Student Solutions Manual for. Design of Nonlinear Control Systems with the Highest Derivative in Feedback

Transcription

1 Student Solutions Manual for Design of Nonlinear Control Systems with the Highest Derivative in Feedback World Scientific, 2004 ISBN Valery D. Yurkevich Copyright c 2007 by Valery D. Yurkevich

2 Preface Student Solutions Manual contains complete solutions of 20 % of Exercises from the book Design of Nonlinear Control Systems with the Highest Derivative in Feedback, World Scientific, 2004, (ISBN ). The manual aims to help students understand a new methodology of output controller design for nonlinear systems in presence of unknown external disturbances and varying parameters of the plant. The solutions manual is accompanied by Matlab-Simulink files 1 for calculations and simulations related with Exercises. The program files provide the student a possibility to design the discussed control systems in accordance with the assigned performance specifications of output transients, and make a comparison of simulation results. The distinguishing feature of the discussed throughout design methodology of dynamic output feedback controllers for nonlinear systems is that two-time-scale motions are induced in the closed-loop system. Stability conditions imposed on the fast and slow modes, and a sufficiently large mode separation rate, can ensure that the full-order closedloop system achieves desired properties: the trajectories of the full singularly perturbed system approximate to the trajectories of the reduced model, where the reduced model is identical to the reference model, by that the output transient performances are as desired, and they are insensitive to parameter variations and external disturbances. Robustness of the closed-loop system properties is guaranteed so far as the stability of the fast mode and the sufficiently large mode separation rate are maintained in the closed-loop system. Consequently, the ensuring of the fast mode stability by selection of control law parameters is the problem requiring undivided attention and that constitutes the core of the controller design procedure. In general, the selection of the control law structure as well as selection of controller parameters are not unique, inasmuch as a set of constraints has to be taken into account such as a range of variations for plant parameters and external disturbances, required control accuracy, requirements on load disturbance rejection as well as high frequency measurement noise rejection. Therefore, it would be much more correctly, if the student will take up the solution presented in the manual as an example or draft version of such solution, and then one can make a try to extend the solution by taking into account some additional practical limitations. Overall, the above mentioned book, along with the Student Solutions Manual, as well as accompanying Matlab-Simulink files are an excellent learning aid for advanced study of real-time control system designing and ones may be used in such course as Design of Nonlinear Control Systems, where prerequisites are Linear Systems and Nonlinear Systems. Any comments about the solutions manual (including any errors noticed) can be sent to yurkev@mail.ru or yurkev@ieee.org with the subject heading book. They will be sincerely appreciated. Valery D. Yurkevich 1 A set of Matlab-Simulink files for the Student Solutions Manual can be downloaded from the website yurkev/books.html.

3 Student Solutions Manual for Design of nonlinear control systems... 3 Contents Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Auxiliary Material (The optimal coefficients based on ITAE criterion) Auxiliary Material (Euler polynomials) Auxiliary Material (Describing functions) Auxiliary Material (The Laplace Transform and the Z-Transform) Errata for the book

4 Student Solutions Manual for Design of nonlinear control systems... 4 Chapter 1 Exercise 1.2 The behavior of a dynamical system is described by the equation x (2) + 1.5x (1) + 0.5x + µ{2x 2 + [x (1) ] 2 } 1/2 = 0. (1) Determine the region of µ such that X = 0 is an exponentially stable equilibrium point of the given system. Solution. Denote x 1 = x, x 2 = x (1), and X = [x 1, x 2 ] T. Hence, we have where A = [ Ẋ = AX + µg(x), ], g(x) = [ 0 (2x x 2 2) 1/2 Hence, g(x) X=0 = 0, and so the perturbation g(x) is vanishing at the equilibrium point of the linear nominal system Ẋ = AX. We can find that g(x) 2 = (2x x 2 2) 1/2 (2x x 2 2) 1/2 = 2 X 2. Denote c 5 = 2. Then, from the Lyapunov equation with Q = I, we get P A + A T P = Q (2) P = [ where λ min (P ) = and λmax(p ) = Hence, if the inequalities 0 < µ < λ min (Q) 2λmax(P )c 5 = hold, then X = 0 is an exponentially stable equilibrium point of the system (1). The above results can be obtained by Matlab program e1 2 Lyap.m as well as the initial value problem solution can be found by running e1 2.mdl. ], ]. Exercise 1.4 The behavior of a dynamical system is described by the equations [ ẋ1 µẋ 2 ] = [ ] [ x1 x 2 ]. (3) Obtain and analyze the stability of the slow-motion subsystem (SMS) and the fast-motion subsystem (FMS). Solution.

5 Student Solutions Manual for Design of nonlinear control systems... 5 where From (3), we have A(µ)= Ẋ = A(µ)X, [ 1 1 2/µ 1/µ The characteristic polynomial of the system is given by ( ) 1 det [si A(µ)] = s 2 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 t 0 = t/µ into the system (3), we obtain Take µ = 0. Hence, we get ] d x 1 dt 0 = µ[x 1 x 2 ], d x 2 dt 0 = 2x 1 + x 2. d x 1 dt 0 = 0, = x 1 = const, d x 2 dt 0 = 2x 1 + x 2. By returning to the primary time scale t, the FMS. µ d dt x 2 = 2x 1 + x 2 is obtained, where x 1 = const. The FMS is unstable. Next, consider an equilibrium point of the FMS, that is dx 2 /dt = 0. Hence, 2x 1 +x 2 = 0 = x 2 = 2x 1. As the result, from the equation of the SMS d dt x 1 = x 1 x 2, 0 = 2x 1 + x 2 d dt x 1 = 3x 1 follows. The SMS is unstable too. Finally, plot the phase portrait of the given system by means of Matlab program 2 pplane.m for µ = 0.1 s. 2 Phase Plane Demo for Matlab by John Polking at Rice Univerisity contains the programs dfield.m, dfsolve, pplane.m, and ppsolve.m. The programs can be downloaded from the website as well as the instructions for use.

6 Student Solutions Manual for Design of nonlinear control systems... 6 Chapter 2 Exercise 2.1 Construct the reference model in the form of the 2nd order differential equation given by T n y (n) + a d n 1T n 1 y (n 1) + + a d 1T y (1) + y = r (4) in such a way that the step response parameters of the output meet the requirements t d s 6 s, σ d 0 %. Plot by computer simulation the output response, and determine the steady-state error from the plot for input signals of type 0 and 1. Solution. Take t d s = 6 s and σ d = 0 %, then by 3 ) ( θ d = tan 1 π ln(100/σ d ), ω d = 4, t d s we get θ d = 0 rad, ζ d = 1, and ω d = ω n = rad/s. By selecting the 2 roots s 1 = s 2 = , we obtain the desired characteristic polynomial given by s s Consider the desired transfer function given by G d yr(s) = s s Hence, the reference model in the form of the type 1 system y (2) y (1) y = r (5) follows. Denote e(t) = r(t) y(t). Let r(t) be the input signal of type 0, that is, r(t) = r s 1(t), where r s = const and r s 0. Hence, r(s) = r s /s and we get e s = lim s 0 se(s) = lim s 0 s[1 G d yr(s)] 1 s rs = lim s 0 s s s s rs = 0. Let r(t) be the input signal of type 1, that is, r(t) = r v t 1(t), where r v = const and r v 0. Hence, r(s) = r v /s 2 and we get e v r = e s = lim s 0 se(s) = lim s 0 s[1 G d yr(s)] 1 s 2 rv = lim s 0 s s s rv = rv 3r v. 3 z = tan 1 (x) denotes the arctangent of x, i.e., tan(z) = x.

