Suppressing Exogenous Disturbances in a Discrete-Time Control System As an Optimization Problem

Abstract

This paper proposes a novel approach to suppressing bounded exogenous disturbances in a linear discrete-time control system by a static state- or output-feedback control law. The approach is based on reducing the original problem to a nonconvex matrix optimization problem with the gain matrix as one variable. The latter problem is solved by the gradient method; its convergence is theoretically justified for several important special cases. An example is provided to demonstrate the effectiveness of the iterative procedure proposed.

APPENDIX

Proof of Lemma 1. Consider a sequence of stabilizing controllers {K_j} ∈ $\mathcal{S}$ such that K_j → K ∈ $\partial \mathcal{S}$, i.e., ρ(A + BKC) = 1. In other words, for any ε > 0 there exists a number N = N(ε) such that

$$\left| {\rho (A + B{{K}_{j}}C) - \rho (A + BKC)} \right| = 1 - \rho (A - B{{K}_{j}}C) < \varepsilon $$

for all j $\geqslant$ N(ε).

Let P_j be the solution of Eq. (5) associated with the controller K_j:

$$\frac{1}{{{{\alpha }_{j}}}}(A + B{{K}_{j}}C){{P}_{j}}{{(A + B{{K}_{j}}C)}^{{\text{T}}}} - {{P}_{j}} + \frac{1}{{1 - {{\alpha }_{j}}}}D{{D}^{{\text{T}}}} = 0.$$

Also, let Y_j be the solution of the dual discrete Lyapunov equation

$$\frac{1}{{{{\alpha }_{j}}}}{{(A + B{{K}_{j}}C)}^{{\text{T}}}}{{Y}_{j}}(A + B{{K}_{j}}C) - {{Y}_{j}} + {{C}_{1}}C_{1}^{{\text{T}}} = 0.$$

Using ([6], Lemmas A.1 and A.2) and ([7], Lemma A.1.2), we have

$$f({{L}_{j}}) = \operatorname{tr} {{C}_{1}}{{P}_{j}}C_{1}^{{\text{T}}} + \rho \left\| {{{K}_{j}}} \right\|_{F}^{2}\; \geqslant \;\operatorname{tr} {{P}_{j}}{{C}_{1}}C_{1}^{{\text{T}}} = \operatorname{tr} \left( {{{Y}_{j}}\frac{1}{{1 - {{\alpha }_{j}}}}D{{D}^{{\text{T}}}}} \right)$$

$$ \geqslant \;\frac{1}{{1 - {{\alpha }_{j}}}}{{\lambda }_{{\min }}}({{Y}_{j}})\left\| D \right\|_{F}^{2}\; \geqslant \;\frac{1}{{1 - {{\alpha }_{j}}}}\frac{{{{\lambda }_{{\min }}}({{C}_{1}}C_{1}^{{\text{T}}})}}{{1 - \sigma _{{\min }}^{2}(A + B{{K}_{j}}C)}}\left\| D \right\|_{F}^{2}$$

$$ \geqslant \;\frac{1}{{1 - {{\rho }^{2}}(A + B{{K}_{j}}C)}}\frac{{{{\lambda }_{{\min }}}({{C}_{1}}C_{1}^{{\text{T}}})}}{{1 - \sigma _{{\min }}^{2}(A + B{{K}_{j}}C)}}\left\| D \right\|_{F}^{2}$$

$$ \geqslant \;\frac{1}{\varepsilon }\frac{1}{{1 + \rho (A + B{{K}_{j}}C)}}\frac{{{{\lambda }_{{\min }}}({{C}_{1}}C_{1}^{{\text{T}}})}}{{1 - \sigma _{{\min }}^{2}(A + B{{K}_{j}}C)}}\left\| D \right\|_{F}^{2}\xrightarrow[{\varepsilon \to 0}]{} + \infty $$

since ρ²(A + BK_jC) < α_j < 1.

On the other hand,

$$f({{K}_{j}}) = \operatorname{tr} {{C}_{1}}{{P}_{j}}C_{1}^{{\text{T}}} + \rho \left\| {{{K}_{j}}} \right\|_{F}^{2}\; \geqslant \;\rho \left\| {{{K}_{j}}} \right\|_{F}^{2}\; \geqslant \;\rho {{\left\| {{{K}_{j}}} \right\|}^{2}}\xrightarrow[{\left\| {{{K}_{j}}} \right\| \to + \infty }]{} + \infty .$$

The proof of Lemma 1 is complete.

Proof of Lemma 2. Differentiation with respect to a is performed in accordance with the results of Section 2, with A replaced by A + BKC.

