PID control | studywolf

In the exciting previous post we looked at how to go about generating a Jacobian matrix, which we could use to transformation both from joint angle velocities to end-effector velocities, and from desired end-effector forces into joint angle torques. I briefly mentioned right at the end that using just this force transformation to build your control signal was only appropriate for very simple systems that didn’t have to account for things like arm-link mass or gravity.

In general, however, mass and gravity must be accounted for and cancelled out. The full dynamics of a robot arm are

$\textbf{M}(\textbf{q}) \ddot{\textbf{q}} = (\textbf{u} - \textbf{C}(\textbf{q}, \dot{\textbf{q}}) - \textbf{g}(\textbf{q})) ,$

where $\ddot{\textbf{q}}$ is joint angle acceleration, $\textbf{u}$ is the control signal (specifying torque), $\textbf{C}(\textbf{q}, \dot{\textbf{q}})$ is a function describing the Coriolis and centrifugal effects, $\textbf{g}(\textbf{q})$ is the effect of gravity in joint space, and $\textbf{M}$ is the mass matrix of the system in joint space.

There are a lot of terms involved in the system acceleration, so while the Jacobian can be used to transform forces between coordinate systems it is clear that just setting the control signal $\textbf{u} = \textbf{J}_{ee}^T (\textbf{q})\textbf{F}_\textbf{x}$ is not sufficient, because a lot of the dynamics affecting acceleration aren’t accounted for. In this section an effective PD controller operating in joint space will be developed that will allow for more precise control by cancelling out unwanted acceleration terms. To do this the effects of inertia and gravity need to be calculated.

Accounting for inertia

The fact that systems have mass is a pain in our controller’s side because it introduces inertia into our system, making control of how the system will move at any given point in time more difficult. Mass can be thought of as an object’s unwillingness to respond to applied forces. The heavier something is, the more resistant it is to acceleration, and the force required to move a system along a desired trajectory depends on both the object’s mass and its current acceleration.

To effectively control a system, the system inertia needs to be calculated so that it can be included in the control signal and cancelled out.

Given the robot arm above, operating in the $(x,z)$ plane, with the $y$ axis extending into the picture where the yellow circles represent each links centre-of-mass (COM). The position of each link is COM is defined relative to that link’s reference frame, and the goal is to figure out how much each link’s mass will affect the system dynamics.

The first step is to transform the representation of each of the COM from Cartesian coordinates in the reference frame of their respective arm segments into terms of joint angles, such that the Jacobian for each COM can be calculated.

Let the COM positions relative to each segment’s coordinate frame be

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

The first segment’s COM is already in base coordinates (since the first link and the base share the same coordinate frame), so all that is required is the position of the second link’s COM in the base reference frame, which can be done with the transformation matrix

$^1_0\textbf{T} = \left[ \begin{array}{cccc} cos(q_1) & 0 & -sin(q_1) & L_0 cos(q_0) \\ 0 & 1 & 0 & 0 \\ sin(q_1) & 0 & cos(q_1) & L_0 sin(q_0) \\ 0 & 0 & 0 & 1 \end{array} \right].$

Using $^1_0\textbf{T}$ to transform the $\textrm{com}_1$ gives

$^1_0\textbf{T} \; \textrm{com}_1 = \left[ \begin{array}{cccc} cos(q_1) & 0 & -sin(q_1) & L_0 cos(q_0) \\ 0 & 1 & 0 & 0 \\ sin(q_1) & 0 & cos(q_1) & L_0 sin(q_0) \\ 0 & 0 & 0 & 1 \end{array} \right] \; \; \left[ \begin{array}{c} \frac{1}{4}cos(q_1) \\ 0 \\ \frac{1}{4}sin(q_1) \\ 1 \end{array} \right]$

$^1_0\textbf{T} \; \textrm{com}_1 = \left[ \begin{array}{c} L_0 cos(q_0) + \frac{1}{4}cos(q_0 + q_1) \\ 0 \\ L_0 sin(q_0) + \frac{1}{4} cos(q_0 + q_1) \\ 1 \end{array} \right].$

To see the full computation worked out explicitly please see my previous robot control post.

