Discrete LQ Problems

The trep.discopt module provides functions for solving time-varying discrete LQ problems.

The Linear Quadratic Regulator (LQR) Problem

The LQR problem is to find the input for a linear system that minimizes a quadratic cost. The optimal input turns out to be a feedback law that is independent of the system’s initial condition. Because of this, the LQR problem is a useful tool to automatically calculate a stabilizing feedback controller for a dynamic system. For nonlinear systems, the LQR problem is solved for the linearization of the system about a trajectory to get a locally stabilizing controller.

Problem Statement: Given a discrete linear system Find the control input \(u(k)\) that minimizes a quadratic cost:

\[V(x(k_0), u(\cdot), k_0) = \sum_{k=k_0}^{k_f-1} \left[ x^T(k)Q(k)x(k) + u^T(k)R(k)u(k) \right] + x^T(k_f) Q(k_f) x(k_f)\]

where

\[\begin{split}\begin{align} R(k) &= R^T(k) \geq 0 \ \forall\ k \in \{k_0 \dots (k_f-1)\} \\ Q(k) &= Q^T(k) \geq 0 \ \forall\ k \in \{k_0 \dots k_f\} \\ x(k_0)&\text{ is known.} \\ x(k+1) &= A(k)x(k) + B(k)u(k) \end{align}\end{split}\]

Solution: The optimal control \(u^*(k)\) and optimal cost \(V^*(x(k_0), k_0)\) are

\[\begin{split}\begin{align} u^*(k) &= -\mathcal{K}(k) x(k) \\ V^*(x(k_0), k_0) &= x^T(k_0) P(k_0) x(k_0) \end{align}\end{split}\]

where

\[ \begin{align}\begin{aligned}\mathcal{K}(k) = \Gamma^{-1}(k) B^T(k) P(k+1) A(k)\\\Gamma(k) = R(k) + B^T(k)P(k+1)B(k)\end{aligned}\end{align} \]

and \(P(k+1)\) is a symmetric time varying matrix satisfying a discrete Ricatti-like equation:

\[\begin{split}\begin{align} P(k_f) &= Q(k_f) \\ P(k) &= Q(k) + A^T(k)P(k+1)A(k) - \mathcal{K}^T(k)\Gamma(k)\mathcal{K}(k) \end{align}\end{split}\]
trep.discopt.solve_tv_lqr(A, B, Q, R)
Parameters:
  • A (Sequence of N numpy arrays, shape (nX, nX)) – Linear system dynamics
  • B (Sequence of N numpy arrays, shape (nX, nU)) – Linear system input matrix
  • Q (Function Q(k) returning numpy array, shape (nX, nX)) – Quadratic State Cost
  • R (Function R(k) returning numpy array, shape (nU, nU)) – Quadratic Input Cost
Return type:

named tuple (K, P)

This function solve the time-varying discrete LQR problem for the linear system A, B and costs Q and R.

A is a sequence of the linear system dynamics, A[k].

B is a sequence of the linear system’s input matrix, B[k].

Q is a function Q(k) that returns the state cost matrix at time k. For example, if \(Q(k) = \mathcal{I}\):

Q = lambda k: numpy.eye(nX)

R is a function Q(k) that returns the state cost matrix at time k. For example, if the cost matrices are stored in an array r_costs:

R = lambda k: r_costs[k]

The function returns the optimal feedback law \(\mathcal{K(k)}\) and the solution to the discrete Ricatti equation at k=0, \(P(0)\). K is a sequence of N numpy arrays of shape (nU,nX). P is a single (nX, nX) numpy array.

The Linear Quadratic (LQ) Problem

The LQ problem is to find the input for a linear system that minimizes a cost with linear and quadratic terms. In trep, the LQ problem is a sub-problem for discrete trajectory optimization that is used to calculate the descent direction at each iteration.

