SCIENCE CHINA Information Sciences, Volume 63 , Issue 7 : 172002(2020) https://doi.org/10.1007/s11432-020-2916-8

## Observer-based multi-objective parametric design for spacecraft with super flexible netted antennas

• AcceptedApr 28, 2020
• PublishedJun 10, 2020
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### Acknowledgment

This work was supported by Major Program of National Natural Science Foundation of China (Grant Nos. 61690210, 61690212), Self-Planned Task of State Key Laboratory of Robotics and System (HIT) (Grant No. SKLRS201716A), and National Natural Science Foundation of China (Grant No. 61333003). The authors are very grateful to the anonymous reviewers for their meaningful suggestions and comments.

### Supplement

Appendix

Proof of Theorem protect 3.2

To prove this theorem, the following lemma which was presented in [22] is needed.

Lemma 3. Let the system (3)–(6) satisfy condition (ref co. Then the right coprime polynomial matrices $N\left(~s\right)~$ and $D\left(~s\right)~$ satisfying the following right coprime factorization (RCF): $$\left( sI-A\right) ^{-1}B=N\left( s\right) D^{-1}\left( s\right) \tag{78}$$ are given by \begin{eqnarray}N\left( s\right) &=&\left[ { \begin{array}{c} {\gamma {s^{2}}+{a_{2}}s+{a_{1}}} \\ {-b_{y}\gamma {s^{2}}} \\ {\gamma {s^{3}}+{a_{2}}{s^{2}}+{a_{1}}s} \\ {-b_{y}\gamma {s^{3}}} \end{array} }\right] , \tag{79} \\ D\left( s\right) &=&-{s^{4}}+{I_{y}}{a_{2}}{s^{3}}+{I_{y}}{a_{1}}{s^{2}.} \tag{80} \end{eqnarray}

According to the eigenstructure assignment result in [23], when the system (3)–(6) is controllable, that is, when the condition ( 14) holds, complete parametric forms of the gain matrix $K$ and a corresponding nonsingular matrix $V$ satisfying $$\left( A+BK\right) V=V\Lambda _{c}, \tag{81}$$ where $\Lambda~_{c}$ is shown in (68), can be given by $$\left \{ \begin{array}{l} K=WV^{-1}, \\ V=[ \begin{array}{cccc} \hat{v}_{1} & \hat{v}_{2} & \hat{v}_{3} & \hat{v}_{4} \end{array} ] , \\ W=[ \begin{array}{cccc} \hat{w}_{1} & \hat{w}_{2} & \hat{w}_{3} & \hat{w}_{4} \end{array} ] , \end{array} \right. \tag{82}$$ with \begin{equation*}\left \{ \begin{array}{l} \hat{v}_{1}=N\left( \alpha _{1}+\alpha _{2} \mathrm{i}\right) \left( f_{1}+f_{2} \mathrm{i}\right) , \\ \hat{v}_{2}=N\left( \alpha _{1}-\alpha _{2} \mathrm{i}\right) \left( f_{1}-f_{2} \mathrm{i}\right) , \\ \hat{v}_{3}=N\left( \alpha _{3}\right) f_{3}, \hat{v}_{4}=N\left( \alpha _{4}\right) f_{4}, \end{array} \right.\end{equation*} and \begin{equation*}\left \{ \begin{array}{l} \hat{w}_{1}=D\left( \alpha _{1}+\alpha _{2} \mathrm{i}\right) \left( f_{1}+f_{2} \mathrm{i}\right) , \\ \hat{w}_{2}=D\left( \alpha _{1}-\alpha _{2} \mathrm{i}\right) \left( f_{1}-f_{2} \mathrm{i}\right) , \\ \hat{w}_{3}=D\left( \alpha _{3}\right) f_{3}, \hat{w}_{4}=D\left( \alpha _{4}\right) f_{4}, \end{array} \right.\end{equation*} where $N\left(~s\right)~\in~\mathbb{R}^{4\times~1}[s]$ and $D\left( s\right)~\in~\mathbb{R}[s]$ are a pair of polynomial matrices satisfying the RCF (78), and $f_{i},\alpha~_{i},i=1,2,3,4,$ are parameters satisfying the following constraint: $$\det \left( V\right) =\Delta _{c}\neq 0. \tag{83}$$

It is known from Lemma 3 that such $N\left(~s\right)~$ and $D\left(~s\right)~$ can be given by (79) and (80), respectively. Then, through simple deductions, we can obtain the expression of $\Delta~_{c}$ as shown in Constraint C1.

