Order Preserving Properties of Vehicle Dynamics with Respect to the Driver s Input

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1 Order Preserving Properties of Vehicle Dynamics with Respect to the Driver s Input Mojtaba Forghani and Domitilla Del Vecchio Massachusetts Institute of Technology September 19, Introduction In this report, we consider longitudinal dynamics of a vehicle and prove that its longitudinal position and velocity are order preserving with respect to the driver s (or automatic controller s) input. We also prove that the position is strictly order preserving with respect to the constant term of the acceleration replaced with a linear function of the position and the velocity. This estimation of the acceleration is used for the cases in which no direct measurement of the acceleration is available. 2 System Model Before introducing the model that we are considering, we define the strict and non-strict order preserving properties. Definition 1. (Order preserving) For all w, z R n we have that w z (w < z) if and only if w i z i (w i < z i ) for all i {1, 2,..., n}, in which w i denotes the ith component of w. We denote the piecewise continuous signal on U by S(U) : R + U. For U R m we define the partial order (strict partial order) by component-wise ordering for all times, that is, for all w, z S(U) we have that w z (w < z) provided w(t) z(t) (w(t) < z(t)) for all t R +. The map f : P Q is order preserving (strict order preserving) provided if for x, y P we have x y (x < y), then f(x) f(y) (f(x) < f(y)). Definition 2. (Continuous system) A continuous system is a tuple Σ = (X, U,, O, f, h), with state x X R n, control input u U R m, disturbance input d R q, output y O X, vector field in the form of f : X U X, and output map h : X O. We denote the flow of a system Σ at time t R + by φ(t, x, u, d), with initial condition x X, control input signal u S(U) and disturbance input signal d S( ). We also denote the ith component of the flow by φ i (t, x, u, d). Definition 3. (Control input/output order preserving) A continuous system Σ = (X, U,, O, f, h) is called input/output order preserving (strict input/output order preserving) with respect to the control input, if the map h(φ(t, x,, d)) : U O, for any fixed t, x and d, is order preserving (strict order preserving). 1

2 Definition 4. (Disturbance input/output order preserving) A continuous system Σ = (X, U,, O, f, h) is called input/output order preserving (strict input/output order preserving) with respect to the disturbance input, if the map h(φ(t, x, u, )) : O, for any fixed t, x and u, is order preserving (strict order preserving). We denote the position and the velocity of vehicle by p and v, respectively. We consider two systems Σ 1 and Σ 2. For Σ 1 we assume that the input is the control input u and there is no disturbance. Therefore it is reasonable to represent the flow of Σ 1 by φ 1 (t, x 1, u), where x 1 is the initial state of the system. The deceleration due to the road load (rolling resistance) and the slope of the road are represented by a r and a s, respectively, and the drag coefficient is denoted by D. We also impose a condition that the speed of the vehicle must be non-negative. We use the superscript T to denote the transpose of a vector or matrix, e.g., A T represents the transpose of matrix A. The dynamics of Σ 1 is as the following: x 1 X R 2, where x 1 = (p, v) T, (1) u U R, with U = {u u [u m, u M ]}, (2) where u m is the minimal control input, and u M is the maximal control input, where f 1 : X U X, with ẋ 1 = f 1 (x 1, u), (3) f 1 (x 1, u) = The function f 1 (x 1, u) is also in the following form: [ f 1 (x 1 v, u) = u Dv 2 a r a s { f 1 (x 1, u) if v > if v. (4) ]. (5) Since we cannot measure the acceleration for system Σ 2, we assume the total acceleration of the vehicle to be in the form of a linear function of the position and the velocity. Moreover we assume the input term (the constant term of the linear function), be a disturbance input. Therefore it is reasonable to represent the flow of Σ 2 by φ 2 (t, x 2, d), where x 2 is the initial state of the system. The dynamics of system Σ 2 is in the following form: where x 2 X R 2, where x 2 = (p, v) T, (6) d R, (7) f 2 : X X, with ẋ 2 = f 2 (x 2, d), (8) f 2 (x 2, d) = The function f 2 (x 2, d) is also in the following form: [ f 2 (x 2 v, d) = ap + bv + d { f 2 (x 2, d) if v > if v. (9) ]. (1) The term ap + bv + d is the total acceleration of the vehicle in Σ 2, where a and b are known constants. Since in Σ 2, unlike Σ 1, we cannot measure the acceleration directly, this model can be a good approximation to the acceleration of the vehicle. 2

