ssm for design of nonlinear control systems

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

Copyright c© 2007 by Valery D. Yurkevich

Preface

Student Solutions Manual contains complete solutions of 20 % of Exercises from thebook “Design of Nonlinear Control Systems with the Highest Derivative in Feedback”,World Scientific, 2004, (ISBN 9812388990). The manual aims to help students under-stand a new methodology of output controller design for nonlinear systems in presence ofunknown external disturbances and varying parameters of the plant.

The solutions manual is accompanied by Matlab-Simulink files1 for calculations andsimulations related with Exercises. The program files provide the student a possibilityto design the discussed control systems in accordance with the assigned performancespecifications of output transients, and make a comparison of simulation results.

The distinguishing feature of the discussed throughout design methodology of dy-namic output feedback controllers for nonlinear systems is that two-time-scale motionsare induced in the closed-loop system. Stability conditions imposed on the fast and slowmodes, and a sufficiently large mode separation rate, can ensure that the full-order closed-loop system achieves desired properties: the trajectories of the full singularly perturbedsystem approximate to the trajectories of the reduced model, where the reduced model isidentical 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 stabilityof the fast mode and the sufficiently large mode separation rate are maintained in theclosed-loop system. Consequently, the ensuring of the fast mode stability by selection ofcontrol law parameters is the problem requiring undivided attention and that constitutesthe core of the controller design procedure.

In general, the selection of the control law structure as well as selection of controllerparameters are not unique, inasmuch as a set of constraints has to be taken into accountsuch as a range of variations for plant parameters and external disturbances, requiredcontrol accuracy, requirements on load disturbance rejection as well as high frequencymeasurement noise rejection. Therefore, it would be much more correctly, if the studentwill take up the solution presented in the manual as an example or draft version of suchsolution, and then one can make a try to extend the solution by taking into account someadditional practical limitations.

Overall, the above mentioned book, along with the Student Solutions Manual, as wellas accompanying Matlab-Simulink files are an excellent learning aid for advanced studyof real-time control system designing and ones may be used in such course as “Design ofNonlinear Control Systems”, where prerequisites are “Linear Systems” and “NonlinearSystems”. Any comments about the solutions manual (including any errors noticed) canbe sent to 〈[email protected]〉 or 〈[email protected]〉 with the subject heading 〈book〉. Theywill be sincerely appreciated.

Valery D. Yurkevich

1A set of Matlab-Simulink files for the Student Solutions Manual can be downloaded from the websitehttp://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


Chapter 1

Exercise 1.2 The behavior of a dynamical system is described by the equation

x(2) + 1.5x(1) + 0.5x + µ2x2 + [x(1)]21/2 = 0. (1)

Determine the region of µ such that X = 0 is an exponentially stable equilibrium pointof the given system.

Solution.Denote x1 = x, x2 = x(1), and X = [x1, x2]

T . Hence, we have

X = AX + µg(X),

where

A =

[0 1

−0.5 −1.5

], g(X) =

[0

(2x21 + x2

2)1/2

].

Hence, g(X)|X=0 = 0, and so the perturbation g(X) is vanishing at the equilibrium pointof the linear nominal system X = AX. We can find that

‖g(X)‖2 = (2x21 + x2

2)1/2 ≤ (2x2

1 + 2x22)

1/2 =√

2 ‖X‖2.

Denote c5 =√

2. Then, from the Lyapunov equation

PA + AT P = −Q (2)

with Q = I, we get

P =

[2 11 1

],

where λmin(P ) = 0.382 and λmax(P ) = 2.618. Hence, if the inequalities

0 < µ <λmin(Q)

2λmax(P )c5

= 0.135

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

[x1

µx2

]=

[1 −12 1

] [x1

x2

]. (3)

Obtain and analyze the stability of the slow-motion subsystem (SMS) and the fast-motionsubsystem (FMS).

Solution.


From (3), we haveX = A(µ)X,

where

A(µ)=

[1 −1

2/µ 1/µ

].

The characteristic polynomial of the system is given by

det [sI − A(µ)] = s2 −(

1

µ+ 1

)s +

3

µ.

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

dt0x1 = µ[x1 − x2],

d

dt0x2 = 2x1 + x2.

Take µ = 0. Hence, we get

d

dt0x1 = 0, =⇒ x1 = const,

d

dt0x2 = 2x1 + x2.

By returning to the primary time scale t, the FMS

µd

dtx2 = 2x1 + x2

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

dtx1 = x1 − x2,

0 = 2x1 + x2

the equation of the SMS

d

dtx1 = 3x1

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.

2Phase Plane Demo for Matlab by John Polking at Rice Univerisity contains the programsdfield.m, dfsolve, pplane.m, and ppsolve.m. The programs can be downloaded from the websitehttp://calclab.math.tamu.edu/docs/math308/MATLAB-pplane/ as well as the instructions for use.


Chapter 2

Exercise 2.1 Construct the reference model in the form of the 2nd order differentialequation given by

T ny(n) + adn−1T

n−1y(n−1) + · · ·+ ad1Ty(1) + y = r (4)

in such a way that the step response parameters of the output meet the requirementstds ≈ 6 s, σd ≈ 0 %. Plot by computer simulation the output response, and determine thesteady-state error from the plot for input signals of type 0 and 1.

Solution.Take tds = 6 s and σd = 0 %, then by3

θd = tan−1

(π

ln(100/σd)

), ωd =

4

tds,

we get θd = 0 rad, ζd = 1, and ωd = ωn = 0.6667 rad/s. By selecting the 2 rootss1 = s2 = −0.6667, we obtain the desired characteristic polynomial given by

s2 + 1.333s + 0.4444.

Consider the desired transfer function given by

Gdyr(s) =

0.4444

s2 + 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 = lims→0

se(s)

= lims→0

s[1−Gdyr(s)]

1

srs

= lims→0

s2 + 1.333s

s2 + 1.333s + 0.4444rs = 0.

Let r(t) be the input signal of type 1, that is, r(t) = rvt 1(t), where rv = const and rv 6= 0.Hence, r(s) = rv/s2 and we get

evr = es = lim

s→0se(s)

= lims→0

s[1−Gdyr(s)]

1

s2rv

= lims→0

s + 1.333

s2 + 1.333s + 0.4444rv

=1.333

0.4444rv ≈ 3rv.

3z = tan−1(x) denotes the arctangent of x, i.e., tan(z) = x.


Run the Matlab program e2 1 Parameters.m in order to calculate the reference modelparameters. Next, run the Simulink program e2 1.mdl to get a plot for the output responseof (5). Then, from inspection of the plot, determine the steady-state error for input signalsof 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 differentialequation 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 determinethe 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

s2 + 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 thates

r = 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)

From (7), we obtain evr = 0 and eacc

r = 1/2, where eaccr is the relative acceleration error

due to the reference input r(t).Run the Matlab program e2 3 Parameters.m to calculate the reference model param-

eters as well as to obtain the step response of the reference model (6). Next, run theSimulink 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


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)

Consider the control lawu = k0[F (X, r)− x(n)] (10)

with real differentiating filter

µqx(q) + dq−1µq−1x(q−1) + · · ·+ d1µx(1) + x = x, X(0) = X0, (11)

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) isassigned by (9). Then, the control law with the highest derivative of the output signal infeedback and an ideal differentiating filter is given by

u = k0[F (x(1), x, r)− x(2)],


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

1 + g(t)k0

[f(x(1), x)− F (x(1), x, r)] (12)

is the SMS equation.Next, let us consider the system given by

µ2x(2) + d1µx(1) + x = x,

where x(1) and x(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 systemequations with the real differentiating filter is given by

x(2) = f(x(1), x) + g(t)k0[F (x(1), x, r)− x(2)],

µ2x(2) + d1µx(1) + x = x. (13)

Denote x(1) = x1 and x(2) = x2. Consider the extended system given by

x(2) = f(x(1), x) + g(t)k0[F (x1, x, r)− x2],

µ2x(2) + d1µx(1) + x = x,

µ2x(2)1 + d1µx

(1)1 + x1 = x(1),

µ2x(2)2 + d1µx

(1)2 + x2 = 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 (x1, x, r)− x2],

µ2x(2) + d1µx(1) + x = x,

µ2x(2)1 + d1µx

(1)1 + x1 = x(1), (14)

µ2x(2)2 + d1µx

(1)2 + [1 + g(t)k0]x2 = f(x(1), x) + g(t)k0F (x1, x, r).

From (14), taking µ = 0, we get the discussed above SMS (12). The FMS of the extendedsystem (14) is given by

µ2x(2) + d1µx(1) + x = x,

µ2x(2)1 + d1µx

(1)1 + x1 = x(1),

µ2x(2)2 + d1µx

(1)2 + [1 + g(t)k0]x2 = f(x(1), x) + g(t)k0F (x1, x, r)


where we assume that f = const, g = const. The behavior of x(2) is described by the lastdifferential equation, where the characteristic equation is given by µ2s2 + d1µs + 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 FMSand 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) withthe desired dynamics given by (9) and the real differentiating filter (11), where k0 = 40and q = 2. Determine the parameters µ and d1 of (11) such that the damping ratioexceeds 0.5 in the FMS and the degree of time-scale separation between FMS and SMSexceeds 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

µ2s2 + d1µs + γ = 0 (15)

is the characteristic equation of the FMS for x(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

s1,2 = − d1

2µ± j

√|d2

1 − 4γ|2µ

=⇒ ζFMS

=d1

2√

γ=⇒

ζminFMS

=d1

2√

γmax, ζmin

FMS= 0.5 =⇒ d1 = 9.4340.

Next, let us find estimates for µ based on different notions for degree of time-scale sepa-ration between FMS and SMS in the closed-loop system. Let us take γ = γmin, then

µ2s2 + d1µs + γmin =⇒ s2 +1

µa

FMS

1 s +1

µ2a

FMS

0


where aFMS

1 = d1, aFMS

0 = γmin = 9, and the state matrix of the FMS is given by

AFMS

= µ−1A22 ,

A22 =

[0 1

−γmin −d1

]=

[0 1−9 −9.4340

].

Since d21 − 4γmin > 0, then from (15) we obtain

s1 = − d1

2µ+

√d2

1 − 4γmin2µ

s2 = − d1

2µ−

√d2

1 − 4γmin2µ

.

Hence, ωminFMS

= |s1|.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

sSMS

1,2 = −0.6± j0.8 =⇒ ωmaxSMS

= 0.6, (aSMS

0 )1/2 = 1.

By solving the Lyapunov equations

PFA22 + AT

22PF= −Q

F,

PSA

S+ AT

SP

S= −Q

S,

where QF

= I and QS

= I, we obtain

PF

=

[1.0541 0.05560.0556 0.0589

], P

S=

[1.4333 0.5

0.5 0.8333

].

Hence,

λmax(PF) = 1.0572, λmin(P

F) = 0.0558,

λmax(PS) = 1.7164, λmin(P

S) = 0.5502.

