A Comparison of Guaranteeing and Kalman Filters

Abstract

We propose a new approach to filtering under arbitrary bounded exogenous disturbances based on reducing this problem to an optimization problem. The approach has a low computational complexity since only Lyapunov equations are solved at each iteration. At the same time, it possesses advantages essential from an engineering-practical point of view, namely, the possibilities to limit the filter matrix and to construct optimal filter matrices separately for each coordinate of the system’s state vector. A gradient method for finding the filter matrix is presented. According to the examples, the proposed recurrence procedure is rather effective and yields quite satisfactory results. This paper continues the series of research works devoted to feedback control design from an optimization perspective.

Appendices

APPENDIX A

Lemma A.1. Let X and Y be the solutions of the dual discrete Lyapunov equations with a Schur matrix A:

$${{A}^{{\text{T}}}}XA - X + W = 0\quad and\quad AY{{A}^{{\text{T}}}} - Y + V = 0.$$

Then tr(XV) = tr(YW).

Proof of Lemma A.1. Indeed, direct calculations give

$$\begin{gathered} \operatorname{tr} (XV) = \operatorname{tr} \left( {X(Y - AY{{A}^{{\text{T}}}})} \right) = \operatorname{tr} (XY) - \operatorname{tr} \left( {XAY{{A}^{{\text{T}}}}} \right) \\ = \operatorname{tr} (YX) - \operatorname{tr} \left( {Y{{A}^{{\text{T}}}}XA} \right) = \operatorname{tr} \left( {Y(X - {{A}^{{\text{T}}}}XA)} \right) = \operatorname{tr} (YW). \\ \end{gathered} $$

The proof of Lemma A.1 is complete.

Lemma A.2. The solution P of the discrete Lyapunov equation

$$AP{{A}^{{\text{T}}}} - P + Q = 0$$

with a Schur matrix A and Q $\succ$ 0 satisfies the lower bounds

$${{\lambda }_{{\max }}}(P)\; \geqslant \;\frac{{{{\lambda }_{{\min }}}(Q)}}{{1 - {{\rho }^{2}}}},\quad {{\lambda }_{{\min }}}(P)\; \geqslant \;\frac{{{{\lambda }_{{\min }}}(Q)}}{{1 - \sigma _{{\min }}^{2}(A)}},$$

(A.1)

where ρ = $\mathop {\max }\limits_i \left| {{{\lambda }_{i}}(A)} \right|$ and σ_min(A) is the smallest singular value of the matrix A.

If Q = DD^T and the pair (A, D) is controllable, then

$${{\lambda }_{{\max }}}(P)\; \geqslant \;\frac{{{{{\left\| {u\text{*}{\kern 1pt} D} \right\|}}^{2}}}}{{1 - {{\rho }^{2}}}} > 0,$$

(A.2)

where

$$u\text{*}{\kern 1pt} A = \lambda u\text{*},\quad \left| \lambda \right| = \rho ,\quad \left\| u \right\| = 1,$$

i.e., u is the left eigenvector of the matrix A corresponding to the eigenvalue λ of the matrix A with the greatest magnitude. The vector u and the value λ can be complex; here, u* denotes the conjugate transpose of u.

Proof of Lemma A.2. The lower bounds (A.1) are well known; for example, see [29]. Let us prove (A.2). The explicit solution of the discrete Lyapunov equation with a Schur matrix A has the form

$$P = \sum\limits_{k = 0}^\infty {{{A}^{k}}D{{D}^{{\text{T}}}}{{{({{A}^{{\text{T}}}})}}^{k}}.} $$

Multiplying this equality by u on the right and by u* on the left, due to u*A^k = λ^ku* and (A^T)^ku = (λ*)^ku, we obtain

$${{\lambda }_{{\max }}}(P)\; \geqslant \;u\text{*}{\kern 1pt} Pu = \sum\limits_{k = 0}^\infty {u\text{*}{\kern 1pt} {{A}^{k}}D{{D}^{{\text{T}}}}{{{({{A}^{{\text{T}}}})}}^{k}}u} = \sum\limits_{k = 0}^\infty {{{{(\lambda \lambda \text{*})}}^{k}}u\text{*}{\kern 1pt} D{{D}^{{\text{T}}}}u} = \frac{{{{{\left\| {u\text{*}{\kern 1pt} D} \right\|}}^{2}}}}{{1 - {{\rho }^{2}}}},$$

where ||u*D|| > 0 by the controllability of the pair (A, D); for example, see [10, Theorem D.1.5]. The proof of Lemma A.2 is complete.

Now, we optimize the function  f(α) and consider the problem

$$\min f(\alpha ),\quad f(\alpha ) = \operatorname{tr} CP{{C}^{{\text{T}}}}$$

subject to the constraint

$$\frac{1}{\alpha }AP{{A}^{{\text{T}}}} - P + \frac{1}{{1 - \alpha }}D{{D}^{{\text{T}}}} = 0$$

for the matrix variables P = P^T ∈ ${{\mathbb{R}}^{{n \times n}}}$ and a scalar parameter 0 < α < 1.

