Modeling and Simulation: Exploring Dynamic System Behaviour Chapter 7 Modelling of Continuous Time Dynamic Systems

Continuous Time Dynamic Models Synopsis Distinctive features Two examples The canonical (standard) form The decomposition problem An M&S project with a CTDS SUI Possible hazards on the road to successful solution of continuous M&S projects

CTDS Models Noteworthy Features CM is formulated as a set of differential equations (ode or pde) possibly augmented with a set of algebraic equations Frequently the SUI has its origins in the “physical world” hence behavior is governed by physical laws (physics, chemistry, heat transfer, etc); i.e., “deep knowledge” Optimization studies are straightforward because (a) random effects rarely present and (b) behavior is ”smooth” Time advance is not a separate issue because it is embedded in the equation solving mechanism

L q΄΄(t) + R q΄(t) + q(t)/C = E(t) Some Examples Electric Circuit (An example of a deductive model formulated from the application of physical laws) L q΄΄(t) + R q΄(t) + q(t)/C = E(t) Initial conditions: q(t0) and q΄(t0) are required

2. Population Dynamics (predator/prey). (An example of an inductive model that is formulated from arguments based on observation and intuition) without interaction: P1΄(t) = – α1 P1(t) P2΄(t) = α2 P2(t)

P1΄(t) = – α1 P1(t) + λ1 P1(t) P2(t) with interaction: P1΄(t) = – α1 P1(t) + λ1 P1(t) P2(t) P2΄(t) = α2 P2(t) – λ2 P1(t) P2(t) (these are the Lotka-Volterra equations )

typical predator/prey behavior

The Canonical Form CTDS Models have the convenient feature of having a standard (general) representation: x΄(t) = f(x(t), u(t), t)   with: x(t0) = x0 and y(t) = g(x(t)) Here x, u and y are vectors of dimension n, r and m respectively

State Variables are often not unique Consider: y΄΄(t) + a1 y΄(t) + a0 y(t) = b1 u΄(t) + b0 u(t) with y(0) = α ; y’(0) = β A possible “decomposition” (what does this mean?) is: x1΄(t) = x2(t) x2΄(t) = - a1 x2(t) - a0 x1(t) + b1 u΄(t) + b0 u(t) y(t) = x1(t) But this can have a problem ----- !!

Consider an alternate decomposition: x1΄(t) = x2(t) x2΄(t) = –a0 x1(t) – a1 x2(t) + u(t)  y(t) = b0 x1(t) + b1 x2(t) Still a possible problem; namely the “initial condition transformation” requires that: b02 –a1 b0 b1 + a0 b12 ≠ 0

Yet another possible decomposition: x1΄(t) = -a0 x2(t) + b0 u(t) x2΄(t) = x1(t) – a1 x2(t) + b1 u(t)   y(t) = x2(t) Here the “initial condition transformation” does not introduce any constraint.

The Safe Ejection Envelope

The Safe Ejection Relationship

Trajectory of Pilot/Seat

Trajectory of Pilot/Seat X(t) = Xp(t) – Xa(t)  X’(t) = Xp’(t) – Xa’(t) ; X(0) = 0   Xa’(t) = Va (constant) ; hence Xa(t) = Va t Xp’(t) = V(t) cos θ(t) Y(t) = Yp(t) – Ya(t) Y’(t) = Yp’(t) – Ya’(t) ; Y(0) = 0 Ya’(t) = 0 (level flight assumption) Yp’(t) = V(t) sin θ(t)

Configuration y vr Өr

Constrained Motion on Rails (Y ≤ Y1)  V(t) = A ; V’(t) = 0 A = [(Vr cos( θr ))2 + (Va – Vr sin( θr ) )2 ]½   θ(t) = B ; θ’(t) = 0 B = tan-1 [Vr cos( θr ) /(Va - Vr sin( θr ) )]

Free Fall Motion (ballistic trajectory) Xp’ (t) = Vx(t) Yp’ (t) = Vy(t) Vx’(t) = – (D/m) cos θ(t) Vy’ (t) = – (D/m) sin θ(t) – g  

But: D(t) = μ V2(t) and μ = ĈD ρ(H). Thus (with some algebraic manipulation): Vx’(t) = – Ψ(t) Vx  Vy’(t) = – Ψ(t) Vy – g   where:Ψ(t) = [ĈD ρ(h + Yp) (Vx2 + Vy2)0.5] / m.

Summary Xp’(t) = Vx(t) ; Xp(0) = 0 Yp’(t) = Vy(t) ; Yp(0) = 0 On Rails: Vx’(t) = 0; Vx(0) = Va – Vr sin θr Vy’(t) = 0; Vy(0) = Va – Vr sin θr Off Rails:  Vx’(t) = – Ψ(t) Vx Vy’(t) = – Ψ(t) Vy – g with Ψ(t) = [ĈD ρ(h + Yp) (Vx2 + Vy2)0.5] / m.

Generating the Envelope Data Va  Vstart h  hstart while (Va < Vlimit) repeat h  h + ∆1 solve ode’s up to t = tT where Xp(tT) = Va tT - BT until (Yp(tT) > HT + Sf) record [Va, h] Va  Va + ∆2 endwhile Plot the collected [Va, h] pairs

Bouncing Ball

Bouncing Ball Ball Dynamics y1(0) =1 y2(0) =V0 sinθ0 x1(0) =0 x2(0) =V0 cosθ0 y1(0) =1 y2(0) =V0 sinθ0

Bouncing Ball Collision Characterization x2(Tc+) = α x2(Tc) y1(Tc+) = 0 y2(Tc+) = - α y2(Tc)