0 = 0. Hence, taking µ → 0
in (109) we get u1 (t) = us1 (t), where us1 (t) is a steady state (more precisely, quasi-steady
state) of the FMS (109) and
us1 = −ay + ar − y − u2
Substitution of us1 into (92) yields the SMS given by
ẏ = −ay + ar,
ż = 5z + 2u2 + 2y − 2a(r − y),
(110)
(111)
where the internal subsystem (111) can be stabilized by applying u2 = kint z. Thus, we
obtain
ẏ = −ay + ar,
ż = [5 + 2kint ]z + 2y − 2a(r − y).
(112)
(113)
From (113), the characteristic polynomial of the internal subsystem
Aint (s) = s − [5 + 2kint ]
follows. Consider, for example, the desired characteristic polynomial given by Adint (s) =
s + aint where aint > 0. Then, from the requirement Aint (s) = Adint (s), we get kint =
−(5 + aint )/2. Take, for example, aint = 1, then kint = −3.
Student Solutions Manual for “Design of nonlinear control systems . . .”
42
Let us take k0 = 10 and a = 1.3333 rad/s. Denote
max = max{a, a } rad/s,
ωSM
int
S
ωFmin
=
MS
d0 + k0
rad/s.
µ
max as a criterion for the degree of time-scale
Let us consider the ratio η2 = ωFmin
/ωSM
MS
S
separation between fast and slow motions in the closed-loop system. Take η2min = 20.
Hence, we obtain
d 0 + k0
min = 20 =⇒
max ≥ η2
µωSM
S
0 + 10
d 0 + k0
=
= 0.375 s.
µ ≤ µmax =
20 · 1.3333
η min ω max
η2 =
2
SM S
As a result, take µ1 = 0.375 s.
The implementation of the control law (101) is similar to the block diagram as shown
in Fig. 15. Run the Matlab program e9 1 Parameters.m to calculate the reference model
and controller parameters. Next, run the Simulink program e9 1.mdl, to get a step
response of the closed-loop system with zero initial conditions.
Exercise 9.2 Consider the system given by
ẋ1 = x1 + x2 + u1 + u2 ,
ẋ2 = x1 + 3x2 − 2u1 ,
y = x1 .
(114)
Stabilize the internal subsystem by insertion of supplementary conditions.
Solution.
From (114), we get
ẏ = x1 + x2 + u1 + u2 .
Hence, we have the relative degree α = 1. Let us insert the supplementary condition for
control variables such that
u1 + u2 = 0.
(115)
Then u2 = −u1 and we get the relative degree α = 2. From (114) and (115), we obtain
ẋ1 = x1 + x2 ,
ẋ2 = x1 + 3x2 − 2u1 ,
y = x1 .
(116)
By taking into account that y = x1 and ẏ = x1 + x2 , the system (116) can by rewritten
in the form
y (2) = 4y (1) − 2y − 2u1 .
Student Solutions Manual for “Design of nonlinear control systems . . .”
43
By following through solution of Exercises 5.9, consider the reference model given by
y (2) = F (y (1) , y, r).
Take tds = 2 s, σ d = 5 %, then by
Ã
d
θ = tan
−1
!
π
,
ln(100/σ d )
ωd =
4
,
tds
we get θd = 0.8092 rad, ζ d = 0.6901, ω d = 2 rad/s and ωn = 2.8981 rad/s. By selecting
the 2 roots s1,2 = −2 ± j2.0974 where Re(s1,2 ) = −ω d = −ωn cos(θd ) and |Im(s1,2 )| =
ωn sin(θd ), we obtain the desired characteristic polynomial s2 + 4s + 8.399. The desired
transfer function is given by
Gdyr (s) =
ad0
8.399
= 2
.
d
d
2
s + a1 s + a0
s + 4s + 8.399
Hence, we get the reference model in the form of the type 1 system
y (2) = −4y (1) − 8.399y + 8.399r =⇒ y (2) = F (y (1) , y, r).
Let us consider the control law given by
(2)
(1)
µ2 u1 + d1 µu1 + d0 u1 = k0 {F (y (1) , y, r) − y (2) }.
Hence, the closed-loop system equations are given by
y (2) = f (y (1) , y) + gu1 ,
(1)
(2)
µ2 u1 + d1 µu1 + d0 u1 = k0 {F (y (1) , y, r) − y (2) },
where f (y (1) , y) = 4y (1) − 2y and g = −2. By following through solution of Exercise 4.2,
we obtain the FMS equation, that is
(2)
(1)
µ2 u1 + d1 µu1 + [d0 + k0 g]u1 = k0 {F (y (1) , y, r) − f (y (1) , y)},
where F = const and f = const during the transients in (116), as well as the SMS
equation, that is
y (2) = F (y (1) , y, r) +
d0
{f (y (1) , y) − F (y (1) , y, r)}.
d0 + k0 g
We have g = −2 < 0. Take d0 > 0 and k0 = 10/g = −5.
From the SMS and reference model equations, it follows that
s2 + ad1 s + ad0
SM S
is the characteristic polynomial of the SMS where a0
µ2 s2 + d1 µs + d0 + k0 g
q
=
ad0 , and
Student Solutions Manual for “Design of nonlinear control systems . . .”
F MS
is the characteristic polynomial of the FMS. Denote a0
44
= d0 + k0 g. Hence, we obtain
√
F MS
(a0 )1/2
d0 + k0 g
η3 =
≥ η3min = 10
SM S 1/2 =
SM S
µ(a0 )
µ(a0 )1/2
q
√
0 − 5 · (−2)
d0 + k0 g
√
=
=⇒ µ ≤
≈ 0.1091 s.
SM S 1/2
10 · 8.399
)
η min (a
0
3
From the characteristic equation of the FMS, we get
q
F MS
s1,2
d21 − 4(d0 + k0 g)
d1
=−
±j
= α ± jβ,
2µ
2µ
where we assume that d21 − 4(d0 + k0 g) < 0. Take ζFmin
= 0.5. Hence, we can find
MS
q
ζF M S = cos(θF M S ) = |α|/ α2 + β 2
d1
= √
≥ ζFmin
= 0.5 =⇒
MS
2 d 0 + k0 g
q
d ≥ 2ζ min d + k g
1
F MS
q
0
0
= 2 · 0.5 0 + (−5) · (−2) ≈ 3.1623.
Take µ = 0.1091 s and d1 = 3.1623.
In conclusion, run the Matlab program e9 2 Parameters.m to calculate the controller
parameters. Next, run the Simulink program e9 2.mdl, to get the step response of the
closed-loop system.
