Lecture 9: Discrete-Time Linear Quadratic Regulator Finite-Horizon Case
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1 Lecture 9: Discrete-Time Linear Quadratic Regulator Finite-Horizon Case Dr. Burak Demirel Faculty of Electrical Engineering and Information Technology, University of Paderborn December 15, 2015
2 2 Previous lectures In the previous lectures, we considered the synthesis problem, solved by pole-placement techniques (i.e., the main design parameter is the location of poles) These techniques are limited to SISO systems.
3 LQR problem: Background Given a discrete-time LTI system x k+1 = Ax k + Bu k y k = Cx k (1) LQR problem: Find the control inputs π = {u 0,, u N 1 } that makes the following criterion as small as possible: where N 1 J(π) = x N Px N + x k Qx k + u k Ru k (2) k=0 Q S 0, P S 0, R S >0, are given state cost, terminal cost, and input cost matrices, respectively.
4 4 Multi-objective interpretation Common form for Q and R: Q = P = C C R = ρi where ρ > 0. Then, the cost is where y k = Cx k. J(π) = N k=0 N 1 y k 2 +ρ k=0 u k 2 ρ gives relative weighting of output and input norm.
5 5 Multi-objective interpretation LQR quadratic cost can be written as: J = J out + ρj in. The term J out = N y k 2 k=0 provides a measure of the output energy, and the term J in = N 1 k=0 u k 2 provides a measurement of the control energy. There are competing objectives; we want both small.
6 Trade-off between conflicting goals π 3 is worse than π 2 on both J out and J in π 1 is better than π 2 in J in but worse in J out J out J out + ρj in = J π 1 π 2 π 3 (Constant) J in When ρ is much larger than 1, the most effective way to decrease J is to employ a small control input, at the expense of a large output. When ρ is much smaller than 1, the most effective way to decrease J is to obtain a very small output, even if this is achieved at the expense of employing a large control input.
7 7 LQR via least-squares LQR can be formulated (and solved) as a least-squares problem X = [x 0,, x N ] is a linear function of x 0 and U = [u 0,, u N 1 ]: 0 x 0 B 0 I u 0. = AB B A. +. x 0 x N.. 0 u N 1 A N 1 B A N 2 A N B B express as X = GU + Hx 0, where G R Nn Nm, H R Nn n
8 8 LQR via least-squares Express linear-quadratic cost as J(U) = diag(q 1/2,, Q 1/2, P 1/2 )(GU + Hx 0 ) 2 + diag(r 1/2,, R 1/2 )U 2 This is only a least-squares problem. This solution method requires solving a least-squares problem with size N(n + m) Nm (see Boyd) Using a naive method (e.g., QR factorization), cost is O(N 3 nm 2 ) (see Boyd)
9 9 Finite-horizon LQR control Theorem Consider the finite horizon LQR problem. The optimal control u k = (R + B S k+1 B) 1 B S k+1 Ax k is a linear function of the state x k. The matrix S k evolves according to the backward Riccati recursion S k = A S k+1 A + Q A S k+1 B(B S k+1 B + R) 1 B S k+1 A with the initial condition S N = P. Furthermore, the optimal control loss is given by J N = x 0 Qx 0.
10 10 Dynamic programming solution gives an efficient, recursive method to solve LQR least-squares problem; cost is O(Nn 3 ) but a less naive approach to solve the LQR least-squares problem will have the same complexity same idea can be applied to many other problems
11 11 Completing the squares Consider the loss function of the form [ ] [ ] [ ] x Qx Q J(x, u) = xu x, (3) u u and minimize with respect to u. Then, there exists a matrix L satisfying Q xu Q u Q u L = Q xu such that the loss function (3) becomes J(x, u) = x (Q x L Q u L)x + (u + Lx) Q u (u + Lx). (4) Since (4) is quadratic in u and both terms are greater or equal zero, (3) is minimized for u = Lx. The minimum is J min = x (Q x L Q u L)x.
12 12 Proof: Via dynamic programming Define the value function V k : R n R by { N 1 } V k = min x u k,,u i Qx i + u i Ru i + x N Px N N 1 i=k subject to x i+1 = Ax i + Bu i for i {k,, N 1}. V k can be interpreted as the loss-to-go from k to N and is a function of x k at time k. We will find that V k is quadratic, i.e., V k (x) = x S k x, where S k S 0 S k can be found recursively, working from k = N the LQR optimal u is easily expressed in terms of S k
13 13 Proof: Via dynamic programming Loss-to-go with no time left is only final state cost: V N = x N S Nx N where S N = P. Now, we will show that V k is quadratic in x k for all k. For k = N 1, we have { } V N 1 = min x u N 1 Qx N 1 + u N 1 Ru N 1 + V N N 1 Using x N = Ax N 1 + Bu N 1 gives V N 1 = min {x u N 1 Qx N 1 + u N 1 Ru N 1 N 1 } + (Ax N 1 + Bu N 1 ) S N (Ax N 1 + Bu N 1 )
14 Proof: Via dynamic programming Then, we get: V N 1 = min u N 1 [ xn 1 u N 1 ] [ Q + A S N A B S N A ] [ ] xn 1 A S N B R + B S N B u N 1 This function is quadratic in u N 1. Completing the squares, we can write: V N 1 = min {x u N 1( A S N A + Q L N 1 (R + ) B S N B)L N 1 xn 1 N 1 } + (u N 1 + L N 1 x N 1 ) (R + B S N B)(u N 1 + L N 1 x N 1 ) (5) where L N 1 = (R + B S N B) 1 B S N A. To minimize (5), we need to select which gives the minimum loss u N 1 = L N 1 x N 1 V N 1 = x N 1 S N 1x N 1
15 DP algorithm for LQR 1. set S N := P 2. for k = N,, 1, compute S k 1 := A S k A + Q A S k B(R + B S k B) 1 B S k A 3. for k = 0,, N 1, compute L k := (R + B S k+1 B) 1 B S k+1 A 4. for k = 0,, N 1, compute u k := L k x k
16 16 LQ-control of the double integrator Consider a double integrator system (with unity sampling interval): x k+1 = y k = [ 1 [ ] [ ] x k + 2 u k ] x k with initial state x 0 = [1 0], horizon N = 20, and weighting matrices: Q = P = C C, R = ρ where ρ > 0.
17 17 LQ-control of the double integrator Optimal trade-off curve of J out and J in : Jout J in Blue circle denotes ρ = 0.3 while red circle denotes ρ = 10.
18 18 LQ-control of the double integrator yk uk ρ = 0.3 ρ = Number of samples [k]
19 Reference 1. S. Boyd, Linear quadratic regulator: Discrete-time finite horizon, Lecture Notes. 2. K. J. Åström and B. Wittenmark, Computer-Controlled Systems, Prentice Hall, D. P. Bertsekas, Dynamic Programming and Optimal Control, Athena Scientific, 2000.
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