Modeling and Analysis of Dynamic Systems

Transcription

1 Modeling and Analysis of Dynamic Systems Dr. Guillaume Ducard Fall 2017 Institute for Dynamic Systems and Control ETH Zurich, Switzerland G. Ducard c 1 / 54

2 Outline 1 G. Ducard c 2 / 54

3 Outline 1 G. Ducard c 3 / 54

4 Figure: G. Ducard c 4 / 54

5 Geostationary orbit around the Earth in the Equatorial plane, exactly the sideral Earth rotational speed: 23h 56min 4.1s, from Earth, satellite is seen at exact same location in sky. Very useful for communication purposes. G. Ducard c 5 / 54

6 Importance of modeling for: understanding the basic properties and limitations of satellite dynamics, the design of an orbit-stabilizing controller. Main steps followed: 1 derivation of the satellite motion, 2 linearization around the orbit trajectory, 3 study stability, observability, controllability. G. Ducard c 6 / 54

7 Problem Definition R M r(t) m F r (t) ϕ(t) F ϕ(t) Notations: R: radius of the Earth [m] M: mass of the Earth [kg] m: mass of the satellite [kg] r(t): dist. Earth center to satellite [m] ϕ(t): orbit angle of the satellite [rad] F r (t) : radial force [N] F ϕ (t): tangential force [N] G. Ducard c 7 / 54

8 Problem Definition m F r (t) R M r(t) ϕ(t) F ϕ(t) Assumptions: No other celestial bodies considered (only Earth gravitational force considered). M m C.O.G located at center of Earth. satellite always remains in the equatorial plane 2 variables are sufficient to describe its position: r(t) and ϕ(t). the attitude (orientation) of the satellite is kept constant (by an inner control system) F r (t) and F ϕ (t) independant. G. Ducard c 8 / 54

9 Outline 1 G. Ducard c 9 / 54

10 Building Up the Step 1: Inputs & Outputs Inputs: radial force F r (t) tangential force F ϕ (t) Outputs: Earth to satellite distance r(t) orbit angle: ϕ(t) G. Ducard c 10 / 54

11 Building Up the Step 2: Energies involved in the satellite motion kinetic energy T T(r,ṙ, ϕ) = 1 2 m ṙ m (r ϕ)2 potential energy V: is due to the Earth Gravitational field G Earth G Earth(r) = GM r 2 u F grav (r) = mg Earth Universal gravitation constant: G = m 3 /(s 2 kg), Earth mass: M = kg, Earth radius: R = m. Satellite mass: m M G. Ducard c 11 / 54

12 Building Up the Step 2: Energy needed to bring the satellite from altitude R to r V(r) = r R r F(ρ)dρ = G M m R ρ 2 dρ r 1 = GM m R ρ 2 dρ [ = GM m 1 ] r ρ R ( 1 = GM m R 1 ), r > R r Remark: Potential energy of satellite: - only a function of the distance r, V(r). - is zero at Earth surface (arbitrary but convenient choice) G. Ducard c 12 / 54

13 V(r) = GM m ( 1 R 1 ), r > R r V/V r/r The potential energy: is 0 on the surface of the earth (arbitrary but convenient choice) and reaches 90% of its maximum value: V = GMm R, at a distance of only 10 times the radius of the Earth. G. Ducard c 13 / 54

14 Minimum Speed to Reach Orbit Question: Which minimum speed should the rocket reach to overcome the Earth gravitation force to place the satellite in its orbit? Assume a rapid acceleration at the start from standstill (take off) to a speed v 0, the energy balance is ( mv2 0 = GM m R 1 ) = ( V grav ) r R r kinetic energy that a satellite must have to reach a certain orbit height r R. ( V grav ) R r is the change in potential energy from dist. R to r. The minimum speed needed to reach altitude r is thus: ( 1 v 0 (r) = 2GM R 1 ) r G. Ducard c 14 / 54

15 Escape Velocity : v esc or v Definition 1 It is the minimum velocity that an object should have, such that without additional acceleration (no propulsion anymore), this object will move away from massive body, without being pulled back. Definition 2 It is the velocity of an object, such that its kinetic energy at a certain point in space A is equal its gravitational energy at that point A. Definition 3 The velocity that is required to completely leave the influence of the gravitation field of the Earth. G. Ducard c 15 / 54