Here we impose more stringent requirements for the problem statement: the matrix C of the system output is supposed to be square and nonsingular. This assumption could be relaxed, but the current goal is to establish the simplest and most obvious results.

Lemma A.3. Assume that A is a Schur matrix, ρ is the spectral radius of A, ρ² < α < 1, the pair (A, D) is controllable, and the matrix C is such that C^TC $\succ$ 0. Then the function f(α) = tr CP(α)C ^T possesses the following properties:

(a) The function f(α) is well-defined, positive, and strongly convex on the interval ρ² < α < 1 and its values tend to infinity at the interval endpoints. Moreover, there exists a constant c > 0 such that

$$f(\alpha )\; \geqslant \;\frac{\alpha }{{(1 - \alpha )(\alpha - {{\rho }^{2}})}}c,\quad {{\rho }^{2}} < \alpha < 1;$$

(A.3)

(b) The function f(α) has the derivative

$$f{\kern 1pt} '(\alpha ) = \operatorname{tr} Y\left( {\frac{1}{{{{{(1 - \alpha )}}^{2}}}}D{{D}^{{\text{T}}}} - \frac{1}{{{{\alpha }^{2}}}}AP{{A}^{{\text{T}}}}} \right),$$

where P and Y are the solutions of the discrete Lyapunov equations

$$\frac{1}{\alpha }AP{{A}^{{\text{T}}}} - P + \frac{1}{{1 - \alpha }}D{{D}^{{\text{T}}}} = 0$$

(A.4)

and

$$\frac{1}{\alpha }{{A}^{{\text{T}}}}YA - Y + {{C}^{{\text{T}}}}C = 0,$$

(A.5)

respectively.

(c) The second derivative of the function f(α) is given by

$$f{\kern 1pt} ''(\alpha ) = 2\operatorname{tr} Y\left( {\frac{1}{{{{{(1 - \alpha )}}^{3}}}}D{{D}^{{\text{T}}}} + \frac{1}{{{{\alpha }^{3}}}}A(P - X){{A}^{{\text{T}}}}} \right),$$

where P, Y, and X satisfy the discrete Lyapunov equations (A.4), (A.5), and

$$\frac{1}{\alpha }AX{{A}^{{\text{T}}}} - X + \frac{1}{{{{{(1 - \alpha )}}^{2}}}}D{{D}^{{\text{T}}}} - \frac{1}{{{{\alpha }^{2}}}}AP{{A}^{{\text{T}}}} = 0,$$

respectively. Moreover,  f ''(α*) > 0 and f ''(α) is monotonically increasing on the left and right of α*.

Proof of Lemma A.3. (a) Equation (6) can be written as

$$\left( {\frac{1}{{\sqrt \alpha }}A} \right)P{{\left( {\frac{1}{{\sqrt \alpha }}A} \right)}^{{\text{T}}}} - P = - \frac{1}{{1 - \alpha }}D{{D}^{{\text{T}}}};$$

according to [10, Lemma 1.2.6], there exists a unique solution if and only if $\frac{1}{{\sqrt \alpha }}A$ is a Schur matrix: $\left| {{{\lambda }_{i}}\left( {\frac{1}{{\sqrt \alpha }}A} \right)} \right|$ < 1, i.e., under the condition ρ² < α < 1.

We estimate the value f(α) = tr CP(α)C ^T using Lemma A.2 with obvious changes:

$$\begin{gathered} f(\alpha ) = \operatorname{tr} CP(\alpha ){{C}^{{\text{T}}}}\; \geqslant \;{{\lambda }_{{\min }}}({{C}^{{\text{T}}}}C){{\lambda }_{{\max }}}(P(\alpha )) \\ \geqslant \;\frac{{{{{\left\| {u\text{*}{\kern 1pt} D} \right\|}}^{2}}{{\lambda }_{{\min }}}({{C}^{{\text{T}}}}C)}}{{(1 - \alpha )\left( {1 - {{\rho }^{2}}\left( {A{\text{/}}\sqrt \alpha } \right)} \right)}} = \frac{\alpha }{{(1 - \alpha )(\alpha - {{\rho }^{2}})}}{{\left\| {u\text{*}{\kern 1pt} D} \right\|}^{2}}{{\lambda }_{{\min }}}({{C}^{{\text{T}}}}C). \\ \end{gathered} $$

Here, u has the same meaning as in Lemma A.2 and the value ||u*D||² is positive by the controllability of the pair (A/$\sqrt \alpha$, D). (It follows from the controllability of (A, D).)

