**7. Appendix**

It is noted that in this appendix all time arguments of either vector-valued or matrix-valued time functions will be dropped for the simplicity of expression. They can be easily distin‐ guished by their contents.

**Proof**: (T1)⇒(T2). We need to show that if the conditions (10) and (11) in (T1) hold, then (O1) and (O2), which are equivalent to (T2), hold. Let quadratic Lyapunov function be

$$V(\mathbf{x}, \tilde{\mathbf{x}}, \boldsymbol{\epsilon}) = \mathbf{x}^T P\_1 \mathbf{x} + \tilde{\mathbf{x}}^T P\_2 \tilde{\mathbf{x}} + \boldsymbol{\epsilon}^T P\_3^{-1} \boldsymbol{\epsilon}\_{\boldsymbol{\epsilon}}$$

with *P*1≻0, *P*2≻0, and *P*3≻0. Then, the performance index (9) can be written as

$$J\_{\infty} = \int\_{0}^{\infty} \|z\|^2 dt = \int\_{0}^{\infty} \left[ z^T z + \frac{d}{dt} V(x, \pounds, \epsilon) \right] dt - V(x(\infty), \pounds(\infty), \epsilon(\infty)), \tag{39}$$

for all states satisfying (1) and (3) with initial states (*x*(0), *x*˜(0))=(0, 0), and (0)=0. In view of (8), the first integrand in (39) is

$$\mathbf{z}^T \mathbf{z} = \mathbf{x}^T \mathbf{D}^T \mathbf{D} \mathbf{x} \tag{40}$$

The second integrand in (39) is

$$\frac{d}{dt}V(\mathbf{x}, \tilde{\mathbf{x}}, \boldsymbol{\varepsilon}) = \frac{d}{dt}\left(\mathbf{x}^T P\_1 \mathbf{x} + \tilde{\mathbf{x}}^T P\_2 \tilde{\mathbf{x}} + \boldsymbol{\varepsilon}^T P\_3^{-1} \boldsymbol{\varepsilon}\right). \tag{41}$$

The right-hand side of equality (41) can be reorganized by using the closed-loop system (8), and thus, the first term is

$$\begin{aligned} &\frac{d}{dt}\mathbf{x}^T P\_1 \mathbf{x} \\ &= \mathbf{x}^T \left( (FA + B\_1 \mathbf{K})^T P\_1 + P\_1 (FA + B\_1 \mathbf{K}) \right) \mathbf{x} \\ &+ \bar{\mathbf{x}}^T (B\_1 \mathbf{K})^T P\_1 \mathbf{x} + \mathbf{x}^T P\_1 B\_1 \mathbf{K} \tilde{\mathbf{x}} \\ &+ \bar{\mathbf{x}}^T B\_2^T P\_1 \mathbf{x} + \mathbf{x}^T P\_1 B\_2 \mathbf{w} \end{aligned} \tag{42}$$

Completing the square of (42), we have

$$\begin{split} &\frac{d}{dt}\mathbf{x}^{T}P\_{1}\mathbf{x} \\ &= \mathbf{x}^{T}\left((FA+B\_{1}K)^{T}P\_{1} + P\_{1}(FA+B\_{1}K)\right)\mathbf{x} \\ &+ \tilde{\mathbf{x}}^{T}(B\_{1}K)^{T}P\_{1}\mathbf{x} + \mathbf{x}^{T}P\_{1}B\_{1}K\tilde{\mathbf{x}} \\ &+ \gamma^{2}w^{T}w + \gamma^{-2}xP\_{1}B\_{2}^{T}B\_{2}^{T}P\_{1}\mathbf{x} \\ &- (w-\gamma^{-2}B\_{2}^{T}P\_{1}\mathbf{x})^{T}\gamma^{2}(w-\gamma^{-2}B\_{2}^{T}P\_{1}\mathbf{x}). \end{split} \tag{43}$$

