MIT16_30F10_rec12

MIT16_30F10_rec12 - Recitation 12 16.30/31 Estimation and...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Recitation 12 16.30/31: Estimation and Control of Aerospace Systems November 29, 2010 Consider a space telescope, and let d be a unit vector aligned with the telescope's line of sight. It is desired to point the telescope towards a star, in direction d0 . The dynamics of the spacecraft are described by the equations of motion J + J = , where is the angular velocity of the spacecraft (in body axes), J is its inertia tensor, and is the control torque (in body axes). In body axes, the star's direction is not fixed, but rotates as d0= - d0 . Consider the control law = -k + d d0 . 1. Find the equilibrium points for the spacecraft, under the given control law. 2. Consider the following function: 1 1 V ( , d0 ) = J + |d - d0 |2 . 2 2 Is it a good candidate for a Lyapunov function? Study the stability of the equilibrium point(s). It may be interesting for you to simulate the system in Matlab, and look at the trajectories of the "star vector" d0 . Notice that the control law is the sum of a "derivative" term k and a d0 . How do trajectories change as you choose different values of k? "proportional" term d Solution: First of all, let us write the dynamics of the system, under the given control law: J = - J - k + d d0 , d0 = - d0 . From the second equation, we see that equilibrium points must be such that d0 = 0, i.e., must 0 , or = cd0 , for some scalar c. Rewriting the first equation using this condition, be parallel to d we find J = -c2 d0 Jd0 - ckd0 + d d0 . From the above equation, we see that since -c2 d0 Jd0 + d d0 is always orthogonal to d0 , can be zero only if c = 0, i.e., if = 0. Moreover, we require that d d0 = 0, i.e., d0 = d. Summarizing, we have two equilibria, one with the spacecraft pointing directly at the star, and one with the spacecraft pointing in the opposite direction. The given function V is indeed a good candidate to study the stability of the "good" equilibrium d0 = d: is it always non-negative, and is equal to zero only at the equilibrium point (d0 , ) = (d, 0). Let us study its time derivative. We get: V ( , d0 ) = J + (d - d0 ) d0 = (- J - k + d d0 ) + (d - d0 ) (- d0 ). Using the following property of the triple vector product u (v w) = w (u v) = v (w u), and remembering that the cross product of two parallel vector is zero, we an simplify the expression , of V ( d0 ) as V ( , d0 ) = -k| |2 , i.e., V is negative semi-definite. As a consequence, Lyapunov's theorem only tells us that the desired equilibrium is stable in the sense of Lyapunov. Asymptotic stability can be proven using Lasalle's theorem, since: The set in which V = 0 is the set for which ( d0 ) = (0, ), i.e., any state in which the angular , velocity is zero--with arbitrary attitude. However, in case d0 = d, we would have J = 0, thus showing that the largest invariant set in which V = 0 is just the equilibrium ( d0 ) = (0, 0). , 2 MIT OpenCourseWare http://ocw.mit.edu 16.30 / 16.31 Feedback Control Systems Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
View Full Document

Ask a homework question - tutors are online