## Improving neural models by compensating for discrete rather than continuous filter dynamics when simulating on digital systems

This is going to be a pretty niche post, but there is some great work by Aaron Voelker from my old lab that has inspired me to do a post. The work is from an upcoming paper, which is all up on Aaron’s GitHub. It applies to building neural models using the Neural Engineering Framework (NEF). There’s a bunch of material on the NEF out there already, (e.g. the book How to Build a Brain by Dr. Chris Eliasmith, an online intro, and you can also check out Nengo, which is neural model development software with some good tutorials on the NEF) so I’m going to assume you already know the basics of the NEF for this post.

Additionally, this is applicable to simulating these models on digital systems, which, probably, most of you are. If you’re not, however! Then use standard NEF methods.

And then last note before starting, these methods are most relevant for systems with fast dynamics (relative to simulation time). If your system dynamics are pretty slow, you can likely get away with the continuous time solution if you resist change and learning. And we’ll see this in the example point attractor system at the end of the post! But even for slowly evolving systems, I would still recommend at least skipping to the end and seeing how to use the library shortcuts when coding your own models. The example code is also all up on my GitHub.

NEF modeling with continuous lowpass filter dynamics

Basic state space equations for linear time-invariant (LTI) systems (i.e. dynamics can be captured with a matrix and the matrices don’t change over time) are:

$\dot{\textbf{x}}(t) = \textbf{A}\textbf{x}(t) + \textbf{B}\textbf{u}(t)$

$\textbf{y}(t) = \textbf{C}\textbf{x}(t) + \textbf{D}\textbf{u}(t)$

where

• $\textbf{x}$ is the system state,
• $\textbf{y}$ is the system output,
• $\textbf{u}$ is the system input,
• $\textbf{A}$ is called the state matrix,
• $\textbf{B}$ is called the input matrix,
• $\textbf{C}$ is called the output matrix, and
• $\textbf{D}$ is called the feedthrough matrix,

and the system diagram looks like this:

and the transfer function, which is written in Laplace space and captures the system output over system input, for the system is

$\textbf{F}(s) = \frac{\textbf{Y}(s)}{\textbf{U}(s)} = \textbf{C}(s\textbf{I} - \textbf{A})^{-1} \textbf{B} + \textbf{D}$

where $s$ is the Laplace variable.

Now, because it’s a neural system we don’t have a perfect integrator in the middle, we instead have a synaptic filter, $H(s)$, giving:

So our goal is: given some synaptic filter $H(s)$, we want to generate some modified transfer function, $\textbf{F}'$, such that $\textbf{F}'(H(s))$ has the same dynamics as our desired system, $\textbf{F}(s)$. In other words, find an $\textbf{F}'$ such that

$\textbf{F}'\left(\frac{1}{H(s)}\right) = \textbf{F}(s).$

Alrighty. Let’s do that.

The transfer function for our neural system is

$\textbf{F}(H(s)) = \frac{\textbf{Y}(s)}{\textbf{U}(s)} = \textbf{C}(H(s)^{-1}\textbf{I} - \textbf{A})^{-1} \textbf{B} + \textbf{D}.$

The effect of the synapse is well captured by a lowpass filter, $H(s) = \frac{1}{\tau s + 1}$, making our equation

$\textbf{F}(H(s)) = \frac{\textbf{Y}(s)}{\textbf{U}(s)} = \textbf{C}((\tau s + 1)\textbf{I} - \textbf{A})^{-1} \textbf{B} + \textbf{D},$

$\textbf{F}(H(s)) = \frac{\textbf{Y}(s)}{\textbf{U}(s)} = \textbf{C}(\tau s \textbf{I} + \textbf{I} - \textbf{A})^{-1} \textbf{B} + \textbf{D}.$

To get this into a form where we can start to modify the system state matrices to compensate for the filter effects, we have to isolate $s\textbf{I}$. To do that, we can do the following basic math jujitsu

$\textbf{F}(H(s)) = \frac{\textbf{Y}(s)}{\textbf{U}(s)} = \textbf{C}(\tau (s \textbf{I} + \frac{1}{\tau}(\textbf{I} - \textbf{A}))^{-1} \textbf{B} + \textbf{D}.$

$\textbf{F}(H(s)) = \frac{\textbf{Y}(s)}{\textbf{U}(s)} = \textbf{C}(s \textbf{I} + \frac{1}{\tau}(\textbf{I} - \textbf{A})^{-1} \frac{1}{\tau}\textbf{B} + \textbf{D}.$

OK. Now, we can make $\textbf{F}'$ by substituting for our $\textbf{A}$ and $\textbf{B}$ matrices with

$\textbf{A}' = \tau\textbf{A} + \textbf{I}$

$\textbf{B}' = \tau\textbf{B}$

then we get

$\textbf{F}'(H(s)) = \frac{\textbf{Y}(s)}{\textbf{U}(s)} = \textbf{C}(s \textbf{I} + \frac{1}{\tau}(\textbf{I} - \textbf{A}')^{-1} \frac{1}{\tau}\textbf{B}' + \textbf{D}.$

$= \textbf{C}(s \textbf{I} + \frac{1}{\tau}(\textbf{I} - (\tau\textbf{A} + \textbf{I}))^{-1} \frac{1}{\tau}(\tau\textbf{B}) + \textbf{D}.$

$= \textbf{C}(s \textbf{I} + \frac{1}{\tau}(\tau\textbf{A}))^{-1}\textbf{B} + \textbf{D}.$

$= \textbf{C}(s \textbf{I} + \textbf{A})^{-1}\textbf{B} + \textbf{D}.$

and voila! We have created an $\textbf{F}'$ such that $\textbf{F}'(H(s)) = \textbf{F}(s)$. Said another way, we have created a system $(\textbf{A}', \textbf{B}', \textbf{C}'=\textbf{C}, \textbf{D}'=\textbf{D})$ that compensates for the synaptic filtering effects to achieve our desired system dynamics!

So, to compensate for the continuous lowpass filter, we use $\textbf{A}' = \tau \textbf{A} + \textbf{I}$ and $\textbf{B}' = \tau \textbf{B}$ when implementing our model and we’re golden.

And so that’s what we’ve been doing for a long time when building our models. Assuming a continuous lowpass filter and going along our merry way. Aaron, however, shrewdly noticed that computers are digital, and thusly that the standard NEF methods are not a fully accurate way of compensating for the filter that is actually being applied in simulation.

To convert our continuous system state equations to discrete state equations we need to make two changes: 1) the first is a variable change to denote the that we’re in discrete time, and we’ll use $z$ instead of $s$, and 2) we need to calculate the discrete version our system, i.e. transform $(\textbf{A}, \textbf{B}, \textbf{C}, \textbf{D}) \rightarrow (\textbf{A}_d, \textbf{B}_d, \textbf{C}_d, \textbf{D}_d)$.

The first step is easy, the second step more complicated. To discretize the system we’ll use the zero-order hold (ZOH) method (also referred to as discretization assuming zero-order hold).

Zero-order hold discretization

Zero-order hold (ZOH) systems simply hold their input over a specified amount of time. The use of ZOH here is that during discretization we assume the input control signal stays constant until the next sample time.

There are good write ups on the derivation of the discretization both on wikipedia and in these course notes from Purdue. I mostly followed the wikipedia derivation, but there were a few steps that get glossed over, so I thought I’d just write it out fully here and hopefully save someone some pain. Also for just a general intro I found these slides from Paul Oh at Drexel University really helpful.

OK. First we’ll solve an LTI system, and then we’ll discretize it.

So, you’ve got yourself a continuous LTI system

$\dot{\textbf{x}}(t) = \textbf{A}\textbf{x}(t) + \textbf{B}\textbf{u}(t)$

and you want to solve for $\textbf{x}(t)$. Rearranging things to put all the $\textbf{x}$ on one side gives

$\dot{\textbf{x}}(t) - \textbf{A}\textbf{x}(t) = \textbf{B}\textbf{u}(t).$

Looking through our identity library to find something that might help us here (after a long and grueling search) we come across:

$\frac{\partial}{\partial t} \textrm{e}^{\textbf{A}t} = \textbf{A} \textrm{e}^{\textbf{A}t} = \textrm{e}^{\textbf{A}t} \textbf{A}.$

We now left multiply our system by $\textrm{e}^{-\textbf{A}t}$ (note the negative in the exponent)

$\textrm{e}^{-\textbf{A}t}\dot{\textbf{x}}(t) - \textrm{e}^{-\textbf{A}t}\textbf{A}\textbf{x}(t) = \textrm{e}^{-\textbf{A}t}\textbf{B}\textbf{u}(t).$

Looking at this carefully, we identify the left-hand side as the result of a chain rule, so we can rewrite it as

$\frac{\partial}{\partial t} (\textrm{e}^{-\textbf{A}t}\textbf{x}(t)) = \textrm{e}^{-\textbf{A}t}\textbf{B}\textbf{u}(t).$

From here we integrate both sides, giving

$\textrm{e}^{-\textbf{A}t}\textbf{x}(t) - \textrm{e}^0\textbf{x}(0) = \int_0^t\textrm{e}^{-\textbf{A}\tau}\textbf{B}\textbf{u}(\tau) d \tau,$

$\textrm{e}^{-\textbf{A}t}\textbf{x}(t) = \int_0^t\textrm{e}^{-\textbf{A}\tau}\textbf{B}\textbf{u}(\tau) d \tau + \textbf{x}(0).$

To isolate the $\textbf{x}(t)$ term on the left-hand side now multiply by $\textrm{e}^{\textbf{A}t}$:

$\textrm{e}^{\textbf{A}t}\textrm{e}^{-\textbf{A}t}\textbf{x}(t) = \textrm{e}^{\textbf{A}t}\int_0^t\textrm{e}^{-\textbf{A}\tau}\textbf{B}\textbf{u}(\tau) d \tau + \textrm{e}^{\textbf{A}t}\textbf{x}(0),$

$\textrm{e}^{\textbf{A}t-\textbf{A}t}\textbf{x}(t) = \textrm{e}^{\textbf{A}t}\int_0^t\textrm{e}^{-\textbf{A}\tau}\textbf{B}\textbf{u}(\tau) d \tau + \textrm{e}^{\textbf{A}t}\textbf{x}(0),$

$\textbf{x}(t) = \textrm{e}^{\textbf{A}t}\int_0^t\textrm{e}^{-\textbf{A}\tau}\textbf{B}\textbf{u}(\tau) d \tau + \textrm{e}^{\textbf{A}t}\textbf{x}(0).$

OK! We solved for $\textbf{x}(t)$.

To discretize our solution we’re going to assume that we’re sampling the system at even intervals, i.e. each sample is at $kT$ for some time step $T$, and that the input $\textbf{u}(t)$ is constant between samples (this is where the ZOH comes in). To simplify our notation as we go, we also define

$\textbf{x}[k] = \textbf{x}(kT).$

So using our new notation, we have

$\textbf{x}[k] = \textrm{e}^{\textbf{A}kT}\textbf{x}(0) + \textrm{e}^{\textbf{A}kT}\int_0^{kT} \textrm{e}^{-\textbf{A}\tau}\textbf{Bu}(\tau) d\tau.$

Now we want to get things back into the form:

$\textbf{x}[k+1] = \textbf{A}_d\textbf{x}[k] + \textbf{B}_d\textbf{u}[k].$

To start, let’s write out the equation for $\textbf{x}[k + 1]$

$\textbf{x}[k+1] = \textrm{e}^{\textbf{A}(k+1)T}\textbf{x}(0) + \textrm{e}^{\textbf{A}(k+1)T}\int_0^{(k+1)T} \textrm{e}^{-\textbf{A}\tau}\textbf{Bu}(\tau) d\tau.$

We want to relate $\textbf{x}[k+1]$ to $\textbf{x}[k]$. Being incredibly clever, we see that we can left multiply $\textbf{x}[k]$ by $\textrm{e}^{\textbf{A}T}$, to get

$\textrm{e}^{\textbf{A}T}\textbf{x}[k] = \textrm{e}^{\textbf{A}(k+1)T}\textbf{x}(0) + \textrm{e}^{\textbf{A}(k+1)T}\int_0^{kT} \textrm{e}^{-\textbf{A}\tau}\textbf{Bu}(\tau) d\tau,$

and can rearrange to solve for a term we saw in $\textbf{x}[k+1]$:

$\textrm{e}^{\textbf{A}(k+1)T}\textbf{x}(0) = \textrm{e}^{\textbf{A}T}\textbf{x}[k] - \textrm{e}^{\textbf{A}(k+1)T}\int_0^{kT} \textrm{e}^{-\textbf{A}\tau}\textbf{Bu}(\tau) d\tau.$

Plugging this in, we get

$\textbf{x}[k+1] = \textrm{e}^{\textbf{A}T}\textbf{x}[k] - \textrm{e}^{\textbf{A}(k+1)T}(\int_0^{kT} \textrm{e}^{-\textbf{A}\tau}\textbf{Bu}(\tau) d\tau + \int_0^{(k+1)T} \textrm{e}^{-\textbf{A}\tau}\textbf{Bu}(\tau) d\tau),$

$\textbf{x}[k+1] = \textrm{e}^{\textbf{A}T}\textbf{x}[k] - \textrm{e}^{\textbf{A}(k+1)T}\int_{kT}^{(k+1)T} \textrm{e}^{-\textbf{A}\tau}\textbf{Bu}(\tau) d\tau.$

OK, we’re getting close.

At this point we’ve got things in the right form, but we can still clean up that second term on the right-hand side quite a bit. First, note that using our starting assumption (that $\textbf{u}(t) \in [k, kT)$ is constant), we can take $\textbf{Bu}(t)$ outside the integral.

