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<h1><a href="index.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> |
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<p><b>Note:</b> These are working notes used for <a
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at MIT</a>. They will be updated throughout the Spring 2020 semester. <a
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<chapter style="counter-reset: chapter 1">
<h1>The Simple Pendulum</h1>
<section><h1>Introduction</h1>
<p> Our goals for this chapter are modest: we'd like to understand the
dynamics of a pendulum.</p>
<p>Why a pendulum? In part, because the dynamics of a majority of our
multi-link robotics manipulators are simply the dynamics of a large number
of coupled pendula. Also, the dynamics of a single pendulum are rich enough
to introduce most of the concepts from nonlinear dynamics that we will use
in this text, but tractable enough for us to (mostly) understand in the next
few pages. </p>
<figure> <img width="30%" src="figures/simple_pend.svg"/>
<figcaption>The simple pendulum</figcaption> </figure>
<p> The Lagrangian derivation of the equations of motion (as described in
the appendix) of the simple pendulum yields: \begin{equation*} m l^2
\ddot\theta(t) + mgl\sin{\theta(t)} = Q. \end{equation*} We'll consider the
case where the generalized force, $Q$, models a damping torque (from
friction) plus a control torque input, $u(t)$: $$Q = -b\dot\theta(t) +
u(t).$$ </p>
</section>
<section class="drake"><h1>Nonlinear dynamics with a
constant
torque</h1>
<p> Let us first consider the dynamics of the pendulum if it is driven in a
particular simple way: a torque which does not vary with time:
\begin{equation} ml^2 \ddot\theta + b\dot\theta + mgl \sin\theta = u_0.
\end{equation} </p>.
<example><h1>Simple Pendulum in Python</h1>
<p>You can experiment with this system in <drake></drake> using</p>
<jupyter no_colab>examples/pendulum/torque_slider_demo.ipynb</jupyter>
</example>
<p> These are relatively simple differential equations, so if I give you
$\theta(0)$ and $\dot\theta(0)$, then you should be able to integrate them
to obtain $\theta(t)$... right? Although it is possible, integrating even
the simplest case ($b = u = 0$) involves elliptic integrals of the first
kind; there is relatively little intuition to be gained here. </p>
<p> This is in stark contrast to the case of linear systems, where much of
our understanding comes from being able to explicitly integrate the
equations. For instance, for a simple linear system we have $$\dot{q} = a q
\quad \rightarrow \quad q(t) = q(0) e^{at},$$ and we can immediately
understand that the long-term behavior of the system is a (stable) decaying
exponential if $a<0$, an (unstable) growing exponential if $a>0$, and that
the system does nothing if $a=0$. Here we are with certainly one of the
simplest nonlinear systems we can imagine, and we can't even solve this
system? </p>
<p> All is not lost. If what we care about is the long-term behavior of the
system, then there are a number of techniques we can apply. In this
chapter, we will start by investigating graphical solution methods. These
methods are described beautifully in a book by Steve
Strogatz<elib>Strogatz94</elib>.</p>
<subsection><h1>The overdamped pendulum</h1>
<p> Let's start by studying a special case -- intuitively when
$b\dot\theta \gg ml^2\ddot\theta$ -- which via <a
href="https://en.wikipedia.org/wiki/Dimensionless_quantity">dimensional
analysis</a> (using the natural frequency $\sqrt{\frac{g}{l}}$ to match
units) occurs when $b \sqrt\frac{l}{g} \gg ml^2$. This is the case of
heavy damping, for instance if the pendulum was moving in molasses. In
this case, the damping term dominates the acceleration term, and we have:
$$ml^2 \ddot\theta + b\dot\theta \approx b\dot\theta = u_0 -
mgl\sin\theta.$$ In other words, in the case of heavy damping, the system
looks approximately first order. This is a general property of
heavily-damped systems, such as fluids at very low Reynolds number. </p>
<p> I'd like to ignore one detail for a moment: the fact that $\theta$
wraps around on itself every $2\pi$. To be clear, let's write the system
without the wrap-around as: \begin{equation}b\dot{x} = u_0 -
mgl\sin{x}.\label{eq:overdamped_pend_ct}\end{equation} Our goal is to
understand the long-term behavior of this system: to find $x(\infty)$
given $x(0)$. Let's start by plotting $\dot{x}$ vs $x$ for the case when
$u_0=0$:</p>
<figure> <img width="70%" src="figures/pend_sinx.svg"/> </figure>
<p> The first thing to notice is that the system has a number of <em>fixed
points</em> or <em>steady states</em>, which occur whenever $\dot{x} = 0$.