It is easy to see that $\hat{v}_{1}$ and $\hat{v}_{2}$ are complex conjugates to each other, so do $\hat{w}_{1}$ and $\hat{w}_{2}.$ Therefore, assume that $$\hat{v}_{1}=\vartheta _{vR}+\vartheta _{vI} \mathrm{i}, \hat{v} _{2}=\vartheta _{vR}-\vartheta _{vI} \mathrm{i}, \tag{84}$$ $$\hat{w}_{1}=\vartheta _{wR}+\vartheta _{wI} \mathrm{i}, \hat{w} _{2}=\vartheta _{wR}-\vartheta _{wI} \mathrm{i}. \tag{85}$$ In view of the first formula in (82), we have $$\hat{w}_{i}=K\hat{v}_{i}, i=1,2,3,4. \tag{86}$$ Substituting (84) and (85) into (86), we obtain the following linear equation: \begin{equation*}W_{0}=KV_{0},\end{equation*} where $$W_{0}=\left[ \begin{array}{cccc} \vartheta _{wR} & \vartheta _{wI} & \hat{w}_{3} & \hat{w}_{4} \end{array} \right] =\left[ \begin{array}{cccc} w_{1} & w_{2} & w_{3} & w_{4} \end{array} \right] , \tag{87}$$ $$V_{0}=\left[ \begin{array}{cccc} \vartheta _{vR} & \vartheta _{vI} & \hat{v}_{3} & \hat{v}_{4} \end{array} \right] =\left[ \begin{array}{cccc} v_{1} & v_{2} & v_{3} & v_{4} \end{array} \right] , \tag{88}$$ with $v_{i},~w_{i},~i=1,2,3,4$ being given by (24)–(27). Obviously, when Eq. (83) holds, $V_{0}$ is also nonsingular. Thus the matrix $K$ can be given by (22). Combining (82), (84) and (88), gives the expression of $V$ shown in (23). Then the proof is completed.

Proof of Theorem protect 3.3

Similarly, to prove Theorem 3.3, the following result obtained in [22] is needed.

Lemma 4. Let the system (3)–(6) satisfy condition ( 14), and then the right coprime polynomial matrices $H\left(~s\right)~$ and $L\left(~s\right)~$ satisfying the following RCF: $$\left( sI-A^{\mathrm{T}}\right) ^{-1}C^{\mathrm{T}}=H\left( s\right) L^{-1}\left( s\right) \tag{89}$$ are given by \begin{eqnarray}H\left( s\right) &=&\left[ { \begin{array}{cc} 1 & 0 \\ 0 & {{a_{1}}b_{y}s} \\ 0 & {-{s^{2}}+{I_{y}}{a_{2}}s+{I_{y}}{a_{1}}} \\ 0 & {{a_{2}}b_{y}s+{a_{1}}b_{y}} \end{array} }\right] , \tag{90} \\ L\left( s\right) &=&\left[ { \begin{array}{cc} s & 0 \\ {-1} & {-{s^{3}}+{I_{y}}{a_{2}}{s^{2}}+{I_{y}}{a_{1}}s} \end{array} }\right] . \tag{91} \end{eqnarray}

