CTMS Example: Frequency Response Control of the Inverted Pendulum Model

Example: Solution to the Inverted Pendulum Problem Using Frequency Response Method

Open-loop Representation
Closed-loop response with no compensation
Closed-loop response with compensation
What happens to the cart's position?

The transfer function of the plant for this problem is given below:

where,

Note: There is a pole/zero cancellation in this transfer function. In previous examples these were removed from the transfer function. However, in this example they will be left in for reasons that will become clear later.

The design criteria (with the pendulum receiving a 1N impulse force from the cart) are:

Settling time of less than 5 seconds.
Pendulum should not move more than 0.05 radians away from the vertical.

To see how this problem was originally set up, consult the inverted pendulum modeling page.

Note: Before trying to work through this problem, it should be noted that this is a complicated problem to solve with the frequency response method. As you will soon find out, this problem has a pole in the right-half-plane, making it unstable. The frequency response method works best when the system is stable in the open loop. For this reason this is not a good example to follow if you are trying to learn how to use frequency response. This problem is for students who want to learn how to solve frequency response problems that are more complicated.

Open-loop Representation

The transfer function found from the Laplace transforms for the output Phi (the pendulum's angle) can be set up using MATLAB by entering the numerator and denominator as vectors. Create an m-file (or a '.m' file located in the same directory as MATLAB) and copy the following text to model the transfer function:

M = .5;
m = 0.2;
b = 0.1;
i = 0.006;
g = 9.8;
l = 0.3;

q = (M+m)*(i+m*l^2)-(m*l)^2;   %simplifies input

num = [m*l/q  0  0];
den = [1  b*(i+m*l^2)/q  -(M+m)*m*g*l/q  -b*m*g*l/q  0];
pendulum = tf(num,den)

Closed-loop response with no compensation

We will now design an inverted pendulum controller for an impulse force using Nyquist diagrams (we cannot use Bode plots because the system is unstable in open loop). Let's begin by looking at the block diagram for this system:

The best way to use MATLAB to solve this problem is to first change the schematic to something that can be modeled much more easily. Rearranging the picture, you can get the following new schematic:

Now we can begin our design. First, we will look at the poles and zeros of this function:

x = roots(num)
y = roots(den)

     x =
     	 0
     	 0

     y =
	 0
	-5.6041
	 5.5651
	-0.1428

As you already know, we have a pole-zero cancellation at the origin, as well as one positive, real pole in the right-half plane. This means that we will need one anti-clockwise encirclement of -1 in order to have a stable closed-loop system (Z = P + N; P = 1, N = -1). The following m-file will be very useful for designing our controller.

function[ ] = pend()

%define TF
num =  [4.5455 0 0];
den = [1.0000 0.1818 -31.1818 -4.4545 0];
pendulum = tf(num,den);
figure(1)

%ask user for controller
numc = input('numc?.........');
denc = input('denc?.........');
k    = input('K?............');

%view compensated system bode
contr = k*tf(numc,denc);
loop = contr*pendulum;
bode(loop)

%view compensated system nyquist
figure(2)
subplot (2,1,1)
nyquist(loop)

%view compensated CL system impulse response
subplot(2,1,2)
sys_cl = feedback(pendulum,contr);
impulse(sys_cl)

With this m-file we will now view the uncompensated system's Nyquist diagram by setting the controller numerator, denominator and gain equal to one. Enter pend at the command prompt, and enter 1 for numc, denc, and K. You should see the following plots in your screen:


	numc?.........1
	denc?.........1
	K?............1

Note: The phase portion of the Bode plot shown in your version of MATLAB may differ from the Bode plot in the tutorials by 360 degrees. Shifting the phase by 360 degrees does not change the relationship between the Bode plot and the dynamic response of the system.

