Assignment 1

Problem 1 - Attitude Control of Satellite

Cube with length $L=0.1$ m. The inertia tensor is given by

$${\mathbf{I}}_g^{CG}={\mathbf{I}}_g^b=m\left[\begin{array}{ccc}{R}_{44}^2& 0& 0\\ 0& R_{55}^2& 0\\ 0& 0& R_{66}^2\end{array}\right]$$

Where $m=10$ kg, and $R_{44}=R_{55}=R_{66}=L/\sqrt{6}$ are the radii of gyration with respect to the CG. The attitude dynamics are given by

$$\begin{array}{r}\dot{q}=T_q(q)\omega \\ I_g\dot{\omega }-S(I_g\omega )\omega =\tau \end{array}$$

Problem 1.1

Skew-Symmetric Matrix

The term $v^x$ is the skew-symmetric form of $v$

$$V=\left[\begin{array}{ccc}0& -c& b\\ c& 0& -a\\ -b& a& 0\end{array}\right]$$

^36cc79 Where

$$v=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$$

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We want to find the equilibrium $\dot{x}=0$, where $q=[\eta ,{ϵ}_1,{ϵ}_2,{ϵ}_3{]}^T=[1,0,0,0{]}^T$, and $\tau =0$

The first equation give us

$$\begin{array}{rl}\dot{q}& =T_q(q)\omega \underset{\text{eq. (2.77)}}{\underbrace{=}}\frac{1}{2}\left[\begin{array}{c}-{ϵ}^T\\ \eta I_3+S(ϵ)\end{array}\right]\omega \end{array}$$

Putting in the values of $q$ we get:

$$\begin{array}{rl}\left[\begin{array}{c}\dot{\eta }\\ \dot{{ϵ}_1}\\ \dot{{ϵ}_2}\\ \dot{{ϵ}_3}\end{array}\right]& =\frac{1}{2}\left[\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{\omega }_1\\ {\omega }_2\\ {\omega }_3\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}0\\ {\omega }_1\\ {\omega }_2\\ {\omega }_3\end{array}\right]\end{array}$$

We can therefore see that $\overrightarrow{\dot{ϵ}}=\vec{\omega }=0$ from $\dot{x}=0$

Putting this into the second equation, we get

$$\begin{array}{r}{I}_g\dot{\omega }-S(I_g\cdot 0)\cdot 0=0\\ \dot{w}=0\end{array}$$

We therefore get that the equilibrium point $x_0$ is

$$\dot{x}=\left[\begin{array}{c}\dot{ϵ}\\ \dot{\omega }\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]=\vec{0}$$

This is when $x_0=0$

We now want to linearize the spacecraft model about $x=x_0$

$$\begin{array}{r}\dot{x}=\left[\begin{array}{c}\dot{ϵ}\\ \dot{\omega }\end{array}\right]=Ax+B\tau \end{array}$$

1.4

$$\begin{array}{rl}\tilde{\mathbf{q}}& =\left[\begin{array}{c}\tilde{\eta }\\ \widetilde{ϵ}\end{array}\right]={\bar{\mathbf{q}}}_d\otimes \mathbf{q}\\ {\mathbf{q}}_1\otimes {\mathbf{q}}_2& =\left[\begin{array}{c}{\eta }_1{\eta }_2-{ϵ}_1^{⊺}{ϵ}_2\\ {\eta }_1{ϵ}_2+{\eta }_2{ϵ}_1+\mathbf{S}({ϵ}_1){ϵ}_2\end{array}\right]\\ & \end{array}$$ $$\begin{array}{rl}\mathbf{S}({ϵ}_1){ϵ}_2& =\left[\begin{array}{ccc}0& -{ϵ}_{13}& {ϵ}_{12}\\ {ϵ}_{13}& 0& -{ϵ}_{11}\\ -{ϵ}_{12}& {ϵ}_{11}& 0\end{array}\right]\left[\begin{array}{c}{ϵ}_{21}\\ {ϵ}_{22}\\ {ϵ}_{23}\end{array}\right]\\ \mathbf{S}({ϵ}_1){ϵ}_2& =\left[\begin{array}{c}-{ϵ}_{13}{ϵ}_{22}+{ϵ}_{12}{ϵ}_{23}\\ {ϵ}_{13}{ϵ}_{21}-{ϵ}_{11}{ϵ}_{23}\\ -{ϵ}_{12}{ϵ}_{21}+{ϵ}_{11}{ϵ}_{22}\end{array}\right]\end{array}$$

