Open-loop Root Locus
Lead Controller
Selecting a Gain
Plotting the Closed-loop Response

The open-loop transfer function of the plant for the ball and beam experiment is given below:

The design criteria for this problem are:

• Settling time less than 3 seconds
• Overshoot less than 5%

To see the derivation of the equations for this problem refer to the ball and beam modeling page. A schematic of the closed loop system with a controller is given below:

Open-loop Root Locus

m = 0.111;
R = 0.015;
g = -9.8;
L = 1.0;
d = 0.03;
J = 9.99e-6;

K = (m*g*d)/(L*(J/R^2+m));   %simplifies input

num = [-K];
den = [1 0 0];

rlocus(num,den)

Now, run the m-file and you should see the following root locus plot:

As you can see the system has two poles at the origin which go off to infinity along the imaginary axes.

The design criteria can also be plotted onto the root locus using the sgrid command. This command generates a grid of constant damping ratio and natural frequency. The damping ratio and natural frequency were found using the following equation, which relates the them to our percent overshoot (PO) and settling time (Ts) requirements:

Note, that the equation with Ts is found by assuming settled is when the response remains within 2% of it's final value. From these equation damping ratio and natural frequency were found to be 0.7 and 1.9 respectively.

sgrid(0.70, 1.9)
axis([-5 5 -2 2])

The area between the two dotted diagnol lines represents locations where the percent overshoot is less than 5%. The area outside the curved line represents locations where the setttling time is less than 3 seconds. Note that no region of the plot falls within the design criteria shown be these lines. To remedy this and bring the root locus into the left-hand plane for stability we will try adding a lead-compensator to the system.

Lead Controller

where, the magnitude of zo is less than the magnitude of po.

Now, let us add the controller to the plant and view the root locus. We will position the zero near the origin to cancel out one of the poles. The pole of our compensator will be placed to the left of the origin to pull the root locus further into the left-hand plane. Add the following lines of Matlab code to your m-file.

zo = 0.01;
po = 5;

numlead = [1 zo];
denlead = [1 po];

numl = conv(num,numlead);
denl = conv(den,denlead);

rlocus(numl,denl)
sgrid(0.70, 1.9)

Now, the branches of the root locus are within our design criteria.

Selecting a Gain

[kc,poles]=rlocfind(numl,denl)

You should see in the Matlab command window something similar to the following (your numbers will be slightly different):

selected_point =
 
  -2.4988+ 1.2493i
 
 
kc =
 
   37.3131
 
 
poles =
 
  -2.4950+ 1.2493i
  -2.4950- 1.2493i
  -0.0101         

Plotting the Closed-loop Response

numl2 = kc*numl;
[numcl,dencl] = cloop(numl2,denl);
t=0:0.01:5;
figure
step(0.25*numcl,dencl,t)

From this plot we see that when a 0.25m step input is given to the system both the settling time and percent overshoot design criteria are met.

Note: A design problem does not necessarily have a unique answer. Using this method (or any other) may result in many different compensators. Try running your m-file several more times selecting a different point each time and study the effect this has on the step response. For practice you may also want to go back to the original open-loop root locus and try to find other ways to add zeros and poles to get a better response.

Root Locus Examples

Cruise Control | Motor Speed | Motor Position | Bus Suspension | Inverted Pendulum | Pitch Controller | Ball & Beam

Ball & Beam Examples

Modeling | PID | Root Locus | Frequency Response | State Space | Digital Control

Tutorials

Basics | Modeling | PID | Root Locus | Frequency Response | State Space | Digital Control | Examples