Problem Statement: Find the control input \(u(k)\) that minimizes the cost:

\[\begin{split}V(x(k_0), u(\cdot), k_0) = \sum_{k=k_0}^{k_f-1} \Bigg[ 2 \begin{bmatrix} q(k) \\ r(k) \end{bmatrix}^T \begin{bmatrix} x(k) \\ u(k) \end{bmatrix} + \begin{bmatrix} x(k) \\ u(k) \end{bmatrix}^T \begin{bmatrix} Q(k) & S(k) \\ S^T(k) & R(k) \end{bmatrix} \begin{bmatrix} x(k) \\ u(k) \end{bmatrix} \Bigg] \\ + 2 q^T(k_f) x(k_f) + x^T(k_f)Q(k_f)x(k_f)\end{split}\]

where

\[\begin{split}\begin{align*} R(k) &= R^T(k) > 0 \ \forall\ k \in \{k_0 \dots (k_f-1)\} \\ Q(k) &= Q^T(k) \geq 0 \ \forall\ k \in \{k_0 \dots k_f\} \\ x(k_0)&\text{ is known.} \\ x(k+1) &= A(k)x(k) + B(k)u(k) \end{align*}\end{split}\]

Solution: The optimal control \(u^*(k)\) and optimal cost \(V^*(x(k_0), k_0)\) are:

\[\begin{split}\begin{align*} u^*(k) &= -\mathcal{K}(k) x(k) - C(k) \\ V^*(x(k_0), k_0) &= x^T(k_0) P(k_0) x(k_0) + 2 b^T(k_0) x(k_0) + c(k_0) \end{align*}\end{split}\]

where:

\[ \begin{align}\begin{aligned}K(k) = \Gamma^{-1}(k) \left[B^T(k)P(k+1)A(k) + S^T(k)\right]\\C(k) = \Gamma^{-1}(k) \left[B^T(k)b(k+1) + r(k) \right]\\\Gamma(k) = \left[ R(k) + B^T(k)P(k+1)B(k) \right]\end{aligned}\end{align} \]

and \(P(k)\), \(b(k)\), and \(c(k)\) are solutions to backwards difference equations:

\[\begin{split}\begin{align*} P(k_f) &= Q(k_f) \\ P(k) &= Q(k) + A^T(k)P(k+1)A(k) - \mathcal{K}^T(k)\Gamma(k)\mathcal{K}(k) \end{align*}\end{split}\]
\[\begin{split}\begin{align*} b(k_f) &= q(k_f) \\ b(k) &= \left[A^T(k) - \mathcal{K}^T(k)B^T(k) \right]b(k+1) + q(k) - \mathcal{K}^T(k)r(k) \end{align*}\end{split}\]
\[\begin{split}\begin{align*} c(k_f) &= 0 \\ c(k) &= c(k+1) - C(k)^T\Gamma(k) C(k) \end{align*}\end{split}\]
trep.discopt.solve_tv_lq(A, B, q, r, Q, S, R)
Parameters:
  • A (Sequence of N numpy arrays, shape (nX, nX)) – Linear system dynamics
  • B (Sequence of N numpy arrays, shape (nX, nU)) – Linear system input matrix
  • q (Sequence of N numpy arrays, shape (nX)) – Linear State Cost
  • r (Sequence of N numpy arrays, shape (nU)) – Linear Input Cost
  • Q (Function Q(k) returning numpy array, shape (nX, nX)) – Quadratic State Cost
  • S (Function S(k) returning numpy array, shape (nX, nU)) – Quadratic Cross Term Cost
  • R (Function R(k) returning numpy array, shape (nU, nU)) – Quadratic Input Cost
Return type:

named tuple (K, C, P, b)

This function solve the time-varying discrete LQ problem for the linear system A, B.

A[k] is a sequence of the linear system dynamics, \(A(k)\).

B[k] is a sequence of the linear system’s input matrix, \(B(k)\).

q[k] is a sequence of the linear state cost, \(q(k)\).

r[k] is a sequence of the linear input cost, \(r(k)\).

Q(k) is a function that returns the quadratic state cost matrix at time k. For example, if \(Q(k) = \mathcal{I}\):

Q = lambda k: numpy.eye(nX)

S(k) is a function that returns the quadratic cross term cost matrix at time k.

R(k) is a function that returns the state cost matrix at time k. For example, if the cost matrices are stored in an array r_costs:

R = lambda k: r_costs[k]

The function returns the optimal feedback law \(\mathcal{K(k)}\), the affine input term \(C(k)\), and the last solution to two of the difference equations, \(P(0)\) and \(b(0)\).

K is a sequence of N numpy arrays of shape (nU,nX).

C is a sequence of N numpy arrays of shape (nU).

P is a single (nX, nX) numpy array.

b is a single (nX) numpy array.