We add the increment ΔK for K in Eq. (5) and denote the corresponding increment of P by ΔP:

$$\frac{1}{\alpha }(A + B(K + \Delta K)C)(P + \Delta P){{(A + B(K + \Delta K)C)}^{{\text{T}}}} - (P + \Delta P) + \frac{1}{{1 - \alpha }}D{{D}^{{\text{T}}}} = 0.$$

Leaving the notation ΔP for the principal part of the increment, we have

$$\begin{gathered} \frac{1}{\alpha }\left( {(A + BKC)P{{{(A + BKC)}}^{{\text{T}}}} + B\Delta KCP{{{(A + BKC)}}^{{\text{T}}}}} \right. \\ \left. { + \;(A + BKC)P{{{(B\Delta KC)}}^{{\text{T}}}} + (A + BKC)\Delta P{{{(A + BKC)}}^{{\text{T}}}}} \right) \\ - \;(P + \Delta P) + \frac{1}{{1 - \alpha }}D{{D}^{{\text{T}}}} = 0. \\ \end{gathered} $$

Subtracting Eq. (5) from this equation gives

$$\begin{gathered} \frac{1}{\alpha }(A + BKC)\Delta P{{(A + BKC)}^{{\text{T}}}} - \Delta P \\ + \;\frac{1}{\alpha }\left( {(A + BKC)P{{{(B\Delta KC)}}^{{\text{T}}}} + B\Delta KCP{{{(A + BKC)}}^{{\text{T}}}}} \right) = 0. \\ \end{gathered} $$

(A.1)

The increment of f(K) is calculated by linearizing the corresponding terms:

$$\begin{gathered} \Delta f(K) = f(K) - f(K + \Delta K) \\ = \operatorname{tr} {{C}_{1}}(P + \Delta P)C_{1}^{{\text{T}}} + \rho \left\| {K + \Delta K} \right\|_{F}^{2} - \left( {\operatorname{tr} {{C}_{1}}PC_{1}^{{\text{T}}} + \rho \left\| K \right\|_{F}^{2}} \right) \\ = \operatorname{tr} {{C}_{1}}\Delta PC_{1}^{{\text{T}}} + \rho \operatorname{tr} {{K}^{{\text{T}}}}\Delta K + \rho \operatorname{tr} {{(\Delta K)}^{{\text{T}}}}K = \operatorname{tr} \Delta PC_{1}^{{\text{T}}}{{C}_{1}} + 2\rho \operatorname{tr} {{K}^{{\text{T}}}}\Delta K. \\ \end{gathered} $$

Due to [6, Lemma A.1], from the dual equations (A.1) and (9) it follows that

$$\begin{gathered} \Delta f(K) = 2\operatorname{tr} Y\frac{1}{\alpha }B\Delta KCP{{(A + BKC)}^{{\text{T}}}} + 2\rho \operatorname{tr} {{K}^{{\text{T}}}}\Delta K \\ = 2\operatorname{tr} \left( {\rho {{K}^{{\text{T}}}} + \frac{1}{\alpha }CP{{{(A + BKC)}}^{{\text{T}}}}YB} \right)\Delta K \\ = 2\left\langle {\rho K + \frac{1}{\alpha }{{B}^{{\text{T}}}}Y(A + BKC)P{{C}^{{\text{T}}}},\Delta K} \right\rangle . \\ \end{gathered} $$

Thus, we arrive at (7). The proof of Lemma 2 is complete.

Proof of Lemma 3. The value

$$\nabla _{K}^{2}f(K,\alpha )[E,E] = \left\langle {\nabla _{K}^{2}f(K,\alpha )[E],E} \right\rangle $$

is calculated by differentiating ${{\nabla }_{K}}f$(K, α)[E] = $\left\langle {{{\nabla }_{K}}f(K,\alpha ),E} \right\rangle$ in the direction E ∈ ${{\mathbb{R}}^{{p \times l}}}$.

For this purpose, linearizing the corresponding terms, we calculate the increment of ${{\nabla }_{K}}f$(K, α)[E] in the direction E:

$$\begin{gathered} \Delta {{\nabla }_{K}}f(K,\alpha )[E] = 2\left( {\rho (K + \delta E) + \frac{1}{\alpha }{{B}^{{\text{T}}}}(Y + \Delta Y)(A + B(K + \delta E)C)(P + \Delta P){{C}^{{\text{T}}}}} \right) \\ - \;2\left( {\rho K + \frac{1}{\alpha }{{B}^{{\text{T}}}}Y(A + BKC)P{{C}^{{\text{T}}}}} \right) \\ = 2\delta \left( {\rho E + \frac{1}{\alpha }{{B}^{{\text{T}}}}(YBECP + Y{\kern 1pt} '(K)[E](A + BKC)P + Y(A + BKC)P{\kern 1pt} '(K)[E]){{C}^{{\text{T}}}}} \right), \\ \end{gathered} $$