7 Student Solutions Manual for Design of nonlinear control systems... 7 Run the Matlab program e2 1 Parameters.m in order to calculate the reference model parameters. Next, run the Simulink program e2 1.mdl to get a plot for the output response of (5). Then, from inspection of the plot, determine the steady-state error for input signals of type 0 and 1, respectively. The simulation results are shown in Fig. 1. Figure 1: Responses of y(t) and e(t) for input signal r(t) of type 0 and 1. Exercise 2.3 Construct the reference model in the form of the 2nd order differential equation as the type 1 system with the following roots of the characteristic polynomial: s 1 = 1+j, s 2 = 1 j. Plot by computer simulation the output response, and determine the steady-state error from the plot for input signals of type 0, 1, and 2. Solution. By selecting the 2 roots s 1,2 = 1±j, we obtain the desired characteristic polynomial s 2 + 2s + 2. Consider the transfer function given by From G d yr(s), the reference model G d yr(s) = 2 s 2 + 2s + 2. y (2) + 2y (1) + 2y = 2r (6) follows. Denote e(t) = r(t) y(t). By the same way as in Exercise 2.1, we obtain that e s r = 0 and e v r = 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 e v r = 0 and e acc r = 1/2, where e acc r 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 parameters as well as to obtain the step response of the reference model (6). Next, run the Simulink program e2 3.mdl in order to get a plot for the output response of (6), and then, from inspection of the plot, determine the steady-state error for input signals of type 0, 1 and 2. Repeat that for (7). The simulation results are shown in Figs Chapter 3

8 Student Solutions Manual for Design of nonlinear control systems... 8 Figure 2: Responses of y(t) and e(t) of the system (6) for input signal r(t) of type 0, 1, and 2. Figure 3: Responses of y(t) and e(t) of the system (7) for input signal r(t) of type 0, 1, and 2. Exercise 3.1 The differential equation of a plant is given by x (2) = x 2 + 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 law with real differentiating filter u = k 0 [F ( ˆX, r) ˆx (n) ] (10) µ q ˆx (q) + d q 1 µ q 1ˆx (q 1) + + d 1 µˆx (1) + ˆx = x, ˆX(0) = ˆX0, (11) where k 0 = 40, q = 2, µ = 0.1 s, d 1 = 3. Determine the fast-motion subsystem (FMS) and slow-motion subsystem (SMS) equations. Perform a numerical simulation. Solution. The differential equation of a plant is given by (8). Hence, n = 2 and x (n) = x (2) is the highest derivative of the output signal. We have that the reference model for x(t) is assigned by (9). Then, the control law with the highest derivative of the output signal in feedback and an ideal differentiating filter is given by u = k 0 [F (x (1), x, r) x (2) ],

9 Student Solutions Manual for Design of nonlinear control systems... 9 where F (x (1), x, r) = 1.2x (1) x + r. As the result, we have the control law given by u = k 0 [ 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 = k 0 [F (x (1), x, r) x (2) ], where f(x (1), x) = x 2 + x (1) and g(t) = 1.2 cos(t). Then x (2) = F (x (1), x, r) + is the SMS equation. Next, let us consider the system given by g(t)k 0 [f(x (1), x) F (x (1), x, r)] (12) µ 2ˆx (2) + d 1 µˆx (1) + ˆx = x, where ˆx (1) and ˆx (2) can be used as the estimates of x (1) and x (2), respectively. Hence, this system can play a role of a real differentiating filter. Then, the closed-loop system equations with the real differentiating filter is given by x (2) = f(x (1), x) + g(t)k 0 [F (ˆx (1), x, r) ˆx (2) ], µ 2ˆx (2) + d 1 µˆx (1) + ˆx = x. (13) Denote ˆx (1) = ˆx 1 and ˆx (2) = ˆx 2. Consider the extended system given by x (2) = f(x (1), x) + g(t)k 0 [F (ˆx 1, x, r) ˆx 2 ], µ 2ˆx (2) + d 1 µˆx (1) + ˆx = x, (2) µ 2ˆx 1 + d 1 µˆx (1) 1 + ˆx 1 = x (1), (2) µ 2ˆx 2 + d 1 µˆx (1) 2 + ˆx 2 = x (2). Substitution of the right member of the first equation into the last one yields x (2) = f(x (1), x) + g(t)k 0 [F (ˆx 1, x, r) ˆx 2 ], µ 2ˆx (2) + d 1 µˆx (1) + ˆx = x, (2) µ 2ˆx 1 + d 1 µˆx (1) 1 + ˆx 1 = x (1), (14) (2) µ 2ˆx 2 + d 1 µˆx (1) 2 + [1 + g(t)k 0 ]ˆx 2 = f(x (1), x) + g(t)k 0 F (ˆx 1, x, r). From (14), taking µ = 0, we get the discussed above SMS (12). The FMS of the extended system (14) is given by µ 2ˆx (2) + d 1 µˆx (1) + ˆx = x, (2) µ 2ˆx 1 + d 1 µˆx (1) 1 + ˆx 1 = x (1), (2) µ 2ˆx 2 + d 1 µˆx (1) 2 + [1 + g(t)k 0 ]ˆx 2 = f(x (1), x) + g(t)k 0 F (ˆx 1, x, r)

10 Student Solutions Manual for Design of nonlinear control systems where we assume that f = const, g = const. The behavior of ˆx (2) is described by the last differential equation, where the characteristic equation is given by µ 2 s 2 + d 1 µs + γ = 0. Note that γ = 1 + g(t)k 0, k 0 = 40, d 1 = 3, and µ = 0.1 s. Hence, γmax = 89, γ min = 9. By taking into account (9), we obtain the degree of time-scale separation between FMS and SMS given by η 3 γ min /µ = 30. Finally, run the Simulink program e3 1.mdl to perform a numerical simulation of the closed-loop system (13). The simulation results are shown in Fig. 4. Figure 4: Simulation results of the closed-loop system (13). Exercise 3.2 A system is given by (8). Consider the control law in the form of (10) with the desired dynamics given by (9) and the real differentiating filter (11), where k 0 = 40 and q = 2. Determine the parameters µ and d 1 of (11) such that the damping ratio exceeds 0.5 in the FMS and the degree of time-scale separation between FMS and SMS exceeds 10. Compare with simulation results. Solution. By following through solution of Exercise 3.1, we get the closed-loop system equations, SMS and FMS equations as well, where is the characteristic equation of the FMS for ˆx (2) where µ 2 s 2 + d 1 µs + γ = 0 (15) γ = 1 + g(t)k 0, g(t) = 1.2 cos(t), k 0 = 40, γmax = 89, γ min = 9. If d 2 1 4γ < 0 when γ = γmax, then from (15) we obtain s 1,2 = d 1 2µ ± j d 2 1 4γ 2µ = ζ F MS = d 1 2 γ = ζ min F MS = d 1 2 γmax, ζmin F MS = 0.5 = d 1 = Next, let us find estimates for µ based on different notions for degree of time-scale separation between FMS and SMS in the closed-loop system. Let us take γ = γ min, then µ 2 s 2 + d 1 µs + γ min = s MS af1 s + 1 MS af µ µ 2 0

