ACS-336 / 6336 Mini Assignment 1 P.Soulatiantork Issued 12:00, 22th Feb 2021. 1 Introduction The aims of this assignment are twofold. Firstly, to derive the ordinary-differential equations that describe the rigid-body dynamics of the helicopter, and secondly, to derive a linearised model of the system. This assignment is worth 10% of the overall module mark. 2 Submission Details Students are requested to submit the completed answer sheet at the back of this handout to ACSE reception by 11:00 on Monday March 1st. Please submit through the BB assessment portal. Late submissions will not be graded and there will be no exceptions other than those that may arise owing to extenuating cir- cumstances. In the event of extenuating circumstances, students must submit an extenuating circumstances form if they have any medical or special circumstances that may have affected their performance on the assignment. You may not collaborate with other students, past or present, and instances of unfair means will be dealt with severely. 3 Feedback Generic feedback on class performance will be presented in lectures. The graded assignments will also be made available to students via the TAs in lab sessions, shortly afterwards. A Derivation of the rigid body dynamics The differential equations that describe the helicopter’s motion about its coordinate axes will be derived via the Euler-Lagrange equation: ∂2L(t) ∂t∂q˙i(t) − ∂L(t) ∂qi(t) = Qi(t), (A.1a) where L(t) ∈ R is the Lagrangian of the system at time t ∈ R. The Lagrangian is defined as the difference between the system’s kinetic energy T (t) ∈ R and potential energy V (t) ∈ R, according to the expression: L(t) := T (t)− V (t). (A.1b) With respect to (A.1a), qi(t) ∈ R are the generalised coordinates of the system, where {i ∈ N | i ≥ 1}, and Qi(t) ∈ R are the generalised forces used to describe non-conservative forces applied to the system. For the helicopter system, the generalised coordinates are defined as follows: q(t)T := [ E(t) Ψ(t) Θ(t) ] , (A.2a) where E, Ψ, and Θ are the elevation, pitch and travel angles. The corresponding velocities are: q˙(t)T := [ E˙(t) Ψ˙(t) Θ˙(t) ] . (A.2b) 1 m2 m1 m1 l3 l1 l2 l2 Figure 1: Schematic diagram of the helicopter, showing the coordinate system employed. Arrows indicate angular rotation in the positive sense. The generalised forces are defined as: Q(t) := l1(Fa(t) + Fb(t)) cos Ψ(t)−kdΨ˙(t) + l2(Fa(t)− Fb(t)) −l1(Fa(t) + Fb(t)) cosE(t) sin Ψ(t) , (A.2c) where Fa,b(t) ∈ R are the fan thrusts, and ks ∈ R, kd ∈ R are the rotational spring and damping coefficients pertaining to rotation about the pitch axis. In the interests of notational convenience, the time dependence of all variables will be dropped henceforth. The derivation of the governing equations proceeds by defining the kinetic energy of the helicopter, shown in Figure 1, about its separate axes. With some minor approximations (that will not be discussed here), the kinetic energy about the elevation axis TE , pitch axis TΨ and travel axis TΘ, is given by: TE = 1 2 JEE˙ 2, JE = 2m1l 2 1 +m2l 2 3, (A.3a) TΨ = 1 2 JΨΨ˙ 2, JΨ = 2m1l 2 2, (A.3b) TΘ = 1 2 JΘΘ˙ 2 cos2E, JΘ = JE , (A.3c) where JE , JΨ and JΘ are the moments of inertia about the elevation, pitch and travel axes, respectively. In turn, these are functions of the lumped fan masses m1 and counterweight mass m2, and the lengths from these masses to the various axes of rotation, l1, l2 and l3. The total kinetic energy is thus given by: T = TE + TΨ + TΘ. (A.3d) The potential energy of the helicopter, with respect to the datum defined at the elevation axis, is the sum of the gravitational potential energy and the potential energy stored in the spring about the pitch axis: V = c sinE + 1 2 ksΨ 2, c = (2m1l1 −m2l3)g, (A.3e) where g is the gravitational acceleration. Inserting (A.3d) and (A.3e) into (A.1b) yields the Lagrangian of our system: L = 1 2 JE ( E˙2 + Θ˙2 cos2E ) + 1 2 JΨΨ˙ 2 − c sinE − 1 2 ksΨ 2. (A.3f) Question (1a) [3 marks]. Having defined the Lagrangian in (A.3f) and the generalised coordinates, velocities and forces in (A.2), solve the Euler-Lagrange equation (A.1a) for 1 ≤ i ≤ 3 and hence obtain expressions for E¨, Ψ¨ and Θ¨. Write your answers in the answer sheet at the back of this handout. 