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<!DOCTYPE html>
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<title>Underactuated Robotics: Linear Quadratic Regulators</title>
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<header>
<h1><a href="underactuated.html" style="text-decoration:none;">Underactuated Robotics</a></h1>
<p data-type="subtitle">Algorithms for Walking, Running, Swimming, Flying, and Manipulation</p>
<p style="font-size: 18px;"><a href="http://people.csail.mit.edu/russt/">Russ Tedrake</a></p>
<p style="font-size: 14px; text-align: right;">
© Russ Tedrake, 2020<br/>
<a href="tocite.html">How to cite these notes</a><br/>
</p>
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<p><b>Note:</b> These are working notes used for <a
href="http://underactuated.csail.mit.edu/Spring2020/">a course being taught
at MIT</a>. They will be updated throughout the Spring 2020 semester. <a
href="https://www.youtube.com/channel/UChfUOAhz7ynELF-s_1LPpWg">Lecture videos are available on YouTube</a>.</p>
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<chapter style="counter-reset: chapter 7"><h1>Linear Quadratic Regulators</h1>
<p>While solving the dynamic programming problem for continuous systems is
very hard in general, there are a few very important special cases where the
solutions are very accessible. Most of these involve variants on the case of
linear dynamics and quadratic cost. The simplest case, called the linear
quadratic regulator (LQR), is formulated as stabilizing a time-invariant
linear system to the origin.</p>
<p>The linear quadratic regulator is likely the most important and influential
result in optimal control theory to date. In this chapter we will derive the
basic algorithm and a variety of useful extensions.</p>
<section><h1>Basic Derivation</h1>
<p>Consider a linear time-invariant system in state-space form, $$\dot{\bx}
= {\bf A}\bx + \bB\bu,$$ with the infinite-horizon cost function given by
$$J = \int_0^\infty \left[ \bx^T {\bf Q} \bx + \bu^T {\bf R} \bu \right] dt,
\quad {\bf Q} = {\bf Q}^T \succeq {\bf 0}, {\bf R} = {\bf R}^T \succ 0.$$
Our goal is to find the optimal cost-to-go function $J^*(\bx)$ which
satisfies the HJB: $$\forall \bx, \quad 0 = \min_\bu \left[ \bx^T {\bf Q}
\bx + \bu^T {\bf R} \bu + \pd{J^*}{\bx} \left( {\bf A}\bx + \bB\bu \right)
\right].$$</p>
<p> There is one important step here -- it is well known that for this
problem the optimal cost-to-go function is quadratic. This is easy to
verify. Let us choose the form: $$J^*(\bx) = \bx^T {\bf S} \bx, \quad {\bf
S} = {\bf S}^T \succeq 0.$$ The gradient of this function is $$\pd{J^*}{\bx}
= 2 \bx^T {\bf S}.$$ </p>
<p> Since we have guaranteed, by construction, that the terms inside the
$\min$ are quadratic and convex (because ${\bf R} \succ 0$), we can take the
minimum explicitly by finding the solution where the gradient of those terms
vanishes: $$\pd{}{\bu} = 2\bu^T {\bf R} + 2 \bx^T {\bf S} \bB = 0.$$ This
yields the optimal policy $$\bu^* = \pi^*(\bx) = - {\bf R}^{-1} \bB^T {\bf
S} \bx = - \bK \bx.$$</p>
<p>Inserting this back into the HJB and simplifying yields $$0 = \bx^T
\left[ {\bf Q} - {\bf S B R}^{-1}\bB^T{\bf S} + 2{\bf SA} \right]\bx.$$ All
of the terms here are symmetric except for the $2{\bf SA}$, but since
$\bx^T{\bf SA}\bx = \bx^T{\bf A}^T{\bf S}\bx$, we can write $$0 = \bx^T
\left[ {\bf Q} - {\bf S B R}^{-1}\bB^T{\bf S} + {\bf SA} + {\bf A}^T{\bf S}
\right]\bx.$$ and since this condition must hold for all $\bx$, it is
sufficient to consider the matrix equation $$0 = {\bf S} {\bf A} + {\bf A}^T
{\bf S} - {\bf S} \bB {\bf R}^{-1} \bB^T {\bf S} + {\bf Q}.$$ This extremely
important equation is a version of the <em>algebraic Riccati equation</em>.