Chapter 10
Exercise 10.1 The system is given by
ẋ = x2 + 4u.
(117)
Find the parameters of the control law in the form given by
uk =
q
X
j=1
d¯j uk−j +
q
X
j=0
āj yk−j +
q
X
b̄j rk−j
(118)
j=0
to meet the following specifications: εr = 0, tds ≈ 3 s, σ d ≈ 0%, q = 1. Determine the
sampling period Ts such that the phase margin of the FMS will meet the requirement
ϕ ≥ 0.7 rad. Compare simulation results of the step output response of the closed-loop
control system with the assignment.
Solution.
Reference model. Consider the reference model given by ẋ = F (x, r). Take tds = 3 s,
d
σ = 0 %, then we get θd = 0 rad, ζ d = 1, a = ω d = ωn = 1.3333 rad/s. By selecting the
Student Solutions Manual for “Design of nonlinear control systems . . .”
45
root s1 = −a, we obtain the desired characteristic polynomial s + a. The desired transfer
function is given by
Gdxr (s) =
a
.
s+a
Hence, we get the reference model given by
ẋ = −ax + ar =⇒ ẋ =
1
[r − x],
T
(119)
where T = 1/a.
Continuous-time controller. We have that x(1) is the highest derivative of the output
signal, then α = 1. As far as the requirement on the high frequency sensor noise attenuation is not specified, then, for simplicity, take q = α = 1 and consider the control law
given by
µu̇ + d0 u = k0 [F (x, r) − ẋ].
(120)
Hence, the control law has the following form:
µu̇ + d0 u = k0 [−ẋ − ax + ar].
(121)
Closed-loop system. The closed-loop system equations are given by
ẋ = x2 + 4u,
µu̇ + d0 u = k0 [F (x, r) − ẋ].
(122)
(123)
Denote f (x) = x2 . From (122) and (123), by following through solution of Exercise 4.2,
we get the FMS given by
µu̇ + [d0 + k0 g]u = k0 {F (x, r) − f (x)},
(124)
where g = gmax = gmin = 4 and F = const, f = const during the transients in (124).
Hence, we obtain that
µs + d0 + k0 g
(125)
is the characteristic polynomial of the FMS.
Suppose the FMS (124) is stable. Then, in order to provide the requirement εr = 0,
we take d0 = 0. Next, taking µ → 0 in (124) we get u(t) = us (t), where us (t) is a steady
state (more precisely, quasi-steady state) of the FMS (124) and
us = uid = g −1 [F (x, r) − f (x)].
Substitution of us into the equation of the plant model (117) yields the SMS, that is the
same as the reference model.
Student Solutions Manual for “Design of nonlinear control systems . . .”
46
Selection of continuous-time controller parameters. Let us consider a simplified version for the gain k0 selection. We have d0 = 0, then the gain k0 can be selected such that
k0 gmin = 10, where gmin = gmax = g = 4. Hence, we get k0 = 5/2.
From (119), we have that the natural frequency of the reference model is given by
ωnd = a = 1.3333 rad/s. Denote
max = ω d = 1.3333 rad/s,
ωSM
n
S
Let us consider the ratio
η2 =
ωFmin
=
MS
d0 + k0 gmin
rad/s.
µ
ωFmin
MS
max
ωSM S
(126)
as a criterion for the degree of time-scale separation between fast and slow motions. Take
η2min = 20. Hence, we obtain
η2 =
d0 + k0 gmin
≥ η2min = 20 =⇒
µω max
SM S
d0 + k0 gmin
0 + 2.5 · 4
µ ≤ µmax =
=
≈ 0.375 s.
min
max
20 · 1.3333
η2 ω
SM S
As a result, take µ = 0.375 s.
From (121), we obtain the block diagram of the controller as shown in Fig. 15 (see
p. 33). Run the Matlab program e10 1 Parameters.m to calculate the continuous-time
controller parameters. Next, run the Simulink program e10 1 Continuous.mdl, to get the
step response of the closed-loop system.
Selection of the sampling period. From (117) we get the following pseudo-continuoustime model:
ẋ(t) = x2 (t) + 4u(t − τ ),
(127)
where τ = Ts /2 and Ts is the sampling period.
Closed-loop system. In accordance with (120) and (127), the closed-loop system equations are given by
ẋ(t) = f (x(t)) + gu(t − τ ),
µu̇(t) + d0 u(t) = k0 [F (x(t), r(t)) − ẋ(t)].
From the above equations, we get the FMS given by
µu̇(t) + d0 u(t) + k0 gu(t − τ ) = k0 {F (x(t), r(t)) − f (x(t))},
(128)
where F = const and f = const during the transients in (128). The block diagram
representation of the FMS (128) is shown in Fig. 16.
The corresponding transfer function of the open-loop FMS with time delay is given
by
O
GF M S (s) =
k0 g exp (−τ s)
,
D(µs)
(129)
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47
Figure 16: Block diagram of the FMS (128) with delay τ , where F = const, f = const.
where D(µs) = µs + d0 . From (129) and the condition given by
O
|GF M S (jωc , µ)| = 1,
we get
|D(jµωc )| = k0 g =⇒ |jµωc + d0 | = k0 g =⇒
q
ωc =
k02 g 2 − d20
µ
(130)
where ωc is the crossover frequency on the Nyquist plot of (129).
Denote by ϕ the value of the phase margin of the FMS (128). Then, by inspection of
the Nyquist for (129), we get that the requirement
ϕ ≥ ϕd > 0
(131)
Ts ≤ 2[π − ϕd − ArgD(jµωc )]/ωc
(132)
holds if the inequality
is satisfied. Finally, from (130) and (132), by taking into account d0 = 0, we get
Ts ≤
(π − 2ϕd )µ
k0 g
(133)
where ϕd < π/2.
Take, for example, ϕd = 0.7 rad (that is about 40 degrees of arc), k0 g = 10, and
µ = 0.375 s. Then, by (133), we get Ts = 0.1742 s.
Digital realization of continuous-time controller. We have the continuous control law
given by
µu̇ + d0 u = k0 [−ẋ − ax + ar].
(134)
From (134), we obtain
2 z−1
=⇒
Ts z + 1
[(2µ + d0 Ts )z + (d0 Ts − 2µ)]u(z) =
k0 [−(2 + aTs )z + (2 − aTs )]x(z) + k0 aTs [z + 1]r(z) =⇒
(2µ + d0 Ts )uk+1 + (d0 Ts − 2µ)uk =
−k0 (2 + aTs )xk+1 + k0 (2 − aTs )uk + k0 aTs rk+1 + k0 aTs rk .