11 Student Solutions Manual for Design of nonlinear control systems where a F MS 1 = d 1, a F MS 0 = γ min = 9, and the state matrix of the FMS is given by A 22 = [ A F MS = µ 1 A 22, ] 0 1 γ min d 1 = [ Since d 2 1 4γ min > 0, then from (15) we obtain s 1 = d 1 d 2 2µ + 1 4γ min s 2 = d 1 d 2 2µ 2µ 1 4γ min. 2µ Hence, ω min F MS = s 1. From the reference model, we get A S = is the state matrix of the SMS, where [ s s + 1 = 0 is the characteristic equation of the SMS. Hence, we obtain s SMS 1,2 = 0.6 ± j0.8 = ω max SMS = 0.6, (a SMS 0 ) 1/2 = 1. By solving the Lyapunov equations P F A 22 + A T 22P F = Q F, ] P S A S + A T S P S = Q S, where Q F = I and Q S = I, we obtain [ ] P F =, P S = Hence, ] [ λmax(p F ) = , λ min (P F ) = , λmax(p S ) = , λ min (P S ) = Finally, we get the following estimates for µ based on the various notions for degree of time-scale separation between FMS and SMS in the closed-loop system, that are: η 1 = λ min (P S ) µλmax(p F ), η 1 = 10 = µ = 0.052, η 2 = ωmin F MS ω max SMS = d 1 2µω max SMS η 3 = (af MS 0 ) 1/2 µ(a SMS 0 ) 1/2 =, η 2 = 10 = µ = , γmin, η 3 = 10 = µ = 0.3. µ. ].

12 Student Solutions Manual for Design of nonlinear control systems Figure 5: Simulation results of the closed-loop system (13) for d 1 = and µ = s. Run the Matlab program e3 2 Parameters.m to calculate d 1 and the above estimates for µ based on such criteria as η 1, η 2, and η 3. Next, run the Simulink program e3 2.mdl, to get the step response of the closed-loop system for d 1 = and µ = s. The simulation results are shown in Fig. 5. Chapter 4 Exercise 4.2 The differential equation of a plant is while that of the reference model is Construct the control law in the form of x (2) = x + x x (1) + {2 + sin(t)}u, (16) x (2) = 3.2x (1) x + 3.2r (1) + r. µ q u (q) + d q 1 µ q 1 u (q 1) + + d 1 µu (1) + d 0 u = k 0 T n { T n x (n) a d n 1T n 1 x (n 1) a d 1T x (1) x + b d ρτ ρ r (ρ) + b d ρ 1τ ρ 1 r (ρ 1) + + b d 1τr (1) + r}. (17) where q = 3. Determine the FMS and SMS equations from the closed-loop system equations. Solution. From (16), we have n = 2 and x (2) is the highest derivative of the output signal, where x (2) = f(x (1), x) + g(t)u (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 µ 3 u (3) + d 2 µ 2 u (2) + d 1 µu (1) + d 0 u = k 0 {F (x (1), x, r (1), r) x (2) },

13 Student Solutions Manual for Design of nonlinear control systems that is µ 3 u (3) + d 2 µ 2 u (2) + d 1 µu (1) + d 0 u = k 0 { 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, µ 3 u (3) + d 2 µ 2 u (2) + d 1 µu (1) + d 0 u = k 0 {F (x (1), x, r (1), r) x (2) }. Denote x 1 = x, x 2 = x (1), u 1 = u, u 2 = µu (1), and u 3 = µ 2 u (2). From the above closed-loop system equations, we obtain d dt x 1 = x 2, d dt x 2 = f(x 1, x 2 ) + g(t)u 1, µ d dt u 1 = u 2, µ d dt u 2 = u 3, µ d dt u 3 = d 0 u 1 d 1 u 2 d 2 u 3 + k 0 {F (x 2, x 1, r (1), r) d dt x 2 }. Substitution of the right member of the second equation into the last one yields the closed-loop system equations in the following form: d dt x 1 = x 2, d dt x 2 = f(x 1, x 2 ) + g(t)u 1, µ d dt u 1 = u 2, µ d dt u 2 = u 3, (19) µ d dt u 3 = {d 0 + k 0 g(t)}u 1 d 1 u 2 d 2 u 3 +k 0 { F (x2, x 1, r (1), r) f(x 1, x 2 ) }. In order to find the FMS equations, let us introduce the new fast time scale t 0 = t/µ into the closed-loop system equations given by (19). We obtain d x 1 dt 0 = µx 2, d x 2 dt 0 = µ{f(x 1, x 2 ) + g(t)u 1 }, d u 1 dt 0 = u 2,

14 Student Solutions Manual for Design of nonlinear control systems d u 2 dt 0 = u 3, d u 3 dt 0 = {d 0 + k 0 g(t)}u 1 d 1 u 2 d 2 u 3 +k 0 { F (x2, x 1, r (1), r) f(x 1, x 2 ) }. If µ 0, then we get the FMS equations in the new time scale t 0, that is d x 1 dt 0 = 0, d x 2 dt 0 = 0, d u 1 dt 0 = u 2, d u 2 dt 0 = u 3, d u 3 dt 0 = {d 0 + k 0 g(t)}u 1 d 1 u 2 d 2 u 3 +k 0 { F (x2, x 1, r (1), r) f(x 1, x 2 ) }. Then, returning to the primary time scale t = µt 0, we obtain the following FMS equations: x 1 = const, x 2 = const, µ d dt u 1 = u 2, µ d dt u 2 = u 3, (20) These equations may be rewritten as µ d dt u 3 = {d 0 + k 0 g(t)}u 1 d 1 u 2 d 2 u 3 +k 0 { F (x2, x 1, r (1), r) f(x 1, x 2 ) }. µ 3 u (3) + d 2 µ 2 u (2) + d 1 µu (1) + {d 0 + k 0 g(t)}u = k 0 {F (x 2, x 1, r (1), r) f(x 1, x 2 )}, (21) where x 1 = const, x 2 = 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: ẋ 1 = x 2, ẋ 2 = F (x 2, x 1, r (1), r) d 0 + d 0 + k 0 g(t) {f(x 1, x 2 ) F (x 2, x 1, r (1), r)}. (22)