Note that it is quadratic in ${\bf S}$, making its solution non-trivial, but
it is well known that the equation has a single positive-definite solution
if and only if the system is controllable and there are good numerical
methods for finding that solution, even in high-dimensional problems. Both
the optimal policy and optimal cost-to-go function are available from
<drake></drake> by calling <code> (K,S) =
LinearQuadraticRegulator(A,B,Q,R)</code>.</p>
<p>If the appearance of the quadratic form of the cost-to-go seemed
mysterious, consider that the solution to the linear system $\dot\bx = (\bA
- \bB\bK)\bx$ takes the form $\bx(t) = e^{(\bA-\bB\bK)t}\bx(0)$, and try
inserting this back into the integral cost function. You'll see that the
cost takes the form $J=\bx^T(0) {\bf S} \bx(0)$.</p>
<p>It is worth examining the form of the optimal policy more closely. Since
the value function represents cost-to-go, it would be sensible to move down
this landscape as quickly as possible. Indeed, $-{\bf S}\bx$ is in the
direction of steepest descent of the value function. However, not all
directions are possible to achieve in state-space. $-\bB^T {\bf S} \bx$
represents precisely the projection of the steepest descent onto the control
space, and is the steepest descent achievable with the control inputs $\bu$.
Finally, the pre-scaling by the matrix ${\bf R}^{-1}$ biases the direction
of descent to account for relative weightings that we have placed on the
different control inputs. Note that although this interpretation is
straight-forward, the slope that we are descending (in the value function,
${\bf S}$) is a complicated function of the dynamics and cost.</p>
<example><h1>LQR for the Double Integrator</h1>
<p>Now can use LQR to reproduce our <a
href="dp.html#hjb_double_integrator">HJB example</a> from the previous
chapter:</p>
<pysrcinclude>double_integrator/lqr.py</pysrcinclude>
<p>As in the hand-derived example, our numerical solution returns $${\bf
K} = [ 1, \sqrt{3} ], \qquad{\bf S} = \begin{bmatrix} \sqrt{3} & 1 \\ 1 &
\sqrt{3} \end{bmatrix}.$$</p>
</example>
<subsection><h1>Local stabilization of nonlinear systems</h1>
<p>LQR is extremely relevant to us even though our primary interest is in
nonlinear dynamics, because it can provide a local approximation of the
optimal control solution for the nonlinear system. Given the nonlinear
system $\dot{\bx} = f(\bx,\bu)$, and a stabilizable operating point,
$(\bx_0,\bu_0)$, with $f(\bx_0,\bu_0) = 0.$ We can define a relative
coordinate system $$\bar\bx = \bx - \bx_0, \quad \bar\bu = \bu - \bu_0,$$
and observe that $$\dot{\bar\bx} = \dot{\bx} = f(\bx,\bu),$$ which we can
approximate with a first-order Taylor expansion to $$\dot{\bar\bx} \approx
f(\bx_0,\bu_0) + \pd{f(\bx_0,\bu_0)}{\bx} (\bx - \bx_0) +
\pd{f(\bx_0,\bu_0)}{\bu} (\bu - \bu_0) = {\bf A}\bar{\bx} +
\bB\bar\bu.$$</p>
<p>Similarly, we can define a quadratic cost function in the error
coordinates, or take a (positive-definite) second-order approximation of a
nonlinear cost function about the operating point (linear and constant
terms in the cost function can be easily incorporated into the derivation
by parameterizing a full quadratic form for $J^*$, as seen in the Linear
Quadratic Tracking derivation below).</p>
<p>The resulting controller takes the form $\bar\bu^* = -{\bf K}\bar\bx$
or $$\bu^* = \bu_0 - {\bf K} (\bx - \bx_0).$$ For convenience,
<drake></drake> allows you to call <code>controller =