Now that we have the COM positions in terms of joint angles, we can find the Jacobians for each point through our Jacobian equation:

$\textbf{J} = \frac{\partial \textbf{x}}{\partial \textbf{q}}$ .

Using this for each link gives us:

$\textbf{J}_0 = \left[ \begin{array}{cc} -\frac{1}{2}sin(q_0) & 0 \\ 0 & 0 \\ \frac{1}{2} cos(q_0) & 0 \\ 0 & 0 \\ 1 & 0 \\ 0 & 0 \end{array} \right]$
$\textbf{J}_1 = \left[ \begin{array}{cc} -L_0sin(q_0) -\frac{1}{4}sin(\theta_0 + q_1) & -\frac{1}{4} sin(q_0 + \theta_1) \\ 0 & 0 \\ L_0 cos(q_0) + \frac{1}{4}cos(q_0 + q_1) & \frac{1}{4} cos(q_0 +q_1) \\ 0 & 0 \\ 1 & 1 \\ 0 & 0 \end{array} \right]$ .

Kinetic energy

The total energy of a system can be calculated as a sum of the energy introduced from each source. The Jacobians just derived will be used to calculate the kinetic energy each link generates during motion. Each link’s kinetic energy will be calculated and summed to get the total energy introduced into the system by the mass and configuration of each link.

Kinetic energy (KE) is one half of mass times velocity squared:

$\textrm{KE} = \frac{1}{2} \; \dot{\textbf{x}}^T \textbf{M}_\textbf{x}(\textbf{q}) \; \dot{\textbf{x}},$

where $\textbf{M}_\textbf{x}$ is the mass matrix of the system, with the subscript $\textbf{x}$ denoting that it is defined in Cartesian space, and $\dot{\textbf{x}}$ is a velocity vector, where $\dot{\textbf{x}}$ is of the form

$\dot{\textbf{x}} = \left[ \begin{array}{c} \dot{x} \\ \dot{y} \\ \dot{z} \\ \dot{\omega_x} \\ \dot{\omega_y} \\ \dot{\omega_z} \end{array} \right],$

and the mass matrix is structured

$\textbf{M}_{\textbf{x}_i} (\textbf{q})= \left[ \begin{array}{cccccc} m_i & 0 & 0 & 0 & 0 & 0 \\ 0 & m_i & 0 & 0 & 0 & 0 \\ 0 & 0 & m_i & 0 & 0 & 0 \\ 0 & 0 & 0 & I_{xx} & I_{xy} & I_{xz} \\ 0 & 0 & 0 & I_{yx} & I_{yy} & I_{yz} \\ 0 & 0 & 0 & I_{zx} & I_{zy} & I_{zz} \end{array} \right],$

where $m_i$ is the mass of COM $i$ , and the $I_{ij}$ terms are the moments of inertia, which define the object’s resistance to change in angular velocity about the axes, the same way that the mass element defines the object’s resistance to changes in linear velocity.

As mentioned above, the mass matrix for the COM of each link is defined in Cartesian coordinates in its respective arm segment’s reference frame. The effects of mass need to be found in joint angle space, however, because that is where the controller operates. Looking at the summation of the KE introduced by each COM:

$\textrm{KE} = \frac{1}{2} \; \Sigma_{i=0}^n ( \dot{\textbf{x}}_i^T \textbf{M}_{\textbf{x}_i}(\textbf{q}) \; \dot{\textbf{x}}_i),$

and substituting in $\dot{\textbf{x}} = \textbf{J} \; \dot{\textbf{q}}$ ,

$\textrm{KE}_i \ \frac{1}{2} \; \Sigma_{i=0}^n (\dot{\textbf{q}}^T \; \textbf{J}_i^T \textbf{M}_{\textbf{x}_i}(\textbf{q})\textbf{J}_i \; \dot{\textbf{q}}),$

and moving the $\dot{\textbf{q}}$ terms outside the summation,

$\textrm{KE}_i = \frac{1}{2} \; \dot{\textbf{q}}^T \; \Sigma_{i=0}^n (\textbf{J}_i^T \textbf{M}_{\textbf{x}_i}(\textbf{q}) \textbf{J}_i) \; \dot{\textbf{q}}.$

