Single Inverted Pendulum¶

First we need the model of the single inverted pendulum. Friedland works it mostly out for us in Chapter 2.

This looks reasonable for hanging down at least.

(The nonlinear state equations, which come from (2E.3) for $\ddot y$ and $\ddot \theta$ are figured out below. )

This is with the pendulum driven by a force. In problem 2.2 which you did in Homework 2, we have the model when driven by an electric motor, which is what we have with our hardware. Here is the solution from Joe Dinius.

Problem 2.1¶

The cart pendulum equations-of-motion are:

$$ \ddot y + \frac{mg}{M} \theta = \frac{f}{M} \\ \ddot \theta - \frac{M+m}{M \ell} g \theta = -\frac{f}{M \ell} $$

$\tau = rf$, $r$ is the torque-to-linear-force factor. For the electric motor:

$$ \tau = -\frac{k^2}{R} \omega + \frac{K}{R}e \\ = rf \\ \implies f = -\frac{k^2}{Rr} \omega + \frac{K}{Rr}e $$

The linear acceleration, $\ddot y$, is equal to $\ddot y = r \dot \omega \implies \dot y = r \omega$ by Newton's law. This implies: $$ f = -\frac{k^2}{Rr^2} \dot y + \frac{K}{Rr}e $$

Plug in $f$ to the first equation: $$ \ddot y + \frac{mg}{M} \theta = -\frac{k^2}{MRr^2} \dot y + \frac{K}{MRr}e \\ \ddot \theta - \frac{M+m}{M \ell} g \theta = \frac{k^2}{MRr^2 \ell} \dot y - \frac{K}{MRr \ell}e $$

Move state derivatives to the LHS: $$ \ddot y +\frac{k^2}{MRr^2} \dot y + \frac{mg}{M} \theta = \frac{K}{MRr}e \\ \ddot \theta - \frac{k^2}{MRr^2 \ell} \dot y - \frac{M+m}{M \ell} g \theta = - \frac{K}{MRr \ell}e $$

The next step is to take this linearized state model and put it into state form. I am going to put the variable $s$ into the equations where $s=1$ for the pendulum up, and $s= -1$ for the pendulum down.

$$\begin{bmatrix} \dot y \\ \dot \theta \\ \ddot {y} \\ \ddot {\theta} \end{bmatrix} = \begin{bmatrix} 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 1 \\ 0 && -mg/M && -k^2/(MRr^2)&& 0 \\ 0 && s(M+m)g/(Ml) && sk^2/(MRr^2l) && 0 \end{bmatrix} \begin{bmatrix} y \\ \theta \\ \dot{y} \\ \dot {\theta} \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ K/(MRr) \\ -sK/(MRrl) \end{bmatrix} e $$

so $$\mathbf A = \begin{bmatrix} 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 1 \\ 0 && -mg/M && -k^2/(MRr^2)&& 0 \\ 0 && s(M+m)g/(Ml) && sk^2/(MRr^2l) && 0 \end{bmatrix}$$ and $$\mathbf B = \begin{bmatrix} 0 \\ 0 \\ K/(MRr) \\ -sK/(MRrl) \end{bmatrix}$$

For an ideal motor, the torque constant is equal to the back electromotive force constant, so $k = K$. The applied voltage is $e$. The mass of the cart is $M$; the resistance of the motor is $R$; the radius of the wheels (or pulley in our case) is $r$; and the mass of the pendulum is $m$, with $g$ for gravity. For our pendulum, we might need to re-write these in terms of their moments of inertia sometime.

For now let's assume all the states are available, so $$ \mathbf C = \begin{bmatrix} 1 && 0 && 0 && 0 \\ 0 && 1 && 0 && 0 \\ 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 1 \end{bmatrix}$$

Let's see if our equations are good by simulating the system to see if it works like we expect.

Unstable Pendulum (Standing up)¶

Let's begin with an investigation when the pendulum is standing up, not hanging down. The model allows this with $s=1$.