Now we show that the function f(α) = tr CP(α)C ^T is strictly convex on the interval (ρ², 1). According to [10, Lemma 1.2.6], the solution of Eq. (A.4) can be explicitly represented as

$$P(\alpha ) = \sum\limits_{k = 0}^\infty {{{{\left( {\frac{1}{{\sqrt \alpha }}A} \right)}}^{k}}\frac{1}{{1 - \alpha }}D{{D}^{{\text{T}}}}{{{\left( {\frac{1}{{\sqrt \alpha }}{{A}^{{\text{T}}}}} \right)}}^{k}}} = \sum\limits_{k = 0}^\infty {\underbrace {\frac{1}{{(1 - \alpha ){{\alpha }^{k}}}}}_{g(\alpha ,k)}\underbrace {{{A}^{k}}D{{D}^{{\text{T}}}}{{{({{A}^{{\text{T}}}})}}^{k}}}_{{{H}_{k}}}.} $$

But H_k $\succ$ 0 and g(α, k) > 0 for 0 < α < 1; therefore, on the interval (ρ², 1) we have

$$P(\alpha ) = \sum\limits_{k = 0}^\infty {g(\alpha ,k){{H}_{k}}} \succ 0$$

and

$$f(\alpha ) = \operatorname{tr} P(\alpha ){{C}^{{\text{T}}}}C > 0.$$

Direct calculations give

$$g{\kern 1pt} '(\alpha ,k) = \left( {\frac{1}{{1 - \alpha }} - \frac{k}{\alpha }} \right)g(\alpha ,k),$$

$$g{\kern 1pt} ''(\alpha ,k) = \left( {{{{\left( {\frac{1}{{1 - \alpha }} - \frac{k}{\alpha }} \right)}}^{2}} + \frac{1}{{{{{(1 - \alpha )}}^{2}}}} + \frac{k}{{{{\alpha }^{2}}}}} \right)g(\alpha ,k)\; \geqslant \;\frac{1}{{{{{(1 - \alpha )}}^{2}}}}g(\alpha ,k).$$

(Here, differentiation is performed with respect to α.) As a result,

$$f{\kern 1pt} ''(\alpha ) = \sum\limits_{k = 0}^\infty {g{\kern 1pt} ''(\alpha ,k)\operatorname{tr} C{{H}_{k}}{{C}^{{\text{T}}}}} \; \geqslant \;\frac{1}{{{{{(1 - \alpha )}}^{2}}}}f(\alpha )\; \geqslant \;\frac{1}{{{{{(1 - {{\rho }^{2}})}}^{2}}}}f(\alpha \text{*}) > 0.$$

Thus, the second derivative of the function f(α) is positive and tends to infinity at the endpoints of the interval (ρ², 1).

Next, with direct calculations of the fourth derivative, we obtain

$${{g}^{{(IV)}}}(\alpha ,k) = \sum\limits_{k = 0}^\infty {\frac{{k(k + 1)(k + 2)(k + 3)}}{{{{\alpha }^{{k + 4}}}}}} + \frac{{24}}{{{{{(1 - \alpha )}}^{4}}}}\; \geqslant \;\frac{{24}}{{{{{(1 - \alpha )}}^{4}}}},$$

so

$$\begin{gathered} {{f}^{{(IV)}}}(\alpha ) = \sum\limits_{k = 0}^\infty {{{g}^{{(IV)}}}(\alpha ,k)\operatorname{tr} C{{H}_{k}}{{C}^{{\text{T}}}}} \\ \geqslant \;\frac{{24}}{{{{{(1 - \alpha )}}^{4}}}}\sum\limits_{k = 0}^\infty {\operatorname{tr} C{{H}_{k}}{{C}^{{\text{T}}}}} > \frac{{24}}{{{{{(1 - {{\rho }^{2}})}}^{4}}}}\sum\limits_{k = 0}^\infty {\operatorname{tr} C{{H}_{k}}{{C}^{{\text{T}}}}} > 0, \\ \end{gathered} $$

i.e., the second derivative f ''(α) is convex and grows at the interval endpoints.

(b) Let us derive the formula for the derivative of  f(α). In Eq. (A.4), the solution P is a function of α. We differentiate this equation, interpreting P ' as the derivative with respect to α:

$$\frac{1}{\alpha }AP{\kern 1pt} '{\kern 1pt} {{A}^{{\text{T}}}} - P{\kern 1pt} '\; + \frac{1}{{{{{(1 - \alpha )}}^{2}}}}D{{D}^{{\text{T}}}} - \frac{1}{{{{\alpha }^{2}}}}AP{{A}^{{\text{T}}}} = 0.$$

(A.6)

Applying Lemma A.1 to the dual Eqs. (A.6) and (A.5) finally yields

$$f{\kern 1pt} '(\alpha ) = \operatorname{tr} CP{\kern 1pt} '{\kern 1pt} {{C}^{{\text{T}}}} = \operatorname{tr} P{\kern 1pt} '{\kern 1pt} {{C}^{{\text{T}}}}C = \operatorname{tr} Y\left( {\frac{1}{{{{{(1 - \alpha )}}^{2}}}}D{{D}^{{\text{T}}}} - \frac{1}{{{{\alpha }^{2}}}}AP{{A}^{{\text{T}}}}} \right).$$

(c) The desired expression for the second derivative of  f(α) can be established by analogy. Differentiating Eq. (A.6) with respect to α, we have

$$\frac{1}{\alpha }AP{\kern 1pt} ''{\kern 1pt} {{A}^{{\text{T}}}} - P{\kern 1pt} ''\; + \frac{2}{{{{{(1 - \alpha )}}^{3}}}}D{{D}^{{\text{T}}}} + \frac{2}{{{{\alpha }^{3}}}}AP{{A}^{{\text{T}}}} - \frac{2}{{{{\alpha }^{3}}}}AP{\kern 1pt} '{\kern 1pt} {{A}^{{\text{T}}}} = 0.$$