Similarly, the second term of (41) is

$$\begin{split} & \frac{d}{dt} \mathfrak{X}^{T} P\_{2} \mathfrak{X} \\ &= \mathfrak{x}^{T} \left( (A + LC)^{T} P\_{2} + P\_{2} (A + LC) \right) \mathfrak{X} \\ &+ \mathfrak{y}^{T} diag[\mathfrak{e}]^{T} L^{T} P\_{2} \mathfrak{x} + \mathfrak{x}^{T} P\_{2} L diag[\mathfrak{e}] \mathfrak{Y} \\ &+ \mathfrak{x}^{T} A^{T} (I - F)^{T} P\_{2} \mathfrak{x} + \mathfrak{x}^{T} P\_{2} (I - F) A \ge \end{split} \tag{44}$$

Applying completing the square to (44), we obtain

$$\begin{split} & \frac{d}{dt} \tilde{\mathbf{x}}^{T} P\_{2} \tilde{\mathbf{x}} \\ &= -\tilde{\mathbf{x}}^{T} \left( (A + LC)^{T} P\_{2} + P\_{2} (A + LC) \right) \tilde{\mathbf{x}} \\ &+ \mathbf{x}^{T} A^{T} (I - F)^{T} P\_{2} \tilde{\mathbf{x}} + \tilde{\mathbf{x}}^{T} P\_{2} (I - F) A \mathbf{x} \\ &+ \beta^{2} \mathfrak{g}^{T} diag[\mathbf{e}]^{T} diag[\mathbf{c}] \mathfrak{g} + \beta^{-2} \tilde{\mathbf{x}}^{T} P\_{2} LL^{T} P\_{2} \tilde{\mathbf{x}} \\ &- \left( diag[\mathbf{c}] \mathfrak{g} - \beta^{-2} L^{T} P\_{2} \mathfrak{x} \right)^{T} \beta^{2} (diag[\mathbf{c}] \mathfrak{g} - \beta^{-2} L^{T} P\_{2} \mathfrak{x}) . \end{split} \tag{45}$$

Substituting (40), (43), and (45) into (39), we have

(41)

(42)

(43)

(44)

(45)

The right-hand side of equality (41) can be reorganized by using the closed-loop system (8),

and thus, the first term is

20 Robust Control - Theoretical Models and Case Studies

Completing the square of (42), we have

Similarly, the second term of (41) is

Applying completing the square to (44), we obtain

$$\begin{split} J\_{\mathbb{S}^{0}} &= \int\_{0}^{\infty} \left[ \begin{pmatrix} \mathbf{x} \\ \ddot{\mathbf{x}} \end{pmatrix}^{T} \begin{pmatrix} \Pi\_{1}(\mathbf{P}\_{1},\mathbf{K}) & \star \\ (P\_{1}\mathbf{B}\_{1}\mathbf{K})^{T} + P\_{2}(I-F)\mathbf{A} & \Pi\_{2}(\mathbf{P}\_{2},L) \end{pmatrix} \begin{pmatrix} \mathbf{x} \\ \ddot{\mathbf{x}} \end{pmatrix} \right. \\ & \left. - \left( w - \gamma^{-2} B\_{2}^{T} P\_{1} \mathbf{x} \right)^{T} \gamma^{2} (w - \gamma^{-2} B\_{2}^{T} P\_{1} \mathbf{x}) \right. \\ & \left. - \left( \operatorname{diag} [\varepsilon] \boldsymbol{\varrho} - \beta^{-2} L^{T} P\_{2} \mathbf{x} \right)^{T} \beta^{2} (\operatorname{diag} [\varepsilon] \boldsymbol{\varrho} - \beta^{-2} L^{T} P\_{2} \mathbf{x}) \right. \\ & \left. + \gamma^{2} w^{T} w + \beta^{2} \boldsymbol{\varrho}^{T} \operatorname{diag} [\varepsilon]^{T} \operatorname{diag} [\varepsilon] \boldsymbol{\varrho} + \frac{d}{dt} (\mathbf{e}^{T} P\_{3}^{-1} \mathbf{c}) \right] dt \\ & - V(\mathbf{x}(\infty), \ddot{\mathbf{x}}(\infty), \boldsymbol{\varepsilon}(\infty)), \end{split} \tag{46}$$