LinearQuadraticRegulator(system, context, Q, R)</code> on most dynamical
systems (including block diagrams built up of many subsystems); it will
perform the linearization for you.</p>
<example><h1>LQR for Acrobots, Cart-Poles,
and Quadrotors</h1>
<p>LQR provides a very satisfying solution to the canonical "balancing"
problem that we've <a href="acrobot.html">already
described for a number of model systems</a>. Here are some of those
examples, again:</p>
<jupyter>examples/acrobot/lqr.ipynb</jupyter>
<jupyter>examples/cartpole/lqr.ipynb</jupyter>
<jupyter>examples/quadrotor2d/lqr.ipynb</jupyter>
<pysrc>quadrotor/lqr.py</pysrc>
<p>I find it very compelling that the same derivation (and effectively
identical code) can stabilize such a diversity of systems!</p>
</example>
</subsection>
</section> <!-- end basic derivation -->
<section><h1>Finite-horizon formulations</h1>
<p>Recall that the cost-to-go for finite-horizon problems is time-dependent,
and therefore the HJB sufficiency condition requires an additional term for
$\pd{J^*}{t}$. $$ \forall \bx, \forall t\in[t_0,t_f],\quad 0 = \min_\bu
\left[ \ell(\bx,\bu) + \pd{J^*}{\bx}f(\bx,\bu) + \pd{J^*}{t} \right]. $$</p>
<subsection><h1>Finite-horizon LQR</h1>
<p> Consider systems governed by an LTI state-space equation of the form
$$\dot{\bx} = {\bf A}\bx + \bB\bu,$$ and a finite-horizon cost function
with
\begin{gather*} h(\bx) = \bx^T {\bf Q}_f \bx,\quad {\bf Q}_f = {\bf Q}_f^T
\succeq {\bf 0} \\ \ell(\bx,\bu,t) = \bx^T {\bf Q} \bx + \bu^T {\bf R}
\bu, \quad {\bf Q} = {\bf Q}^T \succeq 0, {\bf R}={\bf R}^T \succ 0
\end{gather*}
Writing the HJB, we have $$ 0 = \min_\bu \left[\bx^T {\bf Q} \bx + \bu^T
{\bf R}\bu + \pd{J^*}{\bx} \left({\bf A} \bx + \bB \bu \right) +
\pd{J^*}{t} \right]. $$ Due to the positive definite quadratic form on
$\bu$, we can find the minimum by setting the gradient to zero:
\begin{gather*}
\pd{}{\bu} = 2 \bu^T {\bf R} + \pd{J^*}{\bx} \bB = 0 \\
\bu^* = \pi^*(\bx,t) = - \frac{1}{2}{\bf R}^{-1} \bB^T \pd{J^*}{\bx}^T
\end{gather*}
In order to proceed, we need to investigate a particular form for the
cost-to-go function, $J^*(\bx,t)$. Let's try a solution of the form:
$$J^*(\bx,t) = \bx^T {\bf S}(t) \bx, \quad {\bf S}(t) = {\bf S}^T(t)\succ
{\bf 0}.$$ In this case we have $$\pd{J^*}{\bx} = 2 \bx^T {\bf S}(t),
\quad \pd{J*}{t} = \bx^T \dot{\bf S}(t) \bx,$$ and therefore
\begin{gather*} \bu^* = \pi^*(\bx,t) = - {\bf R}^{-1} \bB^T {\bf S}(t) \bx
\\ 0 = \bx^T \left[ {\bf Q} - {\bf S}(t) \bB {\bf R}^{-1} \bB^T {\bf S}(t)
+ {\bf S}(t) {\bf A} + {\bf A}^T {\bf S}(t) + \dot{\bf S}(t) \right]
\bx.\end{gather*}
Therefore, ${\bf S}(t)$ must satisfy the condition (known as the
continuous-time <em>differential Riccati equation</em>): $$-\dot{\bf S}(t)
= {\bf S}(t) {\bf A} + {\bf A}^T {\bf S}(t) - {\bf S}(t) \bB {\bf R}^{-1}
\bB^T {\bf S}(t) + {\bf Q},$$ and the terminal condition $${\bf S}(T) =
{\bf Q}_f.$$ Since we were able to satisfy the HJB with the minimizing
policy, we have met the sufficiency condition, and have found the optimal
policy and optimal cost-to-go function.</p>
<p>Note that the infinite-horizon LQR solution described in the prequel is
exactly the steady-state solution of this equation, defined by $\dot{\bf
S}(t) = 0$. Indeed, for controllable systems this equation is stable
(backwards in time), and as expected the finite-horizon solution converges
on the infinite-horizon solution as the horizon time limits to infinity.