Steve Brunton has some interesting things to say about how controllable a system is on his video here. I want to check the single pendulum to see how controllable it is. Steve describes how you can tell what directions are more controllable by doing the singular value decomposition of $\mathbf Q$ because Grammian $\mathbf W_t \approx \mathbf {Q Q}^T$.

The second eigenvector is the unstable one. Note its direction. Then look at the least controllable direction. Note its direction. We don't want them in the same direction. To find the angle between them, we find the dot or inner product between them. They both appear to be unit vectors the way python calculates them.

The inner product is small and the length of the vectors that were compared is unity, so that is good! It means they are practically orthogonal. Now let's check to see if we place the poles at the same pole locations we started with, the gain coefficients are all come out very close to zero. This is just a check of our the place algorithm.

Stable Pendulum Simulation (Hanging Down)¶

Remember this is the result of a simulation of a linear model. The question to ask is, "Is this accurate?" To check this out, remember the approximations we made were $cos \theta \approx 1$ and $sin\theta \approx \theta$. For the largest angles we see this works out to be:

This isn't the only source of error in our simulation. There are at least a couple others. One is due to not knowing the actual parameters very precisely. An approximate error analysis can be had by looking at the elements in the $A$ matrix to see how they vary as a function of each error. For example if a coefficient, $a_{ij}=f(m,M,r,R,k,K,l)$ then the error is approximately: $$da_{ij} = \frac {\partial a_{ij}} {\partial m} dm+ \frac {\partial a_{ij}} {\partial M} dM+\frac {\partial a_{ij}} {\partial r} dr+\frac {\partial a_{ij}} {\partial R} dR+\frac {\partial a_{ij}} {\partial k} dk+\frac {\partial a_{ij}} {\partial K} dK+\frac {\partial a_{ij}} {\partial l} dl$$

The only reason it damps out is the resistance of the motor windings. We modeled no friction. The underdamped responses (from complex poles) are expected. There could be a damping effect from friction as well. To tell if we should add that to the model, the thing to do is to actually hang the pendulum down, and see if the decay time of the oscillating wave is consistent with what the linear model predicts.

Not changing the pole locations results in a zero gain matrix as expected.

Let's see if we can change the behavior of the pendulum so it is like it was in honey, dieing out in 1, 2, 3, and 4 seconds. We will place the poles at -1, -2, -3 and -4. Then do the simulation.

Note how the cart moves to keep the natural swinging behavior from happening. The reason it looks a little oscillatory is because some eigenmodes damp out faster and some slower.

Note that the maximum angle is still $0.1^o$, so our previous check of the angles still applies.

It looks like we were successful placing the poles where we wanted them.

I would like to know how much difference friction is making for the pendulum. The simulation above may seem like a good place to start, but it assumes the input voltage is zero (shorted), but in that configuration, the resistance of the motor windings, $R$, causes damping. So it would be nice to have a model where the current through the motor is zero. This configuration would have no damping due to $i^2R$ losses in the motor.

Zero Current Input Model¶

In this case, the torque $\tau$ is zero, because there is no Lorentz force with zero current. This is as if the force was set to zero in the original pendulum model before incorporating the motor. In this case the state model is:

With the s variable included this becomes:

$$\begin{bmatrix} \dot y \\ \dot \theta \\ \ddot {y} \\ \ddot {\theta} \end{bmatrix} = \begin{bmatrix} 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 1 \\ 0 && -mg/M && 0 && 0 \\ 0 && s(M+m)g/(Ml) && 0 && 0 \end{bmatrix} \begin{bmatrix} y \\ \theta \\ \dot{y} \\ \dot {\theta} \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} u $$

This is like setting $k=K=0$ in the model. Let's see if that does not decay, as we think it should. It should also give us a better measurement of the viscous friction in the real system.

We should check the real pendulum and make sure the pendulum period is $T_p = (2\pi)/5.78144 = 1.0868$ seconds, assuming the damping is negligible. If not, we should see what happens to this frequency as we add damping, and adjust our parameters so the period is correct based on what we think may be amiss.

