pole placement in M-file

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cmcm
cmcm 2013-7-20
评论： Sam Chak 2024-3-1
hello every one ... how can i determine (p) in pole placement code K = place(A,B,p)

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Azzi Abdelmalek
Azzi Abdelmalek 2013-7-26
p=eig(A-b*k)

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Shashank Prasanna
Shashank Prasanna 2013-7-20
shahad, place is used to perform pole placement using state feedback. The output K is the feedback gain matrix.
PLACE is not an optimal control methodology and does no come up with good p values. It merely attempts to place at the specified location and generates the gain matrix.
You question seems to be related to: How do I generate this stable optimal p?
There are numerous answers to that question but a popular one is linear quadratic controller (LQR):
http://www.mathworks.com/help/control/ref/lqr.html
The first output argument is the gain matrix K and the 3rd output argument 'e' are you poles which are automatically generated when the optimization is performed.
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cmcm
cmcm 2013-7-26
编辑：cmcm 2013-7-26
and if i have the values of (K) and A and B and i want to find P from that ... is there any way to do that ??? because my qus. was how to find this P matrix ??
Azzi Abdelmalek
Azzi Abdelmalek 2013-8-16
Shashank, What I mean by optimal poles is poles that provide a system optimal performances (acceptable stability margin, and optimal performances to be defined by the user).

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Azzi Abdelmalek
Azzi Abdelmalek 2013-7-20
编辑：Azzi Abdelmalek 2013-7-20
p is the poles vector you have to impose to your system in closed loop.
Example:
A=[-6 -5;1 0];
b=[1;0]
The size of A is 2x2 then the length of the pole vector p should be 2
p=[-5 -10] % the poles should be stables (real(p)<0)
%or
p=[-2+j -2-j]
place(A,b,p)
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cmcm
cmcm 2013-7-20
and the values of these poles how i found it ?? by try and error or how ?
Azzi Abdelmalek
Azzi Abdelmalek 2013-7-20
Like I said the pole to impose should be real or complex numbers with real part obligatory negative. This is enough to get a stable control, for optimal performances, you have to read the effects of poles on the system.

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Nagesh
Nagesh 2024-3-1
Ran in:
A=[-6 -5;1 0];
b=[1;0]
b = 2×1
1 0
p=[-5 -10] % the poles should be stables (real(p)<0)
p = 1×2
-5 -10
%or
p=[-2+j -2-j]
p =
-2.0000 + 1.0000i -2.0000 - 1.0000i
place(A,b,p)
ans = 1×2
-2.0000 0.0000
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Sam Chak
Sam Chak 2024-3-1
Ran in:
Hi @Nagesh
Your response didn't fully address the original question. In order to determine the control gain using the place() command, the user must input the poles that result in the Hurwitz characteristic polynomial. These poles specifically correspond to the closed-loop system and generate the desired response specified by the user. Your answer deserves votes if you could provide a practical example to illustrate this.
Let's consider a simple Double Integrator system. If we need to meet performance specifications such as a percent overshoot of ≤ 9% and a settling time of ≤ 1 second, how would you determine the target poles?
A = [0, 1; 0, 0]; % state matrix
B = [0; 1]; % input matrix
C = [1, 0]; % output matrix
D = 0*C*B; % direct matrix
sys = ss(A, B, C, D)
sys = A = x1 x2 x1 0 1 x2 0 0 B = u1 x1 0 x2 1 C = x1 x2 y1 1 0 D = u1 y1 0 Continuous-time state-space model.
Gp = tf(sys)
Gp = 1 --- s^2 Continuous-time transfer function.

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