</p>
<todo>Example to show how the tvlqr solution converges to the tilqr
solution for the double integrator example, and make the connection back
to the value iteration visualizations that we did in the previous
chapter.</todo>
</subsection>
<subsection><h1>Time-varying LQR</h1>
<p>The derivation above holds even if the dynamics are given by
$$\dot{\bx} = {\bf A}(t)\bx + {\bf B(t)}\bu.$$ Similarly, the cost
functions ${\bf Q}$ and ${\bf R}$ can also be time-varying. This is
quite surprising, as the class of time-varying linear systems is a quite
general class of systems. It requires essentially no assumptions on how
the time-dependence enters, except perhaps that if ${\bf A}$ or $\bB$ is
discontinuous in time then one would have to use the proper techniques to
accurately integrate the differential equation. As we will see in the <a
href="trajopt.html">chapter on trajectory
optimization</a>, one of the most powerful applications involves
linearizing around a nominal trajectory of a nonlinear system and using
LQR to provide a trajectory controller.</p>
</subsection>
<subsection><h1>Linear Quadratic Optimal Tracking</h1>
<p>For completeness, we consider a slightly more general form of the
linear quadratic regulator. The standard LQR derivation attempts to drive
the system to zero. Consider now the problem:
\begin{gather*} \dot{\bx} = {\bf A}\bx + \bB\bu \\ h(\bx) = (\bx -
\bx^d(T))^T {\bf Q}_f (\bx - \bx^d(T)), \quad {\bf Q}_f = {\bf Q}_f^T
\succeq 0 \\ \ell(\bx,\bu,t) = (\bx - \bx^d(t))^T {\bf Q} (\bx - \bx^d(t))
+ (\bu - \bu^d(t))^T {\bf R} (\bu - \bu^d(t)),\\ {\bf Q} = {\bf Q}^T
\succeq 0, {\bf R}={\bf R}^T \succ 0 \end{gather*}
Now, guess a solution
\begin{gather*} J^*(\bx,t) = \bx^T {\bf S_2}(t) \bx + \bx^T {\bf s_1}(t) +
s_0(t), \quad {\bf S_2}(t) = {\bf S_2}^T(t)\succ {\bf 0}. \end{gather*}
In this case, we have $$\pd{J^*}{\bx} = 2 \bx^T {\bf S_2}(t) + {\bf
s_1}^T(t),\quad \pd{J^*}{t} = \bx^T \dot{\bf S_2}(t) \bx + \bx^T \dot{\bf
s_1}(t) + \dot{s_0}(t).$$ Using the HJB, $$ 0 = \min_\bu \left[(\bx -
\bx^d(t))^T {\bf Q} (\bx - \bx^d(t)) + (\bu - \bu^d(t))^T {\bf R} (\bu -
\bu^d(t)) + \pd{J^*}{\bx} \left({\bf A} \bx + \bB \bu \right) +
\pd{J^*}{t} \right], $$ we have
\begin{gather*} \pd{}{\bu} = 2 (\bu - \bu^d(t))^T{\bf R} + (2\bx^T{\bf
S_2}(t) + {\bf s_1}^T(t))\bB = 0,\\ \bu^*(t) = \bu^d(t) - {\bf R}^{-1}
\bB^T\left[{\bf S_2}(t)\bx + \frac{1}{2}{\bf s_1}(t)\right] \end{gather*}
The HJB can be satisfied by integrating backwards
\begin{align*}
-\dot{\bf S_2}(t) =& {\bf Q} - {\bf S_2}(t) \bB {\bf R}^{-1} {\bf
B}^T {\bf S_2}(t) + {\bf S_2}(t) {\bf A} + {\bf A}^T {\bf S_2}(t)\\
-\dot{\bf s_1}(t) =& -2 {\bf Q} \bx^d(t) + \left[{\bf A}^T - {\bf S}_2