In this simple example, the zero-crossings are $x^* = \{..., -\pi, 0, \pi,
2\pi, ...\}$. When the system is in one of these states, it will never
leave that state. If the initial conditions are at a fixed point, we know
that $x(\infty)$ will be at the same fixed point.</p>
<p>Next let's investigate the behavior of the system in the local vicinity
of the fixed points. Examining the fixed point at $x^* = \pi$, if the
system starts just to the right of the fixed point, then $\dot{x}$ is
positive, so the system will move away from the fixed point. If it starts
to the left, then $\dot{x}$ is negative, and the system will move away in
the opposite direction. We'll call fixed-points which have this property
<em>unstable</em>. If we look at the fixed point at $x^* = 0$, then the
story is different: trajectories starting to the right or to the left will
move back towards the fixed point. We will call this fixed point
<em>locally stable</em>. More specifically, we'll distinguish between
multiple types of stability (where $\epsilon$ is used to denote an arbitrary <em>small</em> scalar quantity):
<ul>
<li>Locally <b>stable</b> in the sense of Lyapunov (i.s.L.). A fixed
point, $x^*$ is locally stable i.s.L. if for every $\epsilon > 0$,
I can produce a $\delta > 0$ such that if $\| x(0) - x^* \| < \delta$
then $\forall t$ $\| x(t) - x^*\| < \epsilon$. In words, this means
that for any ball of size $\epsilon$ around the fixed point, I can
create a ball of size $\delta$ which guarantees that if the system is
started inside the $\delta$ ball then it will remain inside the
$\epsilon$ ball for all of time.</li>
<li>Locally <b>attractive</b>. A fixed point is locally attractive if
$x(0) = x^* + \epsilon$ implies that $\lim_{t\rightarrow \infty} x(t) =
x^*$.</li>
<li>Locally <b>asymptotically stable</b>. A fixed point is locally
asymptotically stable if it is locally stable i.s.L. and locally attractive.</li>
<li>Locally <b>exponentially stable</b>. A fixed point is locally
exponentially stable if $x(0) = x^* + \epsilon$ implies that $\| x(t) -
x^* \| < Ce^{-\alpha t}$, for some positive constants $C$ and
$\alpha$.</li>
<li><b>Unstable</b>. A fixed point is unstable if it is not locally stable i.s.L.</li>
</ul>
<p>An initial condition near a fixed point that is stable in the sense of
Lyapunov may never reach the fixed point (but it won't diverge), near an
asymptotically stable fixed point will reach the fixed point as $t
\rightarrow \infty$, and near an exponentially stable fixed point will
reach the fixed point with a bounded rate. An exponentially stable fixed
point is also an asymptotically stable fixed point, but the converse is
not true. Attractivity does not actually imply Lyapunov
stability†<sidenote>† we can't see that in one dimension so
will have to hold that example for a moment</sidenote>, which is why we
requires i.s.L. specifically for the definition of asymptotic stability.
Systems which are stable i.s.L. but not asymptotically stable are easy to
construct (e.g. $\dot{x} = 0$). Interestingly, it is also possible to have
nonlinear systems that converge (or diverge) in finite-time; a so-called
<em>finite-time stability</em>; we will see examples of this later in the
book, but it is a difficult topic to penetrate with graphical analysis.
Rigorous nonlinear system analysis is rich with subtleties and surprises.