From this definition we get:

$$\begin{array}{rl}\tilde{\mathbf{q}}={\bar{\mathbf{q}}}_d\otimes \mathbf{q}& =\left[\begin{array}{c}{\eta }_1{\eta }_2-{ϵ}_{11}{ϵ}_{21}-{ϵ}_{12}{ϵ}_{22}-{ϵ}_{13}{ϵ}_{23}\\ {\eta }_1{ϵ}_{21}+{\eta }_2{ϵ}_{11}-{ϵ}_{13}{ϵ}_{22}+{ϵ}_{12}{ϵ}_{23}\\ {\eta }_1{ϵ}_{22}+{\eta }_2{ϵ}_{12}+{ϵ}_{13}{ϵ}_{21}-{ϵ}_{11}{ϵ}_{23}\\ {\eta }_1{ϵ}_{23}+{\eta }_2{ϵ}_{13}-{ϵ}_{12}{ϵ}_{21}+{ϵ}_{11}{ϵ}_{22}\end{array}\right]{|}_{q_1=q_d,q_2=q}\\ \tilde{\mathbf{q}}& =\left[\begin{array}{c}{\eta }_d^2+{ϵ}_{d_1}{ϵ}_1+{ϵ}_{d_2}{ϵ}_2+{ϵ}_{d_3}{ϵ}_3\\ {\eta }_d{ϵ}_1-\eta {ϵ}_{d_1}+{ϵ}_{d_3}{ϵ}_2-{ϵ}_{d_2}{ϵ}_3\\ {\eta }_d{ϵ}_2-\eta {ϵ}_{d_2}-{ϵ}_{d_3}{ϵ}_1+{ϵ}_{d_1}{ϵ}_3\\ {\eta }_d{ϵ}_3-\eta {ϵ}_{d_3}+{ϵ}_{d_2}{ϵ}_1-{ϵ}_{d_1}{ϵ}_2\end{array}\right]\end{array}$$ $$\begin{array}{r}{\mathbf{q}}_d(t)=\left[\begin{array}{c}\varphi (t)\\ \theta (t)\\ \psi (t)\end{array}\right]=\left[\begin{array}{c}0\\ 15\cos (0.1t)\\ 10\sin (0.05t)\end{array}\right]\end{array}$$

1.7

Using the following Lyapunov function

$$\begin{array}{r}V=\frac{1}{2}{\tilde{\omega }}^T{I}_g\tilde{\omega }+2k_p(1-\tilde{\eta })\end{array}$$

Where $w_d=0$, ${ϵ}_d=\text{constant}$, ${\eta }_d=\text{constant}$, and

$$\dot{\tilde{\eta }}=-\frac{1}{2}{\widetilde{ϵ}}^T\tilde{\omega }$$

a)

The time derivative of $V$ is given by

$$\begin{array}{rl}\dot{V}& =\frac{d}{dt}\left(\frac{1}{2}{\tilde{\omega }}^T{I}_g\tilde{\omega }+2k_p(1-\tilde{\eta })\right)\\ \dot{V}& =\tilde{\omega }{I}_g\dot{\tilde{\omega }}-2k_p\dot{\tilde{\eta }}\end{array}$$

Substituting $\tilde{\omega }=\omega$ since $w_d=0$, and $\dot{\tilde{\eta }}=-\frac{1}{2}{\widetilde{ϵ}}^T\tilde{\omega }$, we get

$$\begin{array}{rl}\dot{V}& =\tilde{\omega }{I}_g\dot{\tilde{\omega }}+k_p{\widetilde{ϵ}}^T\tilde{\omega }\end{array}$$

From equation 2 and 4 in the assignment, we can substitute $I_g\dot{\omega }$ and $\tau$ respectively

$$\begin{array}{rl}\dot{V}& ={\omega }^T(S(I_g\omega )\omega -k_d\omega -k_p\omega )+{\widetilde{ϵ}}^T{k}_p\omega \\ \dot{V}& ={\omega }^{T}S(I_g\omega )\omega -{\omega }^T{k}_d\omega -{\omega }^T{k}_p\widetilde{ϵ}+{\widetilde{ϵ}}^T{k}_p\omega \\ \dot{V}& =0-{\omega }^T{k}_d\omega -{\widetilde{ϵ}}^T{k}_p\omega +{\widetilde{ϵ}}^T{k}_p\omega \\ \dot{V}& ={\omega }^T{k}_d\omega ■\end{array}$$

b)