\bB {\bf R}^{-1} \bB^T \right]{\bf s_1}(t) + 2{\bf
S_2}(t) \bB \bu^d(t)\\
-\dot{s_0}(t) =& \bx^d(t)^T {\bf Q} \bx^d(t) - \frac{1}{4} {\bf
s_1}^T(t) \bB {\bf R}^{-1} \bB^T {\bf s_1}(t) + {\bf
s_1}(t)^T \bB \bu^d(t),
\end{align*}
from the final conditions
\begin{align*}
{\bf S_2}(T) =& {\bf Q}_f\\
{\bf s_1}(T) =& -2 {\bf Q}_f \bx^d(T) \\
s_0(T) =& \left[\bx^d(T)\right]^T {\bf Q}_f \left[ \bx^d(T) \right].
\end{align*}
Notice that the solution for ${\bf S_2}$ is the same as the simpler LQR
derivation, and is symmetric (as we assumed). Note also that $s_0(t)$ has
no effect on control (even indirectly), and so can often be ignored.</p>
<p>A quick observation about the quadratic form, which might be helpful in
debugging. We know that $J(x,t)$ must be uniformly positive. This is
true iff ${\bf S}_2>0$ and $s_0 > \frac{1}{4} {\bf s}_1^T {\bf S}_2^{-1}
{\bf s}_1$, which comes from evaluating the function at $x_{min}$ defined
by $\pd{}{x} = 0$.</p>
<todo>test this on an example, get my notation consistent (s(t)^T vs
s^T(t), etc.</todo>
</subsection>
<subsection><h1>Linear Final Boundary Value Problems</h1>
<p> The finite-horizon LQR formulation can be used to impose a strict
final boundary value condition by setting an infinite ${\bf Q}_f$.
However, integrating the Riccati equation backwards from an infinite
initial condition isn't very practical. To get around this, let us
consider solving for ${\bf P}(t) = {\bf S}(t)^{-1}$. Using the matrix
relation $\frac{d {\bf S}^{-1}}{dt} = - {\bf S}^{-1} \frac{d {\bf S}}{dt}
{\bf S}^{-1}$, we have: $$-\dot{\bf P}(t) = - {\bf P}(t){\bf Q P}(t) +
{\bf B R}^{-1} \bB - {\bf A P}(t) - {\bf P}(t){\bf A}^T,$$ with the final
conditions $${\bf P}(T) = 0.$$ This Riccati equation can be integrated
backwards in time for a solution.</p>
<p> It is very interesting, and powerful, to note that, if one chooses
${\bf Q} = 0$, therefore imposing no position cost on the trajectory
before time $T$, then this inverse Riccati equation becomes a linear ODE
which can be solved explicitly. <todo>% add explicit solution here</todo>
These relationships are used in the derivation of the controllability
Grammian, but here we use them to design a feedback controller.</p>
</subsection>
</section> <!-- end finite horizon -->
<section><h1>Variations and extensions</h1>
<!--
<subsection><h1>Minimum-time LQR</h1>
\begin{gather*}
\dot{\bx} = {\bf A}\bx + \bB\bu \\
h(\bx) = \bx^T {\bf Q}_f \bx,
\quad {\bf Q}_f = {\bf Q}_f^T \succeq 0 \\
\ell(\bx,\bu,t) = 1 + \bx^T {\bf Q} \bx + \bu {\bf R} \bu,\\
{\bf Q} = {\bf
Q}^T \succeq 0, {\bf R}={\bf R}^T \succ 0
\end{gather*}
</subsection>
-->
<subsection id="dt_riccati"><h1>Discrete-time Riccati Equations</h1>