Applying Lemma A.1 to this equation and Eq. (A.5) with X = P', we arrive at

$$f{\kern 1pt} ''(\alpha ) = \operatorname{tr} CP{\kern 1pt} ''{\kern 1pt} {{C}^{{\text{T}}}} = \operatorname{tr} P{\kern 1pt} ''{\kern 1pt} {{C}^{{\text{T}}}}C = 2\operatorname{tr} Y\left( {\frac{1}{{{{{(1 - \alpha )}}^{3}}}}D{{D}^{{\text{T}}}} + \frac{1}{{{{\alpha }^{3}}}}A(P - X){{A}^{{\text{T}}}}} \right).$$

The proof of Lemma A.3 is complete.

Note that the function f(α) and its two derivatives are calculated by solving three discrete Lyapunov equations.

Due to the above properties, this function can be minimized using Newton’s method. We specify an initial approximation ρ²(A) < α₀ < 1, e.g., α₀ = (1 + ρ²(A))/2, and apply the iterative process

$${{\alpha }_{{j + 1}}} = {{\alpha }_{j}} - \frac{{f{\kern 1pt} '({{\alpha }_{j}})}}{{f{\kern 1pt} ''({{\alpha }_{j}})}}.$$

(A.7)

The next theorem ensures the global convergence of the algorithm; it can be proved by analogy with a similar result in [16].

Theorem A.1 [16]. In the method (A.7), we have the upper bounds

$$\left| {{{\alpha }_{j}} - \alpha \text{*}} \right|\;\leqslant \;\frac{{f{\kern 1pt} ''({{\alpha }_{0}})}}{{{{2}^{j}}f{\kern 1pt} ''(\alpha \text{*})}}\left| {{{\alpha }_{0}} - \alpha \text{*}} \right|,\quad \left| {{{\alpha }_{{j + 1}}} - \alpha \text{*}} \right|\;\leqslant \;c{{\left| {{{\alpha }_{j}} - \alpha \text{*}} \right|}^{2}},$$

where c > 0 is some constant (possibly, in explicit form).

The first bound ensures the global convergence of the method (faster than a geometric progression with a coefficient of 1/2); the second bound, the quadratic convergence in the neighborhood of the solution. In practice, it takes 3–4 iterations to obtain a solution with a high accuracy (unless the starting point is too close to the interval endpoints).

Returning to the optimization problem (4)–(5), we minimize the function

$$f(L) = \mathop {\min }\limits_\alpha f(L,\alpha )$$

after a preliminary study of its properties.

Lemma A.4. The function f(L) is well-defined and positive on the set $\mathcal{S}$ of admissible filter matrices.

Indeed, if (A – LC) is a Schur matrix, then ρ(A – LC) < 1 and for ρ²(A – LC) < α < 1, there exists a solution P $\succcurlyeq$ 0 of the discrete Lyapunov equation (5). Thus, a strictly positive function f(L, α) is well-defined and f(L) > 0 due to (A.3). As in the continuous-time case, its domain $\mathcal{S}$ may be nonconvex and disconnected and its boundaries nonsmooth.

Lemma A.5. On the set $\mathcal{S}$ the function f(L) is coercive (i.e., it tends to infinity on the boundary of the domain). Moreover, the following lower bounds are valid:

$$\begin{gathered} f(L)\; \geqslant \;\frac{1}{{1 - {{\rho }^{2}}(A - LC)}}\frac{{{{\lambda }_{{\min }}}({{C}_{1}}C_{1}^{{\text{T}}})}}{{1 - \sigma _{{\min }}^{2}(A - LC)}}\left\| {{{D}_{1}} - L{{D}_{2}}} \right\|_{F}^{2}, \\ f(L)\; \geqslant \;\rho {{\left\| L \right\|}^{2}}. \\ \end{gathered} $$

(A.8)

Proof of Lemma A.5. We consider a sequence {L_j} ⊆ $\mathcal{S}$ of admissible matrices such that L_j → L ∈ $\partial \mathcal{S}$, i.e., ρ(A – LC) = 1. In other words, for any ε > 0 there exists a number N = N(ε) such that

$$\left| {\rho (A - {{L}_{j}}C) - \rho (A - LC)} \right| = 1 - \rho (A - {{L}_{j}}C) < \varepsilon $$

for all j $\geqslant$ N(ε).

Let P_j be the solution of Eq. (5) associated with the filter matrix L_j:

$$\frac{1}{{{{\alpha }_{j}}}}(A - {{L}_{j}}C){{P}_{j}}{{(A - {{L}_{j}}C)}^{{\text{T}}}} - {{P}_{j}} + \frac{1}{{1 - {{\alpha }_{j}}}}({{D}_{1}} - {{L}_{j}}{{D}_{2}}){{({{D}_{1}} - {{L}_{j}}{{D}_{2}})}^{{\text{T}}}} = 0;$$

let Y_j be the solution of its dual discrete Lyapunov equation

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

In view of Lemma A.2, we have

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

$$ = \operatorname{tr} \left( {{{Y}_{j}}\frac{1}{{1 - {{\alpha }_{j}}}}({{D}_{1}} - {{L}_{j}}{{D}_{2}}){{{({{D}_{1}} - {{L}_{j}}{{D}_{2}})}}^{{\text{T}}}}} \right)$$

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

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

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

since ρ²(A – L_jC) < α_j < 1.