15 Student Solutions Manual for Design of nonlinear control systems At the same time, we can find the above SMS by some another way. Suppose the FMS (20) is stable. Taking µ 0 in (21) we get u(t) = u s (t), where u s (t) is a steady state (more precisely, quasi-steady state) of the FMS (20) and u s = k 0 d 0 + k 0 g(t) {F (x 2, x 1, r (1), r) f(x 1, x 2 )}. Substitution of u s into (18) yields the SMS equation given by x (2) = F (x (1), x, r (1), r) d 0 + d 0 + k 0 g(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: ε F = 0.05, ε r = 0.02, ζ F MS 0.5, η 3 20, q = 2. Compare simulation results with the assignment. Note that η 3 is the degree of time-scale separation between stable fast and slow motions defined by MS η 3 = (af 0 ) 1/m µ(a SMS 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) = sin(t). We have n = 2 and x (2) is the highest derivative of the output signal. The reference model is given by x (2) = F (x (1), x, r), where F (x (1), x, r) = 2x (1) x + r. As far as the requirement on the high frequency sensor noise attenuation is not specified, then take q = n = 2. Therefore, consider the control law given by that is µ 2 u (2) + d 1 µu (1) + d 0 u = k 0 {F (x (1), x, r) x (2) }, (24) µ 2 u (2) + d 1 µu (1) + d 0 u = k 0 { 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) µ 2 u (2) + d 1 µu (1) + d 0 u = k 0 {F (x (1), x, r) x (2) }. (27)

16 Student Solutions Manual for Design of nonlinear control systems From the above closed-loop system equations, we get the FMS given by µ 2 u (2) +d 1 µu (1) +[d 0 + k 0 g]u=k 0 {F (x (1), x, r) f(x (1), x)}, (28) where F = const, f = const, and g = const during the transients in (28), as well as the SMS given by x (2) = F (x (1), x, r) + d 0 d 0 + k 0 g(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 = = 22, Fmax = 2x (1) x + r max = = 23, g min = 0.5, gmax = 2.5, e F max = ε F Fmax = = We have g(t) > 0 t. Hence, take k 0 > 0. We have ε r = Hence, take d 0 = 1. From the requirement e F (u s ) e F max = 1.15, we obtain k 0 d 0 g min max F (X, R) f(x, w) Ω X,R,w 1 e F max [ ] = Consider a steady state of the SMS, that is x (2) = x (1) = 0. Hence, we obtain x (2) = F (x (1), x, r) + d 0 d 0 + k 0 g {f(x(1), x) F (x (1), x, r)} = x (2) }{{} =0 From the requirement e s d 0 r k 0 g d 0 = 2x (1) x + r }{{} =e s + d 0 d 0 + k 0 g {x + x x(1) { 2x (1) x + r}} = }{{}}{{} =x s =r e s =e s e s = d 0 k 0 g d 0 r. d 0rmax k 0 g min d 0 e s max = ε r rmax = 0.02, we obtain k 0 [ ] [ ] d0 rmax e s max d = g min = 98.

17 Student Solutions Manual for Design of nonlinear control systems Let us take k 0 = 100. From the SMS and reference model equations, it follows that s 2 + 2s + 1 = 0 is the characteristic equation of the SMS, where a SMS 0 = 1, and µ 2 s 2 + d 1 µs + d 0 + k 0 g = 0 is the characteristic equation of the FMS. Hence, we obtain MS η 3 = (af 0 ) 1/2 µ(a SMS 0 ) = d0 + k 0 g d0 1/2 µ(a SMS 0 ) + k 0 g min η min 1/2 µ(a SMS 0 ) 1/2 3 = 20 d0 + k 0 g min = µ η min 3 (a SMS 0 ) = s. 1/ From the characteristic equation of the FMS, we get s F MS 1,2 = d 1 2µ ± j d 2 1 4(d 0 + k 0 g) 2µ = α ± jβ, where we assume that d 2 1 4(d 0 + k 0 g) < 0 when g = gmax. Hence, we can find ζ F MS = cos(θ F MS ) = α / α 2 + β 2 d 1 = 2 d 0 + k 0 g ζmin = 0.5 = F MS d 1 2ζ min d F MS 0 + k 0 gmax = Take µ = 0.3 s and d 1 = 16. Control law implementation. The discussed control law (25) can be rewritten in the form given by where u (2) + d 1 µ u(1) + d 0 µ 2 u = k 0 µ 2 x(2) k 0a d 1 µ 2 T x(1) k 0 µ 2 T 2 x + k 0 µ 2 T 2 r = u (2) + a 1 u (1) + a 0 u = b 2 x (2) + b 1 x (1) + b 0 x + c 0 r (30) a 1 = d 1 µ, a 0 = d 0 µ 2, b 2 = k 0 µ 2, b 1 = k 0a d 1 µ 2 T, b 0 = k 0 µ 2 T 2, c 0 = k 0 µ 2 T 2.

18 Student Solutions Manual for Design of nonlinear control systems Then, in order to find the block diagram of the discussed control law, from (30), we get u (2) b 2 x (2) + a 1 u (1) b 1 x (1) = a 0 u + b 0 x + c 0 r = }{{} = u 2 u (1) b 2 x (1) + a 1 u b 1 x = u 2 = u (1) b 2 x (1) = u 2 a 1 u + b 1 x = }{{} = u 1 u = u 1 + b 2 x. Hence, we obtain the equations of the controller given by u 1 = u 2 a 1 u + b 1 x, u 2 = a 0 u + b 0 x + c 0 r. (31) u = u 1 + b 2 x. From (31), we obtain the block diagram of the controller as shown in Fig. 6. Figure 6: Block diagram of (30) represented in the form (31). In conclusion, run the Matlab program e5 1 Parameters.m to calculate the controller parameters. Next, run the Simulink program e5 1.mdl, to get the step response of the closed-loop system. 4 It can be verified that the simulation results confirm the analytical calculations. The simulation results are shown in Fig. 7. Note that the control law (25) may be expressed in terms of transfer functions as u(s) = k 0 r(s) k 0(s 2 + 2s + 1) x(s). (32) µ 2 s 2 + d 1 µs + d 0 µ 2 s 2 + d 1 µs + d 0 Take d 0 = 0, then from (32) the conventional PID controller with low-pass filtering { 1 u(s) = τ LP F s + 1 k 2x(s) + 1 } s [r(s) x(s)] sx(s) results, where the low-pass filter (LPF) is given by 1/(τ LP F s + 1) and k = k 0 µd 1, τ LP F = µ d 1. 4 Throughout the simulation the following solver options are used: variable-step, ode113(adams), relative tolerance equals 1e-6.