<p>Essentially all of the results above have a natural correlate for
discrete-time systems. What's more, the discrete time versions tend to be
simpler to think about in the model-predictive control (MPC) setting that
we'll be discussing below and in the next chapters.</p>
<p>Consider the discrete time dynamics: $$\bx[n+1] = {\bf A}\bx[n] + {\bf
B}\bu[n],$$ and we wish to minimize $$\min \sum_{n=0}^{N-1} \bx^T[n] {\bf
Q} \bx[n] + \bu^T[n] {\bf R} \bu[n], \qquad {\bf Q} = {\bf Q}^T \succeq 0,
{\bf R} = {\bf R}^T \succ 0.$$ The cost-to-go is given by $$J(\bx,n-1) =
\min_\bu \bx^T {\bf Q} \bx + \bu^T {\bf R} \bu + J({\bf A}\bx + {\bf
B}\bu, n).$$ If we once again take $$J(\bx,n) = \bx^T {\bf S[n]} \bx,
\quad {\bf S}[n] = {\bf S}^T[n] \succ 0,$$ then we have $$\bu^*[n] = -{\bf
K[n]}\bx[n] = -({\bf R} + {\bf B}^T {\bf S}[n] {\bf B})^{-1} {\bf B}^T
{\bf S}[n] {\bf A} \bx[n],$$ yielding $${\bf S}[n-1] = {\bf Q} + {\bf
A}^T{\bf S}[n]{\bf A} - ({\bf A}^T{\bf S}[n]{\bf B})({\bf R} + {\bf B}^T
{\bf S}[n] {\bf B})^{-1} ({\bf B}^T {\bf S}[n]{\bf A}), \quad {\bf S}[N]
= 0,$$ which is the famous <i>Riccati difference equation</i>. The
infinite-horizon LQR solution is given by the (positive-definite)
fixed-point of this equation: $${\bf S} = {\bf Q} + {\bf A}^T{\bf S}{\bf
A} - ({\bf A}^T{\bf S}{\bf B})({\bf R} + {\bf B}^T {\bf S} {\bf B})^{-1}
({\bf B}^T {\bf S}{\bf A}).$$ Like in the continuous time case, this
equation is so important that we have special numerical recipes for
solving this discrete-time algebraic Riccati equation (DARE). <drake>
</drake> delegates to these numerical methods automatically when you
evaluate the <code>LinearQuadraticRegulator</code> method on a system that
has only discrete state and a single periodic timestep.</p>
<todo>Example where the discrete-time solutions converge to the continuous
time solutions as dt-> 0.</todo>
</subsection>
<subsection><h1>LQR with input and state constraints</h1>
<p>A natural extension for linear optimal control is the consideration of
strict constraints on the inputs or state trajectory. Most common are
linear inequality constraints, such as $\forall n, |\bu[n]| \le 1$ or
$\forall n, \bx[n] \ge -2$ (any linear constraints of the form ${\bf Cx} +
{\bf Du} \le {\bf e}$ can be solved with the same tools). Much is known
about the solution to this problem in the discrete-time case, but it's
computation is signficantly harder than the unconstrained case. Almost
always, we give up on solving for the best control policy in closed form,
and instead solve for the optimal control trajectory $\bu[\cdot]$ from a
particular initial condition $\bx[0]$ over some finite horizon.
