TTK4150 Nonlinear Control Systems Solution 6 Part 2

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1 TTK4150 Nonlinear Control Systems Solution 6 Part 2 Department of Engineering Cybernetics Norwegian University of Science and Technology Fall 2003 Solution 1 Thesystemisgivenby φ = R (φ) ω and J 1 ω 1 = (J 2 J 3 ) ω 2 ω 3 + τ 1 J 2 ω 2 = (J 3 J 1 ) ω 3 ω 1 + τ 2 J 3 ω 3 = (J 1 J 2 ) ω 1 ω 2 + τ 3 where φ =[φ 1 φ 2 φ 3 ] T is the orientation vector, ω =[ω 1 ω 2 ω 3 ] T is a vector of angular velocity along the principal axes, τ =[τ 1 τ 2 τ 3 ] T is a vector of torque inputs applied about the principal axes and J i are the principal moments of inertia. Further, we have that R (φ) =R T (φ) and we assume that R (φ) is non singular in out domain of interest (R (φ) is non singular on D φ but not on R 3 ). 1. When analyzing a system it is often preferable to express it in compact form of vectors and matrixes to reduce the amount of equations one has to work on. When doing this, working with differential equations on a compact form, it is important to locate properties of the various vectors and matrixes involved in order to take advantage of these properties in the analysis. (a) The first equation φ = R (φ) ω 1

2 is already given. Properties connected to this differential equation is that R (φ) =R T (φ) and that R (φ) is non singular on D φ (which we assume contains our domain of interest). Our task is therefore to find M and C (ω). It is easily seen that M = diag {J 1,J 2,J 3 } and it follows that M = M T > 0 Whatremainsistoshowthat C (ω) ω = (J 2 J 3 ) ω 2 ω 3 (J 3 J 1 ) ω 3 ω 1 (J 1 J 2 ) ω 1 ω 2 and that C (ω) = C T (ω). Thisimpliesthat 0 c 1 (ω) c 2 (ω) C (ω) ω = c 1 (ω) 0 c 3 (ω) c 2 (ω) c 3 (ω) 0 and ω 1 ω 2 ω 3 c 1 (ω) ω 2 + c 2 (ω) ω 3 = (J 2 J 3 ) ω 2 ω 3 = J 2 ω 2 ω 3 J 3 ω 2 ω 3 c 1 (ω) ω 1 + c 3 (ω) ω 3 = (J 3 J 1 ) ω 3 ω 1 = J 3 ω 3 ω 1 J 1 ω 3 ω 1 c 2 (ω) ω 1 c 3 (ω) ω 2 = (J 1 J 2 ) ω 1 ω 2 = J 1 ω 1 ω 2 J 2 ω 1 ω 2 where it can be recognized that and c 1 (ω) = k 1 ω 3 c 2 (ω) = k 2 ω 2 c 3 (ω) = k 3 ω 1 k 1 + k 2 = J 2 J 3 k 1 + k 3 = J 3 J 1 k 2 k 3 = J 1 J 2 2

3 whereitisseenthat k 1 = J 3 k 2 = J 2 k 3 = J 1 which gives C (ω) = = 0 k 1 ω 3 k 2 ω 2 k 1 ω 3 0 k 3 ω 1 k 2 ω 2 k 3 ω J 3 ω 3 J 2 ω 2 J 3 ω 3 0 J 1 ω 1 J 2 ω 2 J 1 ω 1 0 (b) Since we are solving the stabilization problem, we are working with constant references. The equilibrium of interest is shifted to the origin according to e 1 = φ φ d e 2 = ω ω d = ω and in order to apply our analyzing tools we will hereafter analyse the differential equations and ė 1 = φ φ d = φ = R (φ) ω = R (φ) e 2 ė 2 = ω Mė 2 = C (e 2 ) e 2 + τ The stability properties shown for e = e T 1 e T T 2 =[00] T will now be the same as the stability properties of the point φ T ω T T = φ T d ω T T d. 3

