Ee363 Homework 7 Solutions
نویسنده
چکیده
1. Gain margin for a linear quadratic regulator. Let K be the optimal state feedback gain for the LQR problem with system ˙ x = Ax + Bu, state cost matrix Q ≥ 0, and input cost matrix R > 0. You can assume that (A, B) is controllable and (Q, A) is observable. We consider the system ˙ x = Ax + Bu, u = αKx, where α > 0. If α = 1, this gives the LQR optimal input, but for α = 1 this input is obviously not LQR optimal, and indeed the closed-loop system ˙ x = (A + αBK)x can even be unstable. Show that for α > 1/2, the closed-loop system is stable. In classical control theory, the ability of a system to remain stable when the input is scaled by any factor in the interval (1/2, ∞) is described as a negative gain margin of 6dB, and an infinite positive gain margin. Hint. Use the obvious quadratic Lyapunov function. Solution: Let P be the positive definite solution of the ARE A T P + P A = K T RK − Q. We will use V (z) = z T P z as our Lyapunov function. With ˙ x = (A + αBK)x, we have ˙ V (z) is quadratic, with quadratic form (A + αBK) T P + P (A + αBK) = A T P + P A − 2αK T RK = −(Q + (2α − 1)K T RK). Since Q ≥ 0 and K T RK ≥ 0 for α ≥ 1/2 we have (Q + (2α − 1)K T RK) ≥ 0. To conclude stability, we now just need to show that (Q + (2α − 1)K T RK, A + αBK) is observable, i.e., there is no v = 0 for which (A + αBK)v = λv and (Q + (2α − 1)K T RK)v = 0. If there were such a v, we would have v T (Q + (2α − 1)K T RK)v = v T Qv + (2α − 1)v T K T RKv = 0, which implies Qv = 0 and Kv = 0. This would give (A + BK)v = Av = λv, Qv = 0, which is not possible since we've assumed (Q, A) observable. 2. Gradient systems. Suppose φ is a scalar valued function on R n. (You can assume …
منابع مشابه
Ee363 Homework 6 Solutions
0 = d dt ‖x(t)‖ = 2x(t) ẋ(t) = 2x(t)Ax(t) = x(t) (A+ A )x(t) for all x(t), which occurs if and only A+A = 0, which is the same as A = −A, i.e., A is skew-symmetric. There are many other ways to see this. For example, the norm of the state will be constant provided the velocity vector is always orthogonal to the position vector, i.e., ẋ(t)x(t) = 0. This also leads us to A + A = 0. Another approa...
متن کاملEe363 Homework 5
pt+1 = αpt + βpt−1 + γpt−2 + wt, yt = pt + vt. Here p is the (scalar) time series we are interested in, and y is the scalar measurement available to us. The process noise w is IID zero mean Gaussian, with variance 1. The sensor noise v is IID Gaussian with zero mean and variance 0.01. Our job is to estimate pt+1, based on knowledge of y0, . . . , yt. We will use the parameter values α = 2.4, β ...
متن کاملEe363 Homework 3 Solutions
1. Solution of a two-point boundary value problem. We consider a linear dynamical system ˙ x = Ax, with x(t) ∈ R n. There is an n-dimensional subspace of solutions of this equation, so to single out one of the trajectories we can impose, roughly speaking, n equations. In the most common situation, we specify x(0) = x 0 , in which case the unique solution is x(t) = e tA x 0. This is called an in...
متن کاملEe363 Homework 8 Solutions
holds for all trajectories of the system, and for all t. Here we interpret u and y as power-conjugate quantities (i.e., quantities whose product gives power) such as voltage and current or force and velocity. The inequality above states that at all times, the total energy delivered to the system since t = 0 is nonnegative, i.e., it is impossible to extract any energy from a passive system. (a) ...
متن کاملذخیره در منابع من
با ذخیره ی این منبع در منابع من، دسترسی به آن را برای استفاده های بعدی آسان تر کنید
عنوان ژورنال:
دوره شماره
صفحات -
تاریخ انتشار 2009