State-Space Matrices

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Aysel Alimirzayeva
Aysel Alimirzayeva 2022-11-3
评论： Aysel Alimirzayeva 2022-11-5
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?

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采纳的回答

Sam Chak
Sam Chak 2022-11-3
Hi @Aysel Alimirzayeva
It's because of negative feedback.
The LQR function only computes .
Making a substitution
  2 个评论
Sam Chak
Sam Chak 2022-11-5
编辑：Sam Chak 2022-11-5
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)

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更多回答（1 个）

Sam Chak
Sam Chak 2022-11-5
Ran in:
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)
  1 个评论
Aysel Alimirzayeva
Aysel Alimirzayeva 2022-11-5
Thank you very much for the detailed further explanation.

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