Based on the eigenstructure assignment theory shown in [23], when $\left(~A^{\mathrm{T}},~C^{\mathrm{T}}\right)~$ is controllable, that is, when the condition (14) holds, complete parametric forms of the gain matrix $L$ and a corresponding nonsingular matrix $T$ satisfying $$T^{\mathrm{T}}\left( A+LC\right) =\Lambda _{o}T^{\mathrm{T}}, \tag{92}$$ where $\Lambda~_{o}$ is shown in (68), can be given by $$\left \{ \begin{array}{l} L=T^{-\mathrm{T}}Z^{\mathrm{T}}, \\ T=\left[ \begin{array}{cccc} \hat{t}_{1} & \hat{t}_{2} & \hat{t}_{3} & \hat{t}_{4} \end{array} \right] , \\ Z=\left[ \begin{array}{cccc} \hat{z}_{1} & \hat{z}_{2} & \hat{z}_{3} & \hat{z}_{4} \end{array} \right] , \end{array} \right. \tag{93}$$ with \begin{equation*}\left \{ \begin{array}{ll} \hat{t}_{1}=H\left( \tilde{\alpha}_{1}+\tilde{\alpha}_{2} \mathrm{i}\right) \left( g_{1}+g_{2} \mathrm{i}\right) , \\ \hat{t}_{2}=H\left( \tilde{\alpha}_{1}-\tilde{\alpha}_{2} \mathrm{i}\right) \left( g_{1}-g_{2} \mathrm{i}\right) , \\ \hat{t}_{3}=H\left( \tilde{\alpha}_{3}\right) g_{3}, \hat{t}_{4}=H\left( \tilde{\alpha}_{4}\right) g_{4}, \end{array} \right.\end{equation*} and \begin{equation*}\left \{ \begin{array}{l} \hat{z}_{1}=L\left( \tilde{\alpha}_{1}+\tilde{\alpha}_{2} \mathrm{i}\right) \left( g_{1}+g_{2} \mathrm{i}\right) , \\ \hat{z}_{2}=L\left( \tilde{\alpha}_{1}-\tilde{\alpha}_{2} \mathrm{i}\right) \left( g_{1}-g_{2} \mathrm{i}\right) , \\ \hat{z}_{3}=L\left( \tilde{\alpha}_{3}\right) g_{3}, \hat{z}_{4}=L\left( \tilde{\alpha}_{4}\right) g_{4}, \end{array} \right.\end{equation*} where $H\left(~s\right)~\in~\mathbb{R}^{4\times~2}[s]$ and $L\left( s\right)~\in~\mathbb{R}^{2\times~2}[s]$ are a pair of polynomial matrices satisfying the RCF (89), and $\tilde{\alpha}_{i}$, $g_{i}=[{\tiny \begin{array}{c} g_{i1}~\\ g_{i2} \end{array}} ]~,~g_{i1},g_{i2}\in~\mathbb{R},~i=1,2,3,4$ are parameters satisfying the following constraint: $$\det \left( T\right) =\Delta _{o}\neq 0. \tag{94}$$ It is known from Lemma 4 that such $H\left(~s\right)~$ and $L\left(~s\right)~$ can be given by (90) and (91), respectively. Then, substituting (90), (91) and (93) into (94), gives the expression of $\Delta~_{o}$ as shown in Constraint C2.

It is easy to see that $\hat{t}_{1}$ and $\hat{t}_{2}$ are complex conjugates to each other, so do $\hat{z}_{1}$ and $\hat{z}_{2}.$ Therefore, assume that $$\hat{t}_{1}=\xi _{tR}+\xi _{tI} \mathrm{i}, \hat{t}_{2}=\xi _{tR}-\xi _{tI} \mathrm{i}, \tag{95}$$ $$\hat{z}_{1}=\xi _{zR}+\xi _{zI} \mathrm{i}, \hat{z}_{2}=\xi _{zR}-\xi _{zI} \mathrm{i}. \tag{96}$$ Considering the first formula in (93), we have $$\hat{z}_{i}=L^{\mathrm{T}}\hat{t}_{i}, i=1,2,3,4. \tag{97}$$ Substituting (95) and (96) into (97), we obtain the following linear equation: \begin{equation*}Z_{0}=L^{\mathrm{T}}T_{0},\end{equation*} where $$Z_{0}=\left[ \begin{array}{cccc} \xi _{zR} & \xi _{zI} & \hat{z}_{3} & \hat{z}_{4} \end{array} \right] =\left[ \begin{array}{cccc} z_{1} & z_{2} & z_{3} & z_{4} \end{array} \right] , \tag{98}$$ $$T_{0}=\left[ \begin{array}{cccc} \xi _{tR} & \xi _{tI} & \hat{t}_{3} & \hat{t}_{4} \end{array} \right] =\left[ \begin{array}{cccc} t_{1} & t_{2} & t_{3} & t_{4} \end{array} \right] , \tag{99}$$ with $t_{i},~z_{i},~i=1,2,3,4$ being given by (36)–(39). Obviously, when Eq. (94) holds, $T_{0}$ is also nonsingular. Thus the matrix $L$ can be given by (34). Combining (93), (95) and (99), gives the expression of $T$ shown in (35). Then the proof is completed.