Finally, we get the following estimates for µ based on the various notions for degree oftime-scale separation between FMS and SMS in the closed-loop system, that are:

η1 =λmin(P

S)

µλmax(PF), η1 = 10 =⇒ µ = 0.052,

η2 =ωmin

FMS

ωmaxSMS

=d1

2µωmaxSMS

, η2 = 10 =⇒ µ = 0.1795,

η3 =(a

FMS

0 )1/2

µ(aSMS

0 )1/2=

√γminµ

, η3 = 10 =⇒ µ = 0.3.


Figure 5: Simulation results of the closed-loop system (13) for d1 = 9.4340 and µ = 0.052s.

Run the Matlab program e3 2 Parameters.m to calculate d1 and the above estimates forµ based on such criteria as η1, η2, and η3. Next, run the Simulink program e3 2.mdl, toget the step response of the closed-loop system for d1 = 9.4340 and µ = 0.052 s. Thesimulation 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

µqu(q) + dq−1µq−1u(q−1) + · · ·+ d1µu(1) + d0u

=k0

T n−T nx(n) − ad

n−1Tn−1x(n−1) − · · · − ad

1Tx(1) − x

+ bdρτ

ρr(ρ) + bdρ−1τ

ρ−1r(ρ−1) + · · ·+ bd1τr(1) + r. (17)

where q = 3. Determine the FMS and SMS equations from the closed-loop system equa-tions.

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 (18)

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

µ3u(3) + d2µ2u(2) + d1µu(1) + d0u = k0F (x(1), x, r(1), r)− x(2),


that is

µ3u(3) + d2µ2u(2) + d1µu(1) + d0u = 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,

µ3u(3) + d2µ2u(2) + d1µu(1) + d0u = k0F (x(1), x, r(1), r)− x(2).

Denote x1 = x, x2 = x(1), u1 = u, u2 = µu(1), and u3 = µ2u(2). From the above closed-loopsystem equations, we obtain

d

dtx1 = x2,

d

dtx2 = f(x1, x2) + g(t)u1,

µd

dtu1 = u2,

µd

dtu2 = u3,

µd

dtu3 = −d0u1 − d1u2 − d2u3 + k0

F (x2, x1, r

(1), r)− d

dtx2

.

Substitution of the right member of the second equation into the last one yields theclosed-loop system equations in the following form:

d

dtx1 = x2,

d

dtx2 = f(x1, x2) + g(t)u1,

µd

dtu1 = u2,

µd

dtu2 = u3, (19)

µd

dtu3 = −d0 + k0g(t)u1 − d1u2 − d2u3

+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 intothe closed-loop system equations given by (19). We obtain

d

dt0x1 = µx2,

d

dt0x2 = µf(x1, x2) + g(t)u1,

d

dt0u1 = u2,


d

dt0u2 = u3,

d

dt0u3 = −d0 + k0g(t)u1 − d1u2 − d2u3

+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

dt0x1 = 0,

d

dt0x2 = 0,

d

dt0u1 = u2,

d

dt0u2 = u3,

d

dt0u3 = −d0 + k0g(t)u1 − d1u2 − d2u3

+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 = const, x2 = const,

µd

dtu1 = u2,

µd

dtu2 = u3, (20)

µd

dtu3 = −d0 + k0g(t)u1 − d1u2 − d2u3

+k0

F (x2, x1, r

(1), r)− f(x1, x2)

.

These equations may be rewritten as

µ3u(3) + d2µ2u(2) + d1µu(1) + d0 + k0g(t)u

= k0F (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:

x1 = x2,

x2 = F (x2, x1, r(1), r)

+d0

d0 + k0g(t)f(x1, x2)− F (x2, x1, r

(1), r). (22)


At the same time, we can find the above SMS by some another way. Suppose theFMS (20) is stable. Taking µ → 0 in (21) we get u(t) = us(t), where us(t) is a steadystate (more precisely, quasi-steady state) of the FMS (20) and

us =k0

d0 + k0g(t)F (x2, x1, r

(1), r)− f(x1, x2).

Substitution of us into (18) yields the SMS equation given by

x(2) = F (x(1), x, r(1), r)

+d0

d0 + k0g(t)f(x, x(1))− F (x(1), x, r(1), r),

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,ζ

FMS≥ 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

η3 =(a

FMS

0 )1/m

µ(aSMS

0 )1/n.

Solution.Consider the system given by (23). Then n = 2 and 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).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 asthe requirement on the high frequency sensor noise attenuation is not specified, then takeq = n = 2. Therefore, consider the control law given by

µ2u(2) + d1µu(1) + d0u = k0F (x(1), x, r)− x(2), (24)

that isµ2u(2) + d1µu(1) + d0u = k0−x(2) − 2x(1) − x + r. (25)

Consider the closed-loop system equations given by

x(2) = f(x(1), x) + g(t)u, (26)

µ2u(2) + d1µu(1) + d0u = k0F (x(1), x, r)− x(2). (27)


From the above closed-loop system equations, we get the FMS given by

µ2u(2)+d1µu(1)+[d0 + k0g]u=k0F (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 theSMS given by

x(2) = F (x(1), x, r) +d0

d0 + k0g(t)f(x(1), x)− F (x(1), x, r). (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,

eFmax = 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)| ≤ eF

max = 1.15, we obtain

|k0| ≥ d0

gmin

maxΩX,R,w

|F (X,R)− f(X, w)|eFmax

− 1

=1

0.5

[23 + 22

1.15− 1

]≈ 90.

Consider a steady state of the SMS, that is x(2) = x(1) = 0. Hence, we obtain

x(2) = F (x(1), x, r) +d0

d0 + k0gf(x(1), x)− F (x(1), x, r) =⇒

x(2)︸︷︷︸=0

= −2x(1) − x + r︸︷︷︸=es

+d0

d0 + k0gx + x|x(1)|︸︷︷︸

=xs=r−es

−−2x(1) − x + r︸︷︷︸=es

=⇒

es = − d0

k0g − d0

r.

From the requirement

|es| ≤∣∣∣∣∣−

d0

k0g − d0

r

∣∣∣∣∣ ≤d0rmax

k0gmin − d0

≤ esmax = εrrmax = 0.02,

we obtain

|k0| ≥[d0rmaxesmax

− d0

]1

gmin=

[1 · 10.02

− 1]

1

0.5= 98.


Let us take k0 = 100.From the SMS and reference model equations, it follows that

s2 + 2s + 1 = 0

is the characteristic equation of the SMS, where aSMS

0 = 1, and

µ2s2 + d1µs + d0 + k0g = 0

is the characteristic equation of the FMS. Hence, we obtain

η3 =(a

FMS

0 )1/2

µ(aSMS

0 )1/2=

√d0 + k0g

µ(aSMS

0 )1/2≥

√d0 + k0gmin

µ(aSMS

0 )1/2≥ ηmin

3 = 20

=⇒ µ ≤√

d0 + k0gmin

ηmin3 (a

SMS

0 )1/2=

√1 + 100 · 0.5

20 · 1 ≈ 0.3571 s.

From the characteristic equation of the FMS, we get

sFMS

1,2 = − d1

2µ± j

√d2

1 − 4(d0 + k0g)

2µ= α± jβ,

where we assume that d21 − 4(d0 + k0g) < 0 when g = gmax. Hence, we can find

ζFMS

= cos(θFMS

) = |α|/√

α2 + β2

=d1

2√

d0 + k0g≥ ζmin

FMS= 0.5 =⇒

d1 ≥ 2ζminFMS

√d0 + k0gmax

= 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

u(2) +d1

µu(1) +

d0

µ2u

= −k0

µ2x(2) − k0a

d1

µ2Tx(1) − k0

µ2T 2x +

k0

µ2T 2r =⇒

u(2) + a1u(1) + a0u = b2x

(2) + b1x(1) + b0x + c0r (30)

where

a1 =d1

µ, a0 =

d0

µ2,

b2 = −k0

µ2, b1 = −k0a

d1

µ2T, b0 = − k0

µ2T 2,

c0 =k0

µ2T 2.


Then, in order to find the block diagram of the discussed control law, from (30), we get

u(2) − b2x(2) + a1u

(1) − b1x(1)

= −a0u + b0x + c0r︸︷︷︸=u2

=⇒

u(1) − b2x(1) + a1u− b1x = u2 =⇒

u(1) − b2x(1) = u2 − a1u + b1x︸︷︷︸

=u1

=⇒

u = u1 + b2x.

Hence, we obtain the equations of the controller given by

u1 = u2 − a1u + b1x,

u2 = −a0u + b0x + c0r. (31)

u = u1 + b2x.

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 controllerparameters. Next, run the Simulink program e5 1.mdl, to get the step response of theclosed-loop system.4 It can be verified that the simulation results confirm the analyticalcalculations. 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

µ2s2 + d1µs + d0

r(s)− k0(s2 + 2s + 1)

µ2s2 + d1µs + d0

x(s). (32)

Take d0 = 0, then from (32) the conventional PID controller with low-pass filtering

u(s) =1

τLPF

s + 1k

−2x(s) +

1

s[r(s)− x(s)]− sx(s)

results, where the low-pass filter (LPF) is given by 1/(τLPF

s + 1) and

k =k0

µd1

, τLPF

=µ

d1

.

4Throughout the simulation the following solver options are used: variable-step, ode113(Adams),relative tolerance equals 1e-6.


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 ≈ 1s, σd ≈ 10 %, ζ

FMS≥ 0.3, and η3 ≥ 10. The additional requirement

|Guns(jω)| ≤ εuns(ω), ∀ ω ≥ ωns

min (34)

should be provided such that εuns(ω) = 103 and ωns

min = 103 rad/s. Compare simulationresults 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 outputsignal. Hence, consider the reference model given by

x(2) = F (x(1), x, r).

Take tds = 1 s, σd = 10 %, then by

θd = tan−1

(π

ln(100/σd)

), ωd =

4

tds,

we get θd = 0.9383 rad, ζd = 0.5912, ωd = 4 rad/s and ωn = 6.7664 rad/s. By selectingthe 2 roots s1,2 = −4 ± j5.4575, where Re(s1,2) = −ωd = −ωn cos(θd) and |Im(s1,2)| =ωn sin(θd), we obtain the desired characteristic polynomial s2 + 8s + 45.78. Hence, thedesired transfer function is given by

Gdxr(s) =

45.78

s2 + 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 (35)

follows.


Control law of the 2-nd order. At the beginning, let us take q = 2. Therefore, thecontrol law will be constructed in the form (24), where the reference model is given by(35). Hence, the control law is

µ2u(2) + d1µu(1) + d0u

= k0−x(2) − 8x(1) − 45.78x + 45.78r. (36)

The closed-loop system equations are given by (26)–(27). Hence, the FMS and SMSequations are given by (28) and (29), respectively.