Closed-loop response with compensation

The system is unstable in closed loop (no encirclements of -1). Our first step will be to add an integrator to cancel the extra zero at the origin (we will then have two poles and two zeros at the origin). Use the pend command again.


	numc?.........1
	denc?.........[1 0]
	K?............1

Notice that the nyquist diagram encircles the -1 point in a clockwise fashion. Now we have two poles in the right-half plane (Z= P + N = 1 + 1). We need to add phase in order to get an anti-clockwise encirclement. We will do this by adding a zero to our controller. For starters, we will place this zero at -1.


	numc?.........[1 1]
	denc?.........[1 0]
	K?............1

as you can see, this wasn't enough phase. The encirclement around -1 is still clockwise. We are going to need to add a second zero.


	numc?.........conv([1 1],[1 1])
	denc?.........[1 0]
	K?............1

We still have one clockwise encirclement of the -1 point. However, if we add some gain, we can make the system stable by shifting the nyquist plot to the left, moving the anti-clockwise circle around -1, so that N = -1.


	numc?.........conv([1 1],[1 1])
	denc?.........[1 0]
	K?............10

As you can see, the system is now stable. We can now concentrate on improving the response. We can modify the poles of the controller in order to do this. We have to keep in mind that small poles (close to the origin) will affect the response at small frequencies, while larger poles (farther from the origin) will affect the response at high frequencies. When designing via frequency response, we are interested in obtaining simple plots, since they will be easier to manipulate in order to achieve our design goal. Therefore, we will use these concepts to 'flatten' the frequency response (bode plot). At the same time, you will notice that the nyquist diagram will take an oval shape.

If we try different combinations of poles and gains, we can get a very reasonable response. (Enter the command axis([0 5 -0.05 0.1]) after the pend command.)


	numc?.........conv([1 1.1],[1 5])
	denc?.........[1 0]
	K?............10

Our response has met our design goals. Feel free to vary the parameters and observe what happens.

What happens to the cart's position?

At the beginning on this solution page, the block diagram for this problem was given. The diagram was not entirely complete. The block representing the the position was left out because that variable was not being controlled. It is interesting though, to see what is happening to the cart's position when the controller for the pendulum's angle is in place. To see this we need to consider the actual system block diagram:

Rearranging a little bit, you get the following block diagram:

The feedback loop represents the controller we have designed for the pendulum. The transfer function from the cart's position to the impulse force, with the frequency response feedback controller which we designed, is given as follows:

Transfer Function

Now that we have the transfer function for the entire system, let's take a look at the response. First we need the transfer function for the cart's position. To get this we need to go back to the Laplace transforms of the system equations and find the transfer function from X(s) to U(s). Below is this transfer function:

where,

For more about the Laplace transform please refer to the inverted pendulum modeling page.

Now, create a new m-file and run it in the command window:

M = .5;
m = 0.2;
b = 0.1;
i = 0.006;
g = 9.8;
l = 0.3;

q = (M+m)*(i+m*l^2)-(m*l)^2;   %simplifies input

num1 = [m*l/q  0  0];
den1 = [1  b*(i+m*l^2)/q  -(M+m)*m*g*l/q  -b*m*g*l/q  0];
G1 = tf(num1,den1);

num2 = [(i+m*l^2)/q  0  -m*g*l/q];
den2 = den1;
G2 = tf(num2,den2);

k = 10;
numc = conv([1 1.1],[1 5]);
denc = [1 0];
contr = tf(k*numc,denc);
sys_cl = feedback(1,contr*G1)*G2;
sys_cl = minreal(sys_cl);
t=0:0.01:100;
impulse(sys_cl,t)

The minreal command tells MATLAB to cancel out all common poles and zeros in the closed loop transfer function; this gives the impulse function better numerical properties. As you can see, the cart moves in the negative direction and stabilizes at about -0.18 meters. This design might work pretty well for the actual controller, assuming that the cart had that much room to move. Keep in mind that this was pure luck. We were not trying to design to stabilize the cart's position, and the fact that we have is a fortunate side effect.

Frequency Response Examples: Cruise Control | Motor Speed | Motor Position | Bus Suspension | Inverted Pendulum | Pitch Controller | Ball & Beam
Inverted Pendulum Examples: Modeling | PID | Root Locus | Frequency Response | State Space | Digital Control | Simulink
Tutorials: MATLAB Basics | MATLAB Modeling | PID Control | Root Locus | Frequency Response | State Space | Digital Control | Simulink Basics | Simulink Modeling | Examples