$\textbf{x}[k+1] = \textrm{e}^{\textbf{A}T}\textbf{x}[k] - \textrm{e}^{\textbf{A}(k+1)T}\int_{kT}^{(k+1)T} \textrm{e}^{-\textbf{A}\tau}d\tau \textbf{Bu}[k].$

Next, bring that $\textrm{e}^{\textbf{A}(k+1)T}$ term inside the integral:

$\textbf{x}[k+1] = \textrm{e}^{\textbf{A}T}\textbf{x}[k] - \int_{kT}^{(k+1)T} \textrm{e}^{\textbf{A}((k+1)T - \tau)}d\tau \textbf{Bu}[k].$

And now we’re going to simplify the integral using variable substitution. Let $v = (k+1)T - \tau$, which means also that $\frac{dv}{d\tau} = -1 \rightarrow d\tau = -dv$. Evaluating the upper and lower bounds of the integral, when $\tau = (k+1)T$ then $v = 0$ and when $\tau = kT$ then $v = T$. With this, we can rewrite our equation:

$\textbf{x}[k+1] = \textrm{e}^{\textbf{A}T}\textbf{x}[k] - \int_T^0 \textrm{e}^{\textbf{A}v}dv \textbf{Bu}[k].$

The astute will notice our integral integrates from T to 0 instead of 0 to T. Fortunately for us, we know $\int_a^b x = -\int_b^a$. We can just swap the bounds and multiply by $-1$, giving:

$\textbf{x}[k+1] = \textrm{e}^{\textbf{A}T}\textbf{x}[k] + \int_0^T \textrm{e}^{\textbf{A}v}dv \textbf{Bu}[k].$

And finally, we can evaluate our integral by recalling that $\frac{d}{dt}\textrm{e}^{\textbf{A}t} = \textbf{A}\textrm{e}^{\textbf{A}t}$ and assuming that $\textbf{A}$ is invertible:

$\textbf{x}[k+1] = \textrm{e}^{\textbf{A}T}\textbf{x}[k] + \textbf{A}^{-1} \textrm{e}^{\textbf{A}v}|^T_{v=0} \textbf{Bu}[k].$

$\textbf{x}[k+1] = \textrm{e}^{\textbf{A}T}\textbf{x}[k] + \textbf{A}^{-1} (\textrm{e}^{\textbf{A}T} - \textrm{e}^0) \textbf{Bu}[k].$

$\textbf{x}[k+1] = \textrm{e}^{\textbf{A}T}\textbf{x}[k] + \textbf{A}^{-1} (\textrm{e}^{\textbf{A}T} - \textbf{I}) \textbf{Bu}[k].$

We did it! The state and input matrices for our digital system are:

$\textbf{A}_d = \textrm{e}^{\textbf{A}T}$

$\textbf{B}_d = \textbf{A}^{-1} (\textrm{e}^{\textbf{A}T} - \textbf{I}) \textbf{B}$

And that’s the hard part of discretization, the rest of the system is easy:
where, fortunately for us

$\textbf{C}_d = \textbf{C},$

$\textbf{D}_d = \textbf{D}.$

This then gives us our discrete system transfer function:

$\textbf{F}(z) = \frac{\textbf{Y}_d(z)}{\textbf{U}_d(z)} = \textbf{C}_d(z\textbf{I} - \textbf{A}_d)^{-1} \textbf{B}_d + \textbf{D}_d.$

NEF modeling with continuous lowpass filter dynamics

Now that we know how to discretize our system, we can look at compensating for the lowpass filter dynamics in discrete time. The equation for the discrete time lowpass filter is

$H(z) = \frac{1-a}{z-a},$

where $a = \textrm{e}^{-\frac{dt}{\tau}}.$

Plugging that in to the discrete transfer fuction gets us

$\textbf{F}(H(z)) = \textbf{C}_d(H(z)^{-1}\textbf{I} - \textbf{A}_d)^{-1} \textbf{B}_d + \textbf{D}_d.$

$\textbf{F}(H(z)) = \textbf{C}_d(\frac{z-a}{1-a}\textbf{I} - \textbf{A}_d)^{-1} \textbf{B}_d + \textbf{D}_d,$

$\textbf{F}(H(z)) = \textbf{C}_d(\frac{z\textbf{I}}{1-a} - \frac{a\textbf{I} - (1-a)\textbf{A}_d}{1-a})^{-1} \textbf{B}_d + \textbf{D}_d,$

$\textbf{F}(H(z)) = \textbf{C}_d(z\textbf{I} - a\textbf{I} - (1-a)\textbf{A}_d)^{-1} (1-a)\textbf{B}_d + \textbf{D}_d.$

and we see that if we choose

$\textbf{A}'_d = \frac{1}{1-a}(\textbf{A} + a\textbf{I}),$

$\textbf{B}'_d = \frac{1}{1-a}\textbf{B},$

then we get

$\textbf{F}'(H(z)) = \textbf{C}_d(z\textbf{I} - a\textbf{I} - (1-a)\textbf{A}_d')^{-1} (1-a)\textbf{B}_d' + \textbf{D}_d,$

$= \textbf{C}_d(z\textbf{I} - a\textbf{I} - (1-a)(\frac{1}{1-a}(\textbf{A} + a\textbf{I})))^{-1} (1-a)(\frac{1}{1-a}\textbf{B}) + \textbf{D}_d,$

$= \textbf{C}_d(z\textbf{I} - a\textbf{I} - \textbf{A} + a\textbf{I})^{-1} \textbf{B}) + \textbf{D}_d,$

$= \textbf{C}_d(z\textbf{I} - \textbf{A})^{-1} \textbf{B}) + \textbf{D}_d,$

And now congratulations are in order. Proper compensation for the discrete lowpass filter dynamics has finally been achieved!

Point attractor example

What difference does this actually make in modelling? Well, everyone likes examples, so let’s have one.

Here are the dynamics for a second-order point attractor system:

$\ddot{x} = \alpha(\beta(x^* - x) - \dot{x})$

with $x, \dot{x},$ and $\ddot{x}$ being the system position, velocity, and acceleration, respectively, $x^*$ is the target position, and $\alpha,$ and $\beta$ are gain values. So the acceleration is just going to be set such that it drives the system towards the target position, and compensates for velocity.

Converting this from a second order system to a first order system we have

$\left [ \begin{array}{c} \dot{x} \\ \ddot{x} \end{array} \right ] = \left [ \begin{array}{cc}0 & 1 \\ -\alpha\beta & -\alpha \end{array} \right] \left [ \begin{array}{c} x \\ \dot{x} \end{array} \right ] + \left [ \begin{array}{c}0 \\ \alpha\beta \end{array} \right] \left [ \begin{array}{c} 0 \\ x^* \end{array} \right ]$

which we’ll rewrite compactly as

$\dot{\textbf{x}} = \textbf{A} \textbf{x} + \textbf{B} \textbf{u}$

OK, we’ve got our state space equation of the dynamical system we want to implement.

Given a simulation time step $dt$, we’ll first calculate the discrete state matrices:

$\textbf{A}_d = \textrm{e}^{\textbf{A}dt},$

$\textbf{B}_d = \textbf{A}^-1 (\textrm{e}^{\textbf{A}dt} - \textbf{I})\textbf{B}.$

Great! Easy. Now we can calculate the state matrices that will compensate for the discrete lowpass filter:

$\textbf{A}_d' = \frac{1}{1-a}(\textbf{A}_d + a\textbf{I}),$

$\textbf{B}_d' = \frac{1}{1-a}\textbf{B}_d,$

where $a = \textrm{e}^{-\frac{dt}{\tau}}$.

Alright! So that’s our system now, a basic point attractor implementation in Nengo 2.3 looks like this:

tau = 0.1  # synaptic time constant

# the A matrix for our point attractor
A = np.array([[0.0, 1.0],
[-alpha*beta, -alpha]])

# the B matrix for our point attractor
B = np.array([[0.0], [alpha*beta]])

# account for discrete lowpass filter
a = np.exp(-dt/tau)
if analog:
A = tau * A + np.eye(2)
B = tau * B
else:
# discretize
Bd = np.dot(np.linalg.inv(A), np.dot((Ad - np.eye(2)), B))
A = 1.0 / (1.0 - a) * (Ad - a * np.eye(2))
B = 1.0 / (1.0 - a) * Bd

net = nengo.Network(label='Point Attractor')
net.config[nengo.Connection].synapse = nengo.Lowpass(tau)

with config, net:
net.ydy = nengo.Ensemble(n_neurons=n_neurons, dimensions=2,
# set it up so neurons are tuned to one dimensions only
encoders=nengo.dists.Choice([[1, 0], [-1, 0], [0, 1], [0, -1]]))
# set up Ax part of point attractor
nengo.Connection(net.ydy, net.ydy, transform=A)

# hook up input
net.input = nengo.Node(size_in=1, size_out=1)
# set up Bu part of point attractor
nengo.Connection(net.input, net.ydy, transform=B)

# hook up output
net.output = nengo.Node(size_in=1, size_out=1)
nengo.Connection(net.ydy[0], net.output, synapse=None)


Note that for calculating Ad we’re using expm which is the matrix exp function from scipy.linalg package. The numpy.exp does an elementwise exp, which is definitely not what we want here, and you will get some confusing bugs if you’re not careful.

Code for implementing and also testing under some different gains is up on my GitHub, and generates the following plots for dt=0.001:

In the above results you can see that the when the gains are low, and thus the system dynamics are slower, that you can’t really tell a difference between the continuous and discrete filter compensation. But! As you get larger gains and faster dynamics, the resulting effects become much more visible.

If you’re building your own system, then I also recommend using the ss2sim function from Aaron’s nengolib library. It automatically handles compensation for any synapses and generates the matrices that account for discrete or continuous implementations automatically. Using the library looks like:

tau = 0.1  # synaptic time constant
synapse = nengo.Lowpass(tau)

# the A matrix for our point attractor
A = np.array([[0.0, 1.0],
[-alpha*beta, -alpha]])

# the B matrix for our point attractor
B = np.array([[0.0], [alpha*beta]])

from nengolib.synapses import ss2sim
C = np.eye(2)
D = np.zeros((2, 2))
linsys = ss2sim((A, B, C, D),
synapse=synapse,
dt=None if analog else dt)
A = linsys.A
B = linsys.B


Conclusions

So there you are! Go forward and model free of error introduced by improperly accounting for discrete simulation! If, like me, you’re doing anything with neural modelling and motor control (i.e. systems with very quickly evolving dynamics), then hopefully you’ve found all this work particularly interesting, as I did.

There’s a ton of extensions and different directions that this work can be and has already been taken, with a bunch of really neat systems developed using this more accurate accounting for synaptic filtering as a base. You can read up on this and applications to modelling time delays and time cells and lots lots more up on Aaron’s GitHub, and hisrecent papers, which are listed on his lab webpage.

## Full body obstacle collision avoidance

PreviouslyÂ I’ve discussed how to avoid obstacles using DMPs in the end-effector trajectory. This is good when you’re controlling a single disconnected point-mass, like a mobile robot navigating around an environment. But if you want to use this system to control a robotic manipulator, then pretty quickly you run into the problem that your end-effector is not a disconnected point-mass moving around the environment. It’s attached to the rest of the arm, and moving such thatÂ the arm segments and jointsÂ also avoid the obstacle is a whole other deal.

I was doing a quick lit scan on methods for control methods for avoiding joint collision with obstacles, and was kind of surprised that there wasn’t really much in the realm of recent discussions about it. There is, however, a 1986 paper from Dr. Oussama Khatib titled Real-time obstacle avoidance for manipulators and mobile robots that pretty much solves this problem short of getting into explicit path planning methods. Which could be why there aren’t papers since then about it. All the same, I couldn’t find any implementations of it online, and there are a few tricky parts, so in this post I’m going to work through the implementation.

Note: The implementation that I’ve worked out here uses spheres to represent the obstacles. This works pretty well in general by just making the sphere large enough to cover whatever obstacle you’re avoiding. But if you want a more precise control around other shapes, you can check out the appendix of Dr. Khatib’s paper, where he works provides the math for cubes and cones as well.

Note: You can find all of the code I use here and everything you need for the VREP implementation up on my GitHub. I’ve precalculated the functions that are required, because generating them is quite computationally intensive. Hopefully the file saving doesn’t get weird between different operating systems, but if it does, try deleting all of the files in the ur5_config folder and running the code again. This will regenerate those files (on my laptop this takes ~4 hours, though, so beware).

The general idea

Since it’s from Dr. Khatib, you might expect that this approach involves task space. And indeed, your possible suspicions are correct! The algorithm is going to work by identifying forces in Cartesian coordinates that will move any point of the arm that’s too close to an obstacle away from it. The algorithm follows the form:

Setup

• Specify obstacle location and size
• Specify a threshold distance to the obstacle

While running

• Find the closest point of each arm segment to obstacles
• If within threshold of obstacle, generate force to apply to that point
• Transform this force into joint torques
• Add directly to the outgoing control signal

Something that you might notice about this is that it’s similar to the addition of the null space controller that we’ve seen before in operational space control. There’s a distinct difference here though, in that the control signal for avoiding obstacles is added directly to the outgoing control signal, and that it’s not filtered (like the null space control signal) such that there’s no guarantee that it won’t affect the movement of the end-effector. In fact, it’s very likely to affect the movement of the end-effector, but that’s also desirable, as not ramming the arm through an obstacle is as important as getting to the target.

OK, let’s walk through these steps one at a time.

Setup

I mentioned that we’re going to treat all of our obstacles as spheres. It’s actually not much harder to do these calculations for cubes too, but this is already going to be long enough so I’m only addressing sphere’s here. This algorithm assumes we have a list of every obstacle’s centre-point and radius.

We want the avoidance response of the system to be a function of the distance to the obstacle, such that the closer the arm is to the obstacle the stronger the response. The function that Dr. Khatib provides is of the following form:

$\textbf{F}_{psp} = \left\{ \begin{array}{cc}\eta (\frac{1.0}{\rho} - \frac{1}{\rho_0}) \frac{1}{\rho^2} \frac{\partial \rho}{\partial \textbf{x}} & \rho \leq \rho_0 \\ \textbf{0} & \rho > \rho_0 \end{array} \right. ,$

where $\rho$ is the distance to the target, $\rho_0$ is the threshold distance to the target at which point the avoidance function activates, $\frac{\partial \rho}{\partial \textbf{x}}$ is the partial derivative of the distance to the target with respect to a given point on the arm, and $\eta$ is a gain term.

This function looks complicated, but it’s actually pretty intuitive. The partial derivative term in the function works simply to point in the opposite direction of the obstacle, in Cartesian coordinates, i.e. tells the system how to get away from the obstacle. The rest of the term is just a gain that starts out at zero when $\rho = \rho_0$, and gets huge as the obstacle nears the object (as $\rho \to 0 \Rightarrow \frac{1}{\rho} \to \infty$). Using $\eta = .2$ and $\rho_0 = .2$ gives us a function that looks like

So you can see that very quickly a very, very strong push away from this obstacle is going to be generated once we enter the threshold distance. But how do we know exactly when we’ve entered the threshold distance?

Find the closest point

We want to avoid the obstacle with our whole body, but it turns out we can reduce the problem to only worrying about the closest point of each arm segment to the obstacle, and move that one point away from the obstacle if threshold distance is hit.

To find the closest point on a given arm segment to the obstacle we’ll do some pretty neat trig. I’ll post the code for it and then discuss it below. In this snippet, p1 and p2 are the beginning and ending $(x,y,z)$ locations of arm segment (which we are assuming is a straight line), and v is the center of the obstacle.

# the vector of our line
vec_line = p2 - p1
# the vector from the obstacle to the first line point
vec_ob_line = v - p1
# calculate the projection normalized by length of arm segment
projection = (np.dot(vec_ob_line, vec_line) /
np.dot(vec_line, vec_line))
if projection < 0:
# then closest point is the start of the segment
closest = p1
elif projection > 1:
# then closest point is the end of the segment
closest = p2
else:
closest = p1 + projection * vec_line


The first thing we do is find the arm segment line, and then line from the obstacle center to the start point of the arm segment. Once we have these, we do:

$\frac{\textbf{v}_\textrm{ob\_line} \; \cdot \; \textbf{v}_\textrm{line}}{\textbf{v}_\textrm{line} \; \cdot \; \textbf{v}_\textrm{line}},$

using the geometric definition of the dot product two vectors, we can rewrite the above as

$\frac{||\textbf{v}_\textrm{ob\_line}|| \; || \textbf{v}_\textrm{line} || \; \textrm{cos}(\theta)}{||\textbf{v}_\textrm{line}||^2} = \frac{||\textbf{v}_\textrm{ob\_line}||} {||\textbf{v}_\textrm{line}||} \textrm{cos}(\theta)$

which reads as the magnitude of vec_ob_line divided by the magnitude of vec_line (I know, these are terrible names, sorry) multiplied by the angle between the two vectors. If the angle between the vectors is < 0 (projection will also be < 0), then right off the bat we know that the start of the arm segment, p1, is the closest point. If the projection value is > 1, then we know that 1) the length from the start of the arm segment to the obstacle is longer than the length of the arm, and 2) the angle is such that the end of the arm segment, p2, is the closest point to the obstacle.

Finally, in the last case we know that the closest point is somewhere along the arm segment. To find where exactly, we do the following

$\textbf{p}_1 + \textrm{projection} \; \textbf{v}_\textrm{line},$

which can be rewritten

$\textbf{p}_1 + \frac{||\textbf{v}_\textrm{ob\_line}||} {||\textbf{v}_\textrm{line}||} \textrm{cos}(\theta) \; \textbf{v}_\textrm{line},$

I find it more intuitive if we rearrange the second term to be

$\textbf{p}_1 + \frac{\textbf{v}_\textrm{line}} {||\textbf{v}_\textrm{line}||} \; ||\textbf{v}_\textrm{ob\_line} || \; \textrm{cos}(\theta).$

So then what this is all doing is starting us off at the beginning of the arm segment, p1, and to that we add this other fun term. The first part of this fun term provides direction normalized to magnitude 1. The second part of this term provides magnitude, specifically the exact distance along vec_line we need to traverse to form reach a right angle (giving us the shortest distance) pointing at the obstacle. This handy figure from the Wikipedia page helps illustrate exactly what’s going on with the second part, where B is be vec_line and A is vec_ob_line:

Armed with this information, we understand how to find the closest point of each arm segment to the obstacle, and we are now ready to generate a force to move the arm in the opposite direction!

Check distance, generate force

To calculate the distance, all we have to do is calculate the Euclidean distance from the closest point of the arm segment to the center of the sphere, and then subtract out the radius of the sphere:

# calculate distance from obstacle vertex to the closest point
dist = np.sqrt(np.sum((v - closest)**2))
# account for size of obstacle
rho = dist - obstacle[3]


Once we have this, we can check it and generate $F_{psp}$ using the equation we defined above. The one part of that equation that wasn’t specified was exactly what $\frac{\partial \rho}{\partial \textbf{x}}$ was. Since it’s just the partial derivative of the distance to the target with respect to the closest point, we can calculate it as the normalized difference between the two points:

drhodx = (v - closest) / rho


Alright! Now we’ve found the closest point, and know the force we want to apply, from here it’s standard operational space procedure.

Transform the force into torques

As we all remember, the equation for transforming a control signal from operational space to involves two terms aside from the desired force. Namely, the Jacobian and the operational space inertia matrix:

$\textbf{u}_\textrm{psp} = \textbf{J}^T_{psp} \textbf{M}_{psp} \textbf{F}_{psp},$

where $\textbf{J}_{psp}$ is the Jacobian for the point of interest, $\textbf{M}_{psp}$ is the operational space inertia matrix for the point of interest, and $\textbf{F}_{psp}$ is the force we defined above.

Calculating the Jacobian for an unspecified point

So the first thing we need to calculate is the Jacobian for this point on the arm. There are a bunch of ways you could go about this, but the way I’m going to do it here is by building on the post where I used SymPy to automate the Jacobian derivation. The way we did that was by defining the transforms from the origin reference frame to the first link, from the first link to the second, etc, until we reached the end-effector. Then, whenever we needed a Jacobian we could string together the transforms to get the transform from the origin to that point on the arm, and take the partial derivative with respect to the joints (using SymPy’s derivative method).

As an example, say we wanted to calculate the Jacobian for the third joint, we would first calculate:

$^3_{\textrm{org}}\textbf{T} = ^0_{\textrm{org}}\textbf{T} \; ^1_0\textbf{T} \; ^2_1\textbf{T} \; ^3_2\textbf{T},$

where $^m_n\textbf{T}$ reads the transform from reference frame $m$ to reference frame $n$.

Once we have this transformation matrix, $^3_\textrm{org}\textbf{T}$, we multiply it by the point of interest in reference frame 3, which, previously, has usually been $\textbf{x} = [0, 0, 0]$. In other words, usually we’re just interested in the origin of reference frame 3. So the Jacobian is just

$\frac{\partial \; ^3_\textrm{org}\textbf{T} \textbf{x}}{\partial \textbf{q}}.$

what if we’re interested in some non-specified point along link 3, though? Well, using SymPy we set make $\textbf{x} = [x_0, x_1, x_2, 1]$ instead of $\textbf{x} = [0, 0, 0, 1]$ (recall the 1 at the end in these vectors is just to make the math work), and make the Jacobian function SymPy generates for us dependent on both $\textbf{q}$ and $\textbf{x}$, rather than just $\textbf{q}$. In code this looks like:

Torg3 = self._calc_T(name="3")
# transform x into world coordinates
Torg3x = sp.simplify(Torg3 * sp.Matrix(x))
J3_func = sp.lambdify(q + x, Torg3)


Now it’s possible to calculate the Jacobian for any point along link 3 just by changing the parameters that we pass into J3_func! Most excellent.

We are getting closer.

NOTE: This parameterization can significantly increase the build time of the function, it took my laptop about 4 hours. To decrease build time you can try commenting out the simplify calls from the code, which might slow down run-time a bit but significantly drops the generation time.

Where is the closest point in that link’s reference frame?

A sneaky problem comes up when calculating the closest point of each arm segment to the object: We’ve calculated the closest point of each arm segment in the origin’s frame of reference, and we need thew relative to each link’s own frame of reference. Fortunately, all we have to do is calculate the inverse transform for the link of interest. For example, the inverse transform of $^3_\textrm{org}\textbf{T}$ transforms a point from the origin’s frame of reference to the reference frame of the 3rd joint.

I go over how to calculate the inverse transform at the end of my post on forward transformation matrices, but to save you from having to go back and look through that, here’s the code to do it:

Torg3 = self._calc_T(name="3")
rotation_inv = Torg3[:3, :3].T
translation_inv = -rotation_inv * Torg3[:3, 3]
Torg3_inv = rotation_inv.row_join(translation_inv).col_join(
sp.Matrix([[0, 0, 0, 1]]))


And now to find the closest point in the coordinates of reference frame 3 we simply

x = np.dot(Torg3_inv, closest)


This x value is what we’re going to plug in as parameters to our J3_func above to find the Jacobian for the closest point on link 3.

Calculate the operational space inertia matrix for the closest point

OK. With the Jacobian for the point of interest we are now able to calculate the operational space inertia matrix. This code I’ve explicitly worked through before, and I’ll show it in the full code below, so I won’t go over it again here.

The whole implementation

You can run an example of all this code controlling the UR5 arm to avoid obstacles in VREP using this code up on my GitHub. The specific code added to implement obstacle avoidance looks like this:

# find the closest point of each link to the obstacle
for ii in range(robot_config.num_joints):
# get the start and end-points of the arm segment
p1 = robot_config.Tx('joint%i' % ii, q=q)
if ii == robot_config.num_joints - 1:
p2 = robot_config.Tx('EE', q=q)
else:
p2 = robot_config.Tx('joint%i' % (ii + 1), q=q)

# calculate minimum distance from arm segment to obstacle
# the vector of our line
vec_line = p2 - p1
# the vector from the obstacle to the first line point
vec_ob_line = v - p1
# calculate the projection normalized by length of arm segment
projection = (np.dot(vec_ob_line, vec_line) /
np.sum((vec_line)**2))
if projection < 0:
# then closest point is the start of the segment
closest = p1
elif projection > 1:
# then closest point is the end of the segment
closest = p2
else:
closest = p1 + projection * vec_line
# calculate distance from obstacle vertex to the closest point
dist = np.sqrt(np.sum((v - closest)**2))
# account for size of obstacle

if rho < threshold:
eta = .02
drhodx = (v - closest) / rho
Fpsp = (eta * (1.0/rho - 1.0/threshold) *
1.0/rho**2 * drhodx)

# get offset of closest point from link's reference frame
T_inv = robot_config.T_inv('link%i' % ii, q=q)
m = np.dot(T_inv, np.hstack([closest, [1]]))[:-1]
# calculate the Jacobian for this point
Jpsp = robot_config.J('link%i' % ii, x=m, q=q)[:3]

# calculate the inertia matrix for the
# point subjected to the potential space
Mxpsp_inv = np.dot(Jpsp,
np.dot(np.linalg.pinv(Mq), Jpsp.T))
svd_u, svd_s, svd_v = np.linalg.svd(Mxpsp_inv)
# cut off singular values that could cause problems
singularity_thresh = .00025
for ii in range(len(svd_s)):
svd_s[ii] = 0 if svd_s[ii] < singularity_thresh else \
1./float(svd_s[ii])
# numpy returns U,S,V.T, so have to transpose both here
Mxpsp = np.dot(svd_v.T, np.dot(np.diag(svd_s), svd_u.T))

u_psp = -np.dot(Jpsp.T, np.dot(Mxpsp, Fpsp))
if rho < .01:
u = u_psp
else:
u += u_psp


The one thing in this code I didn’t talk about is that you can see that if rho < .01 then I set u = u_psp instead of just adding u_psp to u. What this does is basically add in a fail safe take over of the robotic control saying that “if we are about to hit the obstacle forget about everything else and get out of the way!”.

Results

And that’s it! I really enjoy how this looks when it’s running, it’s a really effective algorithm.Â Let’s look at some samples of it in action.

First, in a 2D environment, where it’s real easy to move around the obstacle and see how it changes in response to the new obstacle position. The red circle is the target and the blue circle is the obstacle:

And in 3D in VREP, running the code example that I’ve put up on my GitHub implementing this. The example of it running without obstacle avoidance code is on the left, and running with obstacle avoidance is on the right. It’s kind of hard to see but on the left the robot moves through the far side of the obstacle (the gold sphere) on its way to the target (the red sphere):

And one last example, the arm dodging a moving obstacleÂ on its way to the target.

The implementation is a ton of fun to play around with. It’s a really nifty algorithm, that works quite well, and I haven’t found many alternatives in papers that don’t go into path planning (if you know of some and can share that’d be great!). This post was a bit of a journey, but hopefully you found it useful! I continue to find it impressive how many different neat features like this can come about once you have the operational space control framework in place.

## Using SymPy’s lambdify for generating transform matrices and Jacobians

I’ve been working in VREP with some of their different robot models and testing out force control, and one of the things that becomes pretty important for efficient workflow is to have a streamlined method for setting up the transform matrices and calculating the Jacobians for different robots. You do not want to be working these all out by hand and then writing them in yourself.

A solution that’s been working well for me (and is fully implemented in Python) is to use SymPy to set up the basic transform matrices, from each joint to the next, and then use it’s derivative function to calculate the Jacobian. Once this is calculated you can then use SymPy’s lambdify function to parameterize this, and off you go! In this post I’m going to work through an example for controlling VREP’s UR5 arm using force control. And if you’re just looking for code examples, you can find it all up on my GitHub.

Edit: Updated the code to be much nicer, added saving of calculated functions, and a null space controller.
Edit 2: Removed the simplify calls from the code, not worth the generation time increase (if execution is a great concern can generate the functions using cython).

Setting up the transform matrices

This is the part of this process that is unique to each arm. The goal is to set up a system so that you can specify your transforms for each joint (and to each centre-of-mass (COM) too, of course) and then be on your way. So here’s the UR5, cute thing that it is:

For the UR5, there are 6 joints, and we’re going to be specifying 9 transform matrices: 6 joints and the COM for three arm segments (the remaining arm segment COMs are centered at their respective joint’s reference frame). The joints are all rotary, with 0 and 4 rotating around the z-axis, and the rest all rotating around x.

So first, we’re going to create a class called robot_config. Then, to create our transform matrices in SymPy first we need to set up the variables we’re going to be using:

# set up our joint angle symbols (6th angle doesn't affect any kinematics)
self.q = [sp.Symbol('q%i'%ii) for ii in range(6)]
# segment lengths associated with each joint
self.L = np.array([0.0935, .13453, .4251, .12, .3921, .0935, .0935, .0935])


where self.q is an array storing all our joint angle symbols, and self.L is an array of all of the offsets for each joint and arm segment lengths.

Using these to create the transform matrices for the joints, we get a set up that looks like this:

# transform matrix from origin to joint 0 reference frame
self.T0org = sp.Matrix([[sp.cos(self.q[0]), -sp.sin(self.q[0]), 0, 0],
[sp.sin(self.q[0]), sp.cos(self.q[0]), 0, 0],
[0, 0, 1, self.L[0]],
[0, 0, 0, 1]])

# transform matrix from joint 0 to joint 1 reference frame
self.T10 = sp.Matrix([[1, 0, 0, -L[1]],
[0, sp.cos(-self.q[1] + sp.pi/2),
-sp.sin(-self.q[1] + sp.pi/2), 0],
[0, sp.sin(-self.q[1] + sp.pi/2),
sp.cos(-self.q[1] + sp.pi/2), 0],
[0, 0, 0, 1]])

# transform matrix from joint 1 to joint 2 reference frame
self.T21 = sp.Matrix([[1, 0, 0, 0],
[0, sp.cos(-self.q[2]),
-sp.sin(-self.q[2]), self.L[2]],
[0, sp.sin(-self.q[2]),
sp.cos(-self.q[2]), 0],
[0, 0, 0, 1]])

# transform matrix from joint 2 to joint 3
self.T32 = sp.Matrix([[1, 0, 0, L[3]],
[0, sp.cos(-self.q[3] - sp.pi/2),
-sp.sin(-self.q[3] - sp.pi/2), self.L[4]],
[0, sp.sin(-self.q[3] - sp.pi/2),
sp.cos(-self.q[3] - sp.pi/2), 0],
[0, 0, 0, 1]])

# transform matrix from joint 3 to joint 4
self.T43 = sp.Matrix([[sp.sin(-self.q[4] - sp.pi/2),
sp.cos(-self.q[4] - sp.pi/2), 0, -self.L[5]],
[sp.cos(-self.q[4] - sp.pi/2),
-sp.sin(-self.q[4] - sp.pi/2), 0, 0],
[0, 0, 1, 0],
[0, 0, 0, 1]])

# transform matrix from joint 4 to joint 5
self.T54 = sp.Matrix([[1, 0, 0, 0],
[0, sp.cos(self.q[5]), -sp.sin(self.q[5]), 0],
[0, sp.sin(self.q[5]), sp.cos(self.q[5]), self.L[6]],
[0, 0, 0, 1]])

# transform matrix from joint 5 to end-effector
self.TEE5 = sp.Matrix([[1, 0, 0, self.L[7]],
[0, 1, 0, 0],
[0, 0, 1, 0],
[0, 0, 0, 1]])


You can see a bunch of offsetting of the joint angles by -sp.pi/2 and this is to account for the expected 0 angle (straight along the reference frame’s x-axis) at those joints being different than the 0 angle defined in the VREP simulation (at a 90 degrees from the x-axis). You can find these by either looking at and finding the joints’ 0 position in VREP or byÂ trial-and-error empirical analysis.

Once you have your transforms, then you have to specify how to move from the origin to each reference frame of interest (the joints and COMs). For that, I’ve just set up a simple function with a switch statement:

# point of interest in the reference frame (right at the origin)
self.x = sp.Matrix([0,0,0,1])

def _calc_T(self, name, lambdify=True):
""" Uses Sympy to generate the transform for a joint or link

name string: name of the joint or link, or end-effector
lambdify boolean: if True returns a function to calculate
the transform. If False returns the Sympy
matrix
"""

# check to see if we have our transformation saved in file
if os.path.isfile('%s/%s.T' % (self.config_folder, name)):
Tx = cloudpickle.load(open('%s/%s.T' % (self.config_folder, name),
'rb'))
else:
if name == 'joint0' or name == 'link0':
T = self.T0org
elif name == 'joint1' or name == 'link1':
T = self.T0org * self.T10
elif name == 'joint2':
T = self.T0org * self.T10 * self.T21
T = self.T0org * self.T10 * self.Tl21
elif name == 'joint3':
T = self.T0org * self.T10 * self.T21 * self.T32
T = self.T0org * self.T10 * self.T21 * self.Tl32
elif name == 'joint4' or name == 'link4':
T = self.T0org * self.T10 * self.T21 * self.T32 * self.T43
elif name == 'joint5' or name == 'link5':
T = self.T0org * self.T10 * self.T21 * self.T32 * self.T43 * \
self.T54
elif name == 'link6' or name == 'EE':
T = self.T0org * self.T10 * self.T21 * self.T32 * self.T43 * \
self.T54 * self.TEE5
Tx = T * self.x  # to convert from transform matrix to (x,y,z)

# save to file
cloudpickle.dump(Tx, open('%s/%s.T' % (self.config_folder, name),
'wb'))

if lambdify is False:
return Tx
return sp.lambdify(self.q, Tx)


So the first part is pretty straight forward, create the transform matrix, and then at the end to get the (x,y,z) position we just multiply by a vector we created that represents a point at the origin of the last reference frame. Some of the transform matrices (the ones to the arm segments) I didn’t specify above just to cut down on space.

The second part is where we use this awesome lambify function, which lets us turn the matrix we’ve defined into a function, so that we can pass in joint angles and get back out the resulting (x,y,z) position. There’s also the option to get the straight up SymPy matrix return, in case you need the symbolic form (which we will!).

NOTE: You can also see that there’s a check built in to look for saved files, and to just load those saved files instead of recalculating things if they’re available. This is because calculating some of these matrices and their derivatives takes a long, long time. I used the cloudpickle module to do this because it’s able to easily handle saving a whole bunch of weird things that makes normal pickle sour.

Calculating the Jacobian

So now that we’re able to quickly generate the transform matrix for each point of interest on the UR5, we simply take the derivative of the equation for each (x,y,z) coordinate with respect to each joint angle to generate our Jacobian.

def _calc_J(self, name, lambdify=True):
""" Uses Sympy to generate the Jacobian for a joint or link

name string: name of the joint or link, or end-effector
lambdify boolean: if True returns a function to calculate
the Jacobian. If False returns the Sympy
matrix
"""

# check to see if we have our Jacobian saved in file
if os.path.isfile('%s/%s.J' % (self.config_folder, name)):
(self.config_folder, name), 'rb'))
else:
Tx = self._calc_T(name, lambdify=False)
J = []
# calculate derivative of (x,y,z) wrt to each joint
for ii in range(self.num_joints):
J.append([])
J[ii].append(Tx[0].diff(self.q[ii]))  # dx/dq[ii]
J[ii].append(Tx[1].diff(self.q[ii]))  # dy/dq[ii]
J[ii].append(Tx[2].diff(self.q[ii]))  # dz/dq[ii]


Here we retrieve the Tx vector from our _calc_T function, and then calculate the derivatives. When calculating the Jacobian for the end-effector, this is all we need! Huzzah!

But to calculate the Jacobian for transforming the inertia matrices of each arm segment into joint space we’re going to need the orientation information added to our Jacobian as well. This we know ahead of time, for each joint it’s a 3D vector with a 1 on the axis being rotated around. So we can predefine this:

# orientation part of the Jacobian (compensating for orientations)
self.J_orientation = [
[0, 0, 1], # joint 0 rotates around z axis
[1, 0, 0], # joint 1 rotates around x axis
[1, 0, 0], # joint 2 rotates around x axis
[1, 0, 0], # joint 3 rotates around x axis
[0, 0, 1], # joint 4 rotates around z axis
[1, 0, 0]] # joint 5 rotates around x axis


And then we just fill in the Jacobians for each reference frame with self.J_orientation up to the last joint, and fill in the rest of the Jacobian with zeros. So e.g. when we’re calculating the Jacobian for the arm segment just past the second joint we’ll use the first two rows of self.J_orientation and the rest of the rows will be 0.

So this leads us to the second half of the _calc_J function:

        end_point = name.strip('link').strip('joint')
if end_point != 'EE':
end_point = min(int(end_point) + 1, self.num_joints)
# add on the orientation information up to the last joint
for ii in range(end_point):
J[ii] = J[ii] + self.J_orientation[ii]
# fill in the rest of the joints orientation info with 0
for ii in range(end_point, self.num_joints):
J[ii] = J[ii] + [0, 0, 0]

# save to file
cloudpickle.dump(J, open('%s/%s.J' %
(self.config_folder, name), 'wb'))

J = sp.Matrix(J).T  # correct the orientation of J
if lambdify is False:
return J
return sp.lambdify(self.q, J)


The orientation information is added in, we save the result to file, and a function that takes in the joint angles and outputs the Jacobian is created (unless lambdify == False in which case the SymPy symbolic form is returned.)

Then finally, two wrapper functions are added in to make creating / accessing these functions easier. First, define a couple of dictionaries

# create function dictionaries
self._T = {}  # for transform calculations
self._J = {}  # for Jacobian calculations


and then our wrapper functions look like this

def T(self, name, q):
""" Calculates the transform for a joint or link
name string: name of the joint or link, or end-effector
q np.array: joint angles
"""
# check for function in dictionary
if self._T.get(name, None) is None:
print('Generating transform function for %s'%name)
self._T[name] = self.calc_T(name)
return self._T[name](*q)[:-1].flatten()

def J(self, name, q):
""" Calculates the transform for a joint or link
name string: name of the joint or link, or end-effector
q np.array: joint angles
"""
# check for function in dictionary
if self._J.get(name, None) is None:
print('Generating Jacobian function for %s'%name)
self._J[name] = self.calc_J(name)
return np.array(self._J[name](*q)).T


So how you use this class (all of this is in a class) is to call these T and J functions with the current joint angles. They’ll check to see if the functions have already be created or stored in file, if they haven’t then the T and / or J functions will be created, then our wrappers do a bit of formatting to get them into the proper form (i.e. transposing or cropping), and return you your (x,y,z) or Jacobian!

NOTE: It’s a bit of a misnomer to have the function be called T and actually return to you Tx, but hey this is a blog. Lay off.

Calculating the inertia matrix in joint-space and force of gravity
Now, since we’re here we might as well also calculate the functions for our inertia matrix in joint space and the effect of gravity. So, define a couple more placeholders in our robot_config class to help us:

self._M = []  # placeholder for (x,y,z) inertia matrices
self._Mq = None  # placeholder for joint space inertia matrix function
self._Mq_g = None  # placeholder for joint space gravity term function


and then add in our inertia matrix information (defined in each link’s centre-of-mass (COM) reference frame)

# create the inertia matrices for each link of the ur5
self._M.append(np.diag([1.0, 1.0, 1.0,
self._M.append(np.diag([2.5, 2.5, 2.5,
self._M.append(np.diag([5.7, 5.7, 5.7,
self._M.append(np.diag([3.9, 3.9, 3.9,
self._M.append(np.diag([0.7, 0.7, 0.7,


and then using our equations for calculating the system’s inertia and gravity we create our _calc_Mq and _calc_Mq_g functions

def _calc_Mq(self, lambdify=True):
""" Uses Sympy to generate the inertia matrix in
joint space for the ur5

lambdify boolean: if True returns a function to calculate
the Jacobian. If False returns the Sympy
matrix
"""

# check to see if we have our inertia matrix saved in file
if os.path.isfile('%s/Mq' % self.config_folder):
Mq = cloudpickle.load(open('%s/Mq' % self.config_folder, 'rb'))
else:
# get the Jacobians for each link's COM
J = [self._calc_J('link%s' % ii, lambdify=False)

# transform each inertia matrix into joint space
# sum together the effects of arm segments' inertia on each motor
Mq = sp.zeros(self.num_joints)
Mq += J[ii].T * self._M[ii] * J[ii]

# save to file
cloudpickle.dump(Mq, open('%s/Mq' % self.config_folder, 'wb'))

if lambdify is False:
return Mq
return sp.lambdify(self.q, Mq)

def _calc_Mq_g(self, lambdify=True):
""" Uses Sympy to generate the force of gravity in
joint space for the ur5

lambdify boolean: if True returns a function to calculate
the Jacobian. If False returns the Sympy
matrix
"""

# check to see if we have our gravity term saved in file
if os.path.isfile('%s/Mq_g' % self.config_folder):
'rb'))
else:
# get the Jacobians for each link's COM
J = [self._calc_J('link%s' % ii, lambdify=False)

# transform each inertia matrix into joint space and
# sum together the effects of arm segments' inertia on each motor
Mq_g = sp.zeros(self.num_joints, 1)
for ii in range(self.num_joints):
Mq_g += J[ii].T * self._M[ii] * self.gravity

# save to file
cloudpickle.dump(Mq_g, open('%s/Mq_g' % self.config_folder,
'wb'))

if lambdify is False:
return Mq_g
return sp.lambdify(self.q, Mq_g)


and wrapper functions

def Mq(self, q):
""" Calculates the joint space inertia matrix for the ur5

q np.array: joint angles
"""
# check for function in dictionary
if self._Mq is None:
print('Generating inertia matrix function')
self._Mq = self._calc_Mq()
return np.array(self._Mq(*q))

def Mq_g(self, q):
""" Calculates the force of gravity in joint space for the ur5

q np.array: joint angles
"""
# check for function in dictionary
if self._Mq_g is None:
print('Generating gravity effects function')
self._Mq_g = self._calc_Mq_g()
return np.array(self._Mq_g(*q)).flatten()


and we’re all set!

Putting it all together

Now we have nice clean code to generate everything we need for our controller. Using the controller developed in this post as a base, we can replace those calculations with the following nice compact code (which also includes a secondary null-space controller to keep the arm near resting joint angles):

# calculate position of the end-effector
# derived in the ur5 calc_TnJ class
xyz = robot_config.T('EE', q)
# calculate the Jacobian for the end effector
JEE = robot_config.J('EE', q)
# calculate the inertia matrix in joint space
Mq = robot_config.Mq(q)
# calculate the effect of gravity in joint space
Mq_g = robot_config.Mq_g(q)

# convert the mass compensation into end effector space
Mx_inv = np.dot(JEE, np.dot(np.linalg.inv(Mq), JEE.T))
svd_u, svd_s, svd_v = np.linalg.svd(Mx_inv)
# cut off any singular values that could cause control problems
singularity_thresh = .00025
for i in range(len(svd_s)):
svd_s[i] = 0 if svd_s[i] < singularity_thresh else \
1./float(svd_s[i])
# numpy returns U,S,V.T, so have to transpose both here
Mx = np.dot(svd_v.T, np.dot(np.diag(svd_s), svd_u.T))

kp = 100
kv = np.sqrt(kp)
# calculate desired force in (x,y,z) space
u_xyz = np.dot(Mx, target_xyz - xyz)
# transform into joint space, add vel and gravity compensation
u = (kp * np.dot(JEE.T, u_xyz) - np.dot(Mq, kv * dq) - Mq_g)

# calculate our secondary control signal
# calculated desired joint angle acceleration
q_des = (((robot_config.rest_angles - q) + np.pi) %
(np.pi*2) - np.pi)
u_null = np.dot(Mq, (kp * q_des - kv * dq))

# calculate the null space filter
Jdyn_inv = np.dot(Mx, np.dot(JEE, np.linalg.inv(Mq)))
null_filter = (np.eye(robot_config.num_joints) -
np.dot(JEE.T, Jdyn_inv))
u_null_filtered = np.dot(null_filter, u_null)

u += u_null_filtered


And there you go!

You can see all of this code up on my GitHub, along a full example controlling a UR5 VREP model though reaching to a series of targets. It looks something pretty much like this (slightly better because this gif was made before adding in the null space controller):

Overhead of using lambdify instead of hard-coding

This was a big question that I had, because when I’m running simulations the time step is on the order of a few milliseconds, with the controller code called at every time step. So I reaaaally can’t afford a crazy overhead for using lambdify.

To test this, I used the handy Python timeit, which requires a bit awkward setup, but quite nicely calls the function a whole bunch of times (1,000,000 by default) and accounts for various other things going on that could affect the execution time.

I tested two sample functions, one simpler than the other. Here’s the code for setting up and testing the first function:

import timeit
import seaborn

# Test 1 ----------------------------------------------------------------------
print('\nTest function 1: ')
time_sympy1 = timeit.timeit(
stmt = 'f(np.random.random(), np.random.random())',
setup = 'import numpy as np;\
import sympy as sp;\
q0 = sp.Symbol("q0");\
l0 = sp.Symbol("l0");\
a = sp.cos(q0) * l0;\
f = sp.lambdify((q0, l0), a, "numpy")')
print('Sympy lambdify function 1 time: ', time_sympy1)

time_hardcoded1 = timeit.timeit(
stmt = 'np.cos(np.random.random())*np.random.random()',
setup = 'import numpy as np')
print('Hard coded function 1 time: ', time_hardcoded1)


Pretty simple, a bit of a pain in the sympy setup, but other than that not bad at all. The second function I tested was just a random collection of cos and sin calls that resemble what gets computed in a Jacobian:

l1*np.sin(q0 - l0*np.sin(q1)*np.cos(q2) - l2*np.sin(q2) - l0*np.sin(q1) + q0*l0)*np.cos(q0) + l2*np.sin(q0)


And here’s the results:

So it’s slower for sure, but again this is the difference in time after 1,000,000 function calls, so until some big optimization needs to be done using the SymPy lambdify function is definitely worth the slight gain in execution time for the insane convenience.

The full code for the timing tests here are also up on my GitHub.

## The iterative Linear Quadratic Regulator algorithm

A few months ago I posted on Linear Quadratic Regulators (LQRs) for control of non-linear systems using finite-differences. The gist of it was at every time step linearize the dynamics, quadratize (it could be a word) the cost function around the current point in state space and compute your feedback gain off of that, as though the dynamics were both linear and consistent (i.e. didn’t change in different states). And that was pretty cool because you didn’t need all the equations of motion and inertia matrices etc to generate a control signal. You could just use the simulation you had, sample it a bunch to estimate the dynamics and value function, and go off of that.

The LQR, however, operates with maverick disregard for changes in the future. Careless of the consequences, it optimizes assuming the linear dynamics approximated at the current time step hold for all time. It would be really great to have an algorithm that was able to plan out and optimize a sequence, mindful of the changing dynamics of the system.

This is exactly the iterative Linear Quadratic Regulator method (iLQR) was designed for. iLQR is an extension of LQR control, and the idea here is basically to optimize a whole control sequenceÂ rather than just theÂ control signal for the current point in time. The basic flow of the algorithm is:

1. Initialize with initial state $x_0$ and initial control sequence $\textbf{U} = [u_{t_0}, u_{t_1}, ..., u_{t_{N-1}}]$.
2. Do aÂ forward pass, i.e. simulate the system using $(x_0, \textbf{U})$ to get the trajectory through state space, $\textbf{X}$, that results from applying the control sequence $\textbf{U}$ starting in $x_0$.
3. Do a backward pass, estimate the value function and dynamics for each $(\textbf{x}, \textbf{u})$ in the state-space and control signal trajectories.
4. Calculate an updated control signal $\hat{\textbf{U}}$ and evaluate cost of trajectory resulting from $(x_0, \hat{\textbf{U}})$.
1. IfÂ $|(\textrm{cost}(x_0, \hat{\textbf{U}}) - \textrm{cost}(x_0, \textbf{U})| < \textrm{threshold}$ then we've converged and exit.
2. If $\textrm{cost}(x_0, \hat{\textbf{U}}) < \textrm{cost}(x_0, \textbf{U})$, then set $\textbf{U} = \hat{\textbf{U}}$, and change the update size to be more aggressive. Go back to step 2.
3. If $\textrm{cost}(x_0, \hat{\textbf{U}}) \geq$ $\textrm{cost}(x_0, \textbf{U})$ change the update size to be more modest. Go back to step 3.

There are a bunch of descriptions of iLQR, and it also goes by names like ‘the sequential linear quadratic algorithm’. The paper that I’m going to be working off of is by Yuval Tassa out of Emo Todorov’s lab, called Control-limited differential dynamic programming. And the Python implementation of this can be found up on my githubÂ in my Control repo. Also, a big thank you to Dr. Emo Todorov who provided Matlab code for the iLQG algorithm, which was super helpful.

Defining things

So let’s dive in. Formally defining things, we have our system $\textbf{x}$, and dynamics described with the function $\textbf{f}$, such that

$\textbf{x}_{t+1} = \textbf{f}(\textbf{x}_t, \textbf{u}_t),$

where $\textbf{u}$ is the input control signal. The trajectory $\{\textbf{X}, \textbf{U}\}$ is the sequence of states $\textbf{X} = \{\textbf{x}_0, \textbf{x}_1, ..., \textbf{x}_N\}$ that result from applying the control sequence $\textbf{U} = \{\textbf{u}_0, \textbf{u}_1, ..., \textbf{u}_{N-1}\}$ starting in the initial state $\textbf{x}_0$.

Now we need to define all of our cost related equations, so we know exactly what we’re dealing with.

Define the total cost function $J$, which is the sum of the immediate cost, $\ell$, from each state in the trajectory plus the final cost, $\ell_f$:

$J(\textbf{x}_0, \textbf{U}) = \sum\limits^{N-1}\limits_{t=0} \ell(\textbf{x}_t, \textbf{u}_t) + \ell_f(\textbf{x}_N).$

Letting $\textbf{U}_t = \{\textbf{u}_t, \textbf{u}_{t+1}, ..., \textbf{U}_{N-1}\}$, we define theÂ cost-to-go as the sum of costs from time $t$ to $N$:

$J_t(\textbf{x}, \textbf{U}_t) = \sum\limits^{N-1}\limits_{i=t} \ell(\textbf{x}_i, \textbf{u}_i) + \ell_f(\textbf{x}_N).$

The value function $V$ at time $t$ isÂ the optimal cost-to-go from a given state:

$V_t(\textbf{x}) = \min\limits_{\textbf{U}_t} J_t(\textbf{x}, \textbf{U}_t),$

where the above equationÂ just says that the optimal cost-to-go is found by using the control sequence $\textbf{U}_t$ that minimizes $J_t$.