Model Verification and Adjustment¶

In this section we devise various ways to verify the model parameters and modify the model as seems best. To do this we use the pendulum hanging down using the models for that developed above to verify/adjust and adjust the parameters for $\theta$, and then we put the pendulum on it's side and make a model to verify/adjust the motor parameters.

To do this we need to understand how the WWU pendulum works and how to operate it. This the the WWU inverted pendulum.

The pendulum motors and shaft angle encoders are controlled by a dedicated digital controller using a Xylinx FPGA that communicates with the user over Ethernet. To control it you use Ralph's code from the ctrlbox.m file below. You have to setup your computer to communicate with the pendulum at address 169.254.0.100. Here are the settings for Ubuntu.

You can check it works as shown below (on linux). If you can ping the pendulum, you can communicate with it.

There are two physical pendula. They have slightly different characteristics, and so the constants are different for each.

The pendulum needs to know where down is, and where the center of the track is as well. To set those you hit the reset button on the controller. Sometimes you cannot ping the pendulum, and if you reset it, you can. Here is a photo showing which button is the reset button.

The ctrlbox.m file is shown below. It contains the octave/MATLAB routines to communicate with the pendulum. Ralph Stirling is the one who wrote them, and he is also the one who wrote the FPGA controller code to communicate with the Wiznet Ethernet device that the controller uses to send the data over Ethernet.

Now we test the pendulum hanging down to see how it matches our model.

Here is a routine to test the pendulum. This helps us understand how things work. You need to install the octave-sockets package if you don't have it already.

The results from this simulation done with the motor open circuited, so the $R$ has no effect on the slow down of the pendulum seem to indicate that the major contribution to the pendulum friction is not viscous friction. It seems to be a constant torque opposing motion, and it seems to drop off linearly with time instead of exponentially. This paper analyzes this kind of motion. Because it has the term $sgn(\dot \theta)$ in the torque because it changes sign when the angular velocity does, it would require changing models when $\dot \theta$ changes sign. I don't want to mess with that at the moment. I suspect the friction due to the cart moving on the rail is a bigger deal based on just feeling it with my fingers. On that one, I'm not quite sure how to measure it, and I think it will likely also be a force with a $sgn(\dot y)$, I'm going to leave that out of my model too. It may be possible to model those forces as proportional to $\dot \theta$ and $\dot y$, but they would give an exponential decay, and at least for the pendulum, it isn't exponential. However, I do note from the paper that the frequency of oscillation is not a function of the friction, and I also note that the simulation and the measurement are a bit different in the frequency of oscillation. I will take that into account by adjusting the length $l$ to the center of mass of the pendulum. It appears that we measure five periods in about $4.7$ seconds, so the angular frequency is $2 \pi 5/4.7= 6.68$ radians per second. Our simulation has nine periods in about $9.8$ seconds, so that gives an angular frequency of $2 \pi 9/9.8 = 5.7703$ which reminds me we can just get it from the eigenvalue, which was $5.78144$. The characteristic polynomial of the $A$ matrix without the motor connected is $(Mls^4+Mgs^2+gms^2)/(Ml)$. This makes it obvious there are two poles at zero, and the remainder of the poles satisfy $(Mls^2+Mg+gm)=0$ and so the poles are at $\pm j\sqrt{g(1+m/M)/l}$ and this agrees with what octave calculated above. I am not quite sure why, but it seems the of moment of inertia is not quite right, so I am going to reduce the length of the rod to get agreement. The factor for which the resonant frequency is wrong is $6.68/5.78 = \sqrt{l_{old}/l_{new}} = \sqrt{0.323/l_{new}}$ so $l_{new}$ is calculated below as 0.242. The analysis below verifies this change fixes things.

To start the pendulum, you need to let it hang down until it isn't moving. Then you push the calibrate button on the FPGA board to tell it that is hanging down. Then you set the initial condition and start your program. The pause statement in octave lets you start it when you want with a single keystroke.

Cart Falling Down¶

The force the motor develops is important. We need to verify it, especially as some of the code from previous years seems to indicate that we might not have a good handle on the scale constant for the motor drive.
If $J$ is the moment of inertia of the system: