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What are the conditions on a linear time invariant system for a PI controller to converge to a specified set point?

Mathematics Asked on December 6, 2021

What are the conditions on a linear time invariant system for a PI controller to converge to a specified set point? Specifically, given the system:

$$begin{array}{l}
y^prime = Ay + Bu \
w = Cy
end{array}$$

where

  • $Ain mathbb{R}^{mtimes m}$ – state dynamics
  • $Bin mathbb{R}^{mtimes n}$ – control dynamics
  • $Cin mathbb{R}^{ntimes m}$ – observation dynamics
  • $y in mathbb{R}^m$ – state
  • $u in mathbb{R}^n$ – control
  • $w in mathbb{R}^n$ – observations

I’d like to drive the system $y$ to an observable point $bar{w}$ using a PI controller, but don’t understand the conditions necessary on $A$, $B$, and $C$ to make this possible. From what I can tell, we should have:

$$
u = k_p (bar{w}-w) + k_i int_0^t (bar{w}-w)
$$

which gives

$$begin{array}{l}
y^prime = Ay + B(k_p (bar{w}-w) + k_i int_0^t (bar{w}-w)) \
w = Cy
end{array}$$

or

$$begin{array}{l}
y^prime = Ay + B(k_p (bar{w}-Cy) + k_i int_0^t (bar{w}-Cy))
end{array}$$

which, after taking the derivative and regrouping terms gives

$$begin{array}{l}
y^{primeprime} = (A-k_pBC)y^prime -k_iBCy + k_i Bbar{w}
end{array}$$

This can be written as the first order system

$$
begin{bmatrix}
y\z
end{bmatrix}^prime =
begin{bmatrix}
0 & I\
-k_iBC & A-k_pBC
end{bmatrix}
begin{bmatrix}
y\
z
end{bmatrix}
+
begin{bmatrix}
0\
k_i Bbar{w}
end{bmatrix}
$$

It seems like everything works fine if $BC$ is invertible, but I’d like to know if there’s a better condition. If $n < m$, then $BC$ is less than full rank and it seems like this means it’s not always possible to find a control, but I’m not sure what the theory states or the words to search for in order to better determine this.

One Answer

Hint.

As a LTI is easily Laplace transformable, the problem can be stated as

$$ cases{ left(sI-Aright)Y= B U\ U = left(k_p I+frac{k_i I}{s}right)E\ E = W_r - W } $$

and putting all together

$$ E = W_r - C Y = W_r - frac 1s Cleft(sI-Aright)^{-1}Bleft(s k_p I+ k_i Iright)E $$

and calling

$$ G = I+frac 1s Cleft(sI-Aright)^{-1}Bleft(s k_p I+ k_i Iright) $$

we have

$$ E = G^{-1}W_r $$

and the error dynamics depend on the zeros from $det(G)$

as an example, considering

$$ A = left( begin{array}{cc} 1 & 2 \ -3 & 4 \ end{array} right), B = left( begin{array}{c} 1 \ 0 \ end{array} right), C = left( begin{array}{cc} 1 & 1 \ end{array} right) $$

we have

$$ det(G) = frac{2 k_i s-14 k_i+2 k_p s^2-14 k_p s+s^3-5 s^2+10 s}{s left(s^2-5 s+10right)} $$

and the finite zeros are given by

$$ (2s-14) k_i+(2s^2-14s) k_p+s^3-5 s^2+10 s = 0 $$

Those zeros now are continuously dependent on $k_p,k_i$ and should be located inside the left complex plane to attain stability. This can be handled using the Routh-Hurwitz criterion. Additionally, to have asymptotic null error we should verify also that once stable, the error dynamics should obey

$$ lim_{sto 0}s E =0 $$

Answered by Cesareo on December 6, 2021

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