Defining

$\textbf{M}(\textbf{q}) = \Sigma_{i=0}^n \; \textbf{J}_i^T(\textbf{q}) \textbf{M}_{\textbf{x}_i}(\textbf{q}) \; \textbf{J}_i(\textbf{q}),$

gives

$\textrm{KE} = \frac{1}{2} \; \dot{\textbf{q}}^T \; \textbf{M}(\textbf{q}) \; \dot{\textbf{q}},$

which is the equation for calculating kinetic energy in joint space. Thus, $\textbf{M}(\textbf{q})$ denotes the inertia matrix in joint space.

Now that we’ve successfully calculated the mass matrix of the system in joint space, we can incorporate it into our control signal and cancel out its effects on the system dynamics! On to the next problem!

Accounting for gravity

With the forces of inertia accounted for, we can now address the problem of gravity. To compensate for gravity the concept of conservation of energy (i.e. the work done by gravity is the same in all coordinate systems) will once again be pulled out. The controller operates by applying torque on joints, so it is necessary to be able to calculate the effect of gravity in joint space to cancel it out.

While the effect of gravity in joint space isn’t obvious, it is quite easily defined in Cartesian coordinates in the base frame of reference. Here, the work done by gravity is simply the summation of the distance each link’s center of mass has moved multiplied by the force of gravity. Where the force of gravity in Cartesian space is the mass of the object multiplied by -9.8m/s $^2$ along the $z$ axis, the equation for the work done by gravity is written:

$\textbf{W}_g = \Sigma^n_{i=0} (\textbf{F}_{g_i}^T \dot{\textbf{x}}_i),$

where $\textbf{F}_{g_i}$ is the force of gravity on the $i$ th arm segment. Because of the conservation of energy, the equation for work is equivalent when calculated in joint space, substituting into the above equation with the equation for work:

$\textbf{F}_\textbf{q}^T \dot{\textbf{q}} = \Sigma^n_{i=0} (\textbf{F}_{g_i}^T \dot{\textbf{x}}_i),$

and then substitute in using $\dot{\textbf{x}}_i = \textbf{J}_i(\textbf{q}) \; \dot{\textbf{q}}$ ,

$\textbf{F}_\textbf{q}^T \dot{\textbf{q}} = \Sigma^n_{i=0} (\textbf{F}_{g_i}^T \textbf{J}_i(\textbf{q}) \; \dot{\textbf{q}}),$

and cancelling out the $\dot{\textbf{q}}$ terms on both sides,

$\textbf{F}_\textbf{q}^T = \Sigma^n_{i=0} (\textbf{F}_{g_i}^T \textbf{J}_i(\textbf{q})),$

$\textbf{F}_\textbf{q} = \Sigma^n_{i=0} (\textbf{J}_i^T(\textbf{q}) \textbf{F}_{g_i}) = \textbf{g}(\textbf{q}),$

which says that to find the effect of gravity in joint space simply multiply the mass of each link by its Jacobian, multiplied by the force of gravity in $(x,y,z)$ space, and sum over each link. This summation gives the total effect of the gravity on the system.

Making a PD controller in joint space

We are now able to account for the energy in the system caused by inertia and gravity, great! Let’s use this to build a simple PD controller in joint space. Control should be very straight forward because once we cancel out the effects of gravity and inertia then we can almost pretend that the system behaves linearly. This means that we can also treat control of each of the joints independently, since their movements no longer affect one another. So in our control system we’re actually going to have a PD controller for each joint.

The above-mentioned nonlinearity that’s left in the system dynamics is due to the Coriolis and centrifugal effects. Now, these can be accounted for, but they require highly accurate model of the moments of inertia. If the moments are incorrect then the controller can actually introduce instability into the system, so it’s better if we just don’t address them.

Rewriting the system dynamics presented at the very top, in terms of acceleration gives

$\ddot{\textbf{q}} = \textbf{M}^{-1}(\textbf{q}) (\textbf{u} - \textbf{C}(\textbf{q}, \dot{\textbf{q}}) - \textbf{g}(\textbf{q})).$

Ideally, the control signal would be constructed

$\textbf{u} = (\textbf{M}(\textbf{q}) \; \ddot{\textbf{q}}_\textrm{des} + \textbf{C}(\textbf{q}, \dot{\textbf{q}}) + \textbf{g}(\textbf{q})),$

where $\ddot{\textbf{q}}_\textrm{des}$ is the desired acceleration of the system. This would result in system acceleration

$\ddot{\textbf{q}} = \textbf{M}^{-1}(\textbf{q})((\textbf{M}(\textbf{q}) \; \ddot{\textbf{q}}_\textrm{des} + \textbf{C}(\textbf{q}, \dot{\textbf{q}}) + \textbf{g}(\textbf{q})) - \textbf{C}(\textbf{q}, \dot{\textbf{q}}) - \textbf{g}(\textbf{q})),$

$\ddot{\textbf{q}} = \textbf{M}^{-1}(\textbf{q}) \textbf{M}(\textbf{q}) \; \ddot{\textbf{q}}_\textrm{des} ,$

$\ddot{\textbf{q}} = \ddot{\textbf{q}}_\textrm{des},$

which would be ideal. As mentioned, because the Coriolis and centrifugal effects are tricky to account for we’ll leave them out, so the instead the control signal is

$\textbf{u} = (\textbf{M}(\textbf{q}) \; \ddot{\textbf{q}}_\textrm{des} + \textbf{g}(\textbf{q})).$

Using a standard PD control formula to generate the desired acceleration:

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

where $k_p$ and $k_v$ are our gain values, and the control signal has been fully defined:

$\textbf{u} = (\textbf{M}(\textbf{q}) \; (k_p \; (\textbf{q}_{\textrm{des}} - \textbf{q}) + k_v \; (\dot{\textbf{q}}_{\textrm{des}} - \dot{\textbf{q}})) + \textbf{g}(\textbf{q})),$

and we’ve successfully build an effective PD controller in joint space!

Conclusions

Here we looked at building a PD controller that operates in the joint space of a robotic arm that can cancel out the effects of inertia and gravity. By cancelling out the effects of inertia, we can treat control of each of the joints independently, effectively orthogonalizing their control. This makes PD control super easy, we just set up a simple controller for each joint. Also a neat thing is that all of the required calculations can be performed with algorithms of linear complexity, so it’s not a problem to do all of this super fast.

One of the finer points was that we ignored the Coriolis and centrifugal effects on the robot’s dynamics. This is because in the mass matrix model of the moments of inertia are notoriously hard to accurately capture on actual robots. Often you go based off of a CAD model of your robot and then have to do some fine-tuning by hand. So they will be unaccounted for in our control signal, but most of the time as long as you have a very short feedback loop you’ll be fine.

I am really enjoying working through this, as things build on each other so well here and we’re starting to be able to do some really interesting things with the relatively forward transformation matrices and Jacobians that we learned how to build in the previous posts. This was for a very simple robot, but excitingly the next step after this is moving on to operational space control! At last. From there, we’ll go on to look at more complex robotic situations where things like configuration redundancy are introduced and it’s not quite so straightforward.

studywolf

a blog for things I encounter while coding and researching neuroscience, motor control, and learning

Tag Archives: PID control

Robot control part 3: Accounting for mass and gravity