19 Student Solutions Manual for Design of nonlinear control systems Figure 7: Simulation results of the closed-loop system given by (23) and (31) for k 0 = 100, d 1 = 15, d 0 = 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 ε r = 0, t d s 1 s, σ d 10 %, ζ F MS 0.3, and η The additional requirement G uns (jω) ε uns (ω), ω ω ns min (34) should be provided such that ε uns (ω) = 10 3 and ω n s min = 103 rad/s. Compare simulation results with the assignment. Solution. Reference model. From (33), we have x (2) = f(x, x (1) )+gu, where f(x, x (1) ) = 2x (1) +x and g = g min = g max = 2. We have n = 2 and x (2) is the highest derivative of the output signal. Hence, consider the reference model given by Take t d s = 1 s, σ d = 10 %, then by x (2) = F (x (1), x, r). ( ) θ d = tan 1 π, ω d = 4, ln(100/σ d ) t d s we get θ d = rad, ζ d = , ω d = 4 rad/s and ω n = rad/s. By selecting the 2 roots s 1,2 = 4 ± j5.4575, where Re(s 1,2 ) = ω d = ω n cos(θ d ) and Im(s 1,2 ) = ω n sin(θ d ), we obtain the desired characteristic polynomial s 2 + 8s Hence, the desired transfer function is given by G d xr(s) = s 2 + 8s and, from the above, the reference model in the form of the type 1 system follows. x (2) = 8x (1) 45.78x r (35)

20 Student Solutions Manual for Design of nonlinear control systems Control law of the 2-nd order. At the beginning, let us take q = 2. Therefore, the control law will be constructed in the form (24), where the reference model is given by (35). Hence, the control law is µ 2 u (2) + d 1 µu (1) + d 0 u = k 0 { x (2) 8x (1) 45.78x r}. (36) The closed-loop system equations are given by (26) (27). Hence, the FMS and SMS equations are given by (28) and (29), respectively. Selection of control law parameters, when q = 2. The control law parameters k 0, d 0, d 1, µ can be selected by following through solution of Exercise 5.1, if d 0 = 1. Let us consider a simplified version for the gain k 0 selection. In order to provide the requirement ε r = 0, take d 0 = 0. Then the gain k 0 can be selected such that k 0 g min = 10. Hence, we get k 0 = 5. From (28), (29), and (35), we have that s 2 + 8s = 0 is the characteristic equation of the SMS, where a SMS 0 = 45.78, and µ 2 s 2 + d 1 µs + d 0 + k 0 g = 0 is the characteristic equation of the FMS. Hence, we obtain MS η 3 = (af 0 ) 1/2 µ(a SMS 0 ) = d0 + k 0 g d0 1/2 µ(a SMS 0 ) + k 0 g min η min 1/2 µ(a SMS 0 ) 1/2 3 = 10 d0 + k 0 g min = µ η min 3 (a SMS 0 ) = s. 1/ From the characteristic equation of the FMS, we get s F MS 1,2 = d 1 2µ ± j d 2 1 4(d 0 + k 0 g) 2µ = α ± jβ, where we assume that the FMS is underdamped-stable, that is when g = gmax. Hence, we can find d 2 1 4(d 0 + k 0 g) < 0 ζ F MS = cos(θ F MS ) = α / α 2 + β 2 = d 1 d 1 2ζ min F MS 2 d 0 + k 0 g ζmin F MS = 0.3 = d 0 + k 0 gmax =

21 Student Solutions Manual for Design of nonlinear control systems High-frequency sensor noise attenuation, when q = 2. Let us replace x(t) by y(t) = x(t) + n s (t). Then, from the above, we can obtain that A d (s) G uns (s) = k uns D F MS (s) = d 0 + k 0 g (1/45.78)s 2 + (8/45.78)s + 1 [µ 2 /(d 0 + k 0 g)]s 2 + [d 1 µ/(d 0 + k 0 g)]s + 1 is the input sensitivity function with respect to noise for high frequencies, where the requirement on high-frequency sensor noise attenuation is given by the following inequality: G uns (jω) ε uns = 10 3, ω ω n s min = 103 rad/s. (37) Take µ = s and d 1 = , then G uns (jω min ns ) 2208 (where 20 lg db), or, for the sake of simplicity, we can find the limit given by lim G un ω s (jω) = k , µ q where 20 lg db. Hence, the requirement (37) on high-frequency sensor noise attenuation doesn t hold. The Bode plots of G uns (jω) and G uf (jω) are shown in Fig. 8. Figure 8: The Bode plots of G uns (jω) and G uf (jω). Note, the same conclusion can be obtained by inspection the Bode amplitude plot of G uf (jω), where 1 G uf (s) = k uf D F MS (s), k 0 k uf = d 0 + k 0 g, µ 2 D F MS (s) = d 0 + k 0 g s2 + d 1µ d 0 + k 0 g s + 1.

22 Student Solutions Manual for Design of nonlinear control systems Hence, we get L uf (ω min ns ) = 20 lg G uf(jω min ns ) 53 db and L HF max(ω A n s min ) = 20 lg ε n s 20[n + ϑ] lg ω n s min = 60 db, where ϑ = 0. Hence, the requirement for high-frequency sensor noise attenuation L uf (ω n s min ) A LHF max(ω n s min ) doesn t hold. Run the Matlab program e5 10 A Parameters.m to calculate the reference model parameters, Bode plots of G uns (jω) and G uf (jω), as well as parameters of the controller given by (36), where q = 2. Next, run the Simulink program e5 10 a.mdl, to get a step response as well as ramp response (by using Switch 1) of the closed-loop system. The simulation results are shown in Fig. 9. Figure 9: Simulation results of the closed-loop system (33), (36) for k 0 = 5, d 0 = 0, d 1 = , and µ = s, where y(t) = x(t). Note that, by Switch 2, the type of the reference model can be changed from 1 to 2 in the program e5 10 a.mdl. By Switch 3, add the hign frequiency sensor noise n s (t) to the output y(t) = x(t) + n s (t), where n s (t) = 10 3 sin(10 3 t). The simulation results are shown in Fig. 10. Figure 10: Simulation results of the closed-loop system (33), (36) for k 0 = 5, d 0 = 0, d 1 = , and µ = s in the presence of the noise n s (t), where y(t) = x(t)+n s (t). Control law of the 3-rd order. In order to provide the requirement for high-frequency sensor noise attenuation given by (37), let us take q = 3 and consider the control law given by µ 3 u (3) + d 2 µ 2 u (2) + d 1 µu (1) + d 0 u = k 0 {F (y (1), y, r) y (2) },

23 Student Solutions Manual for Design of nonlinear control systems where y(t) = x(t) + n s (t) and the reference model is the same as (35). Hence, the control law can be rewritten as where µ 3 u (3) +d 2 µ 2 u (2) +d 1 µu (1) +d 0 u=k 0 { y (2) ā d 1y (1) ā d 0y+ā d 0r}, (38) ā d 1 = 8, ā d 0 = Note that the control law (38) may be expressed in terms of transfer functions as u(s) = k 0 ā d 0 k 0 (s 2 + ā d r(s) 1s + ā d 0) x(s). (39) µ 3 s 3 + d 2 µ 2 s 2 + d 1 µs + d 0 µ 3 s 3 + d 2 µ 2 s 2 + d 1 µs + d 0 Take d 0 = 0, then from (39) the conventional PID controller with low-pass filtering u(s) = { } 1 τlpf 2 s2 + a LP F τ LP F s + 1 k ā d 1x(s) + ād 0 [r(s) x(s)] sx(s) s results, where the low-pass filter is given by 1/(τ 2 LP F s2 + a LP F τ LP F s + 1) and k = k 0 µd 1, τ LP F = µ d1, a LP F τ LP F = µd 2 d 1. The FMS characteristic polynomial in the closed-loop system is given by µ 3 s 3 + d 2 µ 2 s 2 + d 1 µs + d 0 + k 0 g. Let us consider the selection of the control law parameters based on Bode amplitude plot of the closed-loop FMS given by L uf (ω) = 20 lg G uf (jω), where D F MS (s) = G uf (s) = k uf 1 D F MS (s), k uf = k 0 d 0 + k 0 g, µ 3 d 0 + k 0 g s3 + d 2µ 2 d 0 + k 0 g s2 + d 1µ s + 1. (40) d 0 + k 0 g By the same way as was shown above, take d 0 = 0 and k 0 = 5. Hence, k uf = 0.5. Then, let us perform G d uf(s) in the corner frequency factored form given by Then, the roots of quadratic factor G d uf(s) = k uf 1 [T 2 1 s 2 + 2ζ 1 T 1 s + 1][T 2 s + 1]. (41) T 2 1 s 2 + 2ζ 1 T 1 s + 1