16 Escape Velocity How to compute the escape velocity? v = lim r v 0 (r) = 2GM ( 1 R 1 ) r v = 2G M R m/s 11.2km/s 40300km/h G. Ducard c 16 / 54

17 Building Up the Step 3: Lagrange formalism The dynamics of the satellite are formulated using Lagrange s method: [ ] d L L dt ṙ r = F r and d dt [ ] L L ϕ ϕ = F ϕ r with the Lagrange function L = T V (difference of the kinetic and potential energies): L = 1 2 m ṙ2 + 1 ( 1 2 m (r ϕ)2 GM m R 1 ) r G. Ducard c 17 / 54

18 Building Up the L = 1 2 m ṙ m (r ϕ)2 GM m ( 1 R 1 ) r d dt ( L ṙ L ṙ = mṙ ) = m r L r = mr ϕ2 GM m 1 r 2 L ϕ = mr2 ϕ d dt ( ) L = mr 2 ϕ+2mr ϕṙ ϕ L ϕ = 0 m r = m r ϕ 2 G M m 1 r 2 +F r m r 2 ϕ = 2m r ϕ ṙ +F ϕ r G. Ducard c 18 / 54

19 Building Up the In order to simplify the notations, let s introduce the control accelerations yields u r = F r m, respectively u ϕ = F ϕ m r = r ϕ 2 G M 1 r 2 +u r ϕ = 2 ϕṙ 1 r + 1 r u ϕ G. Ducard c 19 / 54

20 Building Up the Circular Orbit For geostationary conditions, the periodic reference orbit must be chosen to be: circular with constant angular velocity, i.e., u r = 0, r = 0, ṙ = 0, r = r 0 u ϕ = 0, ϕ = 0, ϕ = ω 0, ϕ = ω 0 t What are the satellite turn rate ω 0, and the turn radius r 0? G. Ducard c 20 / 54

21 Building Up the : Circular Orbit Orbit turn rate: ω 0 The satellite turn rate is ω 0 = 2π/day taking 1 sidereal day = 23h 56min 4.1 s, you find: ω 0 = rad/s G. Ducard c 21 / 54

22 Building Up the : Circular Orbit Geostationary orbit radius: r 0, (F centripetal = F gravitational ( ) G M 1/3 r 0 = m ω 2 0 Remarks: r 0 is approximately 6.2 times the radius of the earth. The energy required is more than 80% of the escape energy. The resulting tangential speed v ϕ = r 0 ω 0 = 3.06m/s 10800km/h G. Ducard c 22 / 54

23 How to Launch a Satellite? What is an energy efficient trajectory to place a geostationary satellite to orbit? G. Ducard c 23 / 54

24 Outline 1 G. Ducard c 24 / 54

25 Building Up the State-Space Formulation In order not to have second-order ODE to work with, the system is put into a state space form, such that only 1st order time-derivatives appear. x 1 (t) = r, x 2 (t) = ṙ, u 1 (t) = u r x 3 (t) = ϕ, x 4 (t) = ϕ, u 2 (t) = u ϕ x 1 (t) [ ] x 2 (t) u1 (t) x(t) =, u(t) = x 3 (t) u 2 (t) x 4 (t) G. Ducard c 25 / 54

26 Nonlinear ODE where f(x(t)) = d x(t) = f(x(t),u(t)) dt x 2 (t) x 1 x 2 4 (t) G M/x2 1 (t)+u 1(t) x 4 (t) 2x 2 (t) x 4 (t)/x 1 (t)+u 2 (t)/x 1 (t) G. Ducard c 26 / 54

27 Measurements Equations y(t) = h(x(t)) where [ x1 (t)/r 0 h(x(t)) = x 3 (t) Notice: y 1 (t) is the normalized radius. ] G. Ducard c 27 / 54

28 Model Linearization The model is now linearized around the nominal orbit r 0 [ ] 0 0 x 0 (t) =, u ω 0 t 0 (t) = 0 Notice that in this case the nominal trajectory is not an equilibrium point but a periodic solution of the system equations. ω 0 G. Ducard c 28 / 54