Fortunately, this problem is a convex optimization and we can often solve
it quickly and reliably enough to solve it at every timestep, effectively
turning a motion planning algorithm into a feedback controller; this idea
is famously known as model-predictive control (MPC). We will provide the
details in the <a href="trajopt.html">trajectory
optimization chapter</a>.</p>
<p>We do actually understand what the optimal policy of the
inequality-constrained LQR problem looks like, thanks to work on "explicit
MPC" <elib>Alessio09</elib> -- the optimal policy is now piecewise-linear
(though still continuous), with each piece describe by a polytope, and the
optimal cost-to-go is piecewise-quadratic on the same polytopes.
Unfortunately, the number of pieces grows exponentially with the number of
constraints and the horizon of the problem, making it impractical to
compute for all but very small problems. There are, howeer, a number of
promising approaches to approximate explicit MPC (c.f.
<elib>Marcucci17</elib>).</p>
<p>One important case that does have closed-form solutions is LQR with
linear <i>equality</i> constraints (c.f. <elib>Posa15</elib>, section
IIIb). This is particularly relevant in the case of stabilizing robots
with kinematic constraints such as a closed kinematic chain, which appears
in four-bar linkages or even for the linearization of a biped robot with
two feet on the ground.</p>
<todo>Add the equality-constrained LQR derivation here.</todo>
</subsection>
<subsection><h1>LQR as a convex optimization</h1>
<p>Solving the algebraic Riccati equation is the preferred way of
computing thse LQR solution. But it is helpful to know that one <i>could
</i> also compute it with convex optimization, specifically by formulating
the optimality conditions as a Linear Matrix Inequality (LMI). In
addition to deepening our understanding, this can be useful if we want to
extend the LQR solution or solve for the LQR gains jointly as part of a
bigger optimization.</p>
<p>Derivation coming soon...</p>
</subsection>
<subsection id="sls"><h1>Finite-horizon LQR via least
squares</h1>
<p>We can also obtain the solution to the discrete-time finite-horizon
(including the time-varying or tracking variants) LQR problem using
optimization -- in this case it actually reduces to a simple least-squares
problem. This idea plays an important role in the minimax variants of LQR
(which optimize a worst-case performance), which we will discuss in the
robust control chapter (e.g. <elib>Lofberg03+Sadraddini20</elib>).</p>
<p>First, let us appreciate that the default parameterization is not
convex. Given \begin{gather*} \min \sum_{n=0}^{N-1} \bx^T[n] {\bf Q}
\bx[n] + \bu^T[n] {\bf R} \bu[n], \qquad {\bf Q} = {\bf Q}^T \succeq 0,
{\bf R} = {\bf R}^T \succ 0 \\ \subjto \quad \bx[n+1] = {\bf A} \bx[n] +
{\bf B}\bu[n], \\ \bx[0] = \bx_0 \end{gather*} if we wish to search over
controllers of the form $$\bu[n] = {\bf K}_n \bx[n],$$ then we have
\begin{align*}\bx[1] &= {\bf A}\bx_0 + {\bf B}{\bf K}_0\bx_0, \\ \bx[2] &=
{\bf A}({\bf A} + {\bf BK}_0)\bx_0 + {\bf BK}_1({\bf A} + {\bf BK}_0)\bx_0
\\ \bx[n] &= \left( \prod_{i=0}^{n-1} ({\bf A} + {\bf BK}_i) \right) \bx_0
\end{align*} As you can see, the $\bx[n]$ terms in the cost function
include our decision variables multiplied together -- resulting in a
non-convex objective. The trick is to re-parameterize the decision
variables, and write the feedback in the form: $$\bu[n] = \tilde{\bf K}_n
\bx_0,$$ leading to \begin{align*}\bx[1] &= {\bf A}\bx_0 + {\bf