Moreover, these differences actually matter -- the code that we will write
to stabilize the systems will be subtley different depending on what type
of stability we want, and it can make or break the success of our
methods.</p>
<p> Our graph of $\dot{x}$ vs. $x$ can be used to convince ourselves of
i.s.L. and asymptotic stability by visually inspecting $\dot{x}$ in the
vicinity of a fixed point. Even exponential stability can be inferred if
we can find a negatively-sloped line passing through the equilibrium point
which separates the graph of $f(x)$ from the horizontal axis, since it
implies that the nonlinear system will converge at least as fast as the
linear system represented by the straight line. I will graphically
illustrate unstable fixed points with open circles and stable fixed points
(i.s.L.) with filled circles. </p>
<p>Next, we need to consider what happens to initial conditions which
begin farther from the fixed points. If we think of the dynamics of the
system as a flow on the $x$-axis, then we know that anytime $\dot{x} > 0$,
the flow is moving to the right, and $\dot{x} < 0$, the flow is moving to
the left. If we further annotate our graph with arrows indicating the
direction of the flow, then the entire (long-term) system behavior becomes
clear:</p>
<figure> <img width="70%" src="figures/pend_sinx_annotated.svg"/>
</figure>
<p> For instance, we can see that any initial condition $x(0) \in
(-\pi,\pi)$ will result in $\lim_{t\rightarrow \infty} x(t) = 0$. This
region is called the <em>basin of attraction</em> of the fixed point at
$x^* = 0$. Basins of attraction of two fixed points cannot overlap, and
the manifold separating two basins of attraction is called the
<em>separatrix</em>. Here the unstable fixed points, at $x^* = \{..,
-\pi, \pi, 3\pi, ...\}$ form the separatrix between the basins of
attraction of the stable fixed points. </p>
<p> As these plots demonstrate, the behavior of a first-order one
dimensional system on a line is relatively constrained. The system will
either monotonically approach a fixed-point or monotonically move toward
$\pm \infty$. There are no other possibilities. Oscillations, for
example, are impossible. Graphical analysis is a fantastic analysis tool
for many first-order nonlinear systems (not just pendula); as illustrated
by the following example: </p>
<example><h1>Nonlinear autapse</h1>
<p></p>Consider the following system: \begin{equation} \dot{x} + x =
\tanh(w x), \end{equation} which is plotted below for two values of $w$.
It's convenient to note that $\tanh(z) \approx z$ for small $z$. For $w
\le 1$ the system has only a single fixed point. For $w > 1$ the system
has three fixed points : two stable and one unstable.</p>
<figure> <img width="80%" src="figures/pend_autapse.svg"/>
</figure>
<p>These equations are not arbitrary - they are actually a model for one
of the simplest neural networks, and one of the simplest model of
persistent memory<elib>Seung00</elib>. In the equation $x$ models the
firing rate of a single neuron, which has a feedback connection to
itself. $\tanh$ is the activation (sigmoidal) function of the neuron,
and $w$ is the weight of the synaptic feedback.</p>
<p>Experiment with it for yourself:
<jupyter no_colab>examples/autapse_and_lstm.ipynb</jupyter>
As a bonus, you'll also find a the equations of an <a href="https://en.wikipedia.org/wiki/Long_short-term_memory">LSTM</a> unit that you can also experiment with. See if you can figure it out!
</p>
</example>
<p> One last piece of terminology. In the neuron example, and in many
dynamical systems, the dynamics were parameterized; in this case by a
single parameter, $w$. As we varied $w$, the fixed points of the system
moved around. In fact, if we increase $w$ through $w=1$, something
dramatic happens - the system goes from having one fixed point to having
three fixed points. This is called a <em>bifurcation</em>. This
particular bifurcation is called a pitchfork bifurcation. We often draw
bifurcation diagrams which plot the fixed points of the system as a
function of the parameters, with solid lines indicating stable fixed
points and dashed lines indicating unstable fixed points, as seen in the
figure:</p>
<figure> <todo>bifurcation diagram asymptotes to $x^* = 1$</todo> <img
width="80%" src="figures/pend_autapse_bifurcation.svg"/>
<figcaption>Bifurcation diagram of the nonlinear autapse.</figcaption>
</figure>
<p> Our pendulum equations also have a (saddle-node) bifurcation when we
change the constant torque input, $u_0$. <!-- This is the subject of
exercise~\ref{p:pend_bifurcation}.--> Finally, let's return to the
original equations in $\theta$, instead of in $x$. Only one point to
make: because of the wrap-around, this system will <em>appear</em> to have
oscillations. In fact, the graphical analysis reveals that the pendulum
will turn forever whenever $|u_0| > mgl$, but now you understand
that this is not an oscillation, but an instability with $\theta
\rightarrow \pm \infty$. </p>
</subsection> <!-- end of overdamped pend -->
<subsection id="pend_zero_torque"><h1>The undamped pendulum with zero torque</h1>
<p> Consider again the system $$ml^2 \ddot\theta = u_0 - mgl \sin\theta -
b\dot\theta,$$ this time with $b = 0$. This time the system dynamics are
truly second-order. We can always think of any second-order system as
(coupled) first-order system with twice as many variables. Consider a
general, autonomous (not dependent on time), second-order system,
$$\ddot{q} = f(q,\dot q,u).$$ This system is equivalent to the
two-dimensional first-order system \begin{align*} \dot x_1 =& x_2 \\ \dot
x_2 =& f(x_1,x_2,u), \end{align*} where $x_1 = q$ and $x_2 = \dot q$.