24 Student Solutions Manual for Design of nonlinear control systems are the dominant poles of G d uf(s), where the damping ratio ζ 1 is selected such that ζ 1 = ζ min F MS = 0.3. Take t d s t d s, SMS and t d s, F MS = t d s/η, where η = 10. Then, we can obtain ω 1 4 t d s, F MS = 4η ζ 1 t d s = 4 10 ζ = T 1 = 1 ω 1 = s. Let us calculate a lower bound for T 2 from the condition where L d uf(ω ns min ) = LHF A max(ω ns min ), L HF max(ω A s min ) = 20 lg ε n s 20[n + ϑ] lg ω n s min = 20 lg [2 + 0] lg 10 3 = 60 db. Denote L = ε ns /(ω n s min )n+ϑ = Hence, we get G d uf(jω min ns, T min 2 ) = L = k uf = L = [1 T1 2 [ω n s min ]2 + j2ζ 1 T 1 [ω n s min ]][jt min 2 ω n s min + 1] T min 2 = 1 [k ω n uf /L] 2 1/2 s min (1 [T 1 ω n s min ]2 ) 2 + (2ζ 1 T 1 ω n 1 = s min )2 T min 2 = 1 [ [0.5/10 3 ] 2 ] 1/ (1 [ ] 2 ) 2 +( ) s. The time constant T 2 should be selected such that the inequalities T min 2 T 2 T 1 hold. We see, there is apparent contradiction. Therefore, let us replace the degree of time-scale separation between fast and slow modes η = 10 by η = 8 and redesign the parameter T 1 again. We get T 1 = s. Accordingly, by the same way as above, we obtain T min 2 = s. Hence, the condition T min 2 T 2 T 1 holds and then, we can take T 2 = T min 2 = s. As a result of the above, we obtain D d F MS (s) = [T 2 1 s 2 + 2ζ 1 T 1 s + 1][T 2 s + 1] = s s s + 1. From (40), and by taking into account that d 0 = 0 as well as the requirement D F MS (s) = D d F MS (s),

25 Student Solutions Manual for Design of nonlinear control systems we obtain µ = {d d q k 0 g} 1/ , d 1 = d d 1 [k 0 g] (2)/3 [d d 3 ] 1/ , (42) d 2 = d d 2 [k 0 g] (1)/3 [d d 3 ] 2/ Finally, the control law (38) can be rewritten in the form given by u (3) + d 2 µ u(2) + d 1 µ 2 u(1) + d 0 µ 3 u = k 0 µ 3 y(2) k 0ā d 1 µ 3 T y(1) k 0 µ 3 T 2 y + k 0 µ 3 T 2 r = u (3) +a 2 u (2) +a 1 u (1) +a 0 u = b 2 y (2) +b 1 y (1) +b 0 y+c 0 r. (43) From (43), we can obtain the equations of the controller given by where u 1 = u 2 a 2 u 1 + b 2 y, u 2 = u 3 a 1 u 1 + b 1 y, u 3 = a 0 u 1 + b 0 y + c 0 r, (44) u = u 1, a 2 = d 2 µ, a 1 = d 1 µ 2, a 0 = d 0 µ 3, b 2 = k 0 µ 3, b 1 = k 0ā d 1 µ 3, b 0 = k 0ā 0 µ 3, c 0 = k 0ā 0 µ 3. The Bode plots of G uns (jω) and G uf (jω) are shown in Fig. 11, where the parameters µ, d 1, d 2 are given by (42) with d 0 = 0 and k 0 = 5. In conclusion, run the Matlab program e5 10 B Parameters.m to calculate the reference model parameters, Bode plots of G uns (jω), and G uf (jω), as well as the parameters of the controller given by (44), where q = 3. Next, run the Simulink program e5 10 b.mdl, to get a step response as well as ramp response (by using Switch 1) of the closed-loop system. By Switch 3, add the hign frequiency sensor noise n s (t) to the output, that is y(t) = x(t) + n s (t), where n s (t) = 10 3 sin(10 3 t). The simulation results are shown in Fig. 12. It can be verified that the simulation results confirm the analytical calculations. Note that in the program e5 10 b.mdl, by Switch 2, the type of the reference model can be changed from 1 to 2. Then, instead of (43), we have the controller given by u (3) + a 2 u (2) + a 1 u (1) + a 0 u = b 2 y (2) + b 1 y (1) + b 0 y + c 1 r (1) + c 0 r, (45)

26 Student Solutions Manual for Design of nonlinear control systems Figure 11: The Bode plots of G uns (jω) and G uf (jω). Figure 12: Simulation results of the closed-loop system (33), (44) for k 0 = 5, d 0 = 0, µ = s, d 1 = , d 2 = in the presence of the noise n s (t), where y(t) = x(t) + n s (t). where c 1 = b 1 and c 0 = b 0. Hence, instead of (44), we get u 1 = u 2 a 2 u 1 + b 2 y, u 2 = u 3 a 1 u 1 + b 1 y + c 1 r, u 3 = a 0 u 1 + b 0 y + c 0 r, (46) u = u 1. 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) sin(0.5t) + gu(t τ), (47)

27 Student Solutions Manual for Design of nonlinear control systems Figure 13: Block diagram of (45) represented in the form (46). where g = 1 and the reference model x (2) =F (x (1), x, r) is assigned by The control law has the form x (2) = T 2 { a 1 T x (1) x + r}. (48) µ 2 u (2) + d 1 µu (1) + d 0 u = k 0 {F (x (1), x, r) x (2) }, (49) where T = 1 s, a 1 = 2, k 0 = 10, µ = 0.1 s, d 0 = 0, d 1 = 4. Determine the region of stability for τ of the FMS. Compare with simulation results of the closed-loop system. Solution. The closed-loop system equations are given by x (2) = f(x (1), x, t) + gu(t τ), µ 2 u (2) + d 1 µu (1) + d 0 u = k 0 {F (x (1), x, r) x (2) }, where f(x (1), x, t) = 0.5x(t)x (1) (t) + 0.1x (1) (t) + 0.5x(t) sin(0.5t), F (x (1), x, r) = T 2 { a d 1T x (1) x + r}, g = 1, T = 1 s, a d 1 = 2, k 0 = 10, µ = 0.1 s, d 0 = 0, and d 1 = 4. Hence, from the above equations, we get the FMS given by µ 2 u (2) +d 1 µu (1) +d 0 u+k 0 gu(t τ)=k 0 {F (x (1), x, r) f(x (1), x)}, (50) where F = const and f = const during the transients in (50). representation of the FMS (50) is shown in Fig. 14, where The block diagram D(µs) = µ 2 s 2 + d 1 µs + d 0. (51) Figure 14: Block diagram of the FMS (50) with delay τ, where F = const, f = const.