where definition of Π1(P1) and Π2(P2) are defined as (12) and (13), respectively. Therefore, by eliminating the negative terms from (46), the following inequality is drawn,

$$\begin{split} J\boldsymbol{\omega} &\leq \int\_{0}^{\infty} \biggl( \begin{pmatrix} \boldsymbol{\omega} \\ \boldsymbol{\upxi} \end{pmatrix}^{\top} \begin{pmatrix} \boldsymbol{\Pi}\_{1}(\boldsymbol{P}\_{1},\boldsymbol{K}) & \boldsymbol{\star} \\ (\boldsymbol{P}\_{1}\boldsymbol{B}\_{1}\boldsymbol{K})^{\top} + \boldsymbol{P}\_{2}(\boldsymbol{I}-\boldsymbol{F})\boldsymbol{A} & \boldsymbol{\Pi}\_{2}(\boldsymbol{P}\_{2},\boldsymbol{L}) \end{pmatrix} \begin{pmatrix} \boldsymbol{\upchi} \\ \boldsymbol{\upxi} \end{pmatrix} \\ &+ \gamma^{2} \boldsymbol{w}^{T} \boldsymbol{w} + \beta^{2} \boldsymbol{\mathcal{g}}^{T} \boldsymbol{d} \boldsymbol{a} \boldsymbol{g} [\boldsymbol{\epsilon}]^{T} \boldsymbol{d} \boldsymbol{a} \boldsymbol{g} [\boldsymbol{\epsilon}] \boldsymbol{\mathcal{g}} + \frac{d}{dt} (\boldsymbol{\epsilon}^{T} \boldsymbol{P}\_{3}^{-1} \boldsymbol{\varepsilon}) \Bigg) dt. \end{split} \tag{47}$$

Given that *diag y* ^ <sup>=</sup>*diag <sup>y</sup>* ^ , if (11) of (T1) holds, then it concludes that

$$\mathcal{B}^2 \mathcal{Y}^T \text{diag} [\boldsymbol{\epsilon}]^T \text{diag} [\boldsymbol{\epsilon}] \boldsymbol{\mathcal{Y}} + \frac{d}{dt} (\boldsymbol{\epsilon}^T P\_{\mathfrak{Z}}^{-1} \boldsymbol{\epsilon}) = -\boldsymbol{\epsilon}^T (\mathcal{Q} \boldsymbol{\zeta}) \boldsymbol{\epsilon}.$$

In view of (10) of (T1), we thus find that the inequality (47) is simply

$$J\omega \le \int\_0^\infty \left[\gamma^2 w^T w - \epsilon^T (2\mathcal{Q})\epsilon\right] dt \le \gamma^2 \int\_0^\infty w^T w dt. \tag{48}$$

Therefore, the inequality (48) satisfies the performance index (9), which completes the proof (O2).

To prove that (O1) holds, we use the inequality (10) in (T1) and get the equivalent inequality as follows,

$$\mathcal{P}\mathcal{A} + \mathcal{A}^T \mathcal{P} \prec -\mathcal{P}\mathcal{B}\mathcal{R}^{-1}\mathcal{B}^T \mathcal{P} - \mathcal{D}^T \mathcal{D}\_M$$

where

$$\mathcal{P} = \begin{pmatrix} P\_1 & 0 \\ 0 & P\_2 \end{pmatrix}, \quad \vec{A} = \begin{pmatrix} FA + B\_1K & B\_1K \\ (I - F)A & A + LC \end{pmatrix}, \quad \mathcal{R} = \begin{pmatrix} \gamma^2 & 0 \\ 0 & \beta^2 \end{pmatrix}.$$

$$\mathcal{B} = \begin{pmatrix} B\_2 & 0 \\ 0 & L \end{pmatrix}, \quad \mathcal{D} = \begin{pmatrix} D & 0 \\ 0 & 0 \end{pmatrix}$$

It is concluded, by a standard Lyapunov stability argument, that *Ã*, that is (7), has all eigen‐ values in C<sup>−</sup> , which shows that (O1) holds. This completes the proof of Theorem 1.