Therefore, the graphical depiction of this system is not a line, but a
vector field where the vectors $[\dot x_1, \dot x_2]^T$ are plotted over
the domain $(x_1,x_2)$. This vector field is known as the <em>phase
portrait</em> of the system.</p>
<p> In this section we restrict ourselves to the simplest case when $u_0 =
0$. Let's sketch the phase portrait. First sketch along the
$\theta$-axis. The $x$-component of the vector field here is zero, the
$y$-component is $-\frac{g}{l}\sin\theta.$ As expected, we have fixed
points at $\pm \pi, ...$ Now sketch the rest of the vector field. Can you
tell me which fixed points are stable? Some of them are stable i.s.L.,
none are asymptotically stable.</p>
<figure> <img width="80%" src="figures/pend_undamped_phase.svg"/>
</figure>
<subsubsection><h1>Orbit calculations</h1>
<p> You might wonder how we drew the black contour lines in the figure
above. We could have obtained them by simulating the system
numerically, but those lines can be easily obtained in closed-form.
Directly integrating the equations of motion is difficult, but at least
for the case when $u_0 = 0$, we have some additional physical insight
for this problem that we can take advantage of. The kinetic energy,
$T$, and potential energy, $U$, of the pendulum are given by $$T =
\frac{1}{2}I\dot\theta^2, \quad U = -mgl\cos(\theta),$$ where $I=ml^2$
and the total energy is $E(\theta,\dot\theta) =
T(\dot\theta)+U(\theta)$. The undamped pendulum is a conservative
system: total energy is a constant over system trajectories. Using
conservation of energy, we have:
\begin{gather*} E(\theta(t),\dot\theta(t)) = E(\theta(0),\dot\theta(0))
= E_0 \\ \frac{1}{2} I \dot\theta^2(t) - mgl\cos(\theta(t)) = E_0 \\
\dot\theta(t) = \pm \sqrt{\frac{2}{I}\left[E_0 +
mgl\cos\left(\theta(t)\right)\right]} \end{gather*}
Using this, if you tell me $\theta$ I can determine one of two possible
values for $\dot\theta$, and the solution has all of the richness of the
black contour lines from the plot. This equation has a real solution
when $\cos(\theta) > \cos(\theta_{max})$, where $$\theta_{max} =
\begin{cases} \cos^{-1}\left( -\frac{E_0}{mgl} \right), & E_0 < mgl \\
\pi, & \text{otherwise}. \end{cases}$$ Of course this is just the
intuitive notion that the pendulum will not swing above the height where
the total energy equals the potential energy. As an exercise, you can
verify that differentiating this equation with respect to time indeed
results in the equations of motion.</p>
<p>The particular orbit defined by $E = mgl$ is special -- this is the orbit that visits the (unstable) equilibrium at the upright. It is known as the <a href="https://en.wikipedia.org/wiki/Homoclinic_orbit">homoclinic orbit</a>.</p>
</subsubsection>
<!-- these are correct, but take away from the flow. move to an appendix?
<subsubsection><h1>Trajectory calculations</h1>
<p> For completeness, I'll include what it would take to solve for $\theta(t)$,
even thought it cannot be accomplished using elementary functions. Feel free to
skip this subsection. We begin the integration with
\begin{gather*} \frac{d\theta}{dt} = \sqrt{\frac{2}{I}\left[E +
mgl\cos\left(\theta(t)\right)\right]} \\ \int_{\theta(0)}^{\theta(t)}
\frac{d\theta}{\sqrt{\frac{2}{I}\left[E +
mgl\cos\left(\theta(t)\right)\right]}} = \int_0^t dt' = t \end{gather*}