[µs + d0 ]u(s) = −k0 [s + a]x(s) + ar(s) =⇒ s =
(135)
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From (135), the control law given by the difference equation
uk+1 = d¯1 uk + ā0 yk+1 + ā1 yk + b̄0 rk+1 + b̄1 rk
(136)
d¯0 = 2µ + d0 Ts , d¯1 = (2µ − d0 Ts )/d¯0 ,
ā0 = − k0 (2 + aTs )/d¯0 , ā1 = k0 (2 − aTs )/d¯0 ,
b̄0 = b̄1 = k0 aTs /d¯0 .
(137)
results, where
Implementation of control law. From (136), we obtain
uk+1 − ā0 xk+1 − b̄0 rk+1 = d¯1 uk + ā1 xk + b̄1 rk =⇒
|
{z
}
=ūk+1
uk+1 − ā0 xk+1 − b̄0 rk+1 = ūk+1 =⇒ uk = ā0 xk + b̄0 rk + ūk .
Finally, we get
ūk+1 = d¯1 uk + ā1 xk + b̄1 rk ,
uk = ūk + ā0 xk + b̄0 rk .
(138)
From (138), we obtain the block diagram as shown in Fig. 17.
Figure 17: Block diagram of the control law (136) represented in the form (138).
Run the Simulink program e10 1 Discrete.mdl, to get the step response of the closedloop system with zero initial conditions. Make simulations for ϕd = 0.5 rad and ϕd = 1
rad. Compare the simulation results.
Exercise 10.3 The system is given by
x(2) = x + x|x(1) | + {1.5 + sin(t)}u.
(139)
Find the parameters of the control law (118) to meet the following specifications: εr = 0,
tds ≈ 1 s, σ d ≈ 20%, q = 2. Determine the sampling period Ts such that the phase margin
of the FMS will meet the requirement ϕ ≥ 0.25 rad. Compare simulation results of the
step output response of the closed-loop control system with the assignment.
Solution.
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49
Reference model. From (139), we obtain n = 2, x(2) = f (x(1) , x) + g(t)u where
f (x(1) , x) = x + x|x(1) | and g(t) = 1.5 + sin(t). Hence g ∈ [gmin , gmax ] = [0.5, 2.5].
We have n = 2 and x(2) is the highest derivative of the output signal. Therefore, the
reference model will be constructed in the form
x(2) = F (x(1) , x, r).
Take tds = 1 s, σ d = 20 %, then by
Ã
d
θ = tan
−1
!
π
,
ln(100/σ d )
ωd =
4
,
tds
we get θd = 1.0974 rad, ζ d = 0.4559, ω d = 4 rad/s and ωn = 8.7729 rad/s. By selecting
the 2 roots s1,2 = −4 ± j7.8079, where Re(s1,2 ) = −ω d = −ωn cos(θd ) and |Im(s1,2 )| =
ωn sin(θd ), we obtain the desired characteristic polynomial
(s − s1 )(s − s2 ) = s2 + a1 s + a0 ,
(140)
where a1 = 8, a0 = 76.9637. Hence, we can get the reference model in the form of the
type 1 system, this is
x(2) = −a1 x(1) − a0 x + b0 r,
or type 2 system, this is
x(2) = −a1 x(1) − a0 x + b1 r(1) + b0 r
(141)
where b0 = a0 and b1 = a1 .
Continuous-time controller. Take q = n = 2 Therefore, consider the control law given
by
µ2 u(2) + d1 µu(1) + d0 u
= k0 {−a2 x(2) − a1 x(1) − a0 x + b1 r(1) + b0 r},
(142)
which can be rewritten, for short, as
µ2 u(2) + d1 µu(1) + d0 u = k0 {F (x(1) , x, r) − x(2) }
(143)
where a2 = 1. Hence, the closed-loop system equations are given by
x(2) = f (x(1) , x) + g(t)u,
µ2 u(2) + d1 µu(1) + d0 u = k0 {F (x(1) , x, r) − x(2) }.
From the above closed-loop system equations, we get the FMS given by
µ2 u(2) + d1 µu(1) + {d0 + k0 g}u = k0 {F (x(1) , x, r) − f (x(1) , x)},
(144)
where F = const, f = const, and g = const during the transients in (144), as well as the
SMS given by
x(2) = F (x(1) , x, r) +
d0
{f (x(1) , x) − F (x(1) , x, r)}.
d0 + k0 g(t)
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50
Selection of continuous-time controller parameters. Parameters of the continuoustime controller given by (143) can be found by following through solution of Exercise 5.1.
In order to provide the requirement εr = 0, take d0 = 0 and let us take, for simplicity,
the gain k0 such that the condition k0 gmin = 10 holds. Hence we get k0 = 20.
From the SMS and reference model equations, it follows that the characteristic polySM S
nomial of the SMS is given by (140), where a0 = a0 = 76.9637. and
µ2 s2 + d1 µs + d0 + k0 g
is the characteristic polynomial of the FMS. Take η3min = 10. Then, from the requirement
η3 ≥ η3min , we obtain
F MS
η3 =
1/2
q
√
(a0 )
d0 + k0 g
≥
SM S 1/2 =
SM S
µ(a0 )
µ(a0 )1/2
q
d0 + k0 gmin
SM S
µ(a0
)1/2
≥ η3min = 10
√
d0 + k0 gmin
0 + 20 · 0.5
√
=⇒ µ ≤
=
≈ 0.036 s.
SM
S
10 · 76.9637
η3min (a0 )1/2
From the characteristic equation of the FMS, we get
q
F MS
s1,2 = −
d21 − 4(d0 + k0 g)
d1
±j
= α ± jβ,
2µ
2µ
where we assume that d21 −4(d0 +k0 g) < 0 when g = gmax . Take, for instance, ζFmin
= 0.6.
MS
Then we can find
q
ζF M S = cos(θF M S ) = |α|/ α2 + β 2
d1
= √
≥ ζFmin
= 0.6 =⇒
MS
2 d0 + k0 g
q
d1 ≥ 2ζFmin
d0 + k0 gmax
MS
√
= 2 · 0.6 0 + 20 · 2.5 ≈ 8.4853.
Take µ = 0.036 s and d1 = 8.4853.
From (143), we obtain the block diagram of the controller as shown in Fig. 6 (see
p. 18). Run the Matlab program e10 3 Parameters.m to calculate the controller parameters. Next, run the Simulink program e10 3 Continuous.mdl, to get the step response of
the closed-loop system.