Selection of control law parameters, when q = 2. The control law parameters k0, d0, d1, µcan 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 therequirement εr = 0, take d0 = 0. Then the gain k0 can be selected such that k0gmin = 10.Hence, we get k0 = 5.

From (28), (29), and (35), we have that

s2 + 8s + 45.78 = 0

is the characteristic equation of the SMS, where aSMS

0 = 45.78, and

µ2s2 + d1µs + d0 + k0g = 0

is the characteristic equation of the FMS. Hence, we obtain

η3 =(a

FMS

0 )1/2

µ(aSMS

0 )1/2=

√d0 + k0g

µ(aSMS

0 )1/2≥

√d0 + k0gmin

µ(aSMS

0 )1/2≥ ηmin

3 = 10

=⇒ µ ≤√

d0 + k0gmin

ηmin3 (a

SMS

0 )1/2=

√0 + 5 · 2

10 · 6.766≈ 0.04673 s.


sFMS

1,2 = − d1

2µ± j

√d2

1 − 4(d0 + k0g)

2µ= α± jβ,

where we assume that the FMS is underdamped-stable, that is

d21 − 4(d0 + k0g) < 0

when g = gmax. Hence, we can find

ζFMS

= cos(θFMS

) = |α|/√

α2 + β2

=d1

2√

d0 + k0g≥ ζmin

FMS= 0.3 =⇒

d1 ≥ 2ζminFMS

√d0 + k0gmax = 2 · 0.3√0 + 5 · 2 ≈ 1.8974.


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

Guns(s) = kuns

Ad(s)

DFMS

(s)

=5 · 45.78

d0 + k0g· (1/45.78)s2 + (8/45.78)s + 1

[µ2/(d0 + k0g)]s2 + [d1µ/(d0 + k0g)]s + 1

is the input sensitivity function with respect to noise for high frequencies, where the re-quirement on high-frequency sensor noise attenuation is given by the following inequality:

|Guns(jω)| ≤ εuns = 103, ∀ ω ≥ ωns

min = 103 rad/s. (37)

Take µ = 0.04673 s and d1 = 1.8974, then |Guns(jωns

min)| ≈ 2208 (where 20 lg 2208 ≈66.88 dB), or, for the sake of simplicity, we can find the limit given by

limω→∞ |Guns(jω)| = |k0|

µq≈ 2289,

where 20 lg 2289 ≈ 67.19 dB. Hence, the requirement (37) on high-frequency sensor noiseattenuation 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 ofGuf (jω), where

Guf (s) = kuf1

DFMS

(s),

kuf =k0

d0 + k0g,

DFMS

(s) =µ2

d0 + k0gs2 +

d1µ

d0 + k0gs + 1.


Hence, we getLuf (ω

ns

min) = 20 lg |Guf (jωns

min)| ≈ −53 dB

andL

HFA

max(ωns

min) = 20 lg εns − 20[n + ϑ] lg ωns

min = −60 dB,

where ϑ = 0. Hence, the requirement for high-frequency sensor noise attenuation

Luf (ωns

min) ≤ LHFA

max(ωns

min)

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 controllergiven by (36), where q = 2. Next, run the Simulink program e5 10 a.mdl, to get a stepresponse as well as ramp response (by using Switch 1) of the closed-loop system. Thesimulation 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 2in 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(103t). 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-frequencysensor noise attenuation given by (37), let us take q = 3 and consider the control lawgiven by

µ3u(3) + d2µ2u(2) + d1µu(1) + d0u = k0F (y(1), y, r)− y(2),


where y(t) = x(t) + ns(t) and the reference model is the same as (35). Hence, the controllaw can be rewritten as

µ3u(3)+d2µ2u(2)+d1µu(1)+d0u=k0−y(2)−ad

1y(1)−ad

0y+ad0r, (38)

wheread

1 = 8, ad0 = 45.78.

Note that the control law (38) may be expressed in terms of transfer functions as

u(s) =k0a

d0

µ3s3 + d2µ2s2 + d1µs + d0

r(s)− k0(s2 + ad

1s + ad0)

µ3s3 + d2µ2s2 + d1µs + d0

x(s). (39)

Take d0 = 0, then from (39) the conventional PID controller with low-pass filtering

u(s) =1

τ 2lpfs

2 + aLPF

τLPF

s + 1k

−ad

1x(s) +ad

0

s[r(s)− x(s)]− sx(s)

results, where the low-pass filter is given by 1/(τ 2LPF

s2 + aLPF

τLPF

s + 1) and

k =k0

µd1

, τLPF

=µ√d1

, aLPF

τLPF

=µd2

d1

.

The FMS characteristic polynomial in the closed-loop system is given by

µ3s3 + d2µ2s2 + d1µs + d0 + k0g.

Let us consider the selection of the control law parameters based on Bode amplitude plotof the closed-loop FMS given by

Luf (ω) = 20 lg |Guf (jω)|,

where

Guf (s) = kuf1

DFMS

(s), kuf =

k0

d0 + k0g,

DFMS

(s) =µ3

d0 + k0gs3 +

d2µ2

d0 + k0gs2 +

d1µ

d0 + k0gs + 1. (40)

By the same way as was shown above, take d0 = 0 and k0 = 5. Hence, kuf = 0.5. Then,let us perform Gd

uf (s) in the corner frequency factored form given by

Gduf (s) = kuf

1

[T 21 s2 + 2ζ1T1s + 1][T2s + 1]

. (41)

Then, the roots of quadratic factor

T 21 s2 + 2ζ1T1s + 1


are the dominant poles of Gduf (s), where the damping ratio ζ1 is selected such that

ζ1 = ζminFMS

= 0.3.

Take tds ≈ tds,SMSand tds,FMS

= tds/η, where η = 10. Then, we can obtain

ω1 ≈ 4

tds,FMS

=4η

ζ1tds

=4 · 10

ζ1 · 1 ≈ 133 =⇒ T1 =1

ω1

= 0.0075 s.

Let us calculate a lower bound for T2 from the condition

Lduf (ω

ns

min) = LHFA

max(ωns

min),

where

LHFA

max(ωns

min) = 20 lg εns − 20[n + ϑ] lg ωns

min= 20 lg 103 − 20[2 + 0] lg 103 = −60 dB.

Denote L = εns/(ωns

min)n+ϑ = 10−3. Hence, we get

|Gduf (jω

ns

min, Tmin2 )| = L =⇒

kuf

|[1− T 21 [ωns

min]2 + j2ζ1T1[ωns

min]][jTmin2 ωns

min + 1]|= L =⇒

Tmin2 =

1

ωns

min

[kuf/L]2

(1− [T1ωns

min]2)2 + (2ζ1T1ωns

min)2− 1

1/2

=⇒

Tmin2 =

1

103

[[0.5/10−3]2

(1−[0.0075·103]2)2+(2·0.3·0.0075·103)2−1

]1/2

≈ 0.009 s.

The time constant T2 should be selected such that the inequalities Tmin2 ≤ T2 ≤ T1 hold.

We see, there is apparent contradiction. Therefore, let us replace the degree of time-scaleseparation between fast and slow modes η = 10 by η = 8 and redesign the parameterT1 again. We get T1 = 0.0094 s. Accordingly, by the same way as above, we obtain

Tmin2 = 0.0057 s. Hence, the condition Tmin

2 ≤ T2 ≤ T1 holds and then, we can take

T2 = Tmin2 = 0.0057 s.

As a result of the above, we obtain

DdFMS

(s) = [T 21 s2 + 2ζ1T1s + 1][T2s + 1]

= 4.9699 · 10−7s3 + 1.197 · 10−4s2 + 0.0113s + 1.

From (40), and by taking into account that d0 = 0 as well as the requirement

DFMS

(s) = DdFMS

(s),


we obtain

µ = d dq k0g1/3 ≈ 0.0171,

d1 =d d

1 [k0g](2)/3

[d d3 ]1/3

≈ 6.6097, (42)

d2 =d d

2 [k0g](1)/3

[d d3 ]2/3

≈ 4.1101.

Finally, the control law (38) can be rewritten in the form given by

u(3) +d2

µu(2) +

d1

µ2u(1) +

d0

µ3u

= −k0

µ3y(2) − k0a

d1

µ3Ty(1) − k0

µ3T 2y +

k0

µ3T 2r =⇒

u(3)+a2u(2)+a1u

(1)+a0u = b2y(2)+b1y

(1)+b0y+c0r. (43)

From (43), we can obtain the equations of the controller given by

u1 = u2 − a2u1 + b2y,

u2 = u3 − a1u1 + b1y,

u3 = −a0u1 + b0y + c0r, (44)

u = u1,

where

a2 =d2

µ, a1 =

d1

µ2, a0 =

d0

µ3,

b2 = −k0

µ3, b1 = −k0a

d1

µ3, b0 = −k0a0

µ3, c0 =

k0a0

µ3.

The Bode plots of Guns(jω) and Guf (jω) are shown in Fig. 11, where the parametersµ, 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 refer-ence model parameters, Bode plots of Guns(jω), and Guf (jω), as well as the parametersof 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-loopsystem. By Switch 3, add the hign frequiency sensor noise ns(t) to the output, that isy(t) = x(t) + ns(t), where ns(t) = 10−3 sin(103t). The simulation results are shown inFig. 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 modelcan be changed from 1 to 2. Then, instead of (43), we have the controller given by

u(3) + a2u(2) + a1u

(1) + a0u

= b2y(2) + b1y

(1) + b0y + c1r(1) + c0r, (45)


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

u1 = u2 − a2u1 + b2y,

u2 = u3 − a1u1 + b1y + c1r,

u3 = −a0u1 + b0y + c0r, (46)

u = u1.

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)



where g = 1 and the reference model x(2) =F (x(1), x, r) is assigned by

x(2) = T−2−a1Tx(1) − x + r. (48)

The control law has the form

µ2u(2) + d1µu(1) + d0u = k0F (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 ofstability 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),

µ2u(2) + d1µu(1) + d0u = k0F (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−ad

1Tx(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

µ2u(2)+d1µu(1)+d0u+k0gu(t−τ)=k0F (x(1), x, r)−f(x(1), x), (50)

where F = const and f = const during the transients in (50). The block diagramrepresentation of the FMS (50) is shown in Fig. 14, where

D(µs) = µ2s2 + d1µs + d0. (51)

Figure 14: Block diagram of the FMS (50) with delay τ , where F = const, f = const.


By the Nyquist stability criterion, the FMS (50) is marginally stable if ωc 6= 0 existssuch that the condition

k0ge−jτmωc

D(jµωc)= −1 + j0, (52)

holds, where ωc is the crossover frequency and τm is an upper bound for delay τ . Thevalue τm determines the region of stability for τ of the FMS.

From (52) and by taking into account the condition d0 = 0, we get∣∣∣∣∣

k0ge−jτmωc

jµωc(jµωc + d1)

∣∣∣∣∣ = 1 =⇒

µ4ω4c + d2

1µ2ω2

c − k20g

2 = 0 =⇒ (53)

y = ω2c > 0, µ4y2 + d2

1µ2y − k2

0g2 = 0 =⇒

y =−d2

1 +√

d41 + 4k2

0g2

2µ2

=−42 +

√44 + 4 · 102 · 12

2 · 0.12≈ 481 =⇒ ωc =

√y ≈ 22 rad/s.

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

u(2) +d1

µu(1) +

d0

µ2u

= −k0

µ2x(2) − k0a

d1

µ2Tx(1) − k0

µ2T 2x +

k0

µ2T 2r =⇒

u(2) + a1u(1) + a0u = b2x

(2) + b1x(1) + b0x + c0r,

where

a1 =d1

µ, a0 =

d0

µ2, b2 = −k0

µ2, b1 = −k0a

d1

µ2T, b0 = − k0

µ2T 2, c0 =

k0

µ2T 2.

Run the Matlab program e6 1 Parameters.m to calculate the region of stability for τ ofthe FMS and the controller parameters. Next, run the Simulink program e6 1.mdl, to getthe step response of the closed-loop system.

Exercise 6.4 Determine the phase margin and gain margin of the FMS based on theinput data of Exercise 6.1 for the time delay τ = 0.3τm, where τ = τm corresponds to themarginally stable FMS.

5y = tan−1(x) denotes the arctangent of x, i.e., tan(y) = x.


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) isgiven by

PM = π − ArgD(jµωc)− τωc. (55)

From (51) and d0 = 0, we get

ArgD(jµωc) =π

2+ tan−1(µωc/d1). (56)

Hence, we obtain

PM =π

2− tan−1(µωc/d1)− τωc. (57)

In order to find the gain margin (GM) of the FMS, consider the equation

Arg[GO

FMS(jωπ)] = −π. (58)

From (58), we get

π

2− tan−1(µωπ/d1)− τωπ = 0, (59)

where τ, τm, ωπ and ωc are calculated by joint numerical resolution of (53), (54), and (59).Finally, we can find that

lπ = |GO

FMS(jωπ)| =

∣∣∣∣∣k0ge−jτωπ

jµωπ(jµωπ + d1)

∣∣∣∣∣

=

∣∣∣∣∣k0g

jµωπ(jµωπ + d1)

∣∣∣∣∣

=k0g√

(µωπ)4 + (d1µωπ)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, PM = 0.7486 rad, GM ≈ 2.9685.

Next, run the Simulink program e6 4.mdl, to get the step response of the closed-loopsystem where τ = 0.0146 s.

Chapter 7

Exercise 7.1 The differential equations of a plant model are given by

x1 = x1 + x2,

x2 = x1 + x2 + x3 + u, (60)

x3 = 2x1 − x2 + 2x3 + a u,

y = x1.


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

y = x1 + x2 =⇒ y = x1 + x2 =⇒ y = 2x1 + 2x2 + x3 + u.

Hence, the system (60) is invertible and the relative degree is given by α = 2. Let usintroduce a state-space transformation defined by

y1 = y = x1,

y2 = y = x1 + x2, (61)

z = x3

From (61), we have

x1 = y1,

x2 = y2 − y1, (62)

x3 = z.

By the change of variables (62), we get the normal form of (60) given by

y1 = y2,

y2 = 2y2 + z + u, (63)

z = y1 + y2 + 2z + a u,

y = y1.

Let the desired stable output behavior is defined by y = F (y, y, r), which can be rewrittenas y1 = F (y1, y1, r) or y1 = y2, y2 = F (y2, y1, r). In order to find the equations of theinternal subsystems, take F (y2, y1, r) = 2y2+z+u. Hence, we obtain the inverse dynamicssolution given by

uid = F (y2, y1, r)− 2y2 − z.

Substitution of u = uid into (63) yields

y1 = y2,

y2 = F (y2, y1, r), (64)

z = (2− a)z + y1 + (1− 2a)y2 + aF (y2, y1, r),

y = y1,

where z = (2−a)z+y1+(1−2a)y2+aF (y2, y1, r) is the equation of the internal subsystemand y2, y1, r are treated as bounded external disturbances of the internal subsystem. Ifa = 1, then 2 − a > 0 and the unique equilibrium point of the internal subsystem isunstable. If a = 3, then 2 − a < 0. Hence, the solutions of the internal subsystem arebounded when variables y2, y1, r are bounded, that is the bounded-input-bounded-state


(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

z = (2− a)z + r,

where the unique equilibrium point of the degenerated subsystem is exponentially stablewhen 2− a < 0.

Exercise 7.10 Verify the invertibility and internal stability of the system

x1 = x21 − x3

2 + u,

x2 = |x2| − u, (65)

y = x1,

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, σd ≈ 0%. Run a computer simulationof the closed-loop system with zero initial conditions. Compare simulation results of theoutput 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

y = y2 − z3 + u,

z = |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 y = F (y, r). Take F (y, r) =

y2 − z3 + u. Hence, we obtain the inverse dynamics solution given by

uid = F (y, r)− y2 + z3.

Substitution of u = uid into (66) yields

y = F (y, r), (67)

z = f(z) + φ(y, r),

where φ(y, r) = y2 − F (y, r) and f(z) = |z| − z3.Denote c = φ(y, r) where c is an arbitrary real number. Take z = 0 in the internal

subsystem given byz = f(z) + c. (68)

Hence, from f(z) + c = 0, depending on value c, it can be up to three equilibrium pointsof 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) + d0u = k0[F (y, r)− y(1)],


where the reference model isy(1) = F (y, r).

Take tds = 3 s, σd = 0 %, then we get θd = 0 rad, ζd = 1, a = ωd = 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) + d0u = k0[−y(1) − ay + ar]. (70)

Closed-loop system. The closed-loop system equations are given by

y = y2 − z3 + u,

z = |z| − u, (71)

µu + d0u = k0[F (y, r)− y].

Denote f(y, z) = y2 − z3. From (71), by following through solution of Exercise 4.2, weobtain the FMS equation, that is

µu + [d0 + k0]u = k0[F (y, r)− f(y, z)],

where F = const, f = const during the transients in (72), and

µs + d0 + k0g = 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 steadystate (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 k0gmin = 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 byωd

n = 1.3333 rad/s. Denote

ωmaxSMS

= ωdn = 1.3333 rad/s, ωmin

FMS=

d0 + k0gminµ

rad/s.


Then, without the taking into account the rate of dynamics of the internal subsystem, letus consider the ratio

η2 =ωmin

FMS

ωmaxSMS

(72)

as a criterion for the degree of time-scale separation between fast and slow motions. Take

ηmin2 = 20. Hence, we obtain

η2 =d0 + k0gmin

µωmaxSMS

≥ ηmin2 = 20 =⇒

µ ≤ µmax =d0 + k0gminηmin

2 ωmaxSMS

=0 + 10 · 120 · 1.3333

≈ 0.375 s.

As a result, take µ = 0.375 s.Control law implementation. Finally, the control law (70) can be rewritten in the

form given by

u(1) +d0

µu = −k0

µy(1) − k0

µTy +

k0

µTr =⇒

u(1) + a0u = b1y(1) + b0y + c0r =⇒ u(1) − b1y

(1) = −a0u + b0y + c0r︸︷︷︸=u1

, (73)

where T = 1/a. From (73), we obtain the equations of the controller given by

u1 = −a0u + b0y + c0r, (74)

u = u1 + b1y,

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


Matlab program e7 10 Parameters.m to calculate the reference model parameter T , as


well as the parameters k0, µ of the controller given by (70). Next, run the Simulinkprogram e7 10.mdl, to get a step response of the closed-loop system with zero initialconditions.

Chapter 8

Exercise 8.1 Verify the invertibility and internal stability of the system given by

x1 = x1 + 3x2 + u1 + u2,

x2 = x1 + x2 + u1 − 2u2, (75)

y1 = x1 + x2, y2 = −x1 + 2x2.

Assume that the inequalities |xj(t)| ≤ 2 ∀ j and |r(t)| ≤ 1 hold for all t ∈ [0,∞). Findthe control law of the form

µqii u

(qi)i + di,qi−1µ

qi−1i u

(qi−1)i + · · ·+ di,1µu

(1)i + di,0ui

= kieFi , Ui(0) = U0

i , i = 1, . . . , p, (76)

u = K0u, u = u1, u2, . . . , upT , u = u1, u2, . . . , upT ,

whereµi > 0, ki > 0, Ui = ui, u

(1)i , . . . , u

(qi−1)i T , qi ≥ αi.

Provide the following requirements: εr1 = 0, εr2 = 0, tds1 ≈ 1 s, σd1 ≈ 0%, tds2 ≈ 3 s,

σd2 ≈ 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 Exer-

cise 7.1, we get

y1 = x1 + x2 =⇒ y1 = 2x1 + 4x1 + 2u1 − u2,

y2 = −x1 + 2x2 =⇒ y2 = x1 − x2 + u1 − 5u2,

where

det G∗ = det

[2 −11 −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,σd

1 = 0 %, then we get θd1 = 0 rad, ζd

1 = 1, a1 = ωd1 = ω1,n = 4 rad/s. By selecting the root

s1 = −a1, we obtain the desired characteristic polynomial s + a1. The desired transferfunction Gd

y1r1(s) = y1(s)/r1(s) is given by

Gdy1r1(s) =

a1

s + a1

.

Hence, we get the reference model for y1 given by

y(1)1 = −a1y1 + a1r1 =⇒ y

(1)1 =

1

T1

[r1 − y1], (77)


where T1 = 1/a1. By the same way as above, take tds2 = 3 s, σd2 = 0 %, then

y(1)2 = −a2y2 + a2r2 =⇒ y

(1)2 =

1

T2

[r2 − y2], (78)

where a2 = 1.3333 and T2 = 1/a2. Hence, the reference model (77)–(78) has be con-structed as

y(1)1 = F (y1, r1), y

(1)2 = F (y2, r2). (79)

Control law . Take q1 = q2 = 1, then the control law is

µ1u(1)1 + d1,0u1 = k1[−y

(1)1 − a1y1 + a1r1], (80)

µ2u(1)2 + d2,0u2 = k2[−y

(1)2 − a2y2 + a2r2], (81)

u1 = k11u1 + k12u2, u2 = k21u1 + k22u2. (82)

Take

K0 = [G∗]−1 =

[k11 k12

k21 k22

]=

[0.5556 −0.11110.1111 −0.2222

].

Hence, the controller of the discussed 2I2O system consists of 2 separate linear controllersgenerating the auxiliary controls u1, u2 and accompanied by the matching matrix K0

where the linear controllers are described by (80) and (81), as well as the matrix K0 isimplemented by (82).