The
integral on the left side of this equation is an (incomplete) elliptic integral
of the first kind. Using the identity: $$\cos(\theta) = 1 - 2
\sin^2(\frac{1}{2}\theta),$$ and manipulating, we have $$t =
\sqrt{\frac{I}{2(E+mgl)}} \int_{\theta(0)}^{\theta(t)} \frac{d\theta}{\sqrt{1 -
k_1^2\sin^2(\frac{\theta}{2})}}, \quad \text{with
}k_1=\sqrt{\frac{2mgl}{E+mgl}}.$$ In terms of the incomplete elliptic integral
function, $$F(\phi,k) = \int_0^\phi \frac{d\theta}{\sqrt{1-k^2\sin^2\theta}},$$
accomplished by a change of variables. If $E <= mgl$, which is the case of
closed-orbits, we use the following change of variables to ensure $ 0 < k < 1 $ :
\begin{gather*}\phi = \sin^{-1}\left[ k_1 \sin\left( \frac{\theta}{2} \right)
\right] \\ \cos(\phi) d\phi = \frac{1}{2} k_1 \cos\left(\frac{\theta}{2}\right)
d\theta = \frac{1}{2} k_1 \sqrt{1 - \frac{\sin^2 (\phi)}{k_1^2}} d\theta
\end{gather*} so we have
\begin{gather*}
t = \frac{1}{k_1}\sqrt{\frac{2I}{(E+mgl)}} \int_{\phi(0)}^{\phi(t)}
\frac{d\phi}{\sqrt{1 - \sin^2(\phi)}} \frac{\cos(\phi)}{\sqrt{1 -
\frac{\sin^2\phi}{k_1^2}}} \\ = \sqrt{\frac{I}{mgl}} \left[
F\left(\phi(t),k_2\right) - F\left(\phi(0),k_2\right) \right],\quad
k_2 = \frac{1}{k_1}.\end{gather*} The inverse of $F$ is given by the
Jacobi elliptic functions (sn,cn,...), yielding:
<!-- http://en.wikipedia.org/wiki/Jacobi%27s_elliptic_functions -->
<!--
\begin{gather*}\sin(\phi(t)) = \text{sn}\left(t
\sqrt{\frac{mgl}{I}} + F\left(\phi(0),k_2\right),k_2 \right) \\
\theta(t) = 2\sin^{-1} \left[ k_2 \text{sn}\left(t
\sqrt{\frac{mgl}{I}} + F\left(\phi(0),k_2\right),k_2 \right) \right]
\end{gather*}
The function $\text{sn}$ used here can be evaluated in MATLAB by
calling $$\text{sn}(u,k) = \text{ellipj}(u,k^2).$$ The function
$F$ is not implemented in MATLAB, but implementations can be
downloaded. (note that $F(0,k) = 0$).</p>
<p>
For the open-orbit case, $E>mgl$, we use $$\phi =
\frac{\theta}{2},\quad \frac{d\phi}{d\theta} = \frac{1}{2},$$ yielding
\begin{gather*}
t = \frac{2I}{E+mgl} \int_{\phi(0)}^{\phi(t)} \frac{d\phi}{\sqrt{1 -
k_1^2 \sin^2(\phi)}} \\
\theta(t) = 2 \tan^{-1} \left[ \frac{ \text{sn} \left( t
\sqrt{\frac{E+mgl}{2I}} + F\left( \frac{\theta(0)}{2}, k_1 \right)
\right) } { \text{cn} \left( t
\sqrt{\frac{E+mgl}{2I}} + F\left( \frac{\theta(0)}{2}, k_1 \right)
\right) }
\right]
\end{gather*}
Notes: Use MATLAB's <code>atan2</code> and <code>unwrap</code> to recover the
complete trajectory.</p>
<!-- primary refs:
http://en.wikipedia.org/wiki/Pendulum_%28mathematics%29
http://kr.cs.ait.ac.th/~radok/math/mat6/calc81.htm
http://books.google.com/books?id=xWrJlTIYl_IC&pg=PA280&lpg=PA280&dq=pendulum+elliptic+integrals&source=web&ots=zas9k5qVc0&sig=Tz-OwdqS84VPM2I_pZbtaspVZNk#PPA280,M1 (Computational Physics: Problem Solving with Computers by Rubin H. Landau, Cristian C. Bordeianu, p.280)
http://en.wikipedia.org/wiki/Binomial_theorem
http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html -->
<!--</section> -->
</subsection> <!-- end of undamped pend -->
<subsection><h1>The undamped pendulum with a constant
torque</h1>
<p> Now what happens if we add a constant torque? If you visualize the
bifurcation diagram, you'll see that the fixed points come together,
towards $q = \frac{\pi}{2}, \frac{5\pi}{2}, ...$, until they disappear.