29 However, introducing the notation x(t) = x 0 (t)+δx(t), u(t) = u 0 (t)+δu(t), y(t) = y 0 (t)+δy(t) yields d dt x 0(t)+ d dt δx(t) = f(x 0(t)+δx(t),u 0 (t)+δu(t)) f(x 0 (t),u 0 (t))+ f δx(t) x + f u x0 (t),u 0 (t) δu(t) x0 (t),u 0 (t) G. Ducard c 29 / 54

30 Since, by construction d dt x 0(t) = f(x 0 (t),u 0 (t)) the linearized system is described by the equation d f δx(t) = δx(t)+ f dt x u x0 (t),u 0 (t) x0 (t),u 0 (t) δu(t) The computation of the Jacobian matrices yields the following results f x = x G M/x x 1 x x 2 x 4 /x 2 1 u 2/x 2 1 2x 4 /x 1 0 2x 2 /x 1 G. Ducard c 30 / 54

31 f u = /x 1 G. Ducard c 31 / 54

32 Along the periodic nominal solution, we get f 3ω r 0 ω 0 x = x0 (t),u 0 (t) ω 0 /r 0 0 0, G. Ducard c 32 / 54

33 f u = x0 (t),u 0 (t) /r 0 Remark: In this special case the resulting system matrices are time invariant. This is not true in general. When linearizing a nonlinear system around a non-constant equilibrium orbit x e (t),u e (t) the resulting system matrices will be known functions of time {A(t),B(t),C(t),D(t)}. G. Ducard c 33 / 54

34 System in standard notation ẋ(t) = A x(t)+b u(t), y(t) = C x(t) with and A = f x x0 (t),u 0 (t), B = f u x0 (t),u 0 (t) [ ] 1/r C = G. Ducard c 34 / 54

35 For the system {A,B,C,0} defined above the following questions will be discussed below: 1 What are the stability properties of the system? 2 Is the system controllable and observable? 3 Which actuators and sensors are most important? 4 What are the main dynamic properties of the system? G. Ducard c 35 / 54

36 System Stability Eigenvalues of the matrix A are the roots of s ω0 2 s 0 2r 0 ω 0 det(s I A) = det 0 0 s 1 0 2ω 0 /r 0 0 s G. Ducard c 36 / 54

37 System Stability det(s I A) = s det s 1 0 3ω 2 0 s 2r 0 ω 0 0 2ω 0 /r 0 s = s [s (s 2 +4ω 2 0) ( 1) ( 3ω 2 0 s) ] = s 2 (s 2 +ω 2 0) G. Ducard c 37 / 54

38 System Stability det(s I A) = s 2 (s 2 +ω 2 0) The roots are {0,0,+jω 0, jω 0 }. The eigenvalue pair ±jω 0 shows that the system includes oscillatory modes with eigenfrequency equal to the angular speed of the satellite: ω 0. The double root in the origin indicates that the system might be unstable. G. Ducard c 38 / 54

39 One way to see that is to analyze the eigenstructure of the system matrix A. The rank of the matrix M = (s I A) s=0 is rank(m)=3. there is only one eigenvector associated with the double eigenvalue s = 0. Conclusion Therefore, the matrix A is cyclic and, as it was shown in the main text, this means that the linearized system is unstable (behavior of two integrators in series). Remark : The same result is obtained when the system s transfer functions will be computed. G. Ducard c 39 / 54

40 Controllability and Observability Recall of the linear system ẋ = ω r 0 ω ω 0 /r x /r 0 u and y = [ 1/r ] x G. Ducard c 40 / 54

41 Controllability matrix R R = [B,A B,A 2 B,A 3 B] = ω /r /r 0 2ω 0 /r This system is completely controllable. In fact the four first columns already are linearly independent (the determinant of that submatrix is 1/r 2 0 ). G. Ducard c 41 / 54

42 Failure Scenario 1: Radial Thruster Failure This yields input matrix b 2 = [0, 0, 0, 1/r 0 ] T. The corresponding controllability matrix has the form 0 0 2ω ω 0 0 2ω 3 0 R 2 = 0 1/r 0 0 4ω 0 /r 0 1/r 0 0 4ω0 2/r 0 0 The determinant of this matrix is 12ω0 4/r2 0, i.e., not zero on the reference orbit. Accordingly, the satellite remains completely controllable even if the radial thruster fails. G. Ducard c 42 / 54