On the other hand,

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

The proof of Lemma A.5 is complete.

We introduce the level set

$${{\mathcal{S}}_{0}} = \left\{ {L \in \mathcal{S}{\kern 1pt} :\;\;f(L)\;\leqslant \;f({{L}_{0}})} \right\}.$$

Obviously, Lemma A.5 implies the following result.

Corollary A.1. For any L₀ ∈ $\mathcal{S}$, the set ${{\mathcal{S}}_{0}}$ is bounded.

On the other hand, the function  f(L) achieves minimum on the set ${{\mathcal{S}}_{0}}$. (This function is continuous by the properties of the solution of the discrete Lyapunov equation and is considered on a compact set.) However, the set ${{\mathcal{S}}_{0}}$ has no common points with the boundary of $\mathcal{S}$ due to (A.8). The function  f(L) is differentiable on ${{\mathcal{S}}_{0}}$; see below. Consequently, we arrive at the following result.

Corollary A.2. There exists a minimum point ${{L}_{*}}$ on the set  $\mathcal{S}$, and the gradient of the function f(L) vanishes at this point.

Let us analyze the properties of the gradient of the function f(L, α).

Lemma A.6. The function f(L, α) is defined on the set of stabilizing L for ρ²(A – LC) < α < 1.

On this admissible set, the function is differentiable, and its gradient is given by

$${{\nabla }_{\alpha }}f(L,\alpha ) = \operatorname{tr} Y\left( {\frac{1}{{{{{(1 - \alpha )}}^{2}}}}({{D}_{1}} - L{{D}_{2}}){{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}} - \frac{1}{{{{\alpha }^{2}}}}(A - LC)P{{{(A - LC)}}^{{\text{T}}}}} \right),$$

(A.9)

$${{\nabla }_{L}}f(L,\alpha ) = 2\left( {\rho L - \frac{1}{\alpha }Y(A - LC)P{{C}^{{\text{T}}}} - \frac{1}{{1 - \alpha }}Y({{D}_{1}} - L{{D}_{2}})D_{2}^{{\text{T}}}} \right),$$

(A.10)

where the matrices P and Y are the solutions of the discrete Lyapunov equations (5) and (8), respectively.

The minimum of f(L, α) is achieved at an inner point of the admissible set and is determined by the conditions

$${{\nabla }_{L}}f(L,\alpha ) = 0,\quad {{\nabla }_{\alpha }}f(L,\alpha ) = 0.$$

In addition, f(L, α) as a function of α is strictly convex on ρ²(A – LC) < α < 1 and achieves minimum at an inner point of this interval.

Proof of Lemma A.6. We have the constrained optimization problem

$$\min f(L,\alpha ),\quad f(L,\alpha ) = \operatorname{tr} {{C}_{1}}PC_{1}^{{\text{T}}} + \rho \left\| L \right\|_{F}^{2}$$

subject to the discrete Lyapunov equation (5) for the matrix P of the invariant ellipsoid.

Following Lemma A.3, differentiation with respect to a is performed using the relations (A.9), (5), and (8). To differentiate with respect to L, we add an increment ΔL and denote by ΔP the corresponding increment of P. As a result, the relation (5) takes the form

$$\begin{gathered} \frac{1}{\alpha }(A - (L + \Delta L)C)(P + \Delta P){{(A - (L + \Delta L)C)}^{{\text{T}}}} - (P + \Delta P) \\ + \;\frac{1}{{1 - \alpha }}({{D}_{1}} - (L + \Delta L){{D}_{2}}){{({{D}_{1}} - (L + \Delta L){{D}_{2}})}^{{\text{T}}}} = 0. \\ \end{gathered} $$

Leaving the notation ΔP for the principal terms of the increment, we obtain

$$\begin{gathered} \frac{1}{\alpha }\left( {(A - LC)P{{{(A - LC)}}^{{\text{T}}}} - \Delta LCP{{{(A - LC)}}^{{\text{T}}}} - (A - LC)P{{{(\Delta LC)}}^{{\text{T}}}} + (A - LC)\Delta P{{{(A - LC)}}^{{\text{T}}}}} \right) \\ - \;(P + \Delta P) + \frac{1}{{1 - \alpha }}\left( {({{D}_{1}} - L{{D}_{2}}){{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}} - \Delta L{{D}_{2}}{{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}} - ({{D}_{1}} - L{{D}_{2}}){{{(\Delta L{{D}_{2}})}}^{{\text{T}}}}} \right) = 0. \\ \end{gathered} $$

Subtracting Eq. (12) from this equation yields

$$\frac{1}{\alpha }(A - LC)\Delta P{{(A - LC)}^{{\text{T}}}}$$

$$ - \;\Delta P - \frac{1}{\alpha }\left( {\Delta LCP{{{(A - LC)}}^{{\text{T}}}} + (A - LC)P{{{(\Delta LC)}}^{{\text{T}}}}} \right)$$

