Plotting the frequency response using the bode command
Adding a two-lead controller
Plotting the closed-loop response

From the main problem, the dynamic equations in state-space form are the following:

For the original problem and the derivation of the above equations and schematic, please refer to the bus modeling page.

We want to design a feedback controller so that when the road disturbance (W) is simulated by a unit step input, the output (X1-X2) has a settling time less than 5 seconds and an overshoot less than 5%. For example, when the bus runs onto a 10 cm high step, the bus body will oscillate within a range of +/- 5 mm and will stop oscillating within 5 seconds.

The system model can be represented in Matlab by creating a new m-file and entering the following commands (refer to the main problem for the details of getting those commands).

m1=2500;
m2=320;
k=10000;
b=140000;

A=[0        1       0      0
   -k/m1   -b/m1   k/m1     b/m1
   0        0       0       1
   k/m2     b/m2    -2*k/m2  -b/m2];

B=[0        0
   1        0
   0        0
   0       k/m2];

C=[1   0   -1   0];

D=[0    0];

[nump,denp]=ss2tf(A, B, C, D, 1);
nump=[nump(3) nump(4) nump(5)];
[num1,den1]=ss2tf(A,B,C,D,2);
num1=[num1(3) num1(4) num1(5)];
numf=num1;
denf=nump;

Plotting the frequency response using the bode command

The main idea of frequency-based design is to use the Bode plot of the open-loop transfer function to estimate the closed-loop response. Adding a controller to the system changes the open-loop Bode plot so that the closed-loop response will also change. Let's first draw the Bode plot for the original open-loop transfer function. Add the following line of code to your m-file and rerun:

bode(nump,denp)

Adding a two-lead controller

From the Bode plot above, we see that the phase curve is concave at about 5 rad/sec. First, we will try to add positive phase around this region, so that the phase will remain above the -180 degree line. Since a large phase margin leads to a small overshoot, we will want to add at least 170 degrees of positive phase at the area near 5 rad/sec. Since one lead controller can add no more than +90 degrees, we will use a two-lead controller:

To obtain T and a, the following steps can be used:

1: Determine the positive phase needed :

Since we want 170 degrees total, we will need 85 degrees from each controller.

In our case this frequency should be 5 rad/sec.

4: Determine T and aT from the following equations, these determine the corner frequencies so that the maximum phase will be added at the desired frequency.

Now let's put our 2-Lead controller into the system and see what the Bode plot looks like. Add the following code to your m-file, and add a % in front of the previous bode command(if there is one):

numc=conv([4.59 1], [4.59 1]);
denc=conv([0.0087 1], [0.0087 1]);
margin(conv(nump,numc),conv(denp,denc))

Since the Bode plot has a limited phase range (-360-0), the above plot is a little deceiving. The plot is equivalent to the following:

w=logspace(-4,4);
[mag,phase,w] = bode(conv(nump,numc),conv(denp,denc),w);
subplot(2,1,1);
semilogx(w,20*log10(mag));
subplot(2,1,2);
semilogx(w,phase);
axis([1e-4, 1e4, -180, 90])

From this plot we see that the concave portion of the phase plot is above -180 degrees now, and the phase margin is large enough for the design criteria. Let's see how the output (the distance X1-X2) responds to a bump on the road (W). Recall that the schematic of the system is:

and the closed-loop transfer function can be derived as following:

To obtain the closed-loop transfer function from W to X1-X2, the following commands can be added into the m-file:

numa=conv(conv(numf,nump),denc);
dena=conv(denf,polyadd(conv(denp,denc),conv(nump,numc)));

Plotting the closed-loop response

Let's see what the step response looks like now. Keep in mind that we are using a 0.1 m high step as the disturbance. To simulate this, simply multiply numa by 0.1. Add the following code into the m-file and rerun it. Don't forget to put % mark in front of all bode and margin commands!

t=0:0.01:10;
step(0.1*numa,dena,t)
axis([0 10 -.01 .01])

From this plot we can see that the settling time is shorter than 3 seconds. This response is now satisfactory and no more design iteration is needed.

Frequency Response Examples

Cruise Control | DC Motor | Bus Suspension | Inverted Pendulum

Bus Suspension Examples

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

Tutorials

Basics | Modeling | Starting | Root Locus | Frequency Response | State Space | Examples