The block diagram representation of the FMS (144) is shown in Fig. 18, where
D(µs) = µ2 s2 + d1 µs + d0 .
The corresponding transfer function of the open-loop FMS is given by
O
GF M S (s) =
k0 g
.
D(µs)
Take d0 = 0 and g = gmax , then from (145) and
O
|GF M S (jωc , µ)| = 1,
(145)
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51
Figure 18: Block diagram of the FMS (144), where F = const, f = const.
where ωc is the crossover frequency on the Nyquist plot of the FMS (144) given that
g = gmax = 2.5. Hence, we get
|D(jµωc )| = k0 gmax =⇒ |jµωc (jµωc + d1 )| = k0 gmax =⇒
2
µ4 ωc4 + d21 µ2 ωc2 − k02 gmax
= 0 =⇒
2
4 2
2 2
2 2
y = ωc > 0, µ y + d1 µ y − k0 gmax = 0 =⇒
q
(146)
2
−d21 + d41 + 4k02 gmax
y=
2µ2
√
−8.48532 + 8.48534 + 4 · 202 · 2.52
=
≈ 19712 =⇒
2
2 · 0.036
√
ωc = y ≈ 140 rad/s.
Hence, by taking into account that d0 = 0, and, by inspection of the Nyquist plot for
(145), we can found the phase margin of the FMS given by (144), this is
ϕF M S = π/2 − tan−1 (µωc /d1 )
≈ π/2 − tan−1 (0.036 · 140/8.4853) ≈ 1.033 rad
when g = gmax = 2.5.
Selection of the sampling period. From (139) we get the following pseudo-continuoustime model:
x(2) (t) = f (x(1) (t), x(t)) + g(t)u(t − τ ),
(147)
where τ = Ts /2, Ts is the sampling period. From the closed-loop system equations given
by (143) and (147), we get the FMS equation given by
µ2 u(2) (t) + d1 µu(1) (t) + d0 u(t) + k0 g(t)u(t − τ )
= k0 {F (x(1) (t), x(t), r(t)) − f (x(1) (t), x(t))},
(148)
where F = const, f = const, and g is the frozen parameter during the transients in (148).
The block diagram representation of the FMS (148) is shown in Fig. 19.
The corresponding transfer function of the open-loop FMS with delay τ = Ts /2 is
given by
O
GF M S (s) =
k0 ge−jτ s
.
D(µs)
(149)
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52
Figure 19: Block diagram of the FMS (148) with delay τ = Ts /2, where F = const,
f = const, and g = const.
Then, by (146), the crossover frequency ωc on the Nyquist plot of the FMS (148) can be
√
found, this is ωc = y ≈ 140 rad/s given that g = gmax = 2.5.
Let us select ϕd such that the inequalities
0 < ϕd < ϕF M S = 1.033 rad
hold. Denote ϕ is the phase margin of the FMS given by (148) Then, by inspection of
the Nyquist plot for (149), the condition
0 < ϕd ≤ ϕ < ϕF M S = 1.033 rad
(150)
holds for all g ∈ [gmin , gmax ] = [0.5, 2.5], if the sampling period Ts is selected such that
the condition
0 < Ts ≤ 2[π − ϕd − ArgD(jµωc )]/ωc
(151)
is satisfied. Take, for instance, ϕd = 0.25 rad. From (151), by taking into account d0 = 0,
we get
0 < Ts ≤ [π − 2ϕd − 2 tan−1 (µωc /d1 )]/ωc
≈ [π − 2 · 0.25 − 2 tan−1 (0.036 · 140/8.4853)]/140 ≈ 0.0112 s.
Take for numerical simulation Ts = 0.0112 s.
Digital realization of continuous-time controller. From (142), we obtain
[µ2 s2 + d1 µs + d0 ]u(s)
= k0 [−a2 s2 − a1 s − a0 ]x(s) + k0 [b1 s + b0 ]r(s).
(152)
Then, from (152), by the Tustin transformation
s = 2(z − 1)/[Ts (z + 1)],
the discrete-time control law in the form of the difference equation
uk = d¯1 uk−1 + d¯2 uk−2 + ā0 xk + ā1 xk−1 + ā2 xk−2
+b̄0 rk + b̄1 rk−1 + b̄2 rk−2
(153)
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53
results, where
d¯0 = 4µ2 + 2µd1 Ts + d0 Ts2 ,
d¯1 = {8µ2 − 2d0 Ts2 }/d¯0 ,
d¯2 = − {4µ2 − 2µd1 Ts + d0 Ts2 }/d¯0 ,
ā0 = − k0 {4a2 + 2a1 Ts + a0 Ts2 }/d¯0 ,
ā1 = 2k0 {4a2 − a0 Ts2 }/d¯0 ,
ā2 = − k0 {4a2 − 2a1 Ts + a0 Ts2 }/d¯0 ,
b̄0 = k0 {2b1 Ts + b0 Ts2 }/d¯0 ,
b̄1 = 2k0 b0 Ts2 /d¯0 ,
b̄2 = − k0 {2b1 Ts − b0 Ts2 }/d¯0 .
(154)
From (153), we obtain
uk+2 −ā0 xk+2 − b̄0 rk+2 − d¯1 uk+1 −ā1 xk+1 − b̄1 rk+1 = |d¯2 uk + ā{z
2 xk + b̄2 rk
}
=u2,k+1
=⇒ uk+1 − ā0 xk+1 − b̄0 rk+1 = u2,k + d¯1 uk + ā1 xk + b̄1 rk =⇒
{z
|
}
=u1,k+1
uk = u1,k + ā0 xk + b̄0 rk
From the above, we get
u1,k+1 = u2,k + d¯1 uk + ā1 xk + b̄1 rk ,
u2,k+1 = d¯2 uk + ā2 xk + b̄2 rk ,
uk = u1,k + ā0 xk + b̄0 rk .
(155)
Then, from (155), the block diagram can be obtained as shown in Fig. 20.
Figure 20: Block diagram of the control law (153).
Run the Simulink program e10 3 Discrete.mdl, to get the step response of the closedloop system with zero initial conditions. Make simulations for ϕd = 0.5 rad and ϕd = 0.175
rad. Compare the simulation results.
Student Solutions Manual for “Design of nonlinear control systems . . .”