3 3 Order Preserving Properties We prove that the flows of the position and the velocity in Σ 1, φ 1 1 (t, x1, u) and φ 1 2 (t, x1, u), respectively, have order preserving property with respect to the control input signal u. Proposition 1. The flows φ 1 1 (t, x1, u) and φ 1 2 (t, x1, u) are order preserving with respect to the control input signal u. Proof. If we consider two different input signals u 1 and u 2, such that u 1 > u 2, then for the velocity of Σ 1 at time t corresponding to these two input signals, with the same initial conditions p 1 () = p 2 () = p() and v 1 () = v 2 () = v(), we have v 1 (t) = u 1 (t) Dv 1 (t) 2 a r a s and v 2 (t) = u 2 (t) Dv2 2(t) a r a s, if both v 1 (t) > and v 2 (t) >. Let the function g(t) := v 1 (t) v 2 (t). At an arbitrary time t we have ġ(t) = v 1 (t) v 2 (t) = (u 1 (t) u 2 (t)) D ( v 2 1(t) v 2 2(t) ). (11) Note that since we have chosen the same initial conditions, we have g() = v 1 () v 2 () =. Because of the continuity of flow of the system with respect to time, if order in state v is not preserved, we must have a time t R + such that g(t ) =, since otherwise for all t R +, either g(t) < or g(t) >. Therefore we can define t := min{t R + g(t) = }. Since ġ( + ) = u 1 ( + ) u 2 ( + ) >, ġ(t ) = u 1 (t ) u 2 (t ) >, and g() = g(t ) =, for the interval t (, t ) we have and similarly ġ( + g(h) g() g(h) ) = lim = lim h + h h + h > since h > : h = h 1 (, t ) s.t. g(h 1 ) >, (12) ġ(t g(t ) g(t + h) g(t + h) ) = lim h t (t = lim > + h) h h since h < : h = h 2 (, t ) s.t. g(h 2 ) <, (13) and because of the continuity of the flow with respect to time, there is a t [h 1, h 2 ] such that g(t) =, which is in contradiction with the initial assumption that t := min{t R + g(t) = }. Therefore there is no such t, and for all t R + such that v 1 (t) > and v 2 (t) > we have either g(t) = v 1 (t) v 2 (t) > or g(t) = v 1 (t) v 2 (t) <. From (12) we conclude that the former is true. We had assumed initially that v 1 (t) > and v 2 (t) >. For a case that for some t R + we have v 1 (t ) = and v 2 (t ) =, we let t 1 := min{t R + v 1 (t) = } and t 2 = min{t R + v 2 (t) = }. Because of the non-negativity of v 1 (t), v 2 (t) and v 1 (t) v 2 (t), we must have t 2 t 1. If an arbitrary time t (, t 2 ), then g(t) = v 1 (t) v 2 (t) > ; if t [ t 2, t 1 ), then v 1 (t) v 2 (t) = v 1 (t) > ; and if t [ t 1, ), then g(t) = v 1 (t) v 2 (t) =. Therefore, in any case the order of the flow of the velocity is preserved with respect to the control input signal. Since p 1 () = p 2 () = p(), then based on equation (5), g(s)ds, which implies that the order preserving property of the flow of p is also satisfied with respect to the control input signal. In Proposition 2 we prove that the position in Σ 2, φ 2 1 (t, x2, d), is strictly order preserving with respect to d. 3

4 Proposition 2. input d. The flow φ 2 1 (t, x2, d) is strictly order preserving with respect to the disturbance Proof. Let x 2 := (p, v ) T be the initial condition, where v >. Since d is not a function of time, then by differentiating v from equations (8) and (1), we have that the velocity in Σ 2, for v(t) >, satisfies the following differential equation: v b v av = where v() = v and v() = ap + bv + d. (14) The above differential equation has the solution in the form v(t) = k 1 e λ 1t + k 2 e λ 2t where, λ 1 =.5(b + b 2 + 4a) and λ 2 =.5(b b 2 + 4a). (15) Since complex and real values of λ 1 and λ 2 reveal different behaviors for v(t), we consider different possible cases and analyze the behaviors of v(t) and p(t) with respect to d for each of them. We divide the problem into three different cases; (1): b 2 + 4a >, (2): b 2 + 4a < and (3): b 2 + 4a =. For each case we consider two input signals d 1 = d 1 and d 2 = d 2 such that d 1 > d 2 and determine the relationship between v 1 (t) and v 2 (t) and then between p 1 (t) and p 2 (t), the velocity and the position at time t corresponding to d 1 and d 2, respectively. Case (1): If b 2 + 4a >, then λ 1 and λ 2 in (15) are real numbers. The solution of (14) then takes the form 1 ( ) v(t) = (v (λ 2 b) ap d) e λ1t (v (λ 1 b) ap d) e λ 2t. (16) If we replace d in equation (16) with d 1 and d 2 in order to obtain their corresponding velocities at time t, represented by v 1 (t) and v 2 (t), respectively, we have v 1 (t) v 2 (t) = d1 d 2 ( e λ 2t e λ 1t ). (17) Equation (17) can become zero only when t =. Therefore because of the continuity of flow of the system with respect to time, for all t R +, either v 1 (t) v 2 (t) > or v 1 (t) v 2 (t) <. To determine which of these two cases holds, we note that in general for any x R {} we have that if x >, then e x 1 > and if x <, then e x 1 <. These two statements together imply that ex 1 x >. Since in Case (1) and we are considering t R +, then ( )t. Therefore we can replace x with ( )t in order to obtain e (λ 2 λ 1 )t 1 ( )t > te λ 1t [ e (λ 2 λ 1 )t 1 ( )t ] > eλ 2t e λ 1t > d 1 d 2 ( ) e λ2t e λ 1t > v 1 (t) > v 2 (t), (18) where we have used the facts that t R +, e λ1t > and d 1 d 2 >. By integrating both sides of (18) to determine the position at time t, we obtain v 1 (s)ds > 4 v 2 (s)ds