(A.11)

$$ - \;\frac{1}{{1 - \alpha }}\left( {\Delta L{{D}_{2}}{{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}} + ({{D}_{1}} - L{{D}_{2}}){{{(\Delta L{{D}_{2}})}}^{{\text{T}}}}} \right) = 0.$$

We calculate the increment of the functional f(L) by linearizing the corresponding values:

$$\Delta f(L) = \operatorname{tr} {{C}_{1}}\Delta PC_{1}^{{\text{T}}} + \rho \operatorname{tr} {{L}^{{\text{T}}}}\Delta L + \rho \operatorname{tr} {{(\Delta L)}^{{\text{T}}}}L = \operatorname{tr} \Delta PC_{1}^{{\text{T}}}{{C}_{1}} + 2\rho \operatorname{tr} {{L}^{{\text{T}}}}\Delta L.$$

By Lemma B.1, from the dual Eqs. (A.11) and (8) we have

$$\Delta f(L) = - 2\operatorname{tr} Y\left( {\frac{1}{\alpha }\Delta LCP{{{(A - LC)}}^{{\text{T}}}} + \frac{1}{{1 - \alpha }}\Delta L{{D}_{2}}{{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}}} \right) + 2\rho \operatorname{tr} {{L}^{{\text{T}}}}\Delta L$$

$$ = 2\operatorname{tr} \left( {\rho {{L}^{{\text{T}}}}\Delta L - \frac{1}{\alpha }CP{{{(A - LC)}}^{{\text{T}}}}Y - \frac{1}{{1 - \alpha }}{{D}_{2}}{{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}}Y} \right)\Delta L$$

$$ = \left\langle {2\left( {\rho L - \frac{1}{\alpha }Y(A - LC)P{{C}^{{\text{T}}}} - \frac{1}{{1 - \alpha }}Y({{D}_{1}} - L{{D}_{2}})D_{2}^{{\text{T}}}} \right),\Delta L} \right\rangle .$$

Thus, the relation (A.10) is derived and the proof of Lemma A.6 is complete.

The gradient of the function  f(L) is not Lipschitz on the set $\mathcal{S}$. But it can be shown to possess this property on the subset ${{\mathcal{S}}_{0}}$ (similar to [16]).

These properties of the minimized function and its derivatives justify the minimization method implemented as Algorithm 1.

APPENDIX B

Lemma B.1 [16]. Let X and Y be the solutions of the dual Lyapunov equations with a Hurwitz matrix A:

$${{A}^{{\text{T}}}}X + XA + W = 0\quad and\quad AY + Y{{A}^{{\text{T}}}} + V = 0.$$

Then tr(XV) = tr(YW).

The properties of the function f(α) established in [16] fully apply to the case under consideration. In particular, the function f(α) is well-defined, positive, and strongly convex on the interval 0 < α < 2σ(A – LC) and its values tend to infinity at the interval endpoints. Moreover, there exists a constant c > 0 such that

$$f(\alpha )\; \geqslant \;\frac{c}{{\alpha (2\sigma - \alpha )}},\quad 0 < \alpha < 2\sigma (A - LC).$$

(B.1)

The function  f(α) can be effectively minimized using Newton’s method. We specify an initial approximation 0 < α₀ < 2σ(A – LC), e.g., α₀ = σ(A – LC), and apply the iterative process

$${{\alpha }_{{j + 1}}} = {{\alpha }_{j}} - \frac{{f{\kern 1pt} '({{\alpha }_{j}})}}{{f{\kern 1pt} ''({{\alpha }_{j}})}},$$

where, according to [16],

$$\begin{gathered} f{\kern 1pt} '(\alpha ) = \operatorname{tr} Y\left( {P - \frac{1}{{{{\alpha }^{2}}}}({{D}_{1}} - L{{D}_{2}}){{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}}} \right), \\ f{\kern 1pt} ''(\alpha ) = 2\operatorname{tr} Y\left( {X + \frac{1}{{{{\alpha }^{3}}}}({{D}_{1}} - L{{D}_{2}}){{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}}} \right), \\ \end{gathered} $$

(B.2)

and P, Y, and X are the solutions of the Lyapunov equations (12), (13), and (14), respectively. Theorem A.1 remains valid as well.

The following lemma is a continuous-time analog of Lemma A.4.

Lemma B.2. The function f(L) is well-defined and positive on the set $\mathcal{S}$ of admissible filter matrices.

Indeed, if (A – LC) is a Hurwitz matrix, then σ(A – LC) > 0 and for 0 < α < 2σ(A – LC), there exists a solution P $\succcurlyeq$ 0 of the Lyapunov equation (12). Thus, a strictly positive function f(L, α) is well-defined and  f(L) > 0 due to (B.1). Its domain $\mathcal{S}$ may be nonconvex and disconnected and its boundaries nonsmooth; see [16].