Fast-motion subsystem and slow-motion subsystem. From the closed-loop systemequations given by (75) and (80)–(82), by following through solution of Exercise 7.10again, we obtain the FMS equations, that are

µ1u(1)1 + [d1,0 + k1]u1 = k1[F (y1, r1)− f1(x1, x2)], (83)

µ2u(1)2 + [d2,0 + k2]u2 = k2[F (y2, r2)− f2(x1, x2)], (84)

where y1 = const, y2 = const, x1 = const, x2 = const, f1 = const, and f2 = const duringthe transients in (83)–(84). Due to K0 = [G∗]−1, the characteristic polynomial of theFMS is factorized, this is

(µ1s + d10 + k1)(µ2s + 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 thesame 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 naturalfrequency of the reference model for y1 is given by ωd

1,n = 4 rad/s. Denote

ωmax1,SMS

= ωd1,n = 4 rad/s,

ωmin1,FMS

=d1,0 + k1

µ1

rad/s.


Then, let us consider the ratio

η2 =ωmin

1,FMS

ωmax1,SMS

(86)

as a criterion for the degree of time-scale separation between fast and slow motions in the

first input-output channal. Take ηmin2 = 20. Hence, we obtain

η2 =d1,0 + k1

µ1ωmax1,SMS

≥ ηmin2 = 20 =⇒

µ1 ≤ µ1,max =d1,0 + k1

ηmin2 ωmax

1,SMS

=0 + 10

20 · 4 = 0.125 s.

As a result, take µ1 = 0.125 s. By the same way, we get

µ2 ≤ µ2,max =d2,0 + k2

ηmin2 ωmax

2,SMS

=0 + 10

20 · 1.3333≈ 0.375 s.

As a result, take µ2 = 0.375 s.Much more conservative selection of µ1, µ2 is to take µi = µ ∀ i, where

µ ≤ µmax =minidi,0 + ki

ηmin2 maxiωmax

i,SMS

=0 + 10

20 · 4 = 0.125 s.

Control law implementation. Finally, the separate controllers given by (80) and (81)can be rewritten as

u(1)i +

di,0

µi

ui = −ki

µi

y(1)i − ki

µiTi

yi +ki

µiTi

ri =⇒

u(1)i + ai,0ui = bi,1y

(1)i + bi,0yi + ci,0ri, (87)

where i = 1, 2. From (87), we obtain

u(1)i,1 = −ai,0ui + bi,0yi + ci,0ri, (88)

ui = ui,1 + bi,1yi, i = 1, 2

where

ai,0 =di,0

µi

, bi,1 = −ki

µi

, bi,0 = − ki

µiTi

, ci,0 =ki

µiTi

.

From (88), the block diagram follows, which is similar to the block diagram as shown inFig. 15 (see p. 33).


Run the Matlab program e8 1 Parameters.m to calculate the reference model param-eters 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 theclosed-loop system with zero initial conditions. Make calculations and simulations for thedegree of time-scale separation between fast and slow motions assigned as η2 = 5 andη2 = 10. Compare the simulation results.

Exercise 8.2 Verify the invertibility and internal stability of the system given by

x1 = x21 + x1x2 + 0.5u1 − [1 + 0.2 sin(t)]u2,

x2 = x1 + sin(x2)− u1 − 2[1 + 0.5 sin(2t)]u2, (89)

y1 = x1 − x2, y2 = x1 + x2.

Assume that the inequalities |xj(t)| ≤ 2 ∀ j and |r(t)| ≤ 1 hold for all t ∈ [0,∞). Findthe control law of the form (76) such that εr1 = 0, εr2 = 0, tds1 ≈ 3 s, σd

1 ≈ 0%, tds2 ≈ 3s, σd

2 ≈ 0%. Compare simulation results for the step response of the closed-loop controlsystem with the assignment.

Solution.Invertibility and internal stability. From (89), we get

y1 = x1 − x2 =⇒y1 = x2

1 − x1 + x1x2 − sin(x2) + 1.5u1 + [1− 0.2 sin(t) + sin(2t)]u2,

y2 = x1 + x2 =⇒y2 = x2

1 + x1 + x1x2 + sin(x2)− 0.5u1 − [3 + 0.25 sin(t) + sin(2t)]u2,

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 degreesof the system (89) are given by α1 = α2 = 1. We have that α1 + α2 = n = 2. Then, theinternal subsystem does not exist.

Reference model. By following through solution of Exercise 8.1, take tds1 = 3 s, σd1 =

0 %, tds2 = 3 s, σd2 = 0 %, then we get the reference model for y1 and y2 given by (77)–(78),

where a1 = a2 = ωd = 1.3333 rad/s. Denote ωmaxSMS

= ωd rad/s.Control law . Take q1 = q2 = 1, then the control law can be constructed in the form

(80)–(82). Denote g12 and g22 as the average values of g12 and g22, respectively, whereg12 = 1 and g22 = −3. Denote

G∗ =

[g11 g12

g21 g22

]=

[1.5 1−0.5 −3

].


Take, for instance,

K0 = [G∗]−1 =1

8

[6 2

−1 −3

].

Fast-motion subsystem. From the closed-loop system equations given by (89) and(80)–(82), we obtain the FMS equation, that is

µu(1) + D0 + K1G∗K0u = K1F −H∗, (90)

where µ = diagµ1, µ2, K1 = diagk1, k2, D0 = diagd10, d20, u = [u1, u2]T , and

F = const, H∗ = const during the transients in the FMS (90). Assume that G∗(t) is thematrix with frozen parameters during the transients in (90).

If K0 = [G∗(t)]−1, then the characteristic polynomial of the FMS is factorized asshown by (85). As far as K0 6= [G∗(t)]−1, then the characteristic equation of the FMScan not be factorized on the two separate equations. Take, for simplicity, d10 = d20 = 0,µ = µ1 = µ2 and k = k1 = k2 = 10, and denote µ = µ/k, then the characteristic equationof the FMS (90) is given by

µ2s2 + a1µs + a2,

wherea1 = 1− 0.125g12 − 0.375g22,

a2 =3

32g12g22 − 30

64g22 − 1

8g12,

andg12 ∈ [−0.144, 2.144], g22 ∈ [−1.82,−4.182], ∀ t ∈ [0,∞).

Let µ > 0, then, it can be verified, that

(−1)µ−1K1G∗K0

is Hurwitz matrix. For example, by Matlab program e8 2 Parameters.m, we get that6

maxt∈[0,∞)

Re λi(−G∗K0) ≤ −0.6628

for all i = 1, 2. Take ωminFMS

= 0.6628 and ηmin2 = 10, then µ can be selected such that the

inequality

µ ≤ µmax =kωmin

FMS

ηmin2 ωmax

SMS

=10 · 0.6628

10 · 1.3333≈ 0.4971 s.

holds. As a result, take µ1 = µ2 = 0.4971 s. Note, the control law implementation of(80)–(82) was discussed in Exercise 8.1.

6Reλi(A) is the real part of the eigenvalue λi of A.


Run the Matlab program e8 2 Parameters.m to calculate the reference model param-eters of the controller given by (80)–(82) for η2 = 10. Next, run the Simulink programe8 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 fastand 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 con-trol variables in the system given by

x1 = x1 + x2 + u1 + u2,

x2 = x1 + 3x2 − 2u1, (91)

y = x1.

Solution.Assume that x1, x2 are measurable state variables. We have y = x1 and denote

z = x2. Hence, from (91), we get

y = y + z + u1 + u2,

z = y + 3z − 2u1.

The reference model can be constructed in the form

y = F (y, r) =⇒ y = −ay + ar. (92)

Exclusion of control variable u1. Take u1 = 0 ∀ t. From (92), we get

y = y + z + u2,

z = y + 3z.

Let assume that the control law is given by

µu2 + d0u2 = k0F (y, r)− y =⇒µu2 + d0u2 = k0−y − ay + ar. (93)

Hence, the closed-loop system equations are given by

y = y + z + u2,

z = y + 3z, (94)

µu2 + d0u2 = k0−y − ay + ar.The closed-loop system equations (94) can be rewritten as

y = y + z + u2,

z = y + 3z, (95)

µu2 + [d0 + k0]u2 = k0−ay + ar − y − z.


By following through solutions of Exercises 4.2 and 4.3, from (95), we obtain the FMSequation, that is

µu2 + [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 us

2(t) is a steady state (more precisely, quasi-steadystate) of the FMS (96) and

us2 = −ay + ar − y − z.

Substitution of us2 into (93) yields the SMS given by

y = −ay + ar, (97)

z = 3z + y, (98)

where the internal subsystem (98) is unstable.Exclusion of control variable u2. Take u2 = 0 ∀ t. From (92), we get

y = y + z + u1, (99)

z = y + 3z − 2u1. (100)

Let assume that the control law is given by

µu1 + d0u1 = k0−y − ay + ar. (101)


y = y + z + u1,

z = y + 3z − 2u1, (102)

µu1 + d0u1 = k0−y − ay + ar.The closed-loop system equations (102) can be rewritten as

y = y + z + u1,

z = y + 3z − 2u1, (103)

µu1 + [d0 + k0]u1 = k0−ay + ar − y − z.From (103), we obtain the FMS equation, that is

µu1 + [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 us

1(t) is a steady state of the FMS (104) andus

1 = −ay + ar − y − z. Substitution of us1 into (99), (100) yields the SMS given by

y = −ay + ar, (105)

z = 5z + 3y + 2a(y − r), (106)


where the internal subsystem (106) is unstable again. Hence, the method of selectiveexclusion of redundant control variables does not allow to obtain the stable internal sub-system.

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 aregiven by

y = y + z + u1 + u2,

z = y + 3z − 2u1. (107)

µu1 + d0u1 = k0−y − ay + ar,

where u2 can be utilized for internal dynamics stabilization. From (107) we obtain

y = y + z + u1 + u2,

z = y + 3z − 2u1, (108)

µu1 + [d0 + k0]u1 = k1−ay + ar − y − u2.

From (108), we obtain the FMS equation, that is

µu1 + [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 us

1(t) is a steady state (more precisely, quasi-steadystate) of the FMS (109) and

us1 = −ay + ar − y − u2

Substitution of us1 into (92) yields the SMS given by

y = −ay + ar, (110)

z = 5z + 2u2 + 2y − 2a(r − y), (111)

where the internal subsystem (111) can be stabilized by applying u2 = kintz. Thus, weobtain

y = −ay + ar, (112)

z = [5 + 2kint]z + 2y − 2a(r − y). (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.


Let us take k0 = 10 and a = 1.3333 rad/s. Denote

ωmaxSMS

= maxa, aint rad/s, ωminFMS

=d0 + k0

µrad/s.

Let us consider the ratio η2 = ωminFMS

/ωmaxSMS

as a criterion for the degree of time-scale

separation between fast and slow motions in the closed-loop system. Take ηmin2 = 20.

Hence, we obtain

η2 =d0 + k0

µωmaxSMS

≥ ηmin2 = 20 =⇒

µ ≤ µmax =d0 + k0

ηmin2 ωmax

SMS

=0 + 10

20 · 1.3333= 0.375 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 modeland controller parameters. Next, run the Simulink program e9 1.mdl, to get a stepresponse of the closed-loop system with zero initial conditions.