At the final time step, $N$, the value function is simply

$V(\textbf{x}_N) = \ell_f(\textbf{x}_N).$

For all preceding time steps, we can write the value function as a function of the immediate cost $\ell(\textbf{x}, \textbf{u})$ and the value function at the next time step:

$V(\textbf{x}) = \min\limits_{\textbf{u}} \left[ \ell(\textbf{x}, \textbf{u}) + V(\textbf{f}(\textbf{x}, \textbf{u})) \right].$

NOTE: In the paper, they use the notation $V'(\textbf{f}(\textbf{x}, \textbf{u}))$ to denote the value function at the next time step, which is redundant sinceÂ $\textbf{x}_{t+1} = \textbf{f}(\textbf{x}_t, \textbf{u}_t)$, but it comes in handy later when they drop the dependencies to simplify notation. So, heads up: $V' = V(\textbf{f}(\textbf{x}, \textbf{u})$.

Forward rollout

The forward rollout consists of two parts. The first part is to simulating things to generate the $(\textbf{X}, \textbf{U}),$ from which we can calculate the overall cost of the trajectory, and find out the path that the arm will take. To improve things though we’ll need a lot of information about the partial derivatives of the system, calculating these is the second part of the forward rollout phase.

To calculate all these partial derivatives we’ll use $(\textbf{X}, \textbf{U})$. For each $(\textbf{x}_t, \textbf{u}_t)$ we’ll calculate the derivatives of $\textbf{f}(\textbf{x}_t, \textbf{u}_t)$ with respect to $\textbf{x}_t$ and $\textbf{u}_t$, which will give us what we need for our linear approximation of the system dynamics.

To get the information we need about the value function, we’ll need the first and second derivatives of $\ell(\textbf{x}_t, \textbf{u}_t)$ and $\ell_f(\textbf{x}_t, \textbf{x}_t)$ with respect to $\textbf{x}_t$ and $\textbf{u}_t$.

So all in all, we need to calculate $\textbf{f}_\textbf{x}$, $\textbf{f}_\textbf{u}$, $\ell_\textbf{x}$, $\ell_\textbf{u}$, $\ell_\textbf{xx}$, $\ell_\textbf{ux}$, $\ell_\textbf{uu}$, where the subscripts denote a partial derivative, so $\ell_\textbf{x}$ is the partial derivative of $\ell$ with respect to $\textbf{x}$, $\ell_\textbf{xx}$ is the second derivative of $\ell$ with respect to $\textbf{x}$, etc. And to calculate all of these partial derivatives, we’re going to use finite differences! Just like in the LQR with finite differences post. Long story short, load up the simulation for every time step, slightly vary one of the parameters, and measure the resulting change.

Once we have all of these, we’re ready to move on to the backward pass.

Backward pass

Now, we started out with an initial trajectory, but that was just a guess. We want our algorithm to take it and then converge to a local minimum. To do this, we’re going to add some perturbing values and use them to minimize the value function. Specifically, we’re going to compute a local solution to our value function using a quadratic Taylor expansion. So let’s define $Q(\delta \textbf{x}, \delta \textbf{u})$ to be the change in our value function at $(\textbf{x}, \textbf{u})$ as a result of small perturbations $(\delta \textbf{x}, \delta \textbf{u})$:

$Q(\delta \textbf{x}, \delta \textbf{u}) = \ell (\textbf{x} + \delta \textbf{x}, \textbf{u} + \delta \textbf{u}) + V(\textbf{f}(\textbf{x} + \delta\textbf{x}, \textbf{u} + \delta \textbf{u})).$

The second-order expansion of $Q$ is given by:

$Q_\textbf{x} = \ell_\textbf{x} + \textbf{f}_\textbf{x}^T V'_\textbf{x},$

$Q_\textbf{u} = \ell_\textbf{u} + \textbf{f}_\textbf{u}^T V'_\textbf{x},$

$Q_\textbf{xx} = \ell_\textbf{xx} + \textbf{f}_\textbf{x}^T V'_\textbf{xx} \textbf{f}_\textbf{x} + V'_\textbf{x} \cdot \textbf{f}_\textbf{xx},$

$Q_\textbf{ux} = \ell_\textbf{ux} + \textbf{f}_\textbf{u}^T V'_\textbf{xx} \textbf{f}_\textbf{x}+ V'_\textbf{x} \cdot \textbf{f}_\textbf{ux},$

$Q_\textbf{uu} = \ell_\textbf{uu} + \textbf{f}_\textbf{u}^T V'_\textbf{xx} \textbf{f}_\textbf{u}+ V'_\textbf{x} \cdot \textbf{f}_\textbf{uu}.$

Remember that $V' = V(\textbf{f}(\textbf{x}, \textbf{u}))$, which is the value function at the next time step. NOTE: All of the second derivatives of $\textbf{f}$ are zero in the systems we’re controlling here, so when we calculate the second derivatives we don’t need to worry about doing any tensor math, yay!

Given the second-order expansion of $Q$, we can to compute the optimal modification to the control signal, $\delta \textbf{u}^*$. This control signal update has two parts, a feedforward term, $\textbf{k}$, and a feedback termÂ $\textbf{K} \delta\textbf{x}$. The optimal update is the $\delta\textbf{u}$ that minimizes the cost of $Q$:

$\delta\textbf{u}^*(\delta \textbf{x}) = \min\limits_{\delta\textbf{u}}Q(\delta\textbf{x}, \delta\textbf{u}) = \textbf{k} + \textbf{K}\delta\textbf{x},$

where $\textbf{k} = -Q^{-1}_\textbf{uu} Q_\textbf{u}$ and $\textbf{K} = -Q^{-1}_\textbf{uu} Q_\textbf{ux}.$

Derivation can be found in this earlier paper by Li and Todorov. By then substituting this policy into the expansion of $Q$Â we get a quadratic model of $V$. They do some mathamagics and come out with:

$V_\textbf{x} = Q_\textbf{x} - \textbf{K}^T Q_\textbf{uu} \textbf{k},$

$V_\textbf{xx} = Q_\textbf{xx} - \textbf{K}^T Q_\textbf{uu} \textbf{K}.$

So now we have all of the terms that we need, and they’re defined in terms of the values at the next time step. We know the value of the value function at the final time step $V_N = \ell_f(\textbf{x}_N)$, and so we’ll simply plug this value in and work backwards in time recursively computing the partial derivatives of $Q$ and $V$.

Calculate control signal update

Once those are all calculated, we can calculate the gain matrices, $\textbf{k}$ and $\textbf{K}$, for our control signal update. Huzzah! Now all that’s left to do is evaluate this new trajectory. So we set up our system

$\hat{\textbf{x}}_0 = \textbf{x}_0,$

$\hat{\textbf{u}}_t = \textbf{u}_t + \textbf{k}_t + \textbf{K}_t (\hat{\textbf{x}}_t - \textbf{x}_t),$

$\hat{\textbf{x}}_{t+1} = \textbf{f}(\hat{\textbf{x}}_t, \hat{\textbf{u}}_t),$

and record the cost.Â Now if the cost of the new trajectory $(\hat{\textbf{X}}, \hat{\textbf{U}})$ is less than the cost of $(\textbf{X}, \textbf{U})$ then we set $\textbf{U} = \hat{\textbf{U}}$ and go do it all again! And when the cost from an update becomes less than a threshold value, call it done. In code this looks like:

if costnew < cost:
sim_new_trajectory = True

if (abs(costnew - cost)/cost) < self.converge_thresh:
break


Of course, another option we need to account for is when costnew > cost. What do we do in this case? Our control update hasn’t worked, do we just exit?

The Levenberg-Marquardt heuristic
No! Phew.

The control signal update in iLQR is calculated in such a way that it can behave like Gauss-Newton optimization (which uses second-order derivative information) or like gradient descent (which only uses first-order derivative information). The is that if the updates are going well, then lets include curvature information in our update to helpÂ optimize things faster. If the updates aren’t going well let’s dial back towards gradient descent, stick to first-order derivative information and use smaller steps. This wizardry is known as the Levenberg-Marquardt heuristic. So how does it work?

Something we skimmed over in the iLQR description was that we need to calculate $Q^{-1}_\textbf{uu}$ to getÂ the $\textbf{k}$ and $\textbf{K}$ matrices. Instead of using np.linalg.pinv or somesuch, we’re going to calculate the inverse ourselves using singular value decomposition, so thatÂ we can regularize it. This will let us do a couple of things. First, we’ll be able to make sure that our estimate of curvature ($Q_\textbf{uu}^{-1}$) stays positive definite, which is important to make sure that weÂ always have a descent direction. Second, we’re going to add a regularization term to the singular values to prevent them from exploding when weÂ take their inverse. Here’s our regularization implemented in Python:

 U, S, V = np.linalg.svd(Quu)
S[S < 0] = 0.0 # no negative values
S += lamb # add regularization term
Quu_inv = np.dot(U, np.dot(np.diag(1.0/S), V.T))


Now, what happens when we change lamb? The singular values represent the magnitude of each of the left and right singular vectors, and by taking their reciprocal we flip the contributions of the vectors. So the ones that were contributing the least now have the largest singular values, and the ones that contributed the most now have the smallest singular values. By adding a regularization term we ensure that the inverted singular values can never be larger than 1/lamb. So essentially we throw out information.

In the case where we’ve got a really good approximation of the system dynamics and value function, we don’t want to do this. We want to use all of the information available because it’s accurate, so make lamb small and get a more accurate inverse. In the case where we have a bad approximation of the dynamics we want to be more conservative, which means not having those large singular values. Smaller singular values give a smaller $Q_\textbf{uu}^{-1}$ estimate, which then gives smaller gain matrices and control signal update, which is what we want to do when our control signal updates are going poorly.

How do you know if they’re going poorly or not, you now surely ask! Clever as always, we’re going to use the result of the previous iteration to update lamb. So adding to the code from just above, the end of our control update loop is going to look like:

lamb = 1.0 # initial value of lambda
...
if costnew < cost:
lamb /= self.lamb_factor
sim_new_trajectory = True

if (abs(costnew - cost)/cost) < self.converge_thresh:
break
else:
lamb *= self.lamb_factor
if lamb > self.max_lamb:
break


And that is pretty much everything! OK let’s see how this runs!

Simulation results

If you want to run this and see for yourself, you can go copy my Control repo, navigate to the main directory, and run

python run.py arm2 reach


or substitute in arm3. If you’re having trouble getting the arm2 simulation to run, try arm2_python, which is a straight Python implementation of the arm dynamics, and should work no sweat for Windows and Mac.

Below you can see results from the iLQR controller controlling the 2 and 3 link arms (click on the figures to see full sized versions, they got distorted a bit in the shrinking to fit on the page), using immediate and final state cost functions defined as:

l = np.sum(u**2)


and

pos_err = np.array([self.arm.x[0] - self.target[0],
self.arm.x[1] - self.target[1]])
l = (wp * np.sum(pos_err**2) + # pos error
wv * np.sum(x[self.arm.DOF:self.arm.DOF*2]**2)) # vel error


where wp and wv are just gain values, x is the state of the system, and self.arm.x is the $(x,y)$ position of the hand. These read as “during movement, penalize large control signals, and at the final state, have a big penalty on not being at the target.”

So let’s give it up for iLQR, this is awesome! How much of a crazy improvement is that over LQR? And with all knowledge of the system through finite differences, and with the full movements in exactly 1 second! (Note: The simulation speeds look different because of my editing to keep the gif sizes small, they both take the same amount of time for each movement.)

Changing cost functions
Something that you may notice is that the control of the 3 link is actually straighter than the 2 link. I thought that this might be just an issue with the gain values, since the scale of movement is smaller for the 2 link arm than the 3 link there might have been less of a penalty for not moving in a straight line, BUT this was wrong. You can crank the gains and still get the same movement. The actual reason is that this is what the cost function specifies, if you look in the code, only $\ell_f(\textbf{x}_N)$ penalizes the distance from the target, and the cost function during movement is strictly to minimize the control signal, i.e. $\ell(\textbf{x}_t, \textbf{u}_t) = \textbf{u}_t^2$.

Well that’s a lot of talk, you say, like the incorrigible antagonist we both know you to be, prove it. Alright, fine! Here’s iLQR running with an updated cost function that includes the end-effector’s distance from the target in the immediate cost:

All that I had to do to get this was change the immediate cost from

l = np.sum(u**2)


to

l = np.sum(u**2)
pos_err = np.array([self.arm.x[0] - self.target[0],
self.arm.x[1] - self.target[1]])
l += (wp * np.sum(pos_err**2) + # pos error
wv * np.sum(x[self.arm.DOF:self.arm.DOF*2]**2)) # vel error


where all I had to do was include the position penalty term from the final state cost into the immediate state cost.

Changing sequence length
In these simulations the system is simulating at .01 time step, and I gave it 100 time steps to reach the target. What if I give it only 50 time steps?

It looks pretty much the same! It’s just now twice as fast, which is of course achieved by using larger control signals, which we don’t see, but dang awesome.

What if we try to make it there in 10 time steps??

OK well that does not look good. So what’s going on in this case? Basically we’ve given the algorithm an impossible task. It can’t make it to the target location in 10 time steps. In the implementation I wrote here, if it hits the end of it’s control sequence and it hasn’t reached the target yet, the control sequence starts over back at t=0. Remember that part of the target state is also velocity, so basically it moves for 10 time steps to try to minimize $(x,y)$ distance, and then slows down to minimize final state cost in the velocity term.

In conclusion

This algorithm has been used in a ton of things, for controlling robots and simulations, and is an important part of guided policy search, which has been used to very successfully train deep networks in control problems. It’s getting really impressive results for controlling the arm models that I’ve built here, and using finite differences should easily generalize to other systems.

iLQR is very computationally expensive, though, so that’s definitely a downside. It’s definitely less expensive if you have the equations of your system, or at least a decent approximation of them, and you don’t need to use finite differences. But you pay for the efficiency with a loss in generality.

There are also a bunch of parameters to play around with that I haven’t explored at all here, like the weights in the cost function penalizing the magnitude of the cost function and the final state position error.Â I showed a basic example of changing the cost function, which hopefully gets across just how easy changing these things out can be when you’re using finite differences, and there’s a lot to play around with there too.

Implementation note
In the Yuval and Todorov paper, they talked about using backtracking line search when generating the control signal. So the algorithm they had when generating the new control signal was actually:

$\hat{\textbf{u}}_t = \hat{\textbf{u}}_t + \alpha\textbf{k}_t + \textbf{K}_t(\hat{\textbf{x}}_t - \textbf{x}_t)$

where $\alpha$ was the backtracking search parameter, which gets set to one initially and then reduced. It’s very possible I didn’t implement it as intended, but I found consistently that $\alpha = 1$ always generated the best results, so it was just adding computation time. So I left it out of my implementation. If anyone has insights on an implementation that improves results, please let me know!

And then finally, another thank you to Dr. Emo Todorov for providing Matlab code for the iLQG algorithm, which was very helpful, especially for getting the Levenberg-Marquardt heuristic implemented properly.

## Operational space control of 6DOF robot arm with spiking cameras part 2: Deriving the Jacobian

In the previous exciting post in this series I outlined the project, which is in the title, and we worked through getting access to the arm through Python. The next step was deriving the Jacobian, and that’s what we’re going to be talking about in this post!