B}\tilde{\bf K}_0\bx_0, \\ \bx[2] &= {\bf A}({\bf A} + {\bf B}\tilde{\bf
K}_0)\bx_0 + {\bf B}\tilde{\bf K}_1 \bx_0 \\ \bx[n] &= \left( {\bf A}^n +
\sum_{i=0}^{n-1}{\bf A}^{n-i-1}{\bf B}\tilde{\bf K}_{i} \right) \bx_0
\end{align*} Now all of the decision variables, $\tilde{\bf K}_i$, appear
linearly in the solution to $\bx[n]$ and therefore (convex) quadratically
in the objective.</p>
<p>We still have an objective function that depends on $\bx_0$, but we
would like to find the optimal $\tilde{\bf K}_i$ <i>for all</i>
$\bx_0$. To achieve this let us evaluate the optimality conditions of this
least squares problem, starting by taking the gradient of the objective
with respect to $\tilde{\bf K}_i$, which is: $$\bx_0 \bx_0^T \left(
\tilde{\bf K}_i^T \left({\bf R} + \sum_{m=i+1}^{N-1} {\bf B}^T ({\bf
A}^{m-i-1})^T {\bf Q A}^{m-i-1} {\bf B}\right) + \sum_{m=i+1}^{N-1} ({\bf
A}^m)^T {\bf Q A}^{m-i-1} {\bf B}\right).$$ We can satisfy this
optimality condition for all $\bx_0$ by solving the <i>linear</i> matrix
equation: $$\tilde{\bf K}_i^T \left({\bf R} + \sum_{m=i+1}^{N-1} {\bf B}^T
({\bf A}^{m-i-1})^T {\bf Q A}^{m-i-1} {\bf B}\right) + \sum_{m=i+1}^{N-1}
({\bf A}^m)^T {\bf Q A}^{m-i-1} {\bf B} = 0.$$ We can always solve for
$\tilde{\bf K}_i$ since it's multiplied by a (symmetric) positive definite
matrix (it is the sum of a positive definite matrix and many positive
semi-definite matrices), which is always invertible.</p>
<p>If you need to recover the original ${\bf K}_i$ parameters, you can
extract them recursively with \begin{align*} \tilde{\bf K}_0 &= {\bf K}_0,
\\ \tilde{\bf K}_n &= {\bf K}_n \prod_{i=0}^{n-1} ({\bf A} + {\bf BK}_i),
\qquad 0 < n \le N-1. \end{align*} But often this is not actually
necessary. In some applications it's enough to know the performance cost
under LQR control, or to handle the response to disturbances explicitly
with the disturbance-based feedback (which I've already promised for the
robust control chapter). Afterall, the problem formulation that we've
written here, which makes no mention of disturbances, assumes the model is
perfect and the controls $\tilde{\bf K}_n \bx_0$ are just as suitable for
deployment as ${\bf K}_n\bx[n]$.</p>
<p>"System-Level Synthesis" (SLS) is the name for an important and
slightly different approach, where one optimizes the <i>closed-loop
response</i> directly<elib>Anderson19</elib>. Although SLS is a very
general tool, for the particular formulation we are considering here it
reduces to creating additional decision variables ${\bf \Phi}_i$, such at
that $$\bx[n] = {\bf \Phi}_n \bx[0],$$ and writing the optimization above
as \begin{gather*} \min_{\tilde{\bf K}_*, {\bf \Phi}_*} \sum_{n=0}^{N-1}
\bx^T[0] \left( {\bf \Phi}_n^T {\bf Q \Phi}_n + \tilde{\bf K}_n^T{\bf R}
\tilde{\bf K}_n \right) \bx[0], \\ \subjto \qquad \forall n, \quad {\bf
\Phi}_{n+1} = {\bf A \Phi}_n + {\bf B}\tilde{\bf K}_n. \end{gather*} </p>
<p>Once again, the algorithms presented here are not as efficient as
solving the Riccati equation if we only want the solution to the simple
case, but they become very powerful if we want to combine the LQR
synthesis with other objectives/constraints. For instance, if we want to
add some sparsity constraints (e.g. enforcing that some elements of
$\tilde{\bf K}_i$ are zero), then we could solve the quadratic
optimization subject to linear equality constraints
<elib>Wang14</elib>.</p>
</subsection>
<todo>
Minimum-time LQR
</todo>
</section>
</chapter>
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