One fixed-point is unstable, and one is stable.</p>
<!-- todo: find a way to add an animation or something in here. do a
python demo in class -->
</section> <!-- undamped constant torque -->
<p>Before we continue, let me now give you the promised example of a system
that is not stable i.s.L., but which attracts all trajectories as time
goes to infinity. We can accomplish this with a very pendulum-like example
(written here in polar coordinates):</p>
<example><h1>Unstable equilibrium point that attracts all trajectories</h1>
<p>Consider the system \begin{align*} \dot{r} &= r(1-r), \\ \dot\theta &= \sin^2
\left( \frac{\theta}{2} \right).\end{align*} This system has two equilibrium points, one at
$r^*=0,\theta^*=0$, and the other at $r^* = 1,\theta^*=0$. The fixed
point at zero is clearly unstable. The fixed point with $r^*=1$ attracts
all other trajectories, but it is not stable by any of our
definitions.</p>
</example>
<p>Take a minute to draw the vector field of this (you can draw each
coordinate independently, if it helps) to make sure you understand. Note
that to wrap-around rotation is convenient but not essential -- we could
have written the same dynamical system in cartesian coordinates without
wrapping. And if this still feels too arbitrary, we will see it happen in
practice when we introduce the energy-shaping swing-up controller for the
pendulum in the next chapter.</p>
<subsection><h1>The damped pendulum</h1>
<p>Now let's add damping back. You can still add torque to move the
fixed points (in the same way).</p>
<figure>
<img width="90%" src="figures/pend_damped_phase.svg"/>
<figcaption>Phase diagram for the damped pendulum</figcaption>
</figure>
<p>With damping, the downright fixed point of the pendulum now becomes
asymptotically stable (in addition to stable i.s.L.). Is it also
exponentially stable? How can we tell? One technique is to linearize the
system at the fixed point. A smooth, time-invariant, nonlinear system
that is locally exponentially stable <i>must</i> have a stable
linearization; we'll discuss linearization more in the next chapter. </p>
<p>Here's a thought exercise. If $u$ is no longer a constant, but a
function $\pi(\theta,\dot{\theta})$, then how would you choose $\pi$ to
stabilize the vertical position. Feedback cancellation is the trivial
solution, for example: $$u = \pi(\theta,\dot{\theta}) = 2mgl\sin\theta.$$
But these plots we've been making tell a different story. How would you
shape the natural dynamics - at each point pick a $u$ from the stack of
phase plots - to stabilize the vertical fixed point <em>with minimal
torque effort</em>? This is exactly the way that I would like you to
think about control system design. And we'll give you your first solution
techniques -- using dynamic programming -- in the next lecture.</p>
</subsection> <!-- damped pend -->
</section> <!-- constant torque -->
<section><h1>The torque-limited simple pendulum</h1>
<p>The simple pendulum is fully actuated. Given enough torque, we can
produce any number of control solutions to stabilize the originally
unstable fixed point at the top (such as designing a feedback controller to
effectively invert gravity).</p>
<p>The problem begins to get interesting (a.k.a. becomes underactuated)
if we impose a torque-limit constraint, $|u|\le u_{max}$. Looking at the
phase portraits again, you can now visualize the control problem. Via
feedback, you are allowed to change the direction of the vector field at
each point, but only by a fixed amount. Clearly, if the maximum torque is
small (smaller than $mgl$), then there are some states which cannot be
driven directly to the goal, but must pump up energy to reach the goal.
Furthermore, if the torque-limit is too severe and the system has damping,
then it may be impossible to swing up to the top. The existence of a
solution, and number of pumps required to reach the top, is a non-trivial
function of the initial conditions and the torque-limits.</p>
<p>Although this system is very simple, obtaining a solution with
general-purpose algorithms requires much of the same reasoning necessary for
controlling much more complex underactuated systems; this problem will be a
work-horse for us as we introduce new algorithms throughout this book.</p>
<subsection id="energy_shaping"><h1>Energy-shaping control</h1>
<p>In the specific case of the pendulum, we can give a satisfying
hand-designed nonlinear controller based on our intuition of pumping
energy into the system. We have already observed that the total energy of
the pendulum is given by $$E = \frac{1}{2} m l^2 \dot\theta^2 -
mgl\cos\theta.$$ To understand how to control the energy, let us examine
how that energy changes with time: \begin{align*} \dot{E} =&
ml^2\dot\theta \ddot\theta + \dot\theta mgl\sin\theta \\ =& \dot\theta
\left[ u - mgl\sin\theta \right] + \dot\theta mgl\sin\theta \\ =&