Alright.
This was a time I was very glad to have a previous post talking about generating transformation matrices, because deriving the Jacobian for a 6DOF arm in 3D space comes off as a little daunting when you’re used to 3DOF in 2D space, and I needed a reminder of the derivation process. The first step here was finding out which motors were what, so I went through and found out how each motor moved with something like the following code:

for ii in range(7):
target_angles = np.zeros(7, dtype='float32')
target_angles[ii] = np.pi / 4.0
rob.move(target_angles)
time.sleep(1)


and I found that the robot is setup in the figures below

this is me trying my hand at making things clearer using Inkscape, hopefully it’s worked. Displayed are the first 6 joints and their angles of rotation, $q_0$ through $q_5$. The 7th joint, $q_6$, opens and closes the gripper, so we’re safe to ignore it in deriving our Jacobian. The arm segment lengths $l_1, l_3,$ and $l_5$ are named based on the nearest joint angles (makes easier reading in the Jacobian derivation).

Find the transformation matrix from end-effector to origin

So first thing’s first, let’s find the transformation matrices. Our first joint, $q_0$, rotates around the $z$ axis, so the rotational part of our transformation matrix $^0_\textrm{org}\textbf{T}$ is

$^0_\textrm{org}\textbf{R} = \left[ \begin{array}{ccc} \textrm{cos}(q_0) & -\textrm{sin}(q_0) & 0 \\ \textrm{sin}(q_0) & \textrm{cos}(q_0) & 0 \\ 0 & 0 & 1 \end{array} \right],$

and $q_0$ and our origin frame of reference are on top of each other so we don’t need to account for translation, so our translation component of $^0_\textrm{org}\textbf{T}$ is

$^0_\textrm{org}\textbf{D} = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right].$

Stacking these together to form our first transformation matrix we have

$^0_\textrm{org}\textbf{T} = \left[ \begin{array}{cc} ^0_\textrm{org}\textbf{R} & ^0_\textrm{org}\textbf{D} \\ 0 & 1 \end{array} \right] = \left[ \begin{array}{cccc} \textrm{cos}(q_0) & -\textrm{sin}(q_0) & 0 & 0\\ \textrm{sin}(q_0) & \textrm{cos}(q_0) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] .$

So now we are able to convert a position in 3D space from to the reference frame of joint $q_0$ back to our origin frame of reference. Let’s keep going.

Joint $q_1$ rotates around the $x$ axis, and there is a translation along the arm segment $l_1$. Our transformation matrix looks like

$^1_0\textbf{T} = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & \textrm{cos}(q_1) & -\textrm{sin}(q_1) & l_1 \textrm{cos}(q_1) \\ 0 & \textrm{sin}(q_1) & \textrm{cos}(q_1) & l_1 \textrm{sin}(q_1) \\ 0 & 0 & 0 & 1 \end{array} \right] .$

Joint $q_2$ also rotates around the $x$ axis, but there is no translation from $q_2$ to $q_3$. So our transformation matrix looks like

$^2_1\textbf{T} = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & \textrm{cos}(q_2) & -\textrm{sin}(q_2) & 0 \\ 0 & \textrm{sin}(q_2) & \textrm{cos}(q_2) & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] .$

The next transformation matrix is a little tricky, because you might be tempted to say that it’s rotating around the $z$ axis, but actually it’s rotating around the $y$ axis. This is determined by where $q_3$ is mounted relative to $q_2$. If it was mounted at 90 degrees from $q_2$ then it would be rotating around the $z$ axis, but it’s not. For translation, there’s a translation along the $y$ axis up to the next joint, so all in all the transformation matrix looks like:

$^3_2\textbf{T} = \left[ \begin{array}{cccc} \textrm{cos}(q_3) & 0 & -\textrm{sin}(q_3) & 0\\ 0 & 1 & 0 & l_3 \\ \textrm{sin}(q_3) & 0 & \textrm{cos}(q_3) & 0\\ 0 & 0 & 0 & 1 \end{array} \right] .$

And then the transformation matrices for coming from $q_4$ to $q_3$ and $q_5$ to $q_4$ are the same as the previous set, so we have

$^4_3\textbf{T} = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & \textrm{cos}(q_4) & -\textrm{sin}(q_4) & 0 \\ 0 & \textrm{sin}(q_4) & \textrm{cos}(q_4) & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] .$

and

$^5_4\textbf{T} = \left[ \begin{array}{cccc} \textrm{cos}(q_5) & 0 & -\textrm{sin}(q_5) & 0 \\ 0 & 1 & 0 & l_5 \\ \textrm{sin}(q_5) & 0 & \textrm{cos}(q_5) & 0\\ 0 & 0 & 0 & 1 \end{array} \right] .$

Alright! Now that we have all of the transformation matrices, we can put them together to get the transformation from end-effector coordinates to our reference frame coordinates!

$^\textrm{ee}_\textrm{org}\textbf{T} = ^0_\textrm{org}\textbf{T} \; ^1_0\textbf{T} \; ^2_1\textbf{T} \; ^3_2\textbf{T} \; ^4_3\textbf{T} \; ^5_4\textbf{T}.$

At this point I went and tested this with some sample points to make sure that everything seemed to be being transformed properly, but we won’t go through that here.

Calculate the derivative of the transform with respect to each joint

The next step in calculating the Jacobian is getting the derivative of $^\textrm{ee}_\textrm{org}T$. This could be a big ol’ headache to do it by hand, OR we could use SymPy, the symbolic computation package for Python. Which is exactly what we’ll do. So after a quick

sudo pip install sympy


I wrote up the following script to perform the derivation for us

import sympy as sp

def calc_transform():
# set up our joint angle symbols (6th angle doesn't affect any kinematics)
q = [sp.Symbol('q0'), sp.Symbol('q1'), sp.Symbol('q2'), sp.Symbol('q3'),
sp.Symbol('q4'), sp.Symbol('q5')]
# set up our arm segment length symbols
l1 = sp.Symbol('l1')
l3 = sp.Symbol('l3')
l5 = sp.Symbol('l5')

Torg0 = sp.Matrix([[sp.cos(q[0]), -sp.sin(q[0]), 0, 0,],
[sp.sin(q[0]), sp.cos(q[0]), 0, 0],
[0, 0, 1, 0],
[0, 0, 0, 1]])

T01 = sp.Matrix([[1, 0, 0, 0],
[0, sp.cos(q[1]), -sp.sin(q[1]), l1*sp.cos(q[1])],
[0, sp.sin(q[1]), sp.cos(q[1]), l1*sp.sin(q[1])],
[0, 0, 0, 1]])

T12 = sp.Matrix([[1, 0, 0, 0],
[0, sp.cos(q[2]), -sp.sin(q[2]), 0],
[0, sp.sin(q[2]), sp.cos(q[2]), 0],
[0, 0, 0, 1]])

T23 = sp.Matrix([[sp.cos(q[3]), 0, sp.sin(q[3]), 0],
[0, 1, 0, l3],
[-sp.sin(q[3]), 0, sp.cos(q[3]), 0],
[0, 0, 0, 1]])

T34 = sp.Matrix([[1, 0, 0, 0],
[0, sp.cos(q[4]), -sp.sin(q[4]), 0],
[0, sp.sin(q[4]), sp.cos(q[4]), 0],
[0, 0, 0, 1]])

T45 = sp.Matrix([[sp.cos(q[5]), 0, sp.sin(q[5]), 0],
[0, 1, 0, l5],
[-sp.sin(q[5]), 0, sp.cos(q[5]), 0],
[0, 0, 0, 1]])

T = Torg0 * T01 * T12 * T23 * T34 * T45

# position of the end-effector relative to joint axes 6 (right at the origin)
x = sp.Matrix([0,0,0,1])

Tx = T * x

for ii in range(6):
print q[ii]
print sp.simplify(Tx[0].diff(q[ii]))
print sp.simplify(Tx[1].diff(q[ii]))
print sp.simplify(Tx[2].diff(q[ii]))


And then consolidated the output using some variable shorthand to write a function that accepts in joint angles and generates the Jacobian:

def calc_jacobian(q):
J = np.zeros((3, 7))

c0 = np.cos(q[0])
s0 = np.sin(q[0])
c1 = np.cos(q[1])
s1 = np.sin(q[1])
c3 = np.cos(q[3])
s3 = np.sin(q[3])
c4 = np.cos(q[4])
s4 = np.sin(q[4])

c12 = np.cos(q[1] + q[2])
s12 = np.sin(q[1] + q[2])

l1 = self.l1
l3 = self.l3
l5 = self.l5

J[0,0] = -l1*c0*c1 - l3*c0*c12 - l5*((s0*s3 - s12*c0*c3)*s4 + c0*c4*c12)
J[1,0] = -l1*s0*c1 - l3*s0*c12 + l5*((s0*s12*c3 + s3*c0)*s4 - s0*c4*c12)
J[2,0] = 0

J[0,1] = (l1*s1 + l3*s12 + l5*(s4*c3*c12 + s12*c4))*s0
J[1,1] = -(l1*s1 + l3*s12 + l5*s4*c3*c12 + l5*s12*c4)*c0
J[2,1] = l1*c1 + l3*c12 - l5*(s4*s12*c3 - c4*c12)

J[0,2] = (l3*s12 + l5*(s4*c3*c12 + s12*c4))*s0
J[1,2] = -(l3*s12 + l5*s4*c3*c12 + l5*s12*c4)*c0
J[2,2] = l3*c12 - l5*(s4*s12*c3 - c4*c12)

J[0,3] = -l5*(s0*s3*s12 - c0*c3)*s4
J[1,3] = l5*(s0*c3 + s3*s12*c0)*s4
J[2,3] = -l5*s3*s4*c12

J[0,4] = l5*((s0*s12*c3 + s3*c0)*c4 + s0*s4*c12)
J[1,4] = l5*((s0*s3 - s12*c0*c3)*c4 - s4*c0*c12)
J[2,4] = -l5*(s4*s12 - c3*c4*c12)

return J


Alright! Now we have our Jacobian! Really the only time consuming part here was calculating our end-effector to origin transformation matrix, generating the Jacobian was super easy using SymPy once we had that.

Hack position control using the Jacobian

Great! So now that we have our Jacobian we’ll be able to translate forces that we want to apply to the end-effector into joint torques that we want to apply to the arm motors. Since we can’t control applied force to the motors though, and have to pass in desired angle positions, we’re going to do a hack approximation. Let’s first transform our forces from end-effector space into a set of joint angle torques:

$\textbf{u} = \textbf{J}^T \; \textbf{u}_\textbf{x}.$

To approximate the control then we’re simply going to take the current set of joint angles (which we know because it’s whatever angles we last told the system to move to) and add a scaled down version of $\textbf{u}$ to approximate applying torque that affects acceleration and then velocity.

$\textbf{q}_\textrm{des} = \textbf{q} + \alpha \; \textbf{u},$

where $\alpha$ is the gain term, I used .001 here because it was nice and slow, so no crazy commands that could break the servos would be sent out before I could react and hit the cancel button.

What we want to do then to implement operational space control here then is find the current $(x,y,z)$ position of the end-effector, calculate the difference between it and the target end-effector position, use that to generate the end-effector control signal $u_x$, get the Jacobian for the current state of the arm using the function above, find the set of joint torques to apply, approximate this control by generating a set of target joint angles to move to, and then repeat this whole loop until we’re within some threshold of the target position. Whew.

So, a lot of steps, but pretty straight forward to implement. The method I wrote to do it looks something like:

def move_to_xyz(self, xyz_d):
"""
np.array xyz_d: 3D target (x_d, y_d, z_d)
"""
count = 0
while (1):
count += 1
# get control signal in 3D space
xyz = self.calc_xyz()
delta_xyz = xyz_d - xyz
ux = self.kp * delta_xyz

# transform to joint space
J = self.calc_jacobian()
u = np.dot(J.T, ux)

# target joint angles are current + uq (scaled)
self.q[...] += u * .001
self.robot.move(np.asarray(self.q.copy(), 'float32'))

if np.sqrt(np.sum(delta_xyz**2)) < .1 or count > 1e4:
break


And that is it! We have successfully hacked together a system that can perform operational space control of a 6DOF robot arm. Here is a very choppy video of it moving around to some target points in a grid on a cube.

So, granted I had to drop a lot of frames from the video to bring it’s size down to something close to reasonable, but still you can see that it moves to target locations super fast!

Alright this is sweet, but we’re not done yet. We don’t want to have to tell the arm where to move ourselves. Instead we’d like the robot to perform target tracking for some target LED we’re moving around, because that’s way more fun and interactive. To do this, we’re going to use spiking cameras! So stay tuned, we’ll talk about what the hell spiking cameras are and how to use them for a super quick-to-setup and foolproof target tracking system in the next exciting post!

## Dynamic movement primitives part 3: Rhythmic movements

So far we’ve looked at using DMPs for discrete movements, but as I briefly mentioned it’s also possible to use them to perform rhythmic movements. In this post we’re going to look at the implementation of rhythmic DMPs and see exactly how it’s done. It’s actually pretty straightforward, which is always nich. We’ll go through everything that has to change and then see a couple of different applications of rhythmic DMPs. Again, all the code for the DMPs and arm control implementations can be found up on my Github.

Obtaining consistent and repeatable basis function activation

We’ll start with the canonical system. In the discrete case our canonical system (which drives the activation of the basis functions) decayed from $1$ to $0$ and then we were all done. In the rhythmic case we want our system to repeat indefinitely, so we need a reliable way of continuously activating the basis functions in the same order. One function that may come to mind that reliably repeats is the cosine function. But how to use the cosine function, exactly?

The idea is that we’ll lay out our basis functions in the range from $0$ to $2\pi$, and we’ll set the canonical system to be ever increasing linearly. Then we’ll use the difference between the canonical system state and each of the center points as the value we pass in to our cosine function. Because the cosine function repeats at $2\pi$ we’ll get a nice repeating spread of basis function activations. We’ll then throw these values through our Gaussian equation (with the gain and bias terms) and we’ll be done! Here’s the equation:

$\psi = \textrm{exp}(h * \textrm{cos}(x - c) - 1),$
where $x$ is the state of the canonical system, $c$ is the basis function center point, and $h$ is a gain term controlling the variance of the Gaussian. And here’s a picture of the activations of three basis functions with centers spread out evenly between $0$ and $2\pi$:

Here’s a picture of the basis function activations on a longer time scale:

As long as our canonical system increases at a steady pace the basis functions will continue to activate in a reliable and repeatable way. As mentioned, to get the canonical system to increase at a steady pace we simply give it linear dynamics:

$\dot{x} = 1$
The placement of the center points of the rhythmic system is a lot easier than in the discrete case because the rhythmic canonical system dynamics are linear.

Other differences between discrete and rhythmic

In the actual implementation of the rhythmic system there are a couple of other differences that you need to be aware of. The first is that there is no diminishing term in the rhythmic system, whereas there is in the discrete case.

The second is how to go about establishing goal states when imitating an path. We’re going to assume that whatever path is passed in to us is going to be a rhythmic pattern, and then goal state is going to be set as the center point of the desired trajectory. So for rhythmic systems the ‘goal’ is actually more like a center point for the trajectory.

And that’s pretty much it! If you look through the code that I have up for the rhythmic DMP subclass you’ll see that only a handful of functions needed to be redefined, the rest is the same! Well that’s great, let’s look at some applications.

Example

I was thinking of different examples that would be interesting to see of the rhythmic DMPs, and after a little consideration the most obvious application of rhythmic DMPs is really to a system that does something like walking. In the spinal cord of animals there are circuits which are known as central pattern generators (CPGs), and when stimulated these have been shown to generate rhythmic, repeated movements (as described in Principles of rhythmic motor pattern generation and other papers). Let’s go ahead and build our own CPG here using DMPs. I don’t have a simulation of a set of legs, however, so instead we’ll look at getting a single 3-link arm to reproduce a pattern of behaviour that you’d expect in a leg that is walking.

To do this, I’m going to need to specify the trajectories for the three joints of the arm, and I’ll be controlling the system in joint space, using the generalized coordinates controller, rather than the operational space controller used in the discrete DMP examples for controlling the end-effector position. So that’s one thing. Another thing is that I’m going to need the kinematics of leg joints during motion. After a quick search of the web I found this presentation from Simon Fraser University which has the following images:

Which is some very useful information! I just need to transfer this into a Numpy vector that I can feed into the system and convert degrees to radians and we should be good to go! To get the information out of the graph, use your favorite data stripper, I used http://arohatgi.info/WebPlotDigitizer/app/.

