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Passivity and Feedback Stabilization - Passivity of Hamiltonian Control Systems | AOE 5344, Study notes of Aerospace Engineering

Material Type: Notes; Professor: Woolsey; Class: Nonlinear Control Mech Sys; Subject: Aerospace and Ocean Engineerin; University: Virginia Polytechnic Institute And State University; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 02/13/2009

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Download Passivity and Feedback Stabilization - Passivity of Hamiltonian Control Systems | AOE 5344 and more Study notes Aerospace Engineering in PDF only on Docsity! Passivity and Feedback Stabilization Passivity of Hamiltonian Control Systems. The terms “supply rate,” “energy storage,” and “dis- sipation” have physical significance for canonical Hamiltonian control systems. Recall that a canonical Hamiltonian control system is one which derives from a scalar function H(q,p) of the n “generalized coordinates” qi and their n “conjugate momenta” pi as follows: q̇ = ∂H ∂p ṗ = − ∂H ∂q + u. Often, the Hamiltonian H takes the form H(q,p) = 1 2 pT M(q)−1p + V (q) where M(q) = MT (q) > 0 is the n × n “inertia tensor” or “kinetic metric” and V represents “potential energy.” Taking the output of this system to be y = ∂H ∂p = q̇, one finds that Ḣ = ∂H ∂q · q̇ + ∂H ∂p · ṗ = ∂H ∂q · ∂H ∂p + ∂H ∂p · ( − ∂H ∂q + u ) = u · y. Thus, a Hamiltonian control system is passive (in fact, lossless) with storage function H. As a simple example, consider a point mass sliding without friction which is connected to a wall through a linear spring. Define q = x, p = mẋ, and H(q, p) = 1 2m p2 + 1 2 kq2. It is easy to verify that q̇ = ∂H ∂p = p m (1) ṗ = − ∂H ∂q + u = −kq + u, (2) where u is the input force. “Generalized” or “non-canonical” Hamiltonian systems of the form ẋ = J(x) ∂H ∂x + u, with J(x) = −J(x)T , are also (lossless) passive systems. Defining y = ∂H ∂x , one sees that Ḣ = ∂H ∂x · ẋ = ∂H ∂x · ( J(x) ∂H ∂x + u ) = u · y, by skew-symmetry of J(x). Passivity and Rigid Body Attitude Stabilization. Consider the rigid body shown in Figure 1. Sup- pose the rigid body is rotating with angular velocity ω, expressed with respect to a body-fixed coordinate 1 x b y b z b Ix y z ! I I Figure 1: A rigid body with torque control. frame. The orientation of the rigid body is given, with respect to inertial reference frame, by the proper rotation matrix R: Ix = Rbx, Iy = Rby, and Iz = Rbz, where Ix, Iy, and Iz are unit vectors defining the inertial coordinate frame and bx, by, and bz are unit vectors defining the body-fixed coordinate frame. Suppose that we wish to control the rigid body to some constant desired attitude R = Rd. To do this, we need some measure of the error between the current attitude and the desired attitude. One such measure is R−Rd. In this case, the system acquires the desired orientation when the “error matrix” goes to zero. However, this choice of error does not exploit the fact that the attitude error is actually a rotation. More mathematically, the relevant “group operation” for the group of proper rotations is not matrix addition, but matrix multiplication. It therefore makes more sense to choose the following measure of attitude error: Re = R T d R. For this measure of error, the system acquires the desired orientation when Re = I. Euler’s Theorem on rotations says that any proper three-dimensional rotation may be represented as a rotation, through some angle θ ∈ [0, 2π), about an axis described by a unit vector v = [v1, v2, v3] T . In our case, one may parameterize Re using θ and v (as opposed, say, to Euler angles), Re(θ,v) = cos θI + (1 − cos θ)vv T − sin θv̂ where v̂ :=   0 −v3 v2 v3 0 −v1 −v2 v1 0   . Thus, the 3 × 3 rotation matrix Re can be parameterized by four numbers (θ and the components of v) and one constraint (‖v‖ = 1). This observation leads to a common representation for proper rotations known as “unit quaternions,” or “Euler parameters.” Define the 4 × 1 unit quaternion vector e = ( e0 ev ) := ( cos ( θ 2 ) v sin ( θ 2 ) ) It is easy to verify that ‖e‖ = 1, hence the name “unit” quaternion. Note that the unit quaternion takes the value e = ed = [1, 0, 0, 0] T when Re = R T d R = I. One may easily verify that Ṙ = Rω̂. (Check, for example, the case where R is parameterized by the Euler angles [φ, θ, ψ]T . You will obtain a familiar expression relating the Euler angle rates to body angular rate 2
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