where

$$\Delta P = P(K + \delta E) - P(K) = \delta P{\kern 1pt} '(K)[E],$$

$$\Delta Y = Y(K + \delta E) - Y(K) = \delta Y{\kern 1pt} '(K)[E].$$

Thus, with P ' = P '(K)[E] and Y  ' = Y  '(K)[E], we have

$$\begin{gathered} \frac{1}{2}\nabla _{K}^{2}f(K,\alpha )[E,E] \\ = \left\langle {\rho E + \frac{1}{\alpha }{{B}^{{\text{T}}}}(YBECP + Y{\kern 1pt} '(A + BKC)P + Y(A + BKC)P{\kern 1pt} '){{C}^{{\text{T}}}},E} \right\rangle . \\ \end{gathered} $$

Furthermore, P = P(K) is the solution of the discrete Lyapunov equation (5). We write it in increments in the direction E:

$$\frac{1}{\alpha }(A + B(K + \delta E)C)(P + \delta P{\kern 1pt} '){{(A + B(K + \delta E)C)}^{{\text{T}}}} - (P + \delta P{\kern 1pt} ') + \frac{1}{{1 - \alpha }}D{{D}^{{\text{T}}}} = 0$$

or

$$\begin{gathered} \frac{1}{\alpha }\left( {(A + BKC)P{{{(A + BKC)}}^{{\text{T}}}} + (A + BKC)\delta P{\kern 1pt} '{{{(A + BKC)}}^{{\text{T}}}}} \right. \\ \left. { + \;(A + BKC)P{{{(B\delta EC)}}^{{\text{T}}}} + B\delta ECP{{{(A + BKC)}}^{{\text{T}}}}} \right) \\ - \;(P + \delta P{\kern 1pt} ') + \frac{1}{{1 - \alpha }}D{{D}^{{\text{T}}}} = 0. \\ \end{gathered} $$

In view of (5), this expression yields Eq. (10).

Similarly, Y = Y(K) is the solution of the discrete Lyapunov equation (9). We write it in increments in the direction E:

$$\frac{1}{\alpha }{{(A + B(K + \delta E)C)}^{{\text{T}}}}(Y + \delta Y{\kern 1pt} ')(A + B(K + \delta E)C) - (Y + \delta Y{\kern 1pt} ') + C_{1}^{{\text{T}}}{{C}_{1}} = 0$$

or

$$\begin{gathered} \frac{1}{\alpha }\left( {{{{(A + BKC)}}^{{\text{T}}}}Y(A + BKC) + {{{(A + BKC)}}^{{\text{T}}}}\delta Y{\kern 1pt} '(A + BKC)} \right. \\ \left. { + \;{{{(A + BKC)}}^{{\text{T}}}}YB\delta EC + {{{(B\delta EC)}}^{{\text{T}}}}Y(A + BKC)} \right) - (Y + \delta Y{\kern 1pt} ') + C_{1}^{{\text{T}}}{{C}_{1}} = 0. \\ \end{gathered} $$

Due to (9), we obtain

$$\begin{gathered} \frac{1}{\alpha }{{(A + BKC)}^{{\text{T}}}}Y{\kern 1pt} '(A + BKC) - Y{\kern 1pt} ' \\ + \;\frac{1}{\alpha }\left( {{{{(A + BKC)}}^{{\text{T}}}}YBEC + {{{(BEC)}}^{{\text{T}}}}Y(A + BKC)} \right) = 0. \\ \end{gathered} $$

(A.2)

From (10) and (A.2) it follows that

$$\operatorname{tr} P{\kern 1pt} '{{(A + BKC)}^{{\text{T}}}}YBEC = \operatorname{tr} Y{\kern 1pt} '{\kern 1pt} BECP{{(A + BKC)}^{{\text{T}}}},$$

so

$$\frac{1}{2}\nabla _{K}^{2}f(K,\alpha )[E,E] = \rho \left\langle {E,E} \right\rangle + \frac{1}{\alpha }\left\langle {{{B}^{{\text{T}}}}YBECP{{C}^{{\text{T}}}},E} \right\rangle + \frac{2}{\alpha }\left\langle {{{B}^{{\text{T}}}}Y(A + BKC)P{\kern 1pt} '{\kern 1pt} {{C}^{{\text{T}}}},E} \right\rangle .$$

The proof of Lemma 3 is complete.