54
Chapter 11
Exercise 11.1 Construct the reference model in the form of the 2nd-order difference
equation
yk =
n
X
adj yk−j +
j=1
n
X
bdj rk−j
(156)
j=1
based on the Z-transform in such a way that the step output response with zero initial
conditions meets the requirements tds ≈ 1 s, σ d ≈ 0 % given that the sampling period is
Ts = 0.05 s. Compare simulation results with the assignment.
Solution.
By following through the solution of Exercise 2.1, let us construct the desired continuoustime transfer function Gdyr (s). Take tds = 1 s, σ d = 0 %, then by
Ã
d
θ = tan
−1
!
π
,
ln(100/σ d )
ωd =
4
,
tds
we get θd = 0 rad, ζ d = 1, ω d = ωn = 4 rad/s. By selecting the 2 roots s1 = s2 = −4, we
obtain the desired characteristic polynomial s2 + 8s + 16. Consider the desired transfer
function given by
Gdyr (s) =
s2
16
16
=
.
+ 8s + 16
(s + 4)2
(157)
Let us denote a = ωn = 4 and expand Gdyr (s)/s as
A
B
C
a2
=
+
+
s(s + a)2
s s+a (s+a)2
(A+B)s2 +(2Aa+Ba+C)s+Aa2
=
.
s(s + a)2
Hence, we obtain A = 1, B = −1, and C = −a. Next, by the Z-transform of Gdyr (s)
preceded by a ZOH, we get the desired pulse transfer function given by
d
Hyr
(z)
(
"
(
"
a2
z−1
=
Z L−1
z
s(s + a)2
#¯
¯
¯
¯
¯
)
t=kTs
B
C
A
z−1
Z L−1
+
+
=
z
s s+a (s+a)2
=
bd1 z
z2
−
+
ad1 z
bd2
− ad2
,
#¯
¯
¯
¯
¯
)
t=kTs
(158)
where
bd1 = 1 − e−aTs − aTs e−aTs ,
bd2 = e−aTs [e−aTs − 1 + aTs ],
ad1 = 2e−aTs , ad2 = −e−2aTs .
(159)
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55
Hence, from (158), we obtain the desired difference equation, this is
yk = ad1 yk−1 + ad2 yk−2 + bd1 rk−1 + bd2 rk−2 .
(160)
Run the Matlab program e11 1 Parameters.m in order to calculate the reference model
parameters. Next, run the Simulink program e11 1.mdl to get a plot for the output
response of (160). Then, from inspection of the plot, determine the steady-state error for
input signals of type 0 and 1, respectively.
Exercise 11.4 Determine by the Z-transform the discrete-time model of the system
y (2) − y = 2u
(161)
preceded by ZOH. Design the control law
uk =
n
X
j=1


dj uk−j + λ0 −yk +
n
X
j=1
adj yk−j +
n
X
j=1


bdj rk−j  .
(162)
to meet the following specifications: tds ≈ 2 s; σ d ≈ 0%; η ≥ 10. Compare simulation
results of the step output response of the closed-loop control system with the assignment.
Solution.
Difference equation of the plant. We have the system (161) preceded by ZOH. Hence,
we have y(s) = Gyu (s)u(s), where
Gyu (s) =
2
2
=
.
s2 − 1
(s + 1)(s − 1)
Let us expand Gyu (s)/s as
2
A
B
C
=
+
+
s(s + 1)(s − 1)
s s+1 s−1
(A+B +C)s2 +(C −B)s−A
=
.
s(s + 1)(s − 1)
We get A = −2, B = 1, and C = 1. Next, by the Z-transform of Gyu (s)/s, we obtain the
pulse transfer function given by
(
"
2
z−1
Z L−1
Hyu (z) =
z
s(s + 1)(s − 1)
b1 z + b2
,
= 2
z − a1 z − a2
#¯
¯
¯
¯
¯
)
t=kTs
(163)
where
b1 = b2 = eTs + e−Ts − 2,
a1 = eTs + e−Ts ,
a2 = −1.
(164)
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56
Hence, from (163), we obtain the difference equation of the plant, this is
yk = a1 yk−1 + a2 yk−2 + b1 rk−1 + b2 rk−2 .
(165)
Reference model. By following through the solution of Exercise 11.1, let us construct the
desired continuous-time transfer function Gdyr (s). Take tds = 2 s, σ d = 0 %, then we get
θd = 0 rad, ζ d = 1, ω d = ωn = 2 rad/s. By selecting the 2 roots s1 = s2 = −2, we obtain
the desired characteristic polynomial s2 + 4s + 4. The desired transfer function can be
constructed in the following form:
Gdyr (s) =
4
4
=
.
s2 + 4s + 4
(s + 2)2
Denote a = ωn = 2. By the Z-transform of Gdyr (s) preceded by a ZOH, we get the desired
pulse transfer function given by (158), (159). Hence, the desired difference equation is
given by (160).
Control law. Consider the control law given by
uk = d1 uk−1 + d2 uk−2
+λ0 {−yk + ad1 yk−1 + ad2 yk−2 + bd1 rk−1 + bd2 rk−2 }.
(166)
The closed-loop system equations have the following form:
yk =
2
X
aj yk−j +
j=1
uk =
2
X
2
X
bj uk−j ,
(167)
j=1


2
X
dj uk−j + λ0 −yk +

j=1
adj yk−j +
j=1
2
X


bdj rk−j .
j=1

(168)
Substitution of (167) into (168) yields
yk =
2
X
aj yk−j +
j=1
uk =
2
X
2
X
bj uk−j ,
(169)
j=1
[dj −λ0 bj ]uk−j +λ0
j=1
2
X
{[adj −aj ]yk−j +bdj rk−j },
(170)
j=1
where the FMS is governed by
uk =
2
X
[dj −λ0 bj ]uk−j +λ0
j=1
2
X
{[adj −aj ]yk−j +bdj rk−j },
(171)
j=1
where yk = const during the transients in the system (171).
The controller parameters λ0 and dj can be selected such that
λ0 = [b1 + b2 ]−1 ,
d1 = λ0 b1 ,
d2 = λ0 b2 ,
(172)
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57
where b1 , b2 are defined by (164). Hence, we get the deadbeat response of the FMS.
Finally, take the sampling period such that
tds
qη
2
=
= 0.1 s.
2 · 10
Ts =
(173)
Then, the parameters of the control law (166) can be found by (159), (164), and (172).
Implementation of control law. The control law (166) can be rewritten as
ũk+2 = d1 ũk+1 +d2 ũk −yk+2 +ad1 yk+1 +ad2 yk +bd1 rk+1 +bd2 rk ,
uk = λ0 ũk .