Exercise 9.2 Consider the system given by

x1 = x1 + x2 + u1 + u2,

x2 = x1 + 3x2 − 2u1, (114)

y = x1.

Stabilize the internal subsystem by insertion of supplementary conditions.Solution.From (114), we get

y = x1 + x2 + u1 + u2.

Hence, we have the relative degree α = 1. Let us insert the supplementary condition forcontrol variables such that

u1 + u2 = 0. (115)

Then u2 = −u1 and we get the relative degree α = 2. From (114) and (115), we obtain

x1 = x1 + x2,

x2 = x1 + 3x2 − 2u1, (116)

y = x1.

By taking into account that y = x1 and y = x1 + x2, the system (116) can by rewrittenin the form

y(2) = 4y(1) − 2y − 2u1.


By following through solution of Exercises 5.9, consider the reference model given by

y(2) = F (y(1), y, r).


θd = tan−1

(π

ln(100/σd)

), ωd =

4

tds,

we get θd = 0.8092 rad, ζd = 0.6901, ωd = 2 rad/s and ωn = 2.8981 rad/s. By selectingthe 2 roots s1,2 = −2 ± j2.0974 where Re(s1,2) = −ωd = −ωn cos(θd) and |Im(s1,2)| =ωn sin(θd), we obtain the desired characteristic polynomial s2 + 4s + 8.399. The desiredtransfer function is given by

Gdyr(s) =

ad0

s2 + ad1s + ad

0

=8.399

s2 + 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

µ2u(2)1 + d1µu

(1)1 + d0u1 = k0F (y(1), y, r)− y(2).


y(2) = f(y(1), y) + gu1,

µ2u(2)1 + d1µu

(1)1 + d0u1 = k0F (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

µ2u(2)1 + d1µu

(1)1 + [d0 + k0g]u1 = k0F (y(1), y, r)− f(y(1), y),

where F = const and f = const during the transients in (116), as well as the SMSequation, that is

y(2) = F (y(1), y, r) +d0

d0 + k0gf(y(1), y)− F (y(1), y, r).

We have g = −2 < 0. Take d0 > 0 and k0 = 10/g = −5.From the SMS and reference model equations, it follows that

s2 + ad1s + ad

0

is the characteristic polynomial of the SMS where aSMS

0 =√

ad0, and

µ2s2 + d1µs + d0 + k0g


is the characteristic polynomial of the FMS. Denote aFMS

0 = d0 + k0g. Hence, we obtain

η3 =(a

FMS

0 )1/2

µ(aSMS

0 )1/2=

√d0 + k0g

µ(aSMS

0 )1/2≥ ηmin

3 = 10

=⇒ µ ≤√

d0 + k0g

ηmin3 (a

SMS

0 )1/2=

√0− 5 · (−2)

10 · √8.399≈ 0.1091 s.


sFMS

1,2 = − d1

2µ± j

√d2

1 − 4(d0 + k0g)

2µ= α± jβ,

where we assume that d21 − 4(d0 + k0g) < 0. Take ζmin

FMS= 0.5. Hence, we can find

ζFMS

= cos(θFMS

) = |α|/√

α2 + β2

=d1

2√

d0 + k0g≥ ζmin

FMS= 0.5 =⇒

d1 ≥ 2ζminFMS

√d0 + k0g

= 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 theclosed-loop system.

Chapter 10

Exercise 10.1 The system is given by

x = x2 + 4u. (117)

Find the parameters of the control law in the form given by

uk =q∑

j=1

djuk−j +q∑

j=0

ajyk−j +q∑

j=0

bjrk−j (118)

to meet the following specifications: εr = 0, tds ≈ 3 s, σd ≈ 0%, q = 1. Determine thesampling period Ts such that the phase margin of the FMS will meet the requirementϕ ≥ 0.7 rad. Compare simulation results of the step output response of the closed-loopcontrol system with the assignment.

Solution.Reference model. Consider the reference model given by x = F (x, r). Take tds = 3 s,

σd = 0 %, then we get θd = 0 rad, ζd = 1, a = ωd = ωn = 1.3333 rad/s. By selecting the


root s1 = −a, we obtain the desired characteristic polynomial s + a. The desired transferfunction is given by

Gdxr(s) =

a

s + a.

Hence, we get the reference model given by

x = −ax + ar =⇒ x =1

T[r − x], (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 atten-uation is not specified, then, for simplicity, take q = α = 1 and consider the control lawgiven by

µu + d0u = k0[F (x, r)− x]. (120)

Hence, the control law has the following form:

µu + d0u = k0[−x− ax + ar]. (121)

Closed-loop system. The closed-loop system equations are given by

x = x2 + 4u, (122)

µu + d0u = k0[F (x, r)− x]. (123)

Denote f(x) = x2. From (122) and (123), by following through solution of Exercise 4.2,we get the FMS given by

µu + [d0 + k0g]u = k0F (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 + k0g (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 steadystate (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 thesame as the reference model.


Selection of continuous-time controller parameters. Let us consider a simplified ver-sion for the gain k0 selection. We have d0 = 0, then the gain k0 can be selected such thatk0gmin = 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 byωd

n = a = 1.3333 rad/s. Denote

ωmaxSMS

= ωdn = 1.3333 rad/s, ωmin

FMS=

d0 + k0gminµ

rad/s.

Let us consider the ratio

η2 =ωmin

FMS

ωmaxSMS

(126)

as a criterion for the degree of time-scale separation between fast and slow motions. Take

ηmin2 = 20. Hence, we obtain

η2 =d0 + k0gmin

µωmaxSMS

≥ ηmin2 = 20 =⇒

µ ≤ µmax =d0 + k0gminηmin

2 ωmaxSMS

=0 + 2.5 · 420 · 1.3333

≈ 0.375 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-timecontroller parameters. Next, run the Simulink program e10 1 Continuous.mdl, to get thestep response of the closed-loop system.

Selection of the sampling period. From (117) we get the following pseudo-continuous-time model:

x(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 equa-

tions are given by

x(t) = f(x(t)) + gu(t− τ),

µu(t) + d0u(t) = k0[F (x(t), r(t))− x(t)].

From the above equations, we get the FMS given by

µu(t) + d0u(t) + k0gu(t− τ) = k0F (x(t), r(t))− f(x(t)), (128)

where F = const and f = const during the transients in (128). The block diagramrepresentation of the FMS (128) is shown in Fig. 16.

The corresponding transfer function of the open-loop FMS with time delay is givenby

GO

FMS(s) =k0g exp (−τs)

D(µs), (129)


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

|GO

FMS(jωc, µ)| = 1,

we get

|D(jµωc)| = k0g =⇒ |jµωc + d0| = k0g =⇒

ωc =

√k2

0g2 − d2

0

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)

holds if the inequalityTs ≤ 2[π − ϕd − ArgD(jµωc)]/ωc (132)

is satisfied. Finally, from (130) and (132), by taking into account d0 = 0, we get

Ts ≤ (π − 2ϕd)µ

k0g(133)

where ϕd < π/2.Take, for example, ϕd = 0.7 rad (that is about 40 degrees of arc), k0g = 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

µu + d0u = k0[−x− ax + ar]. (134)

From (134), we obtain

[µs + d0]u(s) = −k0[s + a]x(s) + ar(s) =⇒ s =2

Ts

z − 1

z + 1=⇒

[(2µ + d0Ts)z + (d0Ts − 2µ)]u(z) =

k0[−(2 + aTs)z + (2− aTs)]x(z) + k0aTs[z + 1]r(z) =⇒(2µ + d0Ts)uk+1 + (d0Ts − 2µ)uk =

−k0(2 + aTs)xk+1 + k0(2− aTs)uk + k0aTsrk+1 + k0aTsrk. (135)


From (135), the control law given by the difference equation

uk+1 = d1uk + a0yk+1 + a1yk + b0rk+1 + b1rk (136)

results, where

d0 = 2µ + d0Ts, d1 = (2µ− d0Ts)/d0,

a0 = − k0(2 + aTs)/d0, a1 = k0(2− aTs)/d0, (137)

b0 = b1 = k0aTs/d0.

Implementation of control law. From (136), we obtain

uk+1 − a0xk+1 − b0rk+1 = d1uk + a1xk + b1rk︸︷︷︸=uk+1

=⇒

uk+1 − a0xk+1 − b0rk+1 = uk+1 =⇒ uk = a0xk + b0rk + uk.

Finally, we get

uk+1 = d1uk + a1xk + b1rk, (138)

uk = uk + a0xk + b0rk.

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 closed-loop system with zero initial conditions. Make simulations for ϕd = 0.5 rad and ϕd = 1rad. 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, σd ≈ 20%, q = 2. Determine the sampling period Ts such that the phase marginof the FMS will meet the requirement ϕ ≥ 0.25 rad. Compare simulation results of thestep output response of the closed-loop control system with the assignment.

Solution.


Reference model. From (139), we obtain n = 2, x(2) = f(x(1), x) + g(t)u wheref(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, thereference model will be constructed in the form

x(2) = F (x(1), x, r).


θd = tan−1

(π

ln(100/σd)

), ωd =

4

tds,

we get θd = 1.0974 rad, ζd = 0.4559, ωd = 4 rad/s and ωn = 8.7729 rad/s. By selectingthe 2 roots s1,2 = −4 ± j7.8079, where Re(s1,2) = −ωd = −ωn cos(θd) and |Im(s1,2)| =ωn sin(θd), we obtain the desired characteristic polynomial

(s− s1)(s− s2) = s2 + a1s + a0, (140)

where a1 = 8, a0 = 76.9637. Hence, we can get the reference model in the form of thetype 1 system, this is

x(2) = −a1x(1) − a0x + b0r,

or type 2 system, this is

x(2) = −a1x(1) − a0x + b1r

(1) + b0r (141)

where b0 = a0 and b1 = a1.Continuous-time controller. Take q = n = 2 Therefore, consider the control law given

by

µ2u(2) + d1µu(1) + d0u

= k0−a2x(2) − a1x

(1) − a0x + b1r(1) + b0r, (142)

which can be rewritten, for short, as

µ2u(2) + d1µu(1) + d0u = k0F (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,

µ2u(2) + d1µu(1) + d0u = k0F (x(1), x, r)− x(2).From the above closed-loop system equations, we get the FMS given by

µ2u(2) + d1µu(1) + d0 + k0gu = k0F (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 theSMS given by

x(2) = F (x(1), x, r) +d0

d0 + k0g(t)f(x(1), x)− F (x(1), x, r).


Selection of continuous-time controller parameters. Parameters of the continuous-time 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 k0gmin = 10 holds. Hence we get k0 = 20.

