State-Space Matrices

Hello.Why do we write A-B*K1 in the example of LQR Control using State-Space Matrices in Matlab?Why is B negative?If anyone knows,can you please explain?

Answer 1

[Sam Chak]

Hi @Aysel Alimirzayeva

It's because of negative feedback.

The LQR function only computes .

Making a substitution

[Aysel Alimirzayeva]

@Sam Chak Than you very much.

[Sam Chak]

You're welcome, @Aysel Alimirzayeva.

Just for a note, this "state-feedback"

sys1 = ss(A-B*K1, B, C, D)

only works if your reference state is 0.

If your system is tracking a non-zero reference state, then a pre-compensator N is required to be placed at the reference input. N is just a constant gain to rescale input so that the output converges to 1 in the step response

sys2 = ss(A-B*K1, N*B, C, D)

Answer 2

[Sam Chak]

Hi @Aysel Alimirzayeva

I added a simple example to show you. If you like this example, consider voting 👍 the Answer. Thanks!

Say, the reference state is 1.

A  = [0 1; 2 3];
B  = [0; 1];
K1 = lqr(A, B, eye(2), 1)

K1 = 1×2

4.2361 7.2979

sys1 = ss(A-B*K1, B, [1 0], 0)

sys1 =
 
  A = 
           x1      x2
   x1       0       1
   x2  -2.236  -4.298
 
  B = 
       u1
   x1   0
   x2   1
 
  C = 
       x1  x2
   y1   1   0
 
  D = 
       u1
   y1   0
 
Continuous-time state-space model.

step(sys1, 20)

The step response shows that the output won't reach 1. Thus, a pre-compensator is needed:

N = 1/dcgain(sys1)      % pre-compensator

N = 2.2361

sys2 = ss(A-B*K1, N*B, [1 0], 0)

sys2 =
 
  A = 
           x1      x2
   x1       0       1
   x2  -2.236  -4.298
 
  B = 
          u1
   x1      0
   x2  2.236
 
  C = 
       x1  x2
   y1   1   0
 
  D = 
       u1
   y1   0
 
Continuous-time state-space model.

step(sys2, 20)

[Aysel Alimirzayeva]

Thank you very much for the detailed further explanation.