(174)
(175)
From (174), we obtain
ũk+2 − d1 ũk+1 + yk+2 − ad1 yk+1 − bd1 rk+1 = d2 ũk + ad2 yk + bd2 rk =⇒
|
{z
}
=u2,k+1
ũk+1 + yk+1 = u2,k + d1 ũk + ad1 yk + bd1 rk =⇒
|
{z
}
=u1,k+1
ũk = −yk + u1,k
Finally, we get
u1,k+1 = u2,k + d1 ũk + ad1 yk + bd1 rk ,
u2,k+1 = d2 ũk + ad2 yk + bd2 rk ,
ũk = u1,k − yk , uk = λ0 ũk .
(176)
From (176), we obtain the block diagram as shown in Fig. 21.
Figure 21: Block diagram of the control law (166) represented in the form (176).
Run the Matlab program e11 4 Parameters.m in order to calculate the reference
model parameters. Next, run the Simulink program e11 4.mdl to get a plot for the
output response of the closed-loop system. Make simulations for η = 7, η = 20. Compare
the simulation results.
Student Solutions Manual for “Design of nonlinear control systems . . .”
58
Chapter 12
Exercise 12.1 Consider the system (60) preceded by ZOH where a = 3. Design the
discrete-time control law given by
uk =
q≥α
X
dj uk−j + λ(Ts )[Fk − yk ],
(177)
j=1
to meet the following specifications: εr = 0, tds ≈ 2 s, σ d ≈ 5%. Run a computer simulation
of the closed-loop system with zero initial conditions. Compare simulation results of the
output response with the assignment for r(t) = 1, ∀ t > 0.
Solution.
By following through solution of Exercise 7.1, the invertibility and internal stability
of the system (60) are verified for a = 3, where the relative degree of (60) is α = 2. From
(63), we have ẏ2 = 2y2 + z + g ? u, where g ? = 1 and g ? is so called a high-frequency gain
of the system (60) (or for linear systems that is called as the α-th Markov parameter mα ,
that is g ? = mα = m2 = 1).
Control law structure. Take, for simplicity, the control law given by (177) with q =
α = 2. Then (177) can be rewritten in the following form
uk = d1 uk−1 + d2 uk−2
+λ0 {−yk + ad1 yk−1 + ad2 yk−2 + bd1 rk−1 + bd2 rk−2 }.
(178)
Sampling period. Take, for instance, η = 20, where η is the required degree of timescale separation between the fast and slow modes in the closed-loop system. Then, the
sampling period Ts can be selected by
tds
qη
2
=
= 0.05 s.
2 · 20
Ts =
(179)
Reference model. By following through the solution of Exercise 11.1, let us construct
the desired continuous-time transfer function Gdyr (s). Take tds = 2 s, σ d = 5 %, then we
get θd = 0.8092 rad, ζ d = 0.6901, ωn = 2.8981 rad/s, ω d = 2 rad/s. By selecting the 2
roots s1,2 = −2 ± j2.0974, we obtain the desired characteristic polynomial s2 + 4s + 8.399.
The desired transfer function can be constructed in the following form:
Gdyr (s) =
8.399
.
s2 + 4s + 8.399
By the Z-transform of Gdyr (s) preceded by a ZOH with the sampling period Ts = 0.05 s,
we get the desired pulse transfer function Gdyr (z) given by
Gdyr (z) =
bd1 z + bd2
z 2 − ad1 z − ad2
(180)
Student Solutions Manual for “Design of nonlinear control systems . . .”
59
where bd1 = 0.009816, bd2 = 0.009182, ad1 = 1.8, and ad2 = −0.8187.
Control law parameters. Consider the Euler polynomial Eα (z) for α = 2, that is
Eα (z) = z + 1 (see p. 62).
Assume that the sampling period Ts is small enough to provide the required degree of
time-scale separation between the fast and slow modes in the closed-loop system. Then,
consider the discrete-time approximate model of the input-output mapping that corresponds to continuous-time linear system (60) preceded by a ZOH with high sampling
rate, that is the following difference equation
yk =
α=2
X
j=1
aα,j yk−j +
Tsα
α=2
X
j=1
g?
²α,j
[uk−j + fk−j ],
α!
(181)
where
yk = y(t)|t=kTs ,
uk = u(t)|t=kTs , fk = f (x1 (t), x3 (t))|t=kTs ,
(z − 1)α = z α − aα,1 z α−1 − aα,2 z α−2 − · · · − aα,α ,
α!
.
aα,j = (−1)j+1
(α − j) ! j !
Denote
B(z) = b1 z α−1 + b2 z α−2 + · · · + bα
Tα
= g ? s Eα (z)
α!
2
T
= g ? s E2 (z) = 0.00125(z + 1).
2
(182)
In order to get the deadbeat response of the FMS, the controller parameters d1 , d2 and
λ0 are selected such that
λ0 = [b1 + b2 ]−1 = 400,
d1 = λ0 b1 = 0.5, d2 = λ0 b2 = 0.5.
Implementation of control law. The implementation of the control law (178) was
discussed in Exercise 11.4, and the block diagram of the control law is as shown in Fig. 21
(see p. 57).
Run the Matlab program e12 1 Parameters.m in order to calculate the control law
parameters. Next, run the Simulink program e12 1.mdl to get a plot for the output
response of the closed-loop system. Redesign controller and make simulations for η = 15,
η = 30, η = 40. Compare the simulation results.
Chapter 13
Exercise 13.1 Prove that ϕn (z) are the eigenfunctions of the system
∂x
∂2x
(z, t) = α2 2 (z, t) + c(t)x(z, t) + w(z, t) + u(z, t),
∂t
∂z
(183)
Student Solutions Manual for “Design of nonlinear control systems . . .”
60
where [∂x(z, t)/∂z]|z=0
= 0, [∂x(z, t)/∂z]|z=1 = 0 are√the boundary conditions, α2 = 1,
√
and ϕn (z) = ϕ0n cos( λn z), λn = n2 π 2 , ϕ00 = 1, ϕ0n = 2 ∀ n = 1, 2, . . ..
Solution.
Consider the homogeneous one-dimensional heat equation7
∂x
∂ 2x
(z, t) = 2 (z, t)
∂t
∂z
(184)
with the boundary conditions given by
[∂x(z, t)/∂z]|z=0 = 0,
[∂x(z, t)/∂z]|z=1 = 0.
(185)
In accordance with the method of separating variables, take
x(z, t) = x(t)ϕ(z).