Proof of Theorem protect 4.2

Let \begin{equation*}V^{-\mathrm{T}}=\left[ \begin{array}{cccc} \tilde{v}_{1} & \tilde{v}_{2} & \tilde{v}_{3} & \tilde{v}_{4} \end{array} \right] , T^{-\mathrm{T}}=\left[ \begin{array}{cccc} \tilde{t}_{1} & \tilde{t}_{2} & \tilde{t}_{3} & \tilde{t}_{4} \end{array} \right].\end{equation*} Then, according to (56), the following relations hold: \begin{equation*}\tilde{v}_{i}=\frac{1}{\Delta _{c}}v_{i}^{\ast }, \tilde{t}_{i}=\frac{1}{ \Delta _{o}}t_{i}^{\ast }, i=1,2,3,4.\end{equation*}

It can be seen that $V$ and $V^{-1}$ are, respectively, the right and the left eigenvector matrices of $A_{c}$. Thus, according to Lemma 1 in Para4 we have \begin{eqnarray*}\frac{\partial \lambda _{i}\left( A_{c}\right) }{\partial \Delta a_{j}} &=& \tilde{v}_{i}^{\mathrm{T}}\frac{\partial A_{c}}{\partial \Delta a_{j}}v_{i}= \tilde{v}_{i}^{\mathrm{T}}A_{j}v_{i}=\frac{1}{\Delta _{c}}\left( v_{i}^{\ast }\right) ^{\mathrm{T}}A_{j}v_{i}, \\ i &=&1,2,3,4, j=1,2. \end{eqnarray*} Similarly, considering that $T^{\mathrm{T}}$ and $T^{-\mathrm{T}}$ are, respectively, the left and the right eigenvector matrices of $A_{o}$, it can be known from Lemma 1 in [26] that \begin{eqnarray*}\frac{\partial \lambda _{i}\left( A_{o}\right) }{\partial \Delta a_{j}} &=&t_{i}^{\mathrm{T}}\frac{\partial A_{o}}{\partial \Delta a_{j}}\tilde{t} _{i}=t_{i}^{\mathrm{T}}A_{j}\tilde{t}_{i}=\frac{1}{\Delta _{o}}t_{i}^{ \mathrm{T}}A_{j}t_{i}^{\ast }, \\ i &=&1,2,3,4, j=1,2, \end{eqnarray*} holds. Then, the proof is completed.

Proof of Theorem protect 4.3

At first, let us discuss the eigenstructure of $A_{z}$ as a preliminary. Let $$T_{z}^{\mathrm{T}}=\left[ \begin{array}{cc} V^{-1}-Q_{\ast }T^{\mathrm{T}} & Q_{\ast }T^{\mathrm{T}} \\ -T^{\mathrm{T}} & T^{\mathrm{T}} \end{array} \right] , \tag{100}$$ and $$V_{z}=\left[ \begin{array}{cc} V & -VQ_{\ast } \\ V & T^{-\mathrm{T}}-VQ_{\ast } \end{array} \right] , \tag{101}$$ where $T$ and $V$ are given by (23) and (35), respectively. It is known from Theorems 3.2 and 3.3 that, when $K$ and $L$ are taken as (22) and (34), respectively, the relations (ref as1 and (92) hold. Thus, in view of (100) and (101), we can verify that $$T_{z}^{\mathrm{T}}V_{z}=I, \tag{102}$$ and $$T_{z}^{\mathrm{T}}A_{z}V_{z}=\left[ \begin{array}{cc} \Lambda _{c} & -\Lambda _{c}Q_{\ast }+Q_{\ast }\Lambda _{o}+V^{-1}BKT^{- \mathrm{T}} \\ 0 & \Lambda _{o} \end{array} \right] . \tag{103}$$ According to the matrix equation theory, there exists a unique solution to the following linear matrix equation with respect to $Q$: $$\Lambda _{c}Q-Q\Lambda _{o}=V^{-1}BKT^{-\mathrm{T}}. \tag{104}$$ With the help of matrix vectorization operations, it can be easily verified that $Q_{\ast~}$ given by (66) is the unique solution of the matrix equation (104). Thus, Eq. (103) can be simplified as $$T_{z}^{\mathrm{T}}A_{z}V_{z}=\Lambda _{z}, \tag{105}$$ where $\Lambda~_{z}$ is given by (68).