Lemma B.3. On the set $\mathcal{S}$ of admissible matrices the function f(L) is coercive (i.e., it tends to infinity on the boundary of the domain). Moreover, the following lower bounds are valid:

$$\begin{gathered} f(L)\; \geqslant \;\frac{{{{\lambda }_{{\min }}}({{C}_{1}}C_{1}^{{\text{T}}})\left\| {{{D}_{1}} - L{{D}_{2}}} \right\|_{F}^{2}}}{{4\sigma (A - LC)\left( {\left\| {A - LC} \right\| + \sigma (A - LC)} \right)}}, \\ f(L)\; \geqslant \;\rho {{\left\| L \right\|}^{2}}. \\ \end{gathered} $$

(B.3)

Proof of Lemma B.3. We consider a sequence {L_j} ⊆ $\mathcal{S}$ of admissible matrices such that L_j → L ∈ $\partial \mathcal{S}$, i.e., σ(A – LC) = 0. In other words, for any ε > 0 there exists a number N = N(ε) such that

$$\left| {\sigma (A - {{L}_{j}}C) - \sigma (A - LC)} \right| = \sigma (A - {{L}_{j}}C) < \varepsilon $$

for all j $\geqslant$ N(ε).

Let P_j be the solution of equation (12) associated with the filter matrix L_j:

$$\left( {A - {{L}_{j}}C + \frac{{{{\alpha }_{j}}}}{2}I} \right){{P}_{j}} + {{P}_{j}}{{\left( {A - {{L}_{j}}C + \frac{{{{\alpha }_{j}}}}{2}I} \right)}^{{\text{T}}}} + \frac{1}{{{{\alpha }_{j}}}}({{D}_{1}} - {{L}_{j}}{{D}_{2}}){{({{D}_{1}} - {{L}_{j}}{{D}_{2}})}^{{\text{T}}}} = 0;$$

let Y_j be the solution of its dual Lyapunov equation

$${{\left( {A - {{L}_{j}}C + \frac{{{{\alpha }_{j}}}}{2}I} \right)}^{{\text{T}}}}{{Y}_{j}} + {{Y}_{j}}\left( {A - {{L}_{j}}C + \frac{{{{\alpha }_{j}}}}{2}I} \right) + {{C}_{1}}C_{1}^{{\text{T}}} = 0.$$

In view of [16, Lemma A.3], we have

$$f({{L}_{j}}) = \operatorname{tr} ({{C}_{1}}{{P}_{j}}C_{1}^{{\text{T}}}) + \rho \left\| {{{L}_{j}}} \right\|_{F}^{2}$$

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

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

$$ \geqslant \;\frac{1}{{4\sigma (A - {{L}_{j}}C)}}\frac{{{{\lambda }_{{\min }}}({{C}_{1}}C_{1}^{{\text{T}}})}}{{\left\| {A - {{L}_{j}}C + \frac{{{{\alpha }_{j}}}}{2}I} \right\|}}\left\| {{{D}_{1}} - {{L}_{j}}{{D}_{2}}} \right\|_{F}^{2}$$

$$ \geqslant \;\frac{{{{\lambda }_{{\min }}}({{C}_{1}}C_{1}^{{\text{T}}})}}{{4\varepsilon \left( {\left\| {A - {{L}_{j}}C} \right\| + \varepsilon } \right)}}\left\| {{{D}_{1}} - {{L}_{j}}{{D}_{2}}} \right\|_{F}^{2}\;\xrightarrow[{\varepsilon \to 0}]{}\; + {\kern 1pt} \infty ,$$

since 0 < α_j < 2σ(A – L_jC) and

$$\left\| {A - {{L}_{j}}C + \frac{{{{\alpha }_{j}}}}{2}I} \right\|\;\leqslant \;\left\| {A - {{L}_{j}}C} \right\| + \frac{{{{\alpha }_{j}}}}{2}.$$

On the other hand,

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

The proof of Lemma B.3 is complete.

We introduce the level set

$${{\mathcal{S}}_{0}} = \left\{ {L \in \mathcal{S}{\kern 1pt} :\;\;f(L)\;\leqslant \;f({{L}_{0}})} \right\}.$$

Obviously, Lemma B.3 implies the following result.

Corollary B.3. For any L₀ ∈ $\mathcal{S}$, the set ${{\mathcal{S}}_{0}}$ is bounded.

On the other hand, the function f(L) achieves minimum on the set ${{\mathcal{S}}_{0}}$. (This function is continuous by the properties of the solution of the Lyapunov equation and is considered on a compact set.) However, the set ${{\mathcal{S}}_{0}}$ has no common points with the boundary of $\mathcal{S}$ due to (B.3). The function f(L) is differentiable on ${{\mathcal{S}}_{0}}$; see below. Consequently, we arrive at the following result.

Corollary B.4. There exists a minimum point ${{L}_{*}}$ on the set $\mathcal{S}$, and the gradient of the function f(L) vanishes at this point.

Let us analyze the properties of the gradient of the function f(L, α).