I also had to do a bit of a shuffle with the trajectories provided to make it look right on the arm. The hip to shoulder joint maps fine, but for the knee to elbow the flexion and extension are reversed, and for the wrist to foot the $0$ angle for the foot is straight out. Once those have been accounted for, though, everything works great. Here is an animation of a single leg walking along, if you use your imagination you can just picture a happy-go-lucky torso attached to this leg whistling while he gets his very human-like gait on:

And there you go! It’s neat how straightforward that was, once you have the data of the system you want to model. It shows the power of DMPs for coordinating multiple degrees of freedom quickly and easily. Also something that’s worth pointing out once again is that this is a force based controller. We were able to go from a kinematic description of the movement we wanted to a reliable system generating torques. It’s another good example of the dexterity of dynamic movement primitives.

You can see the code for everything here up online on my github control repo on the walking_demo branch. To run the code you should run:

python run.py


## Robot control part 7: OSC of a 3-link arm

So we’ve done control for the 2-link arm, and control of the one link arm is trivial (where we control joint angle, or x or y coordinate of the pendulum), so here I’ll just show an implementation of operation space control for a more interesting arm model, the 3-link arm model. The code can all be found up on my Github.

In theory there’s nothing different or more difficult about controlling a 3-link arm versus a 2-link arm. For the inertia matrix, what I ended up doing here is just jacking up all the values of the matrix to about 100, which causes the controller to way over control the arm, and you can see the torques are much larger than they would need to be if we had an accurate inertia matrix. But the result is the same super straight trajectories that we’ve come to expect from operational space control:

It’s a little choppy because I cut out a bunch of frames to keep the gif size down. But you get the point, it works. And quite well!

Because this is also a 3-link arm now and our task space force signal is 2D, we have redundant space of solutions, meaning that the task space control signal can be carried out in a number of ways. In other words, a null space exists for this controller. This means that we can throw another controller in our system to operate inside the null space of the first controller. We’ve already worked through all the math for this, so it’s straightforward to implement.

What kind of null space controller should we put in? Well, you may have noticed in the above animation the arm goes through itself, here’s another example:

Often it’s desirable to avoid this (because of physics or whatever), so what we can do is add a secondary controller that works to keep the arm’s elbow and wrist near some comfortable default angles. When we do this we get the following:

And there you have it! Operational space control of a three link arm with a secondary controller in the null space to try to keep the angles near their default / resting positions.

I also added mouse based control to the arm so it will try to follow your mouse when you move over the figure, which makes it pretty fun to explore the power of the controller. It’s interesting to see where the singularities become an issue, and how having a null space controller that’s operating in joint space can actually come to help the system move through those problem points more smoothly. Check it out! It’s all up on my Github.

## Robot control part 6: Handling singularities

We’re back! Another exciting post about robotic control theory, but don’t worry, it’s short and ends with simulation code. The subject of today’s post is handling singularities.

What is a singularity

This came up recently when I had build this beautiful controller for a simple two link arm that would occasionally go nuts. After looking at it for a while it became obvious this was happening whenever the elbow angle reached or got close to 0 or $\pi$. Here’s an animation:

What’s going on here? Here’s what. The Jacobian has dropped rank and become singular (i.e. non-invertible), and when we try to calculate our mass matrix for operational space

$\textbf{M}_\textbf{x}(\textbf{q}) = (\textbf{J} (\textbf{q}) \; \textbf{M}^{-1} (\textbf{q}) \; \textbf{J}^T(\textbf{q}))^{-1},$

the values explode in the inverse calculation.Â Dropping rank means that the rows of the Jacobian are no longer linearly independent, which means that the matrix can be rotated such that it gives a matrix with a row of zeros.Â This row of zeros is the degenerate direction, and the problems come from trying to send forces in that direction.

To determine when the Jacobian becomes singular its determinant can be examined; if the determinant of the matrix is zero, then it is singular.Â Looking the Jacobian for the end-effector:

$\textbf{J}(\textbf{q}) = \left[ \begin{array}{cc} -L_0 sin(q_0) - L_1 sin(q_0 + q_1) & -L_1 sin(q_0 + q_1) \\ L_0 cos(q_0) + L_1 cos(q_0 + q_1) & L_1 cos(q_0 + q_1) \end{array} \right].$

When $q_1 = 0$ it can be that $sin(q_0 + 0) = sin(q_0)$, so the Jacobian becomes

$\textbf{J}(\textbf{q}) = \left[ \begin{array}{cc} (L_0 + L_1)(-sin(q_0)) & -L_1 sin(q_0) \\ (L_0 + L_1) cos(q_0) & L_1 cos(q_0) \end{array} \right],$

which gives a determinant of

$(L_0 + L_1)(-sin(q_0))(L_1)(cos(q_0)) - (L_1)(-sin(q_0))(L_0 + L_1)(cos(q_0)) = 0.$

Similarly, when $q_1 = \pi$, where $sin(q_0 + \pi) = -sin(q_0)$ and $cos(q_0 + \pi) = -cos(q_0)$, the Jacobian is

$\textbf{J}(\textbf{q}) = \left[ \begin{array}{cc} -(L_0 - L_1) sin(q_0) & L_1 sin(q_0) \\ (L_0 + L_1) cos(q_0) & - L_1 cos(q_0) \end{array} \right].$

Calculating the determinant of this we get

$(L_0 + L_1)(-sin(q_0))(L_1)(-cos(q_0)) - (L_1)(sin(q_0))(L_0 + L_1)(-cos(q_0)) = 0.$

Note that while in these cases the Jacobian is a square matrix in the event that it is not a square matrix, the determinant of $\textbf{J}(\textbf{q})\;\textbf{J}^T(\textbf{q})$ can be found instead.

Fixing the problem

When a singularity is occurring it can be detected, but now it must be handled such that the controller behaves appropriately.Â This can be done by identifying the degenerate dimensions and setting the force in those directions to zero.

First the SVD decomposition of $\textbf{M}_\textbf{x}^{-1}(\textbf{q}) = \textbf{V}\textbf{S}\textbf{U}^T$ is found. To get the inverse of this matrix (i.e. to find $\textbf{M}_\textbf{x}(\textbf{q})$) from the returned $\textbf{V}, \textbf{S}$ and $\textbf{U}$ matrices is a matter of inverting the matrix $\textbf{S}$:

$\textbf{M}_\textbf{x}(\textbf{q}) = \textbf{V} \textbf{S}^{-1} \textbf{U}^T,$

where $\textbf{S}$ is a diagonal matrix of singular values.

Because $\textbf{S}$ is diagonal it is very easy to find its inverse, which is calculated by taking the reciprocal of each of the diagonal elements.

Whenever the system approaches a singularity some of the values of $\textbf{S}$ will start to get very small, and when we take the reciprocal of them we start getting huge numbers, which is where the value explosion comes from.Â Instead of allowing this to happen, a check for approaching the singularity can be implemented, which then sets the singular values entries smaller than the threshold equal to zero, canceling out any forces that would be sent in that direction.

Here’s the code:

Mx_inv = np.dot(JEE, np.dot(np.linalg.inv(Mq), JEE.T))
if abs(np.linalg.det(np.dot(JEE,JEE.T))) > .005**2:
# if we're not near a singularity
Mx = np.linalg.inv(Mx_inv)
else:
# in the case that the robot is entering near singularity
u,s,v = np.linalg.svd(Mx_inv)
for i in range(len(s)):
if s[i] < .005: s[i] = 0
else: s[i] = 1.0/float(s[i])
Mx = np.dot(v, np.dot(np.diag(s), u.T))


And here’s an animation of the controlled arm now that we’ve accounted for movement when near singular configurations:

As always, the code for this can be found up on my Github. The default is to run using a two link arm simulator written in Python. To run, simply download everything and run the run_this.py file.

Everything is also included required to run the MapleSim arm simulator. To do this, go into the TwoLinkArm folder, and run python setup.py build_ext -i. This should compile the arm simulation to a shared object library that Python can now access on your system. To use it, edit the run_this.py file to import from TwoLinkArm/arm_python to TwoLinkArm/arm and you should be good to go!
More details on getting the MapleSim arm to run can be found in this post.

## Robot control part 5: Controlling in the null space

In the last post, I went through how to build an operational space controller. It was surprisingly easy after we’ve worked through all the other posts. But maybe that was a little too easy for you. Maybe you want to do something more interesting like implement more than one controller at the same time. In this post we’ll go through how to work inside the null space of a controller to implement several seperate controllers simultaneously without interference.
Buckle up.

Null space forces

The last example comprises the basics of operational space control; describe the system, calculate the system dynamics, transform desired forces from an operational space to the generalized coordinates, and build the control signal to cancel out the undesired system dynamics.Â Basic operational space control works quite well, but it is not uncommon to have several control goals at once;Â such as move the end-effector to this position’ (primary goal), and keep the elbow raised high’ (secondary goal) in the control of a robot arm.

If the operational space can also serve as generalized coordinates, i.e. if the system state specified in operational space constrains all of the degrees of freedom of the robot, then there is nothing that can be done without affecting the performance of the primary controller.Â In the case of controlling a two-link robot arm this is the case.Â The end-effector Cartesian space (chosen as the operational space) could also be a generalized coordinates system, because a specific $(x,y)$ position fully constrains the position of the arm.

But often when using operational space control for more complex robots this is not the case. In these situations, the forces controlled in operational space have fewer dimensions than the robot has degrees of freedom, and so it is possible to accomplish the primary goal in a number of ways. The null space of this primary controller is the region of state space where there is a redundancy of solutions; the system can move in a number of ways and still not affect the completion of the goals of the primary controller. An example of this is all the different configurations the elbow can be in while a person moves their hand in a straight line. In these situations, a secondary controller can be created to operate in the null space of the primary controller, and the full control signal sent to the system is a sum of the primary control signal and a filtered version of the secondary control signal. In this section the derivation of the null-space filter will be worked through for a system with only a primary and secondary controller, but note that the process can be applied iteratively for systems with further controllers.

The filtering of the secondary control signal means that the secondary controller’s goals will only be accomplished if it is possible to do so without affecting the performance of the first controller. In other words, the secondary controller must operate in the null space of the first controller. Denote the primary operational space control signal, e.g. the control signal defined in the previous post, as $\textbf{u}_{\textbf{x}}$ and the control signal from the secondary controller $\textbf{u}_{\textrm{null}}$.Â Define the force to apply to the system

$\textbf{u} = \textbf{u}_{\textbf{x}} + (\textbf{I} - \textbf{J}_{ee}^T(\textbf{q}) \; \textbf{J}_{ee}^{T+}(\textbf{q})) \textbf{u}_{\textrm{null}},$

where $\textbf{J}_{ee}^{T+}(\textbf{q})$ is the pseudo-inverse of $\textbf{J}_{ee}^T(\textbf{q})$.

Examining the filtering term that was added,

$(\textbf{I} - \textbf{J}_{ee}^T(\textbf{q}) \textbf{J}_{ee}^{T+}(\textbf{q})) \textbf{u}_{\textrm{null}},$

it can be seen that the Jacobian transpose multiplied by its pseudo-inverse will be 1’s all along the diagonal, except in the null space.Â This means that $\textbf{u}_{\textrm{null}}$ is subtracted from itself everywhere that affects the operational space movement and is left to apply any arbitrary control signal in the null space of the primary controller.

Unfortunately, this initial set up does not adequately filter out the effects of forces that might be generated by the secondary controller.Â The Jacobian is defined as a relationship between the velocities of two spaces, and so operating in the null space defined by the Jacobian ensures that no velocities are applied in operational space, but the required filter must also prevent any accelerationsÂ from affecting movement in operational space.Â The standard Jacobian pseudo-inverse null space is a velocity null space, and so a filter built using it will allow forces affecting the system’s acceleration to still get through.Â What is required is a pseudo-inverse Jacobian defined to filter signals through an acceleration null space.

To acquire this acceleration filter, our control signalÂ will be substituted into the equation for acceleration in the operational space, which, after cancelling out gravity effects with the control signal and removing the unmodeled dynamics, gives

$\ddot{\textbf{x}} = \textbf{J}_{ee}(\textbf{q}) \textbf{M}^{-1}(\textbf{q}) [\textbf{J}_{ee}^T(\textbf{q}) \; \textbf{M}_{\textbf{x}_{ee}}(\textbf{q}) \; \ddot{\textbf{x}}_\textrm{des} - (\textbf{I} - \textbf{J}_{ee}^T(\textbf{q})\;\textbf{J}_{ee}^{T+}(\textbf{q}))\;\textbf{u}_{\textrm{null}}].$

Rewriting this to separate the secondary controller into its own term

$\ddot{\textbf{x}} = \textbf{J}_{ee}(\textbf{q}) \textbf{M}^{-1}(\textbf{q}) \textbf{J}_{ee}^T(\textbf{q}) \; \textbf{M}_{\textbf{x}_{ee}}(\textbf{q}) \; \ddot{\textbf{x}}_\textrm{des} - \textbf{J}_{ee}(\textbf{q}) \textbf{M}^{-1}(\textbf{q})[\textbf{I} - \textbf{J}_{ee}^T(\textbf{q})\;\textbf{J}_{ee}^{T+}(\textbf{q})]\;\textbf{u}_{\textrm{null}},$

it becomes clear that to not cause any unwanted movement in operational space the second term must be zero.

There is only one free term left in the second term, and that is the pseudo-inverse.Â There are numerous different pseudo-inverses that can be chosen for a given situation, and here what is required is to engineer a pseudo-inverse such that the term multiplying $\textbf{u}_{\textrm{null}}$ in the above operational space acceleration equation is guaranteed to go to zero.

$\textbf{J}_{ee}(\textbf{q})\textbf{M}^{-1}(\textbf{q}) [\textbf{I} - \textbf{J}_{ee}^T (\textbf{q})\textbf{J}_{ee}^{T+}(\textbf{q})] \textbf{u}_{\textrm{null}} = \textbf{0},$

this needs to be true for all $\textbf{u}_{\textrm{null}}$, so it can be removed,

$\textbf{J}_{ee} (\textbf{q}) \; \textbf{M}^{-1}(\textbf{q}) [\textbf{1} - \textbf{J}_{ee}^T (\textbf{q}) \; \textbf{J}_{ee}^{T+} (\textbf{q})] = \textbf{0},$

$\textbf{J}_{ee}(\textbf{q}) \; \textbf{M}^{-1}(\textbf{q}) = \textbf{J}_{ee}(\textbf{q}) \; \textbf{M}^{-1}(\textbf{q}) \; \textbf{J}_{ee}^T(\textbf{q})\; \textbf{J}_{ee}^{T+}(\textbf{q}),$

substituting in our inertia matrix for operational space, which defines

$\textbf{J}_{ee} (\textbf{q}) \textbf{M}^{-1} (\textbf{q}) = \textbf{M}_{\textbf{x}_{ee}}^{-1} (\textbf{q}) \textbf{J}_{ee}^{T+} (\textbf{q}),$

$\textbf{J}_{ee}^{T+}(\textbf{q}) = \textbf{M}_{\textbf{x}_{ee}} (\textbf{q}) \; \textbf{J}_{ee}(\textbf{q}) \; \textbf{M}^{-1}(\textbf{q}).$

This specific Jacobian inverse was presented in this 1987 paper by Dr. Oussama Khatib and is called the `dynamically consistent generalized inverse’.Â Using this psuedo-inverse guarantees that any signal coming from the secondary controller will not affect movement in the primary controller’s operational space. Just as a side-note, the name ‘pseudo-inverse’ is a bit of misnomer here, since it doesn’t try to produce the identity when multiplied by the original Jacobian transpose, but hey. That’s what they’re calling it.