4 2. A storage function is given by where K p = K T p > 0. V (e) = 1 2 et 2 Me et 1 K p e 1 (a) The storage function may be interpreted as the total energy of the system, where the term 1 2 et 2 Me 2 describes the kinetic energy of the system (related to motion/velocity) and the term 1 2 et 1 K p e 1 counts for the potential energy in the system (related to deviation in position/orientation). (b) To find a suitable output based on passivity, we start by calculating the time derivative of the storage function along the solution of the system V (e) = 1 2ėT 2 Me et 2 Mė ėT 1 K p e et 1 K p ė 1 = e T 2 Mė 2 + e T 1 K p ė 1 = e T 2 (C (e 2 ) e 2 + τ)+e T 1 K p R (φ) e 2 = e T 2 C (e 2 ) e 2 + e T 2 τ + e T 1 K p R (φ) e 2 Since C (e 2 ) is skew symmetric we have that e T 2 C (e 2 ) e 2 =0,this can be seen by e T 2 C (e 2 ) e 2 = 1 2 et 2 C (e 2 ) e et 2 C (e 2 ) e 2 = 1 2 et 2 C (e 2 ) e (C (e 2) e 2 ) T e T 2 T = 1 2 et 2 C (e 2 ) e et 2 C T (e 2 ) e 2 = 1 2 et 2 C (e 2 ) e et 2 C (e 2 ) e 2 = 0 Further, it can be seen that e 2 is a common factor in the remaining term. Exploiting this leads to the following expression for V (e) V (e) = e T 2 τ + e T 1 K p R (φ) e 2 = e T 2 τ + e T 1 K p (R (φ) e2 ) = e T 2 τ +(R (φ) e 2 ) T e T 1 K p = e T 2 τ + e T 2 R T (φ) Kp T e 1 = e T 2 τ + e T 2 R (φ) K p e 1 = e T 2 (τ + R (φ) K p e 1 ) 4

5 This last equation suggests that we may choose to make a control law with e 2 as output, which is the angular velocity of the system. This is justifiedbythefactthatwemaychooseτ to cancel the term R (φ) K p e 1 and leave a fictive control input, making the system passive from the fictive input to e 2. Note that this choice of control input, τ, doesnotimplythatweareonlymakinguseoffeedback from the angular velocity, only that we can achieve some passivity properties with respect to ω as output (we are possibly making use of feedback from the orientation to cancel the term R (φ) K p e 1 ). (c) Since we do not know anything of the sing of R (φ), it is impossible to conclude any passivity properties of the system with ω as output. 3. Usingthesamestoragefunctionasabove,wehavethat V (e) = 1 2 et 2 Me et 1 K p e 1 and V (e) =e T 2 (τ + R (φ) K p e 1 ) Since τ isourcontrolinput,wearefreetochooseitaswewishaslong as it is defined (not infinity). (a) In the case of making the system lossless from v to e 2 we must choose it such that V (e) =e T 2 v This implies that the control input, τ, is taken as τ = R (φ) K p e 1 + v where v is our new fictive input. (b) Since the system is passive with a positive definite storage function V (e) = 1 2 et 2 Me et 1 K p e λ min (M) ke 2 k λ min (K p ) ke 1 k 2 2 we know that the equilibrium φ T d ω T d T is stable (Lemma 6.6). 5

6 (c) In the case of making the system output strictly passive from v to e 2 we must choose it such that V (e) e T 2 v e 2 ρ (e 2 ),e 2 ρ (e 2 ) > 0 e 2 6=0 wherewedesiree 2 ρ (e 2 )=δe T 2 e 2 in order to draw conclusions with respect to stability. This leads to the choice of τ = R (φ) K p e 1 K d e 2 + v, K p = K T p > 0 which results in V (e) = e T 2 (τ + R (φ) K p e 1 ) = e T 2 ( R (φ) K p e 1 K d e 2 + v + R (φ) K p e 1 ) = e T 2 K d e 2 + e T 2 v λ max (K d ) e T 2 e 2 + e T 2 v and the system is output strictly passive from v to ω with δ = λ max (K d ). (d) The control input making the system output strictly passive has added the term K d e 2. This term may be interpreted as adding damping to the system with respect to position since it appears in with respect to the angular velocity ω. In the actual system equations it will appear together with the skew symmetric matrix or Mė 2 = C (e 2 ) e 2 + τ = C (e 2 ) e 2 R (φ) K p e 1 K d e 2 + v = (C (e 2 ) K d ) e {z } 2 R (φ) K p e 1 + v adding damping M ω =(C (ω) K d ) ω R (φ) K p (φ φ d )+v (e) In addition to the stated results due to the property of lossless, we can conclude that the system is finite gain L 2 stable from v to ω with an L 2 gain less than or equal to 1/δ (Lemma 6.5). To conclude with the stronger result of asymptotically stable, we need to show that the system is zero-state observable. This is confirmed by ω 0 e 2 0 and ė 2 0 and ė 1 =0 Mė 2 =(C (e 2 ) K d ) e 2 R (φ) K p e 1 =0 R (φ) K p e 1 =0 e 1 =0 φ D φ 6