28 Student Solutions Manual for Design of nonlinear control systems By the Nyquist stability criterion, the FMS (50) is marginally stable if ω c 0 exists such that the condition k 0 ge jτ mω c = 1 + j0, (52) D(jµω c ) holds, where ω c is the crossover frequency and τ m is an upper bound for delay τ. The value τ m determines the region of stability for τ of the FMS. From (52) and by taking into account the condition d 0 = 0, we get k 0 ge jτmωc jµω c (jµω c + d 1 ) = 1 = µ 4 ωc 4 + d 2 1µ 2 ωc 2 k0g 2 2 = 0 = (53) From (52), we have 5 y = ωc 2 > 0, µ 4 y 2 + d 2 1µ 2 y k0g 2 2 = 0 = y = d2 1 + d k0g 2 2 2µ 2 = = ω c = y 22 rad/s. τ m = [π/2 tan 1 (µω c /d 1 )]/ω c. (54) Hence, the region of stability for τ of the FMS is defined as where 0 < τ < τ m = s. Finally, the discussed control law can be rewritten in the form given by u (2) + d 1 µ u(1) + d 0 µ 2 u = k 0 µ 2 x(2) k 0a d 1 µ 2 T x(1) k 0 µ 2 T 2 x + k 0 µ 2 T 2 r = u (2) + a 1 u (1) + a 0 u = b 2 x (2) + b 1 x (1) + b 0 x + c 0 r, a 1 = d 1 µ, a 0 = d 0 µ 2, b 2 = k 0 µ 2, b 1 = k 0a d 1 µ 2 T, b 0 = k 0 µ 2 T 2, c 0 = k 0 µ 2 T 2. Run the Matlab program e6 1 Parameters.m to calculate the region of stability for τ of the FMS and the controller parameters. Next, run the Simulink program e6 1.mdl, to get the step response of the closed-loop system. Exercise 6.4 Determine the phase margin and gain margin of the FMS based on the input data of Exercise 6.1 for the time delay τ = 0.3τ m, where τ = τ m corresponds to the marginally stable FMS. 5 y = tan 1 (x) denotes the arctangent of x, i.e., tan(y) = x.

29 Student Solutions Manual for Design of nonlinear control systems Solution. By following through the solution of Exercise 6.1, the block diagram representation of the FMS (50) is shown in Fig. 14, where the phase margin (PM) of the FMS (50) is given by P M = π ArgD(jµω c ) τω c. (55) From (51) and d 0 = 0, we get Hence, we obtain ArgD(jµω c ) = π 2 + tan 1 (µω c /d 1 ). (56) P M = π 2 tan 1 (µω c /d 1 ) τω c. (57) In order to find the gain margin (GM) of the FMS, consider the equation Arg[G O F MS (jω π)] = π. (58) From (58), we get π 2 tan 1 (µω π /d 1 ) τω π = 0, (59) where τ, τ m, ω π and ω c are calculated by joint numerical resolution of (53), (54), and (59). Finally, we can find that l π = G O (jω k 0 ge jτωπ F MS π) = jµω π (jµω π + d 1 ) k 0 g = jµω π (jµω π + d 1 ) k 0 g = (µω π ) 4 + (d 1 µω π ) 2 Hence, the gain margin (GM) of the FMS is given by GM = 1/l π. By running the Matlab program e6 4 Parameters.m, we obtain ω c 22 rad/s, ω π 47.7 rad/s, τ m s, τ = 0.3τ m s, P M = rad, GM Next, run the Simulink program e6 4.mdl, to get the step response of the closed-loop system where τ = s. Chapter 7 Exercise 7.1 The differential equations of a plant model are given by ẋ 1 = x 1 + x 2, ẋ 2 = x 1 + x 2 + x 3 + u, (60) ẋ 3 = 2x 1 x 2 + 2x 3 + a u, y = x 1.

30 Student Solutions Manual for Design of nonlinear control systems Verify the invertibility and internal stability of the given system (60), where (a) a = 1, and (b) a = 3. Find the degenerated system. Solution. From (60), we obtain ẏ = x 1 + x 2 = ÿ = ẋ 1 + ẋ 2 = ÿ = 2x 1 + 2x 2 + x 3 + u. Hence, the system (60) is invertible and the relative degree is given by α = 2. Let us introduce a state-space transformation defined by From (61), we have y 1 = y = x 1, y 2 = ẏ = x 1 + x 2, (61) z = x 3 x 1 = y 1, x 2 = y 2 y 1, (62) x 3 = z. By the change of variables (62), we get the normal form of (60) given by ẏ 1 = y 2, ẏ 2 = 2y 2 + z + u, (63) ż = y 1 + y 2 + 2z + a u, y = y 1. Let the desired stable output behavior is defined by ÿ = F (ẏ, y, r), which can be rewritten as ÿ 1 = F (ẏ 1, y 1, r) or ẏ 1 = y 2, ẏ 2 = F (y 2, y 1, r). In order to find the equations of the internal subsystems, take F (y 2, y 1, r) = 2y 2 +z+u. Hence, we obtain the inverse dynamics solution given by u id = F (y 2, y 1, r) 2y 2 z. Substitution of u = u id into (63) yields ẏ 1 = y 2, ẏ 2 = F (y 2, y 1, r), (64) ż = (2 a)z + y 1 + (1 2a)y 2 + af (y 2, y 1, r), y = y 1, where ż = (2 a)z+y 1 +(1 2a)y 2 +af (y 2, y 1, r) is the equation of the internal subsystem and y 2, y 1, r are treated as bounded external disturbances of the internal subsystem. If a = 1, then 2 a > 0 and the unique equilibrium point of the internal subsystem is unstable. If a = 3, then 2 a < 0. Hence, the solutions of the internal subsystem are bounded when variables y 2, y 1, r are bounded, that is the bounded-input-bounded-state