(186)
From (184), we get
∂ 2 x(z, t)/∂z 2 = x(t)[d2 ϕ(z)/dz 2 ].
∂x(z, t)/∂t = ϕ(z)[dx(t)/dt],
(187)
Hence, from (184) and (187), we obtain
dx(t)/dt
d2 ϕ(z)/dz 2
=
,
x(t)
ϕ(z)
(188)
where (188) holds for all t and z. Then
dx(t)/dt
d2 ϕ(z)/dz 2
=
= β,
x(t)
ϕ(z)
(189)
where β = const. Hence, from (189), we obtain
d2 ϕ(z)
− βϕ(z) = 0,
dz 2
dx(t)
− βx(t) = 0,
dt
(190)
(191)
In accordance with (185) and (186), we have
¯
¯
dϕ(z) ¯¯
x(t)
¯
= 0,
dz ¯z=0
dϕ(z) ¯¯
x(t)
¯
= 0,
dz ¯z=1
(192)
for all x(t). Then
¯
dϕ(z) ¯¯
¯
= 0,
dz ¯z=0
(193)
¯
dϕ(z) ¯¯
¯
= 0,
dz ¯z=1
(194)
7
From this point the solution can be done by following, for instance, Chapter 11 of the book: Kreyszig
E. Advanced engineering mathematics, 8th ed., New York: John Wiley & Sons, Inc., 1999.
Student Solutions Manual for “Design of nonlinear control systems . . .”
61
If β > 0, then ϕ(z) = Aeµz + Be−µz is the general solution of (190). From (193) and
(194), it follows that ϕ(z) ≡ 0 is the unique solution of (190), (193), and (194).
If β = 0, then ϕ(z) = az + b is the general solution of (190), where
dϕ(z)
= a.
dz
(195)
From (193), (194), and (195), it follows that b 6= 0 and a = 0. Denote ϕ0 (z) = ϕ00 = b = 1.
Take β = −p2 < 0, p > 0, and from (190), (191) we get
d2 ϕ(z)
+ p2 ϕ(z) = 0,
dz 2
dx(t)
+ p2 x(t) = 0.
dt
(196)
(197)
Then ϕ(z) = A cos(pz) + B sin(pz) is the general solution of (196), where
dϕ(z)
= −pA sin(pz) + pB cos(pz).
dz
(198)
From (193) and (198), it follows that B = 0. Take A 6= 0 since otherwise ϕ(z) ≡ 0. From
(194) and (198), it follows that sin(p) = 0. Hence p = nπ for all n = 1, 2, . . ..
We have Rinfinitely many different solutions given by ϕn (z) = ϕ0n cos(nπz). Let us
z=1 2
assume that z=0
ϕn (z)dz = 1. Then
(ϕ0n )2
√
Z z=1
z=0
2
cos (nπz)dz =
((ϕ0n )2 /(nπ))
Z y=nπ
y=0
cos2 (y)dy = (ϕ0n )2 /2 = 1.
Hence ϕ0n = 2 for all n = 1, 2, . . ..
So, for all n = 0, 1, . . ., we get that ϕn (z) = ϕ0n cos(nπz), λn = [nπ]2 are eigenfunctions
and eigenvalues, respectively, of the boundary value problem consisting of (196), (193),
and (194) (that is called as a Sturm-Liouville problem).
The solution of (197) is given by xn (t) = xn (0)e−λn t for all n = 0, 1, . . .. As a result,
we get that
xn (z, t) = xn (t)ϕn (z) = xn (0)e−λn t ϕ0n cos(nπz),
∀ n = 0, 1, . . .
are the eigenfunctions of the problem (184), (185), corresponding to the eigenvalues λn .
Auxiliary Material (The optimal coefficients based on ITAE criterion)
The optimal coefficients based on ITAE criterion for a step input. The optimal coefficients of the normalized transfer function8
Gdyr (s) =
8
ad0
sn + adn−1 sn−1 + · · · + ad1 s + ad0
(199)
Some additional details concerned with the optimal coefficients of the normalized transfer function
based on ITAE criterion can be found in the following references: Dorf, R.C. and Bishop, R.H. Modern
control systems, 9th ed., Upper Saddle River, NJ: Prentice Hall, 2001; Franklin, G.F., Powell, J.D. and
Emami-Naeini A. Feedback control of dynamic systems, 4th ed., Prentice Hall, 2002; Kuo, B.C. and
Golnaraghi, F. Automatic control systems, 8th ed., New York: John Wiley & Sons, Inc., 2003.
Student Solutions Manual for “Design of nonlinear control systems . . .”
62
based on the integral of time multiplied by absolute error (ITAE) criterion
IT AE =
Z ∞
0
t|e(t)| dt
(200)
for a step input are given by
s + ωn ,
s2 + 1.4ωn s + ωn2 ,
s3 + 1.75ωn s2 + 2.15ωn2 s + ωn3 ,
s4 + 2.1ωn s3 + 3.4ωn2 s2 + 2.7ωn3 s + ωn4 ,
s5 + 2.8ωn s4 + 5.0ωn2 s3 + 5.5ωn3 s2 + 3.4ωn4 s + ωn5 ,
s6 + 3.25ωn s5 + 6.60ωn2 s4 + 8.60ωn3 s3 + 7.45ωn4 s2 + 3.95ωn5 s + ωn6 ,
where ωn is the natural frequency.
The optimal coefficients based on ITAE criterion for a ramp input. The optimal
coefficients of the normalized transfer function
ad1 s + ad0
Gdyr (s) = n
(201)
s + adn−1 sn−1 + · · · + ad1 s + ad0
based on the ITAE criterion (200) for a ramp input are given by
s2 + 3.2ωn s + ωn2 ,
s3 + 1.75ωn s2 + 3.25ωn2 s + ωn3 ,
s4 + 2.41ωn s3 + 4.93ωn2 s2 + 5.14ωn3 s + ωn4 ,
s5 + 2.19ωn s4 + 6.50ωn2 s3 + 6.30ωn3 s2 + 5.24ωn4 s + ωn5 .
Auxiliary Material (Euler polynomials)
E1 (z) = 1,
E2 (z) = z + 1,
E3 (z) = z 2 + 4z + 1,
E4 (z) = z 3 + 11z 2 + 11z + 1,
E5 (z) = z 4 + 26z 3 + 66z 2 + 26z + 1.