2 Figure 2: Side-on view of the helicopter at its operating point. Both the pitch axis and elevation axis are perpendicular with respect to the gravitational vector, and the helicopter is motionless. Question (1b) [1 mark]. Your answers to the previous question will be functions of the fan forces Fa,b. However, the actual inputs to the system are the input voltages Ua,b and so, from a control systems per- spective, it is useful to work in terms of the latter. The relationship between the fan forces and the input voltages can be approximated by affine functions of the form: Fa ≈ αkaUa + β, Fb ≈ αkaUb + β, (A.4) where ka ∈ R is the power amplifier gain, and α, β ∈ R are constants that approximate the voltage-thrust characteristic of the fans. By substituting (A.4) into your answers to Question (1a), obtain expressions for E¨, Ψ¨ and Θ¨ as functions of the input voltages. Write your answers in the answer sheet at the back of this handout. B Derivation of a linear, time-invariant, state-space model Having obtained the governing, nonlinear ordinary differential equations that describe the rigid body me- chanics of the helicopter in the previous section, we next seek to derive a linear, time-invariant, state-space model of the usual form: x˙ = Ax+Bu, y = Cx+Du, (B.1) where x ∈ R6, u ∈ R2 and y ∈ R3 are the state, input and measurement vectors of this linear model. The derivation begins by defining an operating point from which a linearised model will be derived. The operating point in question corresponds to the helicopter hovering in a steady fashion about its horizontal position (depicted in Figure 2), in which case the states and controls assume the following equilibrium values: Ee = Ψe = Θe = E˙e = Ψ˙e = Θ˙e = 0, Ua,e = Ub,e = Ue, (B.2) where Ue is the equilibrium control signal required for steady hover. The states, controls and measurements of our linearised system (B.1) will correspond to perturbations away from this operating point, i.e. x = ψ θ ˙ ψ˙ θ˙ := E − Ee Ψ−Ψe Θ−Θe E˙ − E˙e Ψ˙− Ψ˙e Θ˙− Θ˙e , u = [ ua ub ] := [ Ua − Ua,e Ub − Ub,e ] , y := ψ θ . (B.3) Having defined the states x, controls u and measured outputs y of our linearised system, the elements of the state-space matrices in (B.1) can then be found via the Jacobian linearisation method, whereby: Ai,j = ∂fi(X,U) ∂Xj ∣∣∣∣ Xe,Ue , Bi,j = ∂fi(X,U) ∂Uj ∣∣∣∣ Xe,Ue , Ci,j = ∂Yi(X,U) ∂Xj ∣∣∣∣ Xe,Ue , Di,j = ∂Yi(X,U) ∂Uj ∣∣∣∣ Xe,Ue , (B.4) 3 where the states X, controls U and measured outputs Y of our nonlinear system are defined as: X := E Ψ Θ E˙ Ψ˙ Θ˙ , U := [ Ua Ub ] , Y := EΨ Θ , (B.5a) and the vector of ordinary differential equations that describes how the states of the nonlinear system X evolve as a function of the inputs U , can be defined as follows: f(X,U) = f1(X) f2(X) f3(X) f4(X,U) f5(X,U) f6(X,U) := E˙ Ψ˙ Θ˙ E¨ Ψ¨ Θ¨ (B.5b) Question (2a) [4 marks]. Use the Jacobian method (B.4) together with your answers to Question (1b), where appropriate, to derive the elements of the state-space matrices A,B,C and D for the helicopter linearised about the operating point defined in (B.2). Write your answers in the answer sheet at the back of this handout. As an example, the derivation of A1,1 is shown below: A1,1 = ∂f1(X) ∂X1 ∣∣∣∣ Xe,Ue = ∂E˙ ∂E ∣∣∣∣∣ Xe,Ue = 0|Xe,Ue = 0. Question (2b) [2 marks]. Use your answers to Question (2a) to comment on the local stability, reacha- bility and observability of the helicopter system released at the point of steady, level hover. It is emphasised this analysis is based purely upon the open-loop system (described in this case by your linear state-space model) and that there is no feedback controller present. You should assume the following parameter values in your state-space model: ks = 0.0015 kg m 2s−2, kd = 0.0005 kg m 2s−1, l1 = 0.110 m, l2 = 0.070 m, α = 0.0028 kg m s−2V −1, β = −0.0057 kg m s−2, ka = 1.2, Ue = 5.5 V, JE = 0.0024 kg m 2, JΨ = 4.95×10−4 kg m2, JΘ = 0.0024 kg m 2. Write your comments in the answer sheet at the back of this handout. 4 ACS-336/6336 Mini-Assignment: 1 Answer Sheet Student Registration Number: ................................................ Question (1a) [3 marks] E¨ = Ψ¨ = Θ¨ = Question (1b) [1 mark] E¨ = Ψ¨ = Θ¨ = Question (2a) [4 marks] A = , B = , C = , D = . Question (2b) [2 marks] Comment on stability (100 words maximum): Comment on reachability (50 words maximum): Comment on observability (50 words maximum):
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