7 since R (φ) is non singular on D φ. Hence, no solution can stay identical in ω =0other than the trivial solution φ T (t) ω T (t) T = φ T d 0 T. This implies that we can conclude asymptotic stability of the equilibrium (φ, ω) =(φ d, 0) in the system φ = R (φ) ω M ω = C (ω) ω + τ where the control input is given by τ = R (φ) K p (φ φ d ) K d ω Even though the storage function is positive definite and radially unbounded, we may not conclude global asymptotic stability of the equilibrium. This is due to the fact that R (φ) is only guaranteed to be non singular on D φ. 4. This part is solved by considering properties of feedback systems when both systems possess some passivity properties. Figure 6.11 in Khalil shows the feedback connection applied to draw conclusions. To avoid confusion with respect to notation, the signals e 1 and e 2 in the figure are replaced by z 1 and z 2 respectively. (a) In order to draw conclusions of asymptotic stability (Lyapunov stability) and robustness (L 2 stability), we at least require the two feedback components to be output strictly passive. This leads to the choice of the control input τ = R (φ) K p e 1 K d e 2 + v since this turns the system output strictly passive and zero-state observable with v as input and ω as output. Further, let y 1 and y 2 in the figure represent v and e 2 respectively (system H 2 represents the spacecraft in closed loop with the control law τ (φ, ω, 0) which already is shown asymptotically stable). This implies that u 1 represents measurement noise with respect to the rotational velocity and u 2 represents actuator noise with respect to the control law v. (b) To improve the steady state behavior of the rotational velocity we desire some integral effect with respect to e 2. This can be achieved by using the PID control law h PID (s) =K p β (1 + T is)(1+t d s) (1 + βt i s)(1+αt d s) 7

8 where 0 T d T i, 1 β< and 0 <α 1. By choosing H 1 = h PID (s) we are guaranteed integral effect of the rotational velocity ( e 2 ). Further, the output of H 1 is given by v which wearefreetochoose. TheproposedPIDcontrollawisknown to be output strictly passive and zero state observable. Since both dynamic systems are output strictly passive and zero-state observable, we conclude that the equilibrium (φ, ω) =(φ d, 0) is asymptotically stable. No global results can be concluded since R (φ) is non singular only on D φ. Itisnotpossibletousethesameapproachtocopewithasteady state deviation in the orientation. The approach used is based on a relatively strict passivity property with respect to v and ω. Since we do not have any passivity properties with respect to v and (φ φ d ) we are not able to use the same approach. (c) In order to draw any robustness conclusions with respect to uncertainty in measurements and actuators of the proposed system, we require that both feedback components impose the property of output strictly passive with the output strictly part in the form δy T y. Since we do not know if the PID control law imposes this property, we are not able to draw any conclusions. Solution 2 (Exercise a) in Khalil) The system is given by ẋ 1 = x 1 x 2 ẋ 2 = x 1 + u where we globally stabilize the origin. If we follow the steps from Khalil, the design of u is done in six steps Step one consists of formulating the system to fit a general backstepping form and then use the input to transform the system into an integrator backstepping form. Following the notation of Khalil, it can be recognized that η = x 1 ξ = x 2 f (η) = 0 g (η) = η f a (η, ξ) = η g a (η, ξ) = 1 8

9 and the system is given by η = f (η)+g(η) ξ ξ = f a (η, ξ)+g a (η, ξ) u The system is transformed into a integrator backstepping form by choosing the input as u = 1 g a (η, ξ) (u a f a (η, ξ)) = u a η where u a is considered the input in the integrator backstepping form. Step two consists of designing a control law φ (η) such that the origin of η = f (η)+g(η) φ (η) turns asymptotically stable by using a Lyapunov function. In addition to this we require that φ (0) = 0. Using a quadratic Lyapunov function, the stability of the origin is and the choice of φ (η) is found according to where V (η) = 1 2 η2 V (η) = η η = η (f (η)+g(η) φ (η)) = η 2 φ (η) = η 4 φ (η) = η 2 Step three consists of definingavariablez = ξ φ (η) which is used to analyze the fact that ξ differs from φ (η) in the actual system. Step four consists of dividing the control input according to u a = v + φ Step five consists of choosing the control v according to v = V g (η) kz η = ηη k (ξ φ (η)) = η 2 k ξ + η 2 where k>0. 9