Lemma B.4. The function f(L, α) is defined on the set of stabilizing L for 0 < α < 2σ(A – LC). On this admissible set, the function is differentiable, and its gradient is given by

$$\begin{gathered} {{\nabla }_{\alpha }}f(L,\alpha ) = \operatorname{tr} Y\left( {P - \frac{1}{{{{\alpha }^{2}}}}({{D}_{1}} - L{{D}_{2}}){{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}}} \right), \\ {{\nabla }_{L}}f(L,\alpha ) = 2\left( {\rho L - YP{{C}^{{\text{T}}}} - \frac{1}{\alpha }Y({{D}_{1}} - L{{D}_{2}})D_{2}^{{\text{T}}}} \right), \\ \end{gathered} $$

(B.4)

where the matrices P and Y are the solutions of the Lyapunov equations (12) and (13), respectively.

The minimum of f(L, α) is achieved at an inner point of the admissible set and is determined by the conditions

$${{\nabla }_{L}}f(L,\alpha ) = 0,\quad {{\nabla }_{\alpha }}f(L,\alpha ) = 0.$$

In addition,  f(L, α) as a function of α is strictly convex on 0 < α < 2σ(A – LC) and achieves minimum at an inner point of this interval.

Proof of Lemma B.4. We have the constrained optimization problem

$$\min f(L,\alpha ),\quad f(L,\alpha ) = \operatorname{tr} {{C}_{1}}PC_{1}^{{\text{T}}} + \rho \left\| L \right\|_{F}^{2}$$

subject to the Lyapunov equation (12) for the matrix P of the invariant ellipsoid.

Differentiation with respect to α is performed using the relations (B.2), (12), and (13). To differentiate with respect to L, we add an increment ΔL and denote by ΔP the corresponding increment of P. As a result, the relation (12) takes the form

$$\left( {A - (L + \Delta L)C + \frac{\alpha }{2}I} \right)(P + \Delta P) + (P + \Delta P){{\left( {A - (L + \Delta L)C + \frac{\alpha }{2}I} \right)}^{{\text{T}}}}$$

$$ + \;\frac{1}{\alpha }({{D}_{1}} - (L + \Delta L){{D}_{2}}){{({{D}_{1}} - (L + \Delta L){{D}_{2}})}^{{\text{T}}}} = 0.$$

Leaving the notation ΔP for the principal terms of the increment, we obtain

$$\left( {A - (L + \Delta L)C + \frac{\alpha }{2}I} \right)P + P{{\left( {A - (L + \Delta L)C + \frac{\alpha }{2}I} \right)}^{{\text{T}}}}$$

$$ + \;\left( {A - LC + \frac{\alpha }{2}I} \right)\Delta P + \Delta P{{\left( {A - LC + \frac{\alpha }{2}I} \right)}^{{\text{T}}}}$$

$$ + \;\frac{1}{\alpha }\left( {({{D}_{1}} - L{{D}_{2}}){{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}} - \Delta L{{D}_{2}}{{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}} - ({{D}_{1}} - L{{D}_{2}}){{{(\Delta L{{D}_{2}})}}^{{\text{T}}}}} \right) = 0.$$

Subtracting Eq. (12) from this equation yields

$$\left( {A - LC + \frac{\alpha }{2}I} \right)\Delta P + \Delta P{{\left( {A - LC + \frac{\alpha }{2}I} \right)}^{{\text{T}}}} - \Delta LCP - P{{(\Delta LC)}^{{\text{T}}}}$$

(B.5)

$$ - \,\,\frac{1}{\alpha }\left( {\Delta L{{D}_{2}}{{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}} + ({{D}_{1}} - L{{D}_{2}}){{{(\Delta L{{D}_{2}})}}^{{\text{T}}}}} \right) = 0.$$

(B.6)

We calculate the increment of the functional f(L) by linearizing the corresponding values:

$$\Delta f(L) = \operatorname{tr} {{C}_{1}}\Delta PC_{1}^{{\text{T}}} + \rho \operatorname{tr} {{L}^{{\text{T}}}}\Delta L + \rho \operatorname{tr} {{(\Delta L)}^{{\text{T}}}}L = \operatorname{tr} \Delta PC_{1}^{{\text{T}}}{{C}_{1}} + 2\rho \operatorname{tr} {{L}^{{\text{T}}}}\Delta L.$$

By Lemma B.1, from the dual Eqs. (B.6) and (13) we have

$$\Delta f(L) = - \operatorname{tr} 2Y\left( {\Delta LCP + \frac{1}{\alpha }\Delta L{{D}_{2}}{{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}}} \right) + 2\rho \operatorname{tr} {{L}^{{\text{T}}}}\Delta L$$

$$ = 2\operatorname{tr} \left( {\rho {{L}^{{\text{T}}}}\Delta L - CPY\Delta L - \frac{1}{\alpha }{{D}_{2}}{{{({{D}_{1}} - L{{D}_{2}})}}^{{\text{T}}}}Y\Delta L} \right)$$

$$ = \left\langle {2\left( {\rho L - YP{{C}^{{\text{T}}}} - \frac{1}{\alpha }Y({{D}_{1}} - L{{D}_{2}})D_{2}^{{\text{T}}}} \right),\Delta L} \right\rangle .$$

Thus, the relation (B.4) is derived and the proof of Lemma B.4 is complete.