31 Student Solutions Manual for Design of nonlinear control systems (BIBS) stability of the internal subsystem. Next, from (64), by taking y 1 = r = const (hence, y 2 = 0 and F (y 2, y 1, r) = 0), we find the degenerated system given by ż = (2 a)z + r, where the unique equilibrium point of the degenerated subsystem is exponentially stable when 2 a < 0. Exercise 7.10 Verify the invertibility and internal stability of the system ẋ 1 = x 2 1 x u, ẋ 2 = x 2 u, (65) y = x 1, Assume that the inequalities x 1 (t) 1.5, x 2 (t) 1.5, r(t) 1 hold for all t [0, ). Find the control law such that ε r = 0, t d s 3 s, σ d 0%. Run a computer simulation of the closed-loop system with zero initial conditions. Compare simulation results of the output response with the assignment for r(t) = 1, t > 0. Solution. Invertibility and internal stability. By the change of variables y = x 1 and z = x 2, we get ẏ = y 2 z 3 + u, ż = z u. (66) Hence, the system (66) is invertible and the relative degree is given by α = 1. Let the desired stable output behavior is defined by ẏ = F (y, r). Take F (y, r) = y 2 z 3 + u. Hence, we obtain the inverse dynamics solution given by Substitution of u = u id into (66) yields u id = F (y, r) y 2 + z 3. ẏ = F (y, r), (67) ż = f(z) + φ(y, r), where φ(y, r) = y 2 F (y, r) and f(z) = z z 3. Denote c = φ(y, r) where c is an arbitrary real number. Take ż = 0 in the internal subsystem given by ż = f(z) + c. (68) Hence, from f(z) + c = 0, depending on value c, it can be up to three equilibrium points of the internal subsystem. It can be easily verified, all solutions of the internal subsystem (68) are bounded. Control law. We have α = 1. Hence, y (1) is the highest derivative of the output signal. As far as the requirement on the high frequency sensor noise attenuation is not specified, then, for simplicity, take q = α = 1 and consider the control law given by µu (1) + d 0 u = k 0 [F (y, r) y (1) ],

32 Student Solutions Manual for Design of nonlinear control systems where the reference model is y (1) = F (y, r). Take t d s = 3 s, σ d = 0 %, then we get θ d = 0 rad, ζ d = 1, a = ω d = ω n = rad/s. By selecting the root s 1 = a, we obtain the desired characteristic polynomial s + a. Consider the desired transfer function given by G d yr(s) = Hence, we get the reference model given by Finally, the control law is y (1) = ay + ar. a s + a. (69) µu (1) + d 0 u = k 0 [ y (1) ay + ar]. (70) Closed-loop system. The closed-loop system equations are given by ẏ = y 2 z 3 + u, ż = z u, (71) µ u + d 0 u = k 0 [F (y, r) ẏ]. Denote f(y, z) = y 2 z 3. From (71), by following through solution of Exercise 4.2, we obtain the FMS equation, that is µ u + [d 0 + k 0 ]u = k 0 [F (y, r) f(y, z)], where F = const, f = const during the transients in (72), and µs + d 0 + k 0 g = 0 is the characteristic equation of the FMS, where g = 1. Suppose the FMS (72) is stable. Then, in order to provide the requirement ε r = 0, we take d 0 = 0. Next, taking µ 0 in (72) we get u(t) = u s (t), where u s (t) is a steady state (more precisely, quasi-steady state) of the FMS (72) and u s = u id = F (y, r) f(y, z). Substitution of u s 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 k 0 selection. We have d 0 = 0, then the gain k 0 can be selected such that k 0 g min = 10, where g min = g max = g = 1. Hence, we get k 0 = 10. From (69), we have that the natural frequency of the reference model is given by ω d n = rad/s. Denote ω max = ω d SMS n = rad/s, ω min = d 0 + k 0 g min F MS µ rad/s.

33 Student Solutions Manual for Design of nonlinear control systems Then, without the taking into account the rate of dynamics of the internal subsystem, let us consider the ratio η 2 = ωmin F MS ω max (72) SMS as a criterion for the degree of time-scale separation between fast and slow motions. Take η min 2 = 20. Hence, we obtain η 2 = d 0 + k 0 g min µω max SMS η min 2 = 20 = µ µmax = d 0 + k 0 g min η min 2 ω max SMS = s As a result, take µ = s. Control law implementation. Finally, the control law (70) can be rewritten in the form given by u (1) + d 0 µ u = k 0 µ y(1) k 0 µt y + k 0 µt r = u (1) + a 0 u = b 1 y (1) + b 0 y + c 0 r = u (1) b 1 y (1) = a 0 u + b 0 y + c 0 r, (73) }{{} = u 1 where T = 1/a. From (73), we obtain the equations of the controller given by where u 1 = a 0 u + b 0 y + c 0 r, (74) u = u 1 + b 1 y, a 0 = d 0 µ, b 1 = k 0 µ, b 0 = k 0 µt, c 0 = k 0 µt. From (74), we obtain the block diagram of the controller as shown in Fig. 15. Run the Figure 15: Block diagram of (73) represented in the form (74). Matlab program e7 10 Parameters.m to calculate the reference model parameter T, as

34 Student Solutions Manual for Design of nonlinear control systems well as the parameters k 0, µ of the controller given by (70). Next, run the Simulink program e7 10.mdl, to get a step response of the closed-loop system with zero initial conditions. Exercise 8.1 Chapter 8 Verify the invertibility and internal stability of the system given by ẋ 1 = x 1 + 3x 2 + u 1 + u 2, ẋ 2 = x 1 + x 2 + u 1 2u 2, (75) y 1 = x 1 + x 2, y 2 = x 1 + 2x 2. Assume that the inequalities x j (t) 2 j and r(t) 1 hold for all t [0, ). Find the control law of the form where µ q i i ũ (q i) i + d i,qi 1µ q i 1 i ũ (q i 1) i + + d i,1 µũ (1) i + d i,0 ũ i = k i e F i, Ũ i (0) = Ũ i 0, i = 1,..., p, (76) u = K 0 ũ, u = {u 1, u 2,..., u p } T, ũ = {ũ 1, ũ 2,..., ũ p } T, µ i > 0, k i > 0, Ũ i = {ũ i, ũ (1) i,..., ũ (q i 1) i } T, q i α i. Provide the following requirements: ε r1 = 0, ε r2 = 0, t d s1 1 s, σ1 d 0%, t d s2 3 s, σ2 d 0%. Compare simulation results for the step response of the closed-loop control system with the assignment. Solution. Invertibility and internal stability. From (75), by following through solution of Exercise 7.1, we get where ẏ 1 = ẋ 1 + ẋ 2 = ẏ 1 = 2x 1 + 4x 1 + 2u 1 u 2, ẏ 2 = ẋ 1 + 2ẋ 2 = ẏ 2 = x 1 x 2 + u 1 5u 2, det G = det [ ] = 9 0. Hence, the system (75) is invertible and the relative degrees are given by α 1 = α 2 = 1. We have that α 1 + α 2 = n = 2. Then, the internal subsystem does not exist. Reference model. By following through solution of Exercise 7.10, take t d s1 = 1 s, σ d 1 = 0 %, then we get θ d 1 = 0 rad, ζ d 1 = 1, a 1 = ω d 1 = ω 1,n = 4 rad/s. By selecting the root s 1 = a 1, we obtain the desired characteristic polynomial s + a 1. The desired transfer function G d y1r1(s) = y 1 (s)/r 1 (s) is given by G d y1r1(s) = a 1 s + a 1. Hence, we get the reference model for y 1 given by y (1) 1 = a 1 y 1 + a 1 r 1 = y (1) 1 = 1 T 1 [r 1 y 1 ], (77)