The null space filter cancels out the acceleration effects of forces in operational space from a signal that is being applied as part of the control system.Â But it can also be used to cancel out the effects of any unwanted signal that can be modeled.Â Given some undesirable force signal interfering with the system that can be effectively modeled, a null space filtering term can be implemented to cancel it out.Â The control signal in this case, with one primary operational space controller and a null space filter for the undesired force, looks like:

$\textbf{u} = \textbf{J}^T_{ee}(\textbf{q}) \; \textbf{M}_\textbf{x}(\textbf{q}) \; \ddot{\textbf{x}}_\textrm{des} - \textbf{g}(\textbf{q}) - \textbf{J}^T_{ee}(\textbf{q}) \;\textbf{J}^{T+}_{ee}(\textbf{q}) \; \textbf{u}_{\textrm{undesired force}}.$

We did it! This will now allow a high-priority operational space controller to execute without interference from a secondary controller operating in its null space to complete it’s own set of goals (when possible).

Example:

Given a three link arm (revolute-revolute-revolute) operating in the $(x,z)$ plane, shown below:

this example will construct the control system for a primary controller controlling the end-effector and a secondary controller working to keep the arm near its joint angles’ default resting positions.

Let the system state be $\textbf{q} = [q_0, q_1, q_2]^T$Â with default positions $\textbf{q}^0 = \left[\frac{\pi}{3}, \frac{\pi}{4}, \frac{\pi}{4} \right]^T$.Â The control signal of the secondary controller is the difference between the target state and the current system state

$\textbf{u}_{\textrm{null}} = k_{p_{\textrm{null}}}(\textbf{q}^0 - \textbf{q}),$

where $k_{p_\textrm{null}}$ is a gain term.

Let the centres of mass be

$\textrm{com}_0 = \left[ \begin{array}{c} \frac{1}{2}cos(q_0) \\ 0 \\ \frac{1}{2}sin(q_0) \end{array} \right], \;\;\;\; \textrm{com}_1 = \left[ \begin{array}{c} \frac{1}{4}cos(q_1) \\ 0 \\ \frac{1}{4}sin(q_1) \end{array} \right] \;\;\;\; \textrm{com}_2 = \left[ \begin{array}{c} \frac{1}{2}cos(q_2) \\ 0 \\ \frac{1}{4} sin (q_2) \end{array} \right],$

the Jacobians for the COMs are

$\textbf{J}_0(\textbf{q}) = \left[ \begin{array}{ccc} -\frac{1}{2} sin(q_0) & 0 & 0 \\ 0 & 0 & 0 \\ \frac{1}{2} cos(q_0) & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right],$

$\textbf{J}_1(\textbf{q}) = \left[ \begin{array}{ccc} -L_0sin(q_0) - \frac{1}{4}sin(q_{01}) & -\frac{1}{4}sin(q_{01}) & 0 \\ 0 & 0 & 0 \\ L_0 cos(q_0) + \frac{1}{4} cos(q_{01})& \frac{1}{4} cos(q_{01}) & 0 \\ 0 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right]$

$\textbf{J}_2(\textbf{q}) = \left[ \begin{array}{ccc} -L_0sin(q_0) - L_1sin(q_{01}) - \frac{1}{2}sin(q_{012}) & -L_1sin(q_{01}) - \frac{1}{2}sin(q_{012}) & - \frac{1}{2}sin(q_{012}) \\ 0 & 0 & 0 \\ L_0 cos(q_0) + L_1 cos(q_{01}) + \frac{1}{4}cos(q_{012}) & L_1 cos(q_{01}) + \frac{1}{4} cos(q_{012}) & \frac{1}{4}cos(q_{012}) \\ 0 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & 0 & 0 \end{array} \right].$

The Jacobian for the end-effector of this three link arm is

$\textbf{J}_{ee} = \left[ \begin{array}{ccc} -L_0 sin(q_0) - L_1 sin(q_{01}) - L_2 sin(q_{012}) & - L_1 sin(q_{01}) - L_2 sin(q_{012}) & - L_2 sin(q_{012}) \\ L_0 cos(q_0) + L_1 cos(q_{01}) + L_2 cos(q_{012}) & L_1 cos(q_{01}) + L_2 cos(q_{012}) & L_2 cos(q_{012}), \end{array} \right]$

where $q_{01} = q_0 + q_1$ and $q_{012} = q_0 + q_1 + q_2$.

Taking the control signal developed in Section~\ref{sec:exampleOS}

$\textbf{u} = \textbf{J}^T_{ee}(\textbf{q}) \; \textbf{M}_{\textbf{x}_{ee}}(\textbf{q}) [k_p (\textbf{x}_{\textrm{des}} - \textbf{x}) + k_v (\dot{\textbf{x}}_{\textrm{des}} - \dot{\textbf{x}})] - \textbf{g}(\textbf{q}),$

where $\textbf{M}_{\textbf{x}_{ee}}(\textbf{q})$ was defined in the previous post, and $\textbf{g}(\textbf{q})$ is defined two posts ago, and $k_p$ and $k_v$ are gain terms, usually set such that $k_v = \sqrt{k_p}$, and adding in the null space control signal and filter gives

$\textbf{u} = \textbf{J}^T_{ee}(\textbf{q}) \; \textbf{M}_{\textbf{x}_{ee}}(\textbf{q}) [k_p (\textbf{x}_{\textrm{des}} - \textbf{x}) + k_v (\dot{\textbf{x}}_{\textrm{des}} - \dot{\textbf{x}})] - (\textbf{I} - \textbf{J}^T_{ee}(\textbf{q}) \textbf{J}^{T+}_{ee}(\textbf{q})) \textbf{u}_{\textrm{null}} - \textbf{g}(\textbf{q}),$

where $\textbf{J}^{T+}(\textbf{q})$ is the dynamically consistent generalized inverse defined above,Â and $\textbf{u}_{\textrm{null}}$ is our null space signal!

Conclusions

It’s a lot of math, but when you start to get a feel for it what’s really awesome is that this is it. We’re describing the whole system, and so by working with these equations we can get a super effective controller. Which is pretty cool. Especially in relation to other possible controllers.

Alright! We’ve now worked through all the basic theory for operational space control, it is time to get some implementations going.

## Robot control part 4: Operation space control

In this post we’ll look at operational space control and how to derive the control equations. I’d like to mention again that these posts have all come about as a result of me reading and working through Samir Menon’s operational space control tutorial, where he works through an implementation example on a revolute-prismatic-prismatic robot arm.

Generalized coordinates vs operational space

The term generalized coordinates refers to a characterization of the system that uniquely defines its configuration. For example, if our robot has 7 degrees of freedom, then there are 7 state variables, such that when all these variables are given we can fully account for the position of the robot. In the previous posts of this series we’ve been describing robotic arms in joint space, and for these systems joint space is an example of generalized coordinates. This means that if we know the angles of all of the joints, we can draw out exactly what position that robot is in. An example of a coordinate system that does not uniquely define the configuration of a robotic arm would be one that describes only the $x$ position of the end-effector.

So generalized coordinates tell us everything we need to know about where the robot is, that’s great. The problem with generalized coordinates, though, is that planning trajectories in this space for tasks that we’re interested in performing tends not to be straight forward. For example, if we have a robotic arm, and we want to control the position of the end-effector, it’s not obvious how to control the position of the end-effector by specifying a trajectory for each of the arm’s joints to follow through joint space.

The idea behind operational space control is to abstract away from the generalized coordinates of the system and plan a trajectory in a coordinate system that is directly relevant to the task that we wish to perform. Going back to the common end-effector position control situation, we would like to operate our arm in 3D $(x,y,z)$ Cartesian space. In this space, it’s obvious what trajectory to follow to move the end-effector between two positions (most of the time it will just be a straight line in each dimension). So our goal is to build a control system that lets us specify a trajectory in our task space and will transform this signal into generalized coordinates that it can then send out to the system for execution.

Operational space control of simple robot arm

Alright, we’re going to work through an example. The generalized coordinates for this example is going to be joint space, and the operational space is going to be the end-effector Cartesian coordinates relative to the a reference frame attached to the base. Recycling the robot from the second post in this series, here’s the set up we’ll be working with:

Once again, we’re going to need to find the Jacobians for the end-effector of the robot. Fortunately, we’ve already done this:

$\textbf{J} = \left[ \begin{array}{cc} -L_0 sin(\theta_0) - L_1 sin(\theta_0 + \theta_1) & - L_1 sin(\theta_0 + \theta_1) \\ L_0 cos(\theta_0) + L_1 cos(\theta_0 + \theta_1) & L_1 cos(\theta_0 + \theta_1) \\ 0 & 0 \\ 0 & 0 \\ 0 & 0 \\ 1 & 1 \end{array} \right]$

Great! So now that we have $\textbf{J}$, we can go ahead and transform forces from end-effector (hand) space to joint space as we discussed in the second post:

$\textbf{u} = \textbf{J}_{ee}^T(\textbf{q}) \; \textbf{F}_{\textbf{x}}.$

Rewriting $\textbf{F}_\textbf{x}$ as its component parts

$\textbf{F}_{\textbf{x}} = \textbf{M}_{\textbf{x}_{ee}}(\textbf{q}) \; \ddot{\textbf{x}}_\textrm{des},$

where $\ddot{\textbf{x}}$ is end-effector acceleration, and $\textbf{M}_{\textbf{x}_{ee}(\textbf{q})}$ is the inertia matrix in operational space. Unfortunately, this isn’t just the normal inertia matrix, so let’s take a look here at how to go about deriving it.

Inertia in operational space

Being able to calculate $\textbf{M}(\textbf{q})$ allows inertia to be cancelled out in joint-space by incorporating it into the control signal, but to cancel out the inertia of the system in operational space more work is still required.Â The first step will be calculating the acceleration in operational space.Â This can be found by taking the time derivative of our original Jacobian equation.

$\frac{d}{d t}\dot{\textbf{x}} = \frac{d}{d t} (\textbf{J}_{ee}(\textbf{q}) \; \dot{\textbf{q}}),$

$\ddot{\textbf{x}} = \dot{\textbf{J}}_{ee}(\textbf{q}) \; \dot{\textbf{q}} + \textbf{J}_{ee} (\textbf{q})\; \ddot{\textbf{q}}.$

Substituting in the dynamics of the system, as defined in the previous post, but ignoring the effects of gravity for now, gives:

$\ddot{\textbf{x}} = \dot{\textbf{J}}_{ee}(\textbf{q}) \; \dot{\textbf{q}} + \textbf{J}_{ee} (\textbf{q})\; \textbf{M}^{-1}(\textbf{q}) [ \textbf{u} - \textbf{C}(\textbf{q}, \dot{\textbf{q}})].$

Define the control signal

$\textbf{u} = \textbf{J}_{ee}^T(\textbf{q}) \textbf{F}_\textbf{x},$

where substituting in for $\textbf{F}_\textbf{x}$, the desired end-effector force, gives

$\textbf{u} = \textbf{J}_{ee}^T(\textbf{q})\; \textbf{M}_{\textbf{x}_{ee}}(\textbf{q})\; \ddot{\textbf{x}}_\textrm{des},$

where $\ddot{\textbf{x}}_\textrm{des}$ denotes the desired end-effector acceleration.Â Substituting the above equation into our equation for acceleration in operational space gives

$\ddot{\textbf{x}} = \dot{\textbf{J}}_{ee}(\textbf{q}) \; \dot{\textbf{q}} + \textbf{J}_{ee} (\textbf{q})\; \textbf{M}^{-1}(\textbf{q}) [ \textbf{J}_{ee}^T(\textbf{q})\; \textbf{M}_{\textbf{x}_{ee}}(\textbf{q})\; \ddot{\textbf{x}}_\textrm{des} - \textbf{C}(\textbf{q}, \dot{\textbf{q}})].$

$\ddot{\textbf{x}} = \textbf{J}_{ee}(\textbf{q})\; \textbf{M}^{-1}(\textbf{q}) \; \textbf{J}_{ee}^T(\textbf{q})\; \textbf{M}_{\textbf{x}_{ee}}(\textbf{q})\; \ddot{\textbf{x}}_\textrm{des} + [\dot{\textbf{J}}_{ee}(\textbf{q}) \; \dot{\textbf{q}} - \textbf{J}_{ee}(\textbf{q})\textbf{M}^{-1}(\textbf{q}) \; \textbf{C}(\textbf{q}, \dot{\textbf{q}})],$

the last term is ignored due to the complexity of modeling it, resulting in

$\ddot{\textbf{x}} = \textbf{J}_{ee}(\textbf{q})\; \textbf{M}^{-1}(\textbf{q}) \textbf{J}_{ee}^T(\textbf{q})\; \textbf{M}_{\textbf{x}_{ee}}(\textbf{q})\; \ddot{\textbf{x}}_\textrm{des}.$

At this point, to get the dynamics $\ddot{\textbf{x}}$ to be equal to the desired acceleration $\ddot{\textbf{x}}_\textrm{des}$, the end-effector inertia matrix $\textbf{M}_{\textbf{x}_{ee}}$ needs to be chosen carefully.Â By setting

$\textbf{M}_{\textbf{x}_{ee}}(\textbf{q}) = [\textbf{J}_{ee}(\textbf{q}) \; \textbf{M}^{-1}(\textbf{q}) \; \textbf{J}_{ee}^T(\textbf{q})]^{-1},$

we now get

$\ddot{\textbf{x}} = \textbf{J}_{ee}(\textbf{q})\; \textbf{M}^{-1}(\textbf{q}) \textbf{J}_{ee}^T(\textbf{q})\; [\textbf{J}_{ee}(\textbf{q}) \; \textbf{M}^{-1}(\textbf{q}) \; \textbf{J}_{ee}^T(\textbf{q})]^{-1} \; \ddot{\textbf{x}}_\textrm{des},$

$\ddot{\textbf{x}} = \ddot{\textbf{x}}_\textrm{des}.$

And that’s why and how the inertia matrix in operational space is defined!

The whole signal

Going back to the control signal we were building, let’s now add in a term to cancel the effects of gravity in joint space. This gives

$\textbf{u} = \textbf{J}_{ee}^T(\textbf{q}) \textbf{M}_{\textbf{x}_{ee}}(\textbf{q}) \ddot{\textbf{x}}_\textrm{des} + \textbf{g}(\textbf{q}),$

where $\textbf{g}(\textbf{q})$ is the same as defined in the previous post.Â This controller converts desired end-effector acceleration into torque commands, and compensates for inertia and gravity.

Defining a basic PD controller in operational space

$\ddot{\textbf{x}}_\textrm{des} = k_p (\textbf{x}_{\textrm{des}} - \textbf{x}) + k_v (\dot{\textbf{x}}_{\textrm{des}} - \dot{\textbf{x}}),$

and the full equation for the operational space control signal in joint space is:

$\textbf{u} = \textbf{J}_{ee}^T(\textbf{q}) \; \textbf{M}_{\textbf{x}_{ee}}(\textbf{q}) [k_p (\textbf{x}_{\textrm{des}} - \textbf{x}) + k_v (\dot{\textbf{x}}_{\textrm{des}} - \dot{\textbf{x}})] + \textbf{g}(\textbf{q}).$

Hurray! That was relatively simple. The great thing about this, though, is that it’s the same process for any robot arm! So go out there and start building controllers! Find your robot’s mass matrix and gravity term in generalized coordinates, the Jacobian for the end effector, and you’re in business.

Conclusions

So, this feels a little anticlimactic without an actual simulation / implementation of operational space, but don’t worry! As avid readers (haha) will remember, a while back I worked out how to import some very realistic MapleSim arm simulations into Python for use with some Python controllers. This seems a perfect application opportunity, so that’s next! A good chance to work through writing the controllers for different arms and also a chance to play with controllers operating in null spaces and all the like.

Actual simulation implementations will also be a good chance to play with trying to incorporate those other force terms into the control equation, and get to see the results without worrying about breaking an actual robot. In actual robots a lot of the time you leave out anything where your model might be inaccurate because the last thing to do is falsely compensate for some forces and end up injecting energy into your system, making it unstable.

There’s still some more theory to work through though, so I’d like to do that before I get to implementing simulations. One more theory post, and then we’ll get back to code!