Auxiliary Material (Describing functions)
Let us consider a symmetric nonlinearity9 v = ϕ(z) as shown in Fig. 22. Assume that
the nonlinearity input is z(t), where z(t) = A sin(ωt). Consider the output v(t) of the
nonlinearity represented by its Fourier series
v(t) = b1 sin(ωt) + c1 cos(ωt) +
∞
X
{bk sin(kωt) + ck cos(kωt)}.
k=2
9
The angle θ defines the slope k of the backlash hysteresis, that is k = tan(θ).
(202)
Student Solutions Manual for “Design of nonlinear control systems . . .”
63
In accordance with the describing function method, the nonlinearity v = ϕ(z) can by
replace by its quasi-linear approximation. Hence, we get a sinusoidal describing function10
Gn (j, A) = q1 + jq2 , where q1 = b1 /A and q2 = c1 /A.
Figure 22: Nonsmooth nonlinearities.
Nonlinear function
Describing function Gn (j, A) = q1 + jq2
Saturation
∆ 2
2M  −1 ∆
∆
,
q1 =
sin
+
1−
π∆
A
A
A

µ
s
¶
·
¸

q2 = 0, A ≥ ∆
Relay with dead zone
4M
q1 =
πA
Ideal relay
Hysteresis
s
·
1−
q1 =
4M
q1 =
πA
s
·
1−
¸
∆ 2
, q2 = 0, A ≥ ∆
A
4M
,
πA
¸
∆ 2
4M ∆
, q2 = −
, A≥∆
A
πA2
·
Backlash hysteresis
q1 =
2∆
+2 1 −
A
10
µ
k π
2∆
+ sin−1 1 −
π 2
A
µ
q2 = −
q2 = 0
¶
q
A
∆
−1
A/∆
¶

,
A
− 1)
4k ( ∆
, k = tan(θ), A ≥ ∆
π (A/∆)2
y = sin−1 (x) denotes the inverse sine of x, i.e., sin(y) = x.
Student Solutions Manual for “Design of nonlinear control systems . . .”
Auxiliary Material (The Laplace Transform and the Z-Transform)
f (t)
1
t
t2
2!
t3
3!
e−at
te−at
sin ωt
cos ωt
e−at sin ωt
e−at cos ωt
F (s) = L[f (t)]
F (z) = Z[f (kTs )]
1
s
1
s2
1
s3
1
s4
1
s+a
a
s(s + a)
1
(s + a)2
1
(s + a)3
ω
s2 + ω 2
s
2
s + ω2
ω
(s + a)2 + ω 2
s+a
(s + a)2 + ω 2
b−a
(s + a)(s + b)
s
(s + a)2
a
s2 (s + a)
a2
s(s + a)2
(b − a)s
(s + a)(s + b)
z
z−1
Ts z
(z − 1)2
Ts2 z(z + 1)
2! (z − 1)3
3
Ts z(z 2 + 4z + 1)
3! (z − 1)4
z
, d = e−aTs
z−d
(1 − d)z
(z − 1)(z − d)
zdTs
(z − d)2
z(z + d)dTs2
2! (z − d)3
z sin ωTs
z 2 − 2z(cos ωTs ) + 1
z(z − cos ωTs )
2
z − 2z(cos ωTs ) + 1
zd sin ωTs
2
z − 2zd(cos ωTs ) + d2
z 2 − zd cos ωTs
z 2 − 2zd(cos ωTs ) + d2
(d − c)z
, c = e−bTs
(z − d)(z − c)
z[z − d(1 + aTs )]
(z − d)2
z[(aTs − 1 + d)z + (1 − d − aTs d)]
a(z − 1)2 (z − d)
z[z(1 − d − aTs d) + d2 − d + aTs d]
(z − 1)(z − d)2
z[z(b − a) − (bd − ac)]
(z − d)(z − c)
64
Student Solutions Manual for “Design of nonlinear control systems . . .”
65
Errata for the book
“Design of nonlinear control systems with the highest derivative in feedback”, 2004.
Page
Line
It is printed
Should be printed
30
5 from top
+6.30ωn2 s2 +
+6.30ωn3 s2 +
60
12 from top
V (u)
dt
dV (u)
dt
74
5 from top
uN ID = ϕ(X, R, w)
U1s (t) = U1N ID (t) = [uN ID , 0,
· · · , 0]T , U1N ID = ϕ(X, R, w)
74
14 from top
+µϕ̃(X, R, Ṙ, w, ẇ)
−µϕ̃(X, R, Ṙ, w, ẇ)
91
1 from below
Lduf (ω)
Gduf (s)
92
2 from top
Lduf (ω)
Gduf (s)
92
10 from top
Guf (s)
Gduf (s)
97
4 from below
Luf (ω)
Luns (ω)
102
15 from top
Luns (ω)
Luf (ω)
113
14 from top
x(2) = x(1) +x+5u
x(2) = x(1) +x+5u+w
114
2 from below
184
13 from top
ns
ωmin
= 102
{T2−1 [r2 − r] − r(1) }
ns
ωmin
= 103
(1)
{T2−1 [r2 − x2 ] − x2 }
188
9 from below
µu1 + d10 u =
µu1 + d0 u1 =
188
3 from below
bd
1τ
T2
ad
1
T
200
2 from below
216
4 from top
y1 = x1 +x2 , y2 = x1 +2x2
y1 = x1 , y2 = x2
250
16 from top
ϕ(τ ) ≥ 0.35 rad
ϕ ≥ 0.7 rad
250
9 from below
ϕ(τ ) ≥ 0.2 rad
ϕ ≥ 0.7 rad
251
9 from top
ϕ(τ ) ≥ 0.3 rad
ϕ ≥ 0.7 rad
251
19 from top
ϕ(τ ) ≥ 0.25 rad
ϕ ≥ 0.5 rad
251
12 from below
ϕ(τ ) ≥ 0.25 rad
ϕ ≥ 0.7 rad
251
6 from below
ϕ(τ ) ≥ 0.2 rad
ϕ ≥ 0.7 rad
252
2 from top
ϕ(τ ) ≥ 0.3 rad
ϕ ≥ 0.7 rad
310
2 from below
sampling period
sampling rate
HF A
HF A
(1)
(2)
µ2i ui
(1)
+ di1 µi ui
+ di0 ui
d
(1)
(2)
µ2i ũi
(1)
+ di1 µi ũi
+ di0 ũi
311
11 from below
σ ≈ 20%
σ d ≈ 0%
324
11 from top
ϕ(τ ) ≥ 0.2 rad
ϕ ≥ 0.7 rad
324
7 from below
ϕ(τ ) ≥ 0.3 rad
ϕ ≥ 0.7 rad