10 Step five consists of summing up the control law according to u = u a η = v + φ η = η 2 k ξ + η 2 2η η η = η 2 k ξ + η 2 2ηηξ η = (1 + k +2ξ) η 2 η kξ = (1 + k +2x 2 ) x 2 1 x 1 kx 2 Since V (η) is a radially unbounded Lyapunov function and the results hold globally, we conclude that the system is globally asymptotically stable. Solution 3 (Exercise in Khalil) The system is given by ẋ 1 = x 2 + a + x 1 a 1/3 3 ẋ 2 = x 1 + u As in the previous exercise the system is in the form of (14.53)-(14.54) with Take f = a + x 1 a 1/3 3 g = 1 f a = x 1 g a = 1 φ (x 1 ) = a x 1 a 1/3 3 x1 V = 1 2 x2 1 and use (14.56). Solution 4 (Exercise in Khalil) The system is given by ẋ 1 = x 1 + x 2 ẋ 2 = x 1 x 2 x 1 x 3 + u ẋ 3 = x 1 + x 1 x 2 2x 3 10

11 1. Since we are evaluating the stability of the origin, let z 1 = x 1 and x 2 be considered as input in the first equation where ż 1 = x 1 + x 2 = z 1 + φ 1 + z 2 z 2 = x 2 φ 1 The system is analyzed according to V 1 = 1 2 z2 1 V 1 = z 1 ż 1 = z 1 ( z 1 + φ 1 + z 2 ) = z1 2 + z 1 φ 1 + z 1 z 2 by choosing φ 1 =0we have that the origin of z 1 is "stable" with V 1 = z1 2 + z 1 z 2 The next step consists of including the z 2 dynamics in the analysis and by choosing V 2 = V z2 2 V 2 = V 1 + z 2 ż 2 = z z 1 z 2 + z 2 ³ ẋ 2 φ 1 = z1 2 + z 1 z 2 + z 2 (x 1 x 2 x 1 x 3 + u) = z1 2 + z 1 z 2 + z 2 (z 1 z 2 z 1 x 3 + u) = z1 2 z2 2 + z 2 (2z 1 z 1 x 3 + u) u = 2z 1 + z 1 x 3 = 2x 1 + x 1 x 3 it can be recognized that the origin is asymptotically stable. By further investigation we see that the dynamics is globally exponential stable by our choice of u. This is seen from the Lyapunov function for the system V = 1 x x2 2 = 1 2 k(x 1,x 2 )k 2 2 V = k(x 1,x 2 )k

12 2. We have shown that the two firststatesinthemodelwillconverge exponentially to the origin (even when x 3 is unstable). To show that the overall system is asymptotically stable we must show that the last dynamic equation is asymptotically stale. To do this we apply the theory of cascade system and input-to-state stability, where ẋ 3 = x 1 + x 1 x 2 2x 3 = 2x 3 + z 3 where z 3 = x 1 + x 1 x 2. Due to the exponential stability of x 1 and x 2, we have that kz 3 k = kx 1 + x 1 x 2 k kx 1 k + kx 1 x 2 k = kx 1 k + kx 1 kkx 2 k k 1 kx 1 (0)k e γ 1 (t t0) + k 1 kx 1 (0)k e γ 1 (t t0) k 2 kx 2 (0)k e γ 2 (t t 0) which shows that kz 3 k is exponentially stable (globally asymptotically stable). Using the theory of cascade systems, we can conclude global asymptotically stability of the overall system if x 3 is input-to-state stable. We start by noticing that the origin of x 3 is globally asymptotically stable when z 3 =0, which implies that the dynamics may be ISS. The ISS property is analyzed with a standard quadratic function V 3 = 1 2 x2 3 V 3 = x 3 ẋ 3 = x 3 ( 2x 3 + z 3 ) = 2x x 3 z 3 = 2(1 θ) x 2 3 2θx x 3 z 3 2(1 θ) kx 3 k 2 2 2θ kx 3k kx 3z 3 k = 2(1 θ) kx 3 k 2 2 (2θ kx 3k 2 kz 3 k) kx 3 k 2 2(1 θ) kx 3 k 2 2, 2θ kx 3k 2 kz 3 k which shows that the system is input to state stable (z 3 regarded as input) with W 3 = 2(1 θ) kx 3 k 2 2 ρ = 1 2θ kz 3k 0 < θ < 1 12