5 p + v 1 (s)ds > p + v 2 (s)ds p 1 (t) > p 2 (t). (19) Case (2): If b 2 + 4a <, then λ 1 and λ 2 in (15) are complex numbers. The solution of (14) then takes the form ( ) v(t) = e αt ap + (b α)v + d sin βt + v cos βt, β with α =.5b, and β =.5 (b 2 + 4a). (2) If we replace d in equation (2) with d 1 and d 2 in order to obtain their corresponding velocities at time t, represented by v 1 (t) and v 2 (t), respectively, we have v 1 (t) v 2 (t) = d1 d 2 e αt sin βt. (21) β We observe that in Case (2), unlike Case (1), we cannot guarantee that for all t R +, v 1 (t) v 2 (t). Note that v 1 (t) v 2 (t) = for all t such that sin βt = or alternatively, βt = kπ, for all k Z. The smallest t R + that satisfies sin βt = is t = π β. The velocity at time t corresponding to d 1 and d 2, based on equation (2), is given by v i (t ) = v e αt cos βπ β = v e αt < for i {1, 2}. (22) Since for all t R +, v(t), we must have v i (t ) =, or in other words, for all t [, t ] we have either v 1 (t) v 2 (t) or v 1 (t) v 2 (t). In order to determine which case holds, we note that for all t [, t ] we have β > and sin βt 1. Therefore in any case, for all t [, t ] we have. Also e αt (d 1 d 2 ) >. These two statements along with (21) imply sin βt β that v 1 (t) v 2 (t). Since in (2) we have v 2 () = v > and v 2 (t ) = v e αt <, then because of the continuity of flow of the system with respect to time, there is a t (, t ) such that t := min{t (, t ) v 2 (t) = }. Then we have for all t (, t), v 1 (t) v 2 (t) >. For a t (, t), we have for a t [ t, t ) we have + and for a t [t, ), we have t t (v 1 (s) v 2 (s))ds + (v 1 (s) v 2 (s))ds > ; (23) t (v 1 (s) v 2 (s))ds > (v 1 (s) v 2 (s))ds p 1 (t) p 2 (t) > ; (24) (v 1 (s) v 2 (s))ds + (v 1 (s) v 2 (s))ds = t (v 1 (s) v 2 (s))ds + > p 1 (t) p 2 (t) >. (25) 5

6 Case (3): If b 2 + 4a =, then λ 1 = λ 2 = λ, which is also a real number. The solution of (14) then takes the form v(t) = e λt [v + (ap + (b λ)v + d) t], (26) and for v 1 (t) v 2 (t) we have which implies v 1 (t) v 2 (t) = t(d 1 d 2 )e λt >, (27) (v 1 (s) v 2 (s))ds >. (28) We had assumed initially that v 1 (t) > and v 2 (t) >. In general, we may have a time t such that v 1 (t ) = and v 2 (t ) =. In this case, because of the continuity of flow of the system with respect to time, there are times t 1 and t 2 such that t 1 = sup{t (, t ) v 1 (t) > } and t 2 = sup{t (, t ) v 2 (t) > }. Since we have proved through Cases (1)-(3) that as long as v 1 (t) > and v 2 (t) > we have v 1 (t) v 2 (t) >, then t 2 t 1. For an arbitrary time τ (, t 2 ) Cases (1)-(3) imply that p 1 (τ) p 2 (τ) > ; if τ [ t 2, t 1 ), then and if τ [ t 1, ), then p 1 (τ) p 2 (τ) = and the proof is complete. p 1 (τ) p 2 (τ) = t 2 t 2 τ (v 1 (s) v 2 (s))ds + (v 1 (s) )ds > ; t 2 (29) t 1 τ (v 1 (s) v 2 (s))ds + t 2 (v 1 (s) )ds + ( )ds >, t 1 (3) 6