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 / 21

2 Outline 1 Lecture 4: Modeling Tools for Mechanical Systems 2 G. Ducard c 2 / 21

3 Outline 1 Lecture 4: Modeling Tools for Mechanical Systems 2 G. Ducard c 3 / 21

4 Lagrange: G. Ducard c 4 / 21

5 : Recipe 1 Define inputs and outputs 2 Define the generalized coordinates: q(t) = [q 1 (t),q 2 (t),...,q n (t)] and q(t) = [ q 1 (t), q 2 (t),..., q n (t)] 3 Build the Lagrange function: L(q, q) = T(q, q) U(q) 4 System dynamics equations: { } d L L = Q k, k = 1,...,n dt q k q k Notes: Q k represents the k-th generalized force or torque acting on the k th generalized coordinate variable q k n: number of degrees of freedom in the system always n generalized variables G. Ducard c 5 / 21

6 Outline 1 Lecture 4: Modeling Tools for Mechanical Systems 2 G. Ducard c 6 / 21

7 Lagrange Equations for Constrained Systems Remarks: { } d L L dt q k q k ν µ j α j,k = Q k, k = 1,...,n, (1) j=1 the constraints are included using Lagrange multipliers : µ j j = 1...ν Number of constraints: ν with (ν < n) n may be seen as the number of DOF In the end, we obtain: n+ν coupled equations to be solved for q k and µ j (usually requires computing the time derivative of the constraints, i.e., µ). G. Ducard c 7 / 21

8 Outline 1 Lecture 4: Modeling Tools for Mechanical Systems 2 G. Ducard c 8 / 21

9 Nonlinear m = 1kg y(t) ϕ(t) mg l = 1m u(t) M = 1kg Figure: Pendulum on a cart, u(t) is the force acting on the cart ( input ), y(t) the distance of the cart to an arbitrary but constant origin, and ϕ(t) the angle of the pendulum. G. Ducard c 9 / 21

10 Step 1: Inputs & Outputs Input: force acting on the cart: u(t) Output: angle of the pendulum: ϕ(t) Step 2: System s coordinate variables q 1 = y, q 1 = ẏ q 2 = ϕ, q 2 = ϕ Step 3: Lagrange functions L 1 (t) = T 1 (t) U 1 (t) L 2 (t) = T 2 (t) U 2 (t) L(t) = L 1 (t)+l 2 (t) G. Ducard c 10 / 21

11 Step 4: System s dynamics equations { } d L dt q 1 { d L dt q 2 L q 1 = Q 1 } L q 2 = Q 2 We are looking for dynamic equations of the form: ÿ(t) = f(ϕ(t), ϕ(t),u(t)) ϕ(t) = g(ϕ(t), ϕ(t),u(t)) G. Ducard c 11 / 21

12 Outline 1 Lecture 4: Modeling Tools for Mechanical Systems 2 G. Ducard c 12 / 21

13 Introduction In general they are described by Navier-Stokes equations. For control purposes simpler formulations are necessary, to build networks with building blocks: ducts, compressible nodes, valves, etc. G. Ducard c 13 / 21

14 Water duct in gravitational field v(t) h p 1 A l p 2 v(t) Objective d dt v(t) = f (p 1(t),p 2 (t),v(t),h,ρ,a,l) G. Ducard c 14 / 21

15 Change in momentum: Newton s law d p dt = md v dt = Fpressure + F gravity + F friction = [P 1 A P 2 A] x+ g dm+ F friction The mass m of the fluid in the element of tube of length l is given by m = ρ A l dm = ρ A dl tube G. Ducard c 15 / 21

16 Angle of the duct sinα = dh dl Gravity force g dm = g ( cosα y +sinα x)ρ A dl tube tube [ h = ρ g A 0 cosα sinα dh y + h 0 ] sinα sinα dh x = ρ g A(tanα) 1 h y +ρgah x G. Ducard c 16 / 21

17 Water duct in gravitational field Dynamics along the x axis (because v = v x) with ρ A l dv(t) dt = A (P 1 P 2 )+ρ g A h F friction,x (t) F friction,x (t) = 1 2 ρv2 (t)sign[v(t)] λ(v(t)) A l d Remark: shape factor: l d G. Ducard c 17 / 21

18 Outline 1 Lecture 4: Modeling Tools for Mechanical Systems 2 G. Ducard c 18 / 21

19 V(t) V = 0 V in (t) p(t) k = 1/(σ 0 V 0 ) V out (t) Definitions of compressability Property of a body (solid, liquid, gas, etc.) to deform (to change its volume) under the effect of applied pressure. Defined as: σ 0 = 1 V 0 dv dp V 0 : nominal volume [m 3 ], P : pressure [Pa] σ 0 : compressibility [Pa 1 ] k 0 = 1 σ 0 is called elasticity constant [ Pa m 3]. G. Ducard c 19 / 21

20 V(t) = V(t) V 0 G. Ducard c 20 / 21 Lecture 4: Modeling Tools for Mechanical Systems Compressibility effects V(t) V = 0 V in (t) p(t) k = 1/(σ 0 V 0 ) V out (t) Modeling d dt V(t) = V in(t) V out(t) = A in v in (t) A out v out (t) P(t) = k V(t) = 1 σ 0 V 0 V(t)

21 Next lecture + Upcoming Exercise Next lecture Pelton Turbine Electromagnetic systems Next exercise: Online next Friday Hydro-electric Power plant, Part I G. Ducard c 21 / 21