Then, let us discuss the explicit expression of $\left~\Vert~G_{c}\left( s\right)~\right~\Vert~_{2}.$ Considering (105), the function $\left~\Vert G_{c}\left(~s\right)~\right~\Vert~_{2}$ can be transformed into \begin{equation*}\left \Vert G_{c}\left( s\right) \right \Vert _{2}=\left \Vert C_{z}V_{z}\left( sI-\Lambda _{z}\right) ^{-1}T_{z}^{\mathrm{T}}D_{z}\right \Vert _{2}.\end{equation*} According to Theorems 3.2 and 3.3, when $K$ and $L$ are taken as (22) and (34), respectively, both $A_{c}$ and $A_{o}$ are stable. Then, from (105), we know that $A_{z}$ is also stable. Therefore, it is known from Lemma 4.1 of the previous study 1) that there exist unique symmetric positive definite solutions $P_{1}$ and $P_{2}$ to the following Lyapunov matrix equations: $$\Lambda _{z}P_{1}+P_{1}\Lambda _{z}=-T_{z}^{\mathrm{T}}D_{z}D_{z}^{\mathrm{T} }T_{z}, \tag{106}$$ and $$\Lambda _{z}P_{2}+P_{2}\Lambda _{z}=-V_{z}^{\mathrm{T}}C_{z}^{\mathrm{T} }C_{z}V_{z}, \tag{107}$$ and $\left~\Vert~G_{c}\left(~s\right)~\right~\Vert~_{2}$ can be given by \begin{eqnarray*}\left \Vert G_{dy_{p}}\left( s\right) \right \Vert _{2} =\left( \mathrm{trace }\left( C_{z}V_{z}P_{1}V_{z}^{\mathrm{T}}C_{z}^{\mathrm{T}}\right) \right) ^{ \frac{1}{2}} =\left( \mathrm{trace}\left( D_{z}^{\mathrm{T}}T_{z}P_{2}T_{z}^{\mathrm{T} }D_{z}\right) \right) ^{\frac{1}{2}}. \end{eqnarray*} In view of (100) and (101), with the help of matrix vectorization operations, it can be verified that $P_{1}^{\ast~}$ and $P_{2}^{\ast~}$ which are given by (66) are the unique solutions of the matrix equations (106) and (107), respectively. Thus the result (ref rrr can be obtained. Then the proof is completed.

Duan G-R, Liu G P, Thompson S. Disturbance attenuation in Luenberger function observer designs—a parametric approach. IFAC Proc Vol, 2000, 33: 41–46.

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• Figure 1

Structure of classic PID controller.

• Figure 2

(Color online) The index values $J_{\mathrm{sen}}$ under the three control methods.

• Figure 3

(Color online) (a) Pitch angle; (b) pitch angular velocity; (c) control torque.

• Table 1

Table 1Symbols

 Symbol Meaning $\mathrm{diag}\left(~s_{1},s_{2},\ldots,s_{n}\right)$ $\text{Diagonal~matrix~with~}s_{1},s_{2},\ldots,s_{n}\text{~as~diagonal~elements}$ $\lambda~_{i}\left(~M\right)$ textThe $i$text-th eigenvalue of a matrix $M$ $\mathrm{trace}\left(~M\right)$ textSum of diagonal elements of a matrix $M$ $\mathrm{Blockdiag}\left(~M_{1},M_{2},\ldots,M_{n}\right)$ textBlock diagonal matrix with $M_{1},M_{2},\ldots,M_{n}$text as diagonal elements $\mathrm{vec}(~[ \begin{array}{cccc} \eta~_{1}~&~\eta~_{2}~&~\cdots~&~\eta~_{n} \end{array} ]~)$ $[ \begin{array}{cccc} \eta~_{1}^{\mathrm{T}}~&~\eta~_{2}^{\mathrm{T}}~&~\cdots~&~\eta~_{n}^{\mathrm{T} } \end{array} ]~^{\mathrm{T}}$ $\mathrm{unvec}(~[ \begin{array}{cccc} \eta~_{1}^{\mathrm{T}}~&~\eta~_{2}^{\mathrm{T}}~&~\cdots~&~\eta~_{n}^{\mathrm{T} } \end{array} ]~^{\mathrm{T}})$ $[ \begin{array}{cccc} \eta~_{1}~&~\eta~_{2}~&~\cdots~&~\eta~_{n} \end{array} ]$ $A\otimes~B$ $\text{Kronecker~product~of~}A\text{~and~}B$
• Table 2

Table 2Nominal values of parameters

 Parameter Value Unit $I_{y}$ 20667.25 $\mathrm{ kg\cdot~m}^{2}$ $b$ $-108.88$ $\sqrt{\mathrm{kg}}\mathrm{\cdot~m}$ $\xi$ $0.005$ – $\Lambda~_{y}$ $2\pi~\times~0.151$ –

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