Control design using pole placement
Reference input
The state equations for this problem are:
where
- m=1000 kg
- b=50 N*sec/kg
- u=500 N
- v=velocity
- y=output
The design criteria are:
To see the original problem setup , see Cruise Control Modeling page.
where Recall from the State-Space Tutorial page, we should use the technique called
"pole placement" to obtain the desired output. Poles of a closed-loop system can be found from the characteristic
equation: the determinate of [sI-(A-B*K)] matrix. If desired poles can be placed into the system by designing right
control matrix (K), then the desired output can be obtained. In this tutorial, poles will be chosen first,
then use MATLAB to find the corresponding control matrix (K).
Now, we need to determine where to place poles for our system. Since our [sI-(A-B*K)] matrix is 1x1,
we have only one pole to place.
Let the pole to be at -1.5 (arbitrary). Just as in the State-Space Tutorial, the MATLAB function called
place will be used to find the control matrix K
. Create an new m-file and enter the following commands.
As you can see, the rise time is satisfactory, but the steady-state error is too large.
Once again from the State-Space Tutorial, a scaling factor called
Nbar (the schematic is shown below) should be used to eliminate the steady-state error.
We can use the rscale function to compute the scaling factor. The input is
already multiplied by 500, and we want the steady-state speed to be 10 m/sec, so we need to account
for these factors as well.
Copy the following commands to an m-file and run it in the MATLAB command window.
You should get the step response shown below.
As you can see, the steady-state error has been eliminated. The rise time is less than 5 seconds and the overshoot is,
in fact, zero. All the design requirements are satisfied.
Control design using pole placement
The schematic of a full-state feedback system is shown below.
m = 1000;
b = 50;
t = 0:0.1:10;
u = 500*ones(size(t));
A = [-b/m];
B = [1/m];
C = [1];
D = [0];
cruise = ss(A,B,C,D);
x0 = [0];
p1 = -1.5;
K=place(A,B,[p1])
sys_cl=ss(A-B*K,B,C,D);
lsim(sys_cl,u,t,x0);
Running this m-file in the MATLAB command window should give you the control matrix and the following step response.
Reference input
m=1000;
b=50;
t=0:0.1:10;
u=500*ones(size(t));
A=[-b/m];
B=[1/m];
C=[1];
D=[0];
x0=[0];
p1=-1.5
K=place(A,B,[p1]);
Nbar=rscale(cruise,k)*10/500;
sys_cl=ss(A-B*K,B*Nbar,C,D);
lsim(sys_cl,u,t,x0);