43 Failure Scenario 2: Tangential Thruster Failure This yields b 1 = [0, 1, 0, 0] T and the controllability matrix R 1 = ω ω ω 0 /r ω 0 /r 0 0 2ω 3 0 /r 0 in this case has only three independent columns (the fourth column is equal to ω0 2 times the second column), i.e., In case of tangential-thruster failure the system is no longer completely controllable. G. Ducard c 43 / 54

44 Observability O = C C A C A 2 C A 3 = 1/r /r The system with no faults is completely observable. G. Ducard c 44 / 54

45 Fault Scenario 1: Radial Sensor Failure r meas If the radial sensor fails, the measurement matrix reduces to c 2 = [0, 0, 1, 0] and the resulting observability matrix O 2 has the form O 2 = 0 2ω 0 /r ω0 3/r ω0 2 This matrix is regular (its determinant is 12ω0 4/r2 0 ). Accordingly, the system remains completely observable even if the radial sensor fails. G. Ducard c 45 / 54

46 Fault Scenario 2: Tangential Sensor Failure ϕ meas If the tangential sensor fails, the measurement matrix reduces to c 1 = [1, 0, 0, 0] and the resulting observability matrix O 1 has the form 1/r /r O 1 = 3ω 0 2/r r 0 ω 0 0 ω0 2/r Obviously, this matrix is singular (the third row is zero). Accordingly, the system is not completely observable in this case. G. Ducard c 46 / 54

47 Conclusions on the satellite design This analysis is helpful for the design of the satellite. The main result is: tangential actuator and sensor are more important than radial ones. If radial actuator and sensor fail, the satellite stability is still guaranteed (control reconfiguration). G. Ducard c 47 / 54

48 Transfer Function P(s) = P(s) = [ P11 (s) P 12 (s) P 21 (s) P 22 (s) 1 s 2 +ω 2 0 ] = C(sI A) 1 B 2ω 0 s (s 2 +ω 2 0) 2ω 0 r 0 s (s 2 +ω 2 0 ) s 2 3ω 2 0 r 0 s 2 (s 2 +ω 2 0 ) G. Ducard c 48 / 54

49 Note that even if the system is stabilizable with only the tangential thruster and sensor working, the control problem is difficult, because the corresponding SISO system transfer function P 22 (s) has a non-minimumphase zero at 3 ω 0 (limits the attainable crossover frequency to about 0.85 ω 0. G. Ducard c 49 / 54

50 Remark: The MIMO system has no finite transmission zeros. This can be seen from [ ] (s I A) B Z(s) = C D which turns out to be detz(s) = 1 r 0 Accordingly, no finite s yields a determinant equal to zero and, therefore, no finite zero exists. In other words, the MIMO system P(s) is minimumphase. Compared to the SISO design using only P 22 (s), the design of a suitable controller using both inputs and outputs will yield a much better closed-loop system performance in terms of bandwidth, response times, etc. G. Ducard c 50 / 54

51 Two simulations The main objective is to convey a first impression of the differences between the linear system behavior, as analyzed above, and the nonlinear system behavior. The system is assumed to be in its reference state for all t < 0. At that moment a test signal { d0 0 t < 1 d(t) = 0 1 t is applied. In a first case d 0 is chosen rather small and in a second case very large, in order to clearly show the differences between the linear an the nonlinear system behavior. G. Ducard c 51 / 54

52 r(t) r0 ϕ(t) ω0t Input: u r = d(t), u ϕ = time t/t Input: u r = d(t), u ϕ= time t/t 0 r(t) r0 ϕ(t) ω0t Input: u r = 0, u ϕ = d(t) time t/t 0 Input: u r = 0, u ϕ = d(t) time t/t 0 G. Ducard c 52 / 54

53 r(t) r0 ϕ(t) ω0t Input: u r = d(t), u ϕ = time t/t Input: u r = d(t), u ϕ= time t/t 0 r(t) r0 ϕ(t) ω0t Input: u r = 0, u ϕ = d(t) time t/t Input: u r = 0, u ϕ = d(t) time t/t 0 G. Ducard c 53 / 54

54 Next lecture + Upcoming Exercise Next lecture Zero dynamics Nonlinear systems Thank you for your attention. G. Ducard c 54 / 54