From the SMS and reference model equations, it follows that the characteristic poly-nomial of the SMS is given by (140), where a

SMS

0 = a0 = 76.9637. and

µ2s2 + d1µs + d0 + k0g

is the characteristic polynomial of the FMS. Take ηmin3 = 10. Then, from the requirement

η3 ≥ ηmin3 , we obtain

η3 =(a

FMS

0 )1/2

µ(aSMS

0 )1/2=

√d0 + k0g

µ(aSMS

0 )1/2≥

√d0 + k0gmin

µ(aSMS

0 )1/2≥ ηmin

3 = 10

=⇒ µ ≤√

d0 + k0gmin

ηmin3 (a

SMS

0 )1/2=

√0 + 20 · 0.5

10 · √76.9637≈ 0.036 s.


sFMS

1,2 = − d1

2µ± j

√d2

1 − 4(d0 + k0g)

2µ= α± jβ,

where we assume that d21−4(d0+k0g) < 0 when g = gmax. Take, for instance, ζmin

FMS= 0.6.

Then we can find

ζFMS

= cos(θFMS

) = |α|/√

α2 + β2

=d1

2√

d0 + k0g≥ ζmin

FMS= 0.6 =⇒

d1 ≥ 2ζminFMS

√d0 + k0gmax

= 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 parame-ters. Next, run the Simulink program e10 3 Continuous.mdl, to get the step response ofthe closed-loop system.

The block diagram representation of the FMS (144) is shown in Fig. 18, whereD(µs) = µ2s2 + d1µs + d0.

The corresponding transfer function of the open-loop FMS is given by

GO

FMS(s) =k0g

D(µs). (145)

Take d0 = 0 and g = gmax, then from (145) and

|GO

FMS(jωc, µ)| = 1,


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 thatg = gmax = 2.5. Hence, we get

|D(jµωc)| = k0gmax =⇒ |jµωc(jµωc + d1)| = k0gmax =⇒µ4ω4

c + d21µ

2ω2c − k2

0g2max = 0 =⇒

y = ω2c > 0, µ4y2 + d2

1µ2y − k2

0g2max = 0 =⇒ (146)

y =−d2

1 +√

d41 + 4k2

0g2max

2µ2

=−8.48532 +

√8.48534 + 4 · 202 · 2.52

2 · 0.0362≈ 19712 =⇒

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

ϕFMS

= 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-continuous-

time 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 givenby (143) and (147), we get the FMS equation given by

µ2u(2)(t) + d1µu(1)(t) + d0u(t) + k0g(t)u(t− τ)

= k0F (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

GO

FMS(s) =k0ge−jτs

D(µs). (149)


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 befound, this is ωc =

√y ≈ 140 rad/s given that g = gmax = 2.5.

Let us select ϕd such that the inequalities

0 < ϕd < ϕFMS

= 1.033 rad

hold. Denote ϕ is the phase margin of the FMS given by (148) Then, by inspection ofthe Nyquist plot for (149), the condition

0 < ϕd ≤ ϕ < ϕFMS

= 1.033 rad (150)

holds for all g ∈ [gmin, gmax] = [0.5, 2.5], if the sampling period Ts is selected such thatthe 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

[µ2s2 + d1µs + d0]u(s)

= k0[−a2s2 − a1s− a0]x(s) + k0[b1s + 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 = d1uk−1 + d2uk−2 + a0xk + a1xk−1 + a2xk−2

+b0rk + b1rk−1 + b2rk−2 (153)


results, where

d0 = 4µ2 + 2µd1Ts + d0T2s ,

d1 = 8µ2 − 2d0T2s /d0,

d2 = −4µ2 − 2µd1Ts + d0T2s /d0,

a0 = − k0 4a2 + 2a1Ts + a0T2s /d0,

a1 = 2k0 4a2 − a0T2s /d0, (154)

a2 = − k0 4a2 − 2a1Ts + a0T2s /d0,

b0 = k0 2b1Ts + b0T2s /d0,

b1 = 2k0 b0T2s /d0,

b2 = − k0 2b1Ts − b0T2s /d0.


uk+2−a0xk+2−b0rk+2−d1uk+1−a1xk+1−b1rk+1 = d2uk + a2xk + b2rk︸︷︷︸=u2,k+1

=⇒ uk+1 − a0xk+1 − b0rk+1 = u2,k + d1uk + a1xk + b1rk︸︷︷︸=u1,k+1

=⇒

uk = u1,k + a0xk + b0rk

From the above, we get

u1,k+1 = u2,k + d1uk + a1xk + b1rk,

u2,k+1 = d2uk + a2xk + b2rk, (155)

uk = u1,k + a0xk + b0rk.

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 closed-loop system with zero initial conditions. Make simulations for ϕd = 0.5 rad and ϕd = 0.175rad. Compare the simulation results.


Chapter 11

Exercise 11.1 Construct the reference model in the form of the 2nd-order differenceequation

yk =n∑

j=1

adjyk−j +

n∑

j=1

bdjrk−j (156)

based on the Z-transform in such a way that the step output response with zero initialconditions meets the requirements tds ≈ 1 s, σd ≈ 0 % given that the sampling period isTs = 0.05 s. Compare simulation results with the assignment.

Solution.By following through the solution of Exercise 2.1, let us construct the desired continuous-

time transfer function Gdyr(s). Take tds = 1 s, σd = 0 %, then by

θd = tan−1

(π

ln(100/σd)

), ωd =

4

tds,

we get θd = 0 rad, ζd = 1, ωd = ωn = 4 rad/s. By selecting the 2 roots s1 = s2 = −4, weobtain the desired characteristic polynomial s2 + 8s + 16. Consider the desired transferfunction given by

Gdyr(s) =

16

s2 + 8s + 16=

16

(s + 4)2. (157)

Let us denote a = ωn = 4 and expand Gdyr(s)/s as

a2

s(s + a)2=

A

s+

B

s+a+

C

(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

Hdyr(z) =

z − 1

zZ

L−1

[a2

s(s + a)2

]∣∣∣∣∣t=kTs

=z − 1

zZ

L−1

[A

s+

B

s+a+

C

(s+a)2

]∣∣∣∣∣t=kTs

=bd1z + bd

2

z2 − ad1z − ad

2

, (158)

where

bd1 = 1− e−aTs − aTse

−aTs ,

bd2 = e−aTs [e−aTs − 1 + aTs], (159)

ad1 = 2e−aTs , ad

2 = −e−2aTs .


Hence, from (158), we obtain the desired difference equation, this is

yk = ad1yk−1 + ad

2yk−2 + bd1rk−1 + bd

2rk−2. (160)

Run the Matlab program e11 1 Parameters.m in order to calculate the reference modelparameters. Next, run the Simulink program e11 1.mdl to get a plot for the outputresponse of (160). Then, from inspection of the plot, determine the steady-state error forinput 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∑

j=1

djuk−j + λ0

−yk +

n∑

j=1

adjyk−j +

n∑

j=1

bdjrk−j

. (162)

to meet the following specifications: tds ≈ 2 s; σd ≈ 0%; η ≥ 10. Compare simulationresults 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

s2 − 1=

2

(s + 1)(s− 1).

Let us expand Gyu(s)/s as

2

s(s + 1)(s− 1)=

A

s+

B

s + 1+

C

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 thepulse transfer function given by

Hyu(z) =z − 1

zZ

L−1

[2

s(s + 1)(s− 1)

]∣∣∣∣∣t=kTs

=b1z + b2

z2 − a1z − a2

, (163)

where

b1 = b2 = eTs + e−Ts − 2, a1 = eTs + e−Ts , a2 = −1. (164)


Hence, from (163), we obtain the difference equation of the plant, this is

yk = a1yk−1 + a2yk−2 + b1rk−1 + b2rk−2. (165)

Reference model. By following through the solution of Exercise 11.1, let us construct thedesired continuous-time transfer function Gd

yr(s). Take tds = 2 s, σd = 0 %, then we getθd = 0 rad, ζd = 1, ωd = ωn = 2 rad/s. By selecting the 2 roots s1 = s2 = −2, we obtainthe desired characteristic polynomial s2 + 4s + 4. The desired transfer function can beconstructed in the following form:

Gdyr(s) =

4

s2 + 4s + 4=

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 isgiven by (160).

Control law. Consider the control law given by

uk = d1uk−1 + d2uk−2

+λ0−yk + ad1yk−1 + ad


2rk−2. (166)

The closed-loop system equations have the following form:

yk =2∑

j=1

ajyk−j +2∑

j=1

bjuk−j, (167)

uk =2∑

j=1

djuk−j + λ0

−yk +

2∑

j=1

adjyk−j +

2∑

j=1

bdjrk−j

. (168)

Substitution of (167) into (168) yields

yk =2∑

j=1

ajyk−j +2∑

j=1

bjuk−j, (169)

uk =2∑

j=1

[dj−λ0bj]uk−j+λ0

2∑

j=1

[adj−aj]yk−j+bd

jrk−j, (170)

where the FMS is governed by

uk =2∑

j=1

[dj−λ0bj]uk−j+λ0

2∑

j=1

[adj−aj]yk−j+bd

jrk−j, (171)

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 = λ0b1, d2 = λ0b2, (172)


where b1, b2 are defined by (164). Hence, we get the deadbeat response of the FMS.Finally, take the sampling period such that

Ts =tdsqη

=2

2 · 10= 0.1 s. (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

uk+2 = d1uk+1+d2uk−yk+2+ad1yk+1+ad

2yk+bd1rk+1+bd

2rk, (174)

uk = λ0uk. (175)


uk+2 − d1uk+1 + yk+2 − ad1yk+1 − bd

1rk+1 = d2uk + ad2yk + bd

2rk︸︷︷︸=u2,k+1

=⇒

uk+1 + yk+1 = u2,k + d1uk + ad1yk + bd

1rk︸︷︷︸=u1,k+1

=⇒

uk = −yk + u1,k

Finally, we get

u1,k+1 = u2,k + d1uk + ad1yk + bd

1rk,

u2,k+1 = d2uk + ad2yk + bd

2rk, (176)

uk = u1,k − yk, uk = λ0uk.

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 referencemodel parameters. Next, run the Simulink program e11 4.mdl to get a plot for theoutput response of the closed-loop system. Make simulations for η = 7, η = 20. Comparethe simulation results.


Chapter 12

Exercise 12.1 Consider the system (60) preceded by ZOH where a = 3. Design thediscrete-time control law given by

uk =q≥α∑

j=1

djuk−j + λ(Ts)[Fk − yk], (177)

to meet the following specifications: εr = 0, tds ≈ 2 s, σd ≈ 5%. Run a computer simulationof the closed-loop system with zero initial conditions. Compare simulation results of theoutput 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 y2 = 2y2 + z + g?u, where g? = 1 and g? is so called a high-frequency gainof 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 = d1uk−1 + d2uk−2

+λ0−yk + ad1yk−1 + ad


2rk−2. (178)

Sampling period. Take, for instance, η = 20, where η is the required degree of time-scale separation between the fast and slow modes in the closed-loop system. Then, thesampling period Ts can be selected by

Ts =tdsq η

(179)

=2

2 · 20= 0.05 s.