u\dot\theta. \end{align*} In words, adding energy to the system is simple
- apply torque in the same direction as $\dot\theta$. To remove
energy, apply torque in the opposite direction (e.g., damping).</p>
<p>To swing up the pendulum, even with torque limits, let us use this
observation to drive the system to its homoclinic orbit, and then let the
dynamics of the pendulum carry us to the upright equilibrium. Recall that
the homoclinic orbit has energy $mgl$ -- let's call this our
<i>desired</i> energy:$$E^d = mgl.$$ Furthermore, let's define the
difference between our current energy and this desired energy as
$\tilde{E} = E - E^d$, and note that we still have $$\dot{\tilde{E}} =
\dot{E} = u\dot\theta.$$ Now consider the feedback controller of the form
$$u = -k \dot\theta \tilde{E},\quad k>0.$$ I admit that this looks a bit
strange; it was chosen for the simple reason that it turns the resulting
"error dynamics" into something simple: $$\dot{\tilde{E}} = - k
\dot\theta^2 \tilde{E}.$$ Think about the graphical analysis of this
system if you were to draw $\dot{\tilde{E}}$ vs. $\tilde{E}$ for any fixed
$\dot\theta$ -- it's a straight line separated from the horizontal axis
which would imply an exponential convergence: $\tilde{E} \rightarrow 0.$
This is true for any $\dot\theta$, except for $\dot\theta=0$ (so it will
not actually swing us up from the downright fixed point... but if you
nudge the system just a bit, then it will start pumping energy and will
swing all of the way up). The essential property is that when $E > E^d$,
we should remove energy from the system (damping) and when $E < E^d$, we
should add energy (negative damping). Even if the control actions are
bounded, the convergence is easily preserved.</p>
<p> This is a nonlinear controller that will push all system trajectories
to the unstable equilibrium. But does it make the unstable equilibrium
locally stable? With only this controller, the fixed point is <em>
attractive, but is not stable</em> -- just like our example above. Small
perturbations may cause the system to drive all of the way around the
circle in order to once again return to the unstable equilibrium. For
this reason, to actually balance the system, we'll have to switch to a
different controller once we get near the fixed point (an idea that we'll
discuss more in the next chapter).</p>
<example><h1>Energy Shaping for the Pendulum</h1>
<p>Take a minute to play around with the energy-shaping controller for
swinging up the pendulum</p>
<jupyter>examples/pendulum/energy_shaping.ipynb</jupyter>
<p>Make sure that you take a minute to look at the code which runs during
these examples. Note the somewhat arbitrary threshold for switching to
the balancing controller. We'll give a much more satisfying answer for
that in the <a href="lyapunov.html">chapter on Lyapunov methods</a>.</p>
</example>
<p>There are a few things that are extremely nice about this controller.
Not only is it simple, but it is actually incredibly robust to errors
we might have in our estimate of the model parameters, $m,$ $l,$ and $g.$
In fact, the only place that the model enters our control equation is in
the calculation of $\tilde{E} = \frac{1}{2} m l^2 \dot\theta^2 - mgl(1 +
\cos\theta)$, and the only property of this estimate that impacts
stability is the location of the orbit when $\tilde{E} = 0$, which is
$\frac{1}{2}l\dot\theta^2 = g(\cos\theta + 1).$ This doesn't depend at all
on our estimate of the mass, and only linearly on our estimates of the
length and gravity (and if one cannot measure the length of the pendulum
accurately, then a proper measuring device would make an excellent
investment). This is somewhat amazing; we will develop many
optimization-based algorithms capable of swinging up the pendulum, but it
will take a lot of work to find one that is as insensitive to the model
parameters.</p>
<p>If there is damping in the original system, of course we can cancel
this out, too, using $u = -k \dot\theta \tilde{E} + b\dot\theta.$ And
once again, the controller is quite robust if we have errors in the
estimated damping ratio.</p>
</subsection>
</section>
<section><h1>Exercises</h1>
<exercise><h1>Graphical Analysis</h1>
<figure>
<img width="60%" src="figures/exercises/graphical_analysis.svg"/>
<figcaption>First-order systems with multiple equilibria.</figcaption>
</figure>
Consider the first-order system $$\dot x = \begin{cases} - x^5 + 2 x^3 - x & \text{if} \quad x \leq 1, \\ 0 & \text{if} \quad 1 < x \leq 2, \\ - x + 2 & \text{if} \quad x > 2, \end{cases}$$ whose dynamics is represented in the figure above. Notice that, together with $x^*=-1$ and $x^*=0$, all the points in the closed interval $[1,2]$ are equilibria for this system. For each equilibrium point, determine whether it is unstable, stable i.s.L., asymptotically stable, or exponentially stable.
