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Limitations of Pole Assignment
It is not always obvious what set of poles we should target (especially for large systems),
We do not control explicitly the trade-off between “speed of convergence” and “intensity of the control” (large input values maybe costly or impossible).
Let
\[\dot{x} = A x + Bu\]
where
\(A \in \mathbb{R}^{n\times n}\), \(B \in \mathbb{R}^{m\times n}\) and
\(x(0) = x_0 \in \mathbb{R}^n\) is given.
Find \(u(t)\) that minimizes
\[ J = \int_0^{+\infty} x(t)^t Q x(t) + u(t)^t R u(t) \, dt \]
where:
\(Q \in \mathbb{R}^{n \times n}\) and \(R \in \mathbb{R}^{m\times m}\),
(to be continued …)
\(Q\) and \(R\) are symmetric (\(R^t = R\) and \(Q^t = Q\)),
\(Q\) and \(R\) are positive definite (denoted “\(>0\)”)
\[x^t Q x \geq 0 \, \mbox{ and } \, x^t Q x = 0 \, \mbox{ iff }\, x=0\]
and
\[u^t R u \geq 0 \, \mbox{ and } \, u^t R u = 0 \, \mbox{ iff }\, u=0.\]
If \(x \in \mathbb{R}\) and \(u \in \mathbb{R}\),
\[ J = \int_0^{+\infty} q x(t)^2 + r u(t)^2 \, dt \]
with \(q > 0\) and \(r > 0\).
When we minimize \(J\):
Only the relative values of \(q\) and \(r\) matters.
Large values of \(q\) penalize strongly non-zero states:
\(\Rightarrow\) fast convergence.
Large values of \(r\) penalize strongly non-zero inputs:
\(\Rightarrow\) small input values.
If \(x \in \mathbb{R}^n\) and \(u \in \mathbb{R}^m\) and \(Q\) and \(R\) are diagonal,
\[ Q = \mathrm{diag}(q_1, \cdots, q_n), \; R=\mathrm{diag}(r_1, \cdots, r_m), \]
\[ J = \int_0^{+\infty} \sum_{i} q_i x_i(t)^2 + \sum_j r_j u_j(t)^2 \, dt \]
with \(q_i > 0\) and \(r_j > 0\).
Thus we can control the cost of each component of \(x\) and \(u\) independently.
Assume that \(\dot{x} = A x + Bu\) is controllable.
There is an optimal solution; it is a linear feedback
\[u = - K x\]
The closed-loop dynamics is asymptotically stable.
The gain matrix \(K\) is given by
\[ K = R^{-1} B^t \Pi, \]
where \(\Pi \in \mathbb{R}^{n \times n}\) is the unique matrix such that \(\Pi^t = \Pi\), \(\Pi > 0\) and
\[ \Pi B R^{-1} B^t \Pi - \Pi A - A^t \Pi - Q = 0. \]
Consider the double integrator \(\ddot{x} = u\)
\[ \frac{d}{dt} \left[\begin{array}{c} x \\ \dot{x} \end{array}\right] = \left[\begin{array}{cx} 0 & 1 \\ 0 & 0\end{array}\right] \left[\begin{array}{c} x \\ \dot{x} \end{array}\right] + \left[\begin{array}{c} 0 \\ 1 \end{array}\right] u \]
(in standard form)
It is stable:
Consider the controllable dynamics
\[ \dot{x} = A x + Bu \]
and \(u(t)\) the control that minimizes
\[ J = \int_{0}^{+\infty} x(t)^t Q x(t) + u(t)^t R u(t) \, dt. \]
Let
\[ j(x, u) := x^tQ x + u^t R u. \]
Show that
\[ j(x(t), u(t)) = - \frac{d}{dt} x(t)^t \Pi x(t) \]
What is the value of \(J\)?
We know that \(u = -Kx\) where \(K = R^{-1} B^t \Pi\) and \(\Pi\) is a symmetric solution of
\[ \Pi BR^{-1} B^t \Pi - \Pi A - A^t \Pi - Q = 0. \]
Since \(R\) is symmetric,
\[ \Pi BR^{-1} B^t \Pi = \Pi B(R^{-1})^tR R^{-1} B^t \Pi = K^t R K \]
and thus \[\Pi A + A^t \Pi = K^t R K - Q.\]
Since \(\dot{x} = (A - B K ) x\),
\[ \begin{split} \frac{d}{dt} x^t \Pi x &= x^t (\Pi (A - BK) + (A - BK)^t \Pi) x \\ & = x^t (\Pi A + A^t \Pi - \Pi BK - (BK)^t \Pi) x \\ & = x^t ( K^t R K - Q - K^t R K - K^t R K) x \\ & = x^t (- Q - K^t R K ) x^t \\ & = - x^t Q x - u^t R u \\ & = -j(x, u). \end{split} \]
Since the system is controllable, the optimal control makes the origin of the closed-loop system asymptotically stable. Consequently, \(x(t) \to 0\) when \(t \to +\infty\). Hence,
\[ \begin{split} J &= \int_0^{+\infty} j(x, u) \, dt \\ &= - \int_0^{+\infty} \frac{d}{dt} x^t \Pi x \, dt \\ &= - \left[x^t \Pi x\right]_0^{+\infty} \\ & = x(0)^t \Pi x(0). \end{split} \]