Reference model. By following through the solution of Exercise 11.1, let us constructthe desired continuous-time transfer function Gd

yr(s). Take tds = 2 s, σd = 5 %, then weget θd = 0.8092 rad, ζd = 0.6901, ωn = 2.8981 rad/s, ωd = 2 rad/s. By selecting the 2roots 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) =

bd1z + bd

2

z2 − ad1z − ad

2

(180)


where bd1 = 0.009816, bd

2 = 0.009182, ad1 = 1.8, and ad

2 = −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 corre-sponds to continuous-time linear system (60) preceded by a ZOH with high samplingrate, that is the following difference equation

yk =α=2∑

j=1

aα,jyk−j + Tαs

α=2∑

j=1

g? εα,j

α ![uk−j + fk−j], (181)

where

yk = y(t)|t=kTs, uk = u(t)|t=kTs

, fk = f(x1(t), x3(t))|t=kTs,

(z − 1)α = zα − aα,1zα−1 − aα,2z

α−2 − · · · − aα,α,

aα,j = (−1)j+1 α !

(α− j) ! j !.

Denote

B(z) = b1zα−1 + b2z

α−2 + · · ·+ bα

= g? Tαs

α !Eα(z) (182)

= g? T 2s

2E2(z) = 0.00125(z + 1).

In order to get the deadbeat response of the FMS, the controller parameters d1, d2 andλ0 are selected such that

λ0 = [b1 + b2]−1 = 400,

d1 = λ0b1 = 0.5, d2 = λ0b2 = 0.5.

Implementation of control law. The implementation of the control law (178) wasdiscussed 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 lawparameters. Next, run the Simulink program e12 1.mdl to get a plot for the outputresponse 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

∂t(z, t) = α2∂2x

∂z2(z, t) + c(t)x(z, t) + w(z, t) + u(z, t), (183)


where [∂x(z, t)/∂z]|z=0 = 0, [∂x(z, t)/∂z]|z=1 = 0 are the boundary conditions, α2 = 1,and ϕn(z) = ϕ0

n cos(√

λn z), λn = n2π2, ϕ00 = 1, ϕ0

n =√

2 ∀ n = 1, 2, . . ..Solution.Consider the homogeneous one-dimensional heat equation7

∂x

∂t(z, t) =

∂2x

∂z2(z, t) (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

∂x(z, t)/∂t = ϕ(z)[dx(t)/dt], ∂2x(z, t)/∂z2 = x(t)[d2ϕ(z)/dz2]. (187)

Hence, from (184) and (187), we obtain

dx(t)/dt

x(t)=

d2ϕ(z)/dz2

ϕ(z), (188)

where (188) holds for all t and z. Then

dx(t)/dt

x(t)=

d2ϕ(z)/dz2

ϕ(z)= β, (189)

where β = const. Hence, from (189), we obtain

d2ϕ(z)

dz2− βϕ(z) = 0, (190)

dx(t)

dt− βx(t) = 0, (191)

In accordance with (185) and (186), we have

x(t)dϕ(z)

dz

∣∣∣∣∣z=0

= 0, x(t)dϕ(z)

dz

∣∣∣∣∣z=1

= 0, (192)

for all x(t). Then

dϕ(z)

dz

∣∣∣∣∣z=0

= 0, (193)

dϕ(z)

dz

∣∣∣∣∣z=1

= 0, (194)

7From this point the solution can be done by following, for instance, Chapter 11 of the book: KreyszigE. Advanced engineering mathematics, 8th ed., New York: John Wiley & Sons, Inc., 1999.


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)

dz= a. (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

d2ϕ(z)

dz2+ p2ϕ(z) = 0, (196)

dx(t)

dt+ p2x(t) = 0. (197)

Then ϕ(z) = Acos(pz) + B sin(pz) is the general solution of (196), where

dϕ(z)

dz= −pA sin(pz) + pB cos(pz). (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 infinitely many different solutions given by ϕn(z) = ϕ0n cos(nπz). Let us

assume that∫ z=1z=0 ϕ2

n(z)dz = 1. Then

(ϕ0n)2

∫ z=1

z=0cos2(nπz)dz = ((ϕ0

n)2/(nπ))∫ y=nπ

y=0cos2(y)dy = (ϕ0

n)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−λnt for all n = 0, 1, . . .. As a result,we get that

xn(z, t) = xn(t)ϕn(z) = xn(0)e−λntϕ0n 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 coef-ficients of the normalized transfer function8

Gdyr(s) =

ad0

sn + adn−1s

n−1 + · · ·+ ad1s + ad

0

(199)

8Some additional details concerned with the optimal coefficients of the normalized transfer functionbased on ITAE criterion can be found in the following references: Dorf, R.C. and Bishop, R.H. Moderncontrol systems, 9th ed., Upper Saddle River, NJ: Prentice Hall, 2001; Franklin, G.F., Powell, J.D. andEmami-Naeini A. Feedback control of dynamic systems, 4th ed., Prentice Hall, 2002; Kuo, B.C. andGolnaraghi, F. Automatic control systems, 8th ed., New York: John Wiley & Sons, Inc., 2003.


based on the integral of time multiplied by absolute error (ITAE) criterion

ITAE =∫ ∞

0t|e(t)| dt (200)

for a step input are given by

s + ωn,

s2 + 1.4ωns + ω2n,

s3 + 1.75ωns2 + 2.15ω2

ns + ω3n,

s4 + 2.1ωns3 + 3.4ω2ns

2 + 2.7ω3ns + ω4

n,

s5 + 2.8ωns4 + 5.0ω2ns

3 + 5.5ω3ns2 + 3.4ω4

ns + ω5n,

s6 + 3.25ωns5 + 6.60ω2

ns4 + 8.60ω3ns3 + 7.45ω4

ns2 + 3.95ω5

ns + ω6n,

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

Gdyr(s) =

ad1s + ad

0

sn + adn−1s

n−1 + · · ·+ ad1s + ad

0

(201)

based on the ITAE criterion (200) for a ramp input are given by

s2 + 3.2ωns + ω2n,

s3 + 1.75ωns2 + 3.25ω2

ns + ω3n,

s4 + 2.41ωns3 + 4.93ω2

ns2 + 5.14ω3ns + ω4

n,

s5 + 2.19ωns4 + 6.50ω2

ns3 + 6.30ω3ns

2 + 5.24ω4ns + ω5

n.

Auxiliary Material (Euler polynomials)

E1(z) = 1,

E2(z) = z + 1,

E3(z) = z2 + 4z + 1,

E4(z) = z3 + 11z2 + 11z + 1,

E5(z) = z4 + 26z3 + 66z2 + 26z + 1.

Auxiliary Material (Describing functions)

Let us consider a symmetric nonlinearity9 v = ϕ(z) as shown in Fig. 22. Assume thatthe nonlinearity input is z(t), where z(t) = A sin(ωt). Consider the output v(t) of thenonlinearity represented by its Fourier series

v(t) = b1 sin(ωt) + c1 cos(ωt) +∞∑

k=2

bk sin(kωt) + ck cos(kωt). (202)

9The angle θ defines the slope k of the backlash hysteresis, that is k = tan(θ).


In accordance with the describing function method, the nonlinearity v = ϕ(z) can byreplace 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 q1 =2M

π∆

sin−1

(∆

A

)+

∆

A

√1−

[∆

A

]2 ,

q2 = 0, A ≥ ∆

Relay with dead zone q1 =4M

πA

√1−

[∆

A

]2

, q2 = 0, A ≥ ∆

Ideal relay q1 =4M

πA, q2 = 0

Hysteresis q1 =4M

πA

√1−

[∆

A

]2

, q2 = −4M∆

πA2, A ≥ ∆

Backlash hysteresis q1 =k

π

[π

2+ sin−1

(1− 2∆

A

)

+2(1− 2∆

A

) √A∆− 1

A/∆

,

q2 = −4k

π

(A∆− 1)

(A/∆)2, k = tan(θ), A ≥ ∆

10y = sin−1(x) denotes the inverse sine of x, i.e., sin(y) = x.


Auxiliary Material (The Laplace Transform and the Z-Transform)

f(t) F (s) = L[f(t)] F (z) = Z[f(kTs)]

11

s

z

z − 1

t1

s2

Tsz

(z − 1)2

t2

2!

1

s3

T 2s z(z + 1)

2! (z − 1)3

t3

3!

1

s4

T 3s z(z2 + 4z + 1)

3! (z − 1)4

e−at 1

s + a

z

z − d, d = e−aTs

a

s(s + a)

(1− d)z

(z − 1)(z − d)

te−at 1

(s + a)2

zdTs

(z − d)2

1

(s + a)3

z(z + d)dT 2s

2! (z − d)3

sin ωtω

s2 + ω2

z sin ωTs

z2 − 2z(cos ωTs) + 1

cos ωts

s2 + ω2

z(z − cos ωTs)

z2 − 2z(cos ωTs) + 1

e−at sin ωtω

(s + a)2 + ω2

zd sin ωTs

z2 − 2zd(cos ωTs) + d2

e−at cos ωts + a

(s + a)2 + ω2

z2 − zd cos ωTs

z2 − 2zd(cos ωTs) + d2

b− a

(s + a)(s + b)

(d− c)z

(z − d)(z − c), c = e−bTs

s

(s + a)2

z[z − d(1 + aTs)]

(z − d)2

a

s2(s + a)

z[(aTs − 1 + d)z + (1− d− aTsd)]

a(z − 1)2(z − d)a2

s(s + a)2

z[z(1− d− aTsd) + d2 − d + aTsd]

(z − 1)(z − d)2

(b− a)s

(s + a)(s + b)

z[z(b− a)− (bd− ac)]

(z − d)(z − c)


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 +6.30ω2ns2+ +6.30ω3

ns2+

60 12 from top V (u)dt

dV (u)dt

74 5 from top uNID = ϕ(X, R,w) Us1 (t)=UNID

1 (t)=[uNID, 0,

· · · , 0]T , UNID1 =ϕ(X, R,w)

74 14 from top +µϕ(X, R, R, w, w) −µϕ(X,R, R, w, w)

91 1 from below Lduf (ω) Gd

uf (s)

92 2 from top Lduf (ω) Gd

uf (s)

92 10 from top Guf (s) Gduf (s)

97 4 from below LHF A

uf (ω) LHF A

uns(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 ωns

min = 102 ωns

min = 103

184 13 from top T−12 [r2 − r]− r(1) T−1

2 [r2 − x2]− x(1)2

188 9 from below µu(1)1 + d10u = µu

(1)1 + d0u1 =

188 3 from below bd1τT 2

ad1

T

200 2 from below µ2i u

(2)i + di1µiu

(1)i + di0ui µ2

i u(2)i + di1µiu

(1)i + di0ui

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






310 2 from below sampling period sampling rate

311 11 from below σd ≈ 20% σd ≈ 0%



ssm for design of nonlinear control systems

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