</exercise>
<exercise><h1>Basin of Attraction and Bifurcation Diagram</h1>
Consider the first-order system with polynomial dynamics $$\dot x = f(x) = - x^3 + 4x^2 +11x - 30.$$
<ol type="a">
<li> Sketch the graph of the function $f(x)$, and identify all the equilibrium points. (Feel free to plot this with a tool of your choice to check your work.)</li>
<li> For each equilibrium point, determine whether it is stable (i.s.L.) or unstable. For each one of the stable equilibria, identify the basin of attraction.</li>
<li> Consider an additive term $w$ in the system dynamics: $\dot x = f(x) + w$. As $w$ ranges from $0$ to $\infty$, determine how many stable and unstable equilibrium points the system has. Support your answer with a sketch of the bifurcation diagram for nonnegative values of $w$.</li>
</ol>
</exercise>
<exercise><h1>Enumerate Unstable Equilibria</h1>
We are given a dynamical system $\dot x = f(x)$ with a scalar state $x$. The dynamics $f(x)$ is unknown, but we are given two pieces of information:
<ul>
<li>the function $f(x)$ is continuous;</li>
<li>the system has exactly three stable (i.s.L.) equilibrium points.</li>
</ul>
Identify the minimum $n_{\text{min}}$ and the maximum $n_{\text{max}}$ number of unstable equilibria that the system can have. Support your answer with the sketch of two functions $f_{\text{min}}(x)$ and $f_{\text{max}}(x)$ that verify the requirements above and have, respectively, $n_{\text{min}}$ and $n_{\text{max}}$ unstable equilibria.
</exercise>
<exercise><h1>Attractivity vs Stability</h1>
<figure>
<img width="60%" src="figures/exercises/attractivity_vs_stability.svg"/>
<figcaption>Phase diagram of a two-dimensional system.</figcaption>
</figure>
This figure shows the phase portrait of the dynamical system \begin{align*} \dot x_1 &= x_1 (1-|\bx|) + x_2 \frac{x_1-|\bx|}{2|\bx|}, \\ \dot x_2 &= x_2 (1-|\bx|) - x_1 \frac{x_1-|\bx|}{2|\bx|}, \end{align*} where $|\bx| = \sqrt{x_1^2+x_2^2}$. To help you with the analysis of this system, we set up <a href="https://colab.research.google.com/github/RussTedrake/underactuated/blob/master/exercises/pend/attractivity_vs_stability/attractivity_vs_stability.ipynb" target="_blank">this python notebook</a>. Take your time to understand the code in it, and then answer the following questions.
<ol type="a">
<li>Find all the equilibrium points of this system (no need to look outside the portion of state space depicted above). Use the notebook to double check your answer: simulate the evolution of the system setting the equilibrium points you identified as initial conditions.</li>
<li>Determine whether the equilibria you identified at the previous point are attractive and/or stable i.s.L. Explain briefly your reasoning, and feel free to include some plots generated with the notebook in your written answer.</li>
<li>This dynamical system is a (very) close relative of one of the systems we analyzed in this chapter. Can you guess which one is it? What is the connection between the two? Extra points: support your claim with a mathematical derivation.</li>
</ol>
</exercise>
<exercise><h1>Pendulum with Vibrating Base</h1>
Consider an actuated pendulum whose base (pivot of the rod) is forced to oscillate horizontally according to the harmonic law $h \sin(\omega t)$, where $h$ denotes the amplitude of the oscillation and $\omega$ the frequency. The equation of motion for this system is $$m l^2 \ddot \theta + m g l \sin \theta = m l \omega^2 h \sin (\omega t) \cos \theta + u.$$ (If this equation seems obscure to you, try to derive it using the <a href="multibody.html">Lagrangian approach</a>; but be careful, the kinetic energy $T(\theta, \dot \theta, t)$ of this system depends explicitly on time.) Our goal is to design a time-dependent control law $u = \pi (\theta, \dot \theta, t)$ that makes the pendulum spin at constant velocity $\dot \theta = 1$.
<ol type="a">
<li>Identify a desired closed-loop dynamics $\ddot \theta = f(\dot \theta)$, whose unique equilibrium point is stable and is located in $\dot \theta = 1$.</li>
<li>Use feedback cancellation to design a control law $u = \pi(\theta, \dot \theta, t)$ that makes the closed-loop dynamics coincide with the one chosen at the previous point.</li>
<li><a href="https://colab.research.google.com/github/RussTedrake/underactuated/blob/master/exercises/pend/vibrating_pendulum/vibrating_pendulum.ipynb" target="_blank">In this python notebook</a>, we set up a simulation environment to test your control law. Try to understand the structure of the code: this workflow is quite general, and it could be a good starting point for your future Drake projects. Implement your control law in the dedicated cell, and check your work using the animation and the plots at the end of the notebook.</li>
</ol>
</exercise>
</section>
</chapter>
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