Question 13.7: Consider an axisymmetric prolate satellite with a reaction-j...

First, we should note that because the satellite is axisymmetric \left(I_{1}=I_{2}\right) and no torque is produced about the 3 axis \left(M_{3}=0\right), Euler’s moment equations are reduced to Eqs. (13.75) and (13.76).

\begin{array}{c}{{M_{1}=I_{1} \dot{\omega}_{1}+a\omega_{2}}} & (13.75)\\ {{M_{2}=I_{1}\dot{\omega}_{2}-a\omega_{1}}} & (13.76)\end{array}

Next, let us substitute the nonlinear nutation control laws (13.83) and (13.84) for control torques M_{1} and M_{2} in Eqs. (13.75) and (13.76). The result is

\begin{aligned} & I_{1} \dot{\omega}_{1}=-a \omega_{2}-F z \operatorname{sgn}\left(\omega_{1}\right) & (13.85) \\ & I_{1} \dot{\omega}_{2}=a \omega_{1}-2 F z \operatorname{sgn}\left(\omega_{2}\right) & (13.86) \end{aligned}

where a=\left(I_{3}-I_{2}\right) n and n=\omega_{3}. Because M_{3}=0 and I_{1}=I_{2}, the spin component along the 3 axis is constant, that is,

n=\omega_{3}=\omega \cos \gamma=5.9416  \mathrm{rad} / \mathrm{s}(\text { constant })

Using the 3-axis spin component n and moments of inertia I_{3} and I_{2}, we determine the constant a=-9,013.42 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}. Figure 13.44 shows the closed-loop nutation control system for the axisymmetric satellite. Although it requires carefully tracing the signal paths, the reader should be able to identify the governing equations (13.85) and (13.86) in Figure 13.44. Note also that a relay with dead zone is used in the feedback path instead of a pure “sign” or signum function (again, the dead zone will reduce the control “chatter” when spin components \omega_{1} and \omega_{2} become acceptably small). In an operational setting, the jet selection is simply determined by checking the signs of the feedback from rate gyroscopes mounted along the 1 and 2 axes. For example, if \omega_{1}>0 and \omega_{2}<0, fire the reaction jet \mathrm{b} (for M_{1}=-F z ) and jets a, \mathrm{d} (for M_{2}=+2 F z ). If either spin component is smaller than the threshold of the dead zone, the jets are not fired.

Simulink was used to create and simulate the closed-loop system shown in Figure 13.44. The initial value of the component of the angular velocity vector \boldsymbol{\omega} projected onto the 1-2 plane is

\omega_{12}=\omega \sin \gamma=0.8350  \mathrm{rad} / \mathrm{s}

Suppose the \boldsymbol{\omega}_{12} vector is initially 30^{\circ} (counterclockwise) from the 1 axis (recall for torque-free motion, the \omega_{12} vector will rotate clockwise for a prolate satellite; see Figure 12.9). The initial spin components are \omega_{1}(0)=\omega_{12} \cos \left(30^{\circ}\right)=0.7232  \mathrm{rad} / \mathrm{s}, and \omega_{2}(0)=\omega_{12} \sin \left(30^{\circ}\right)=0.4175  \mathrm{rad} / \mathrm{s}. Each dead zone threshold is set to 0.001  \mathrm{rad} / \mathrm{s} (\approx 0.06  \mathrm{deg} / \mathrm{s}). Figure 13.45 shows the time histories of the spin components \omega_{1}(t) and \omega_{2}(t), respectively. The results of the nutation control effort are apparent in these figures as both off-axis spin components \omega_{1} and \omega_{2} are driven to zero in about 15 \mathrm{~s}.

After 15 \mathrm{~s}, the satellite’s angular velocity vector \boldsymbol{\omega} is essentially aligned with the 3 axis [recall that \omega_{3}=5.9416  \mathrm{rad} / \mathrm{s} (constant) at all times because M_{3}=0 and I_{1}=I_{2}; therefore, the magnitude of the final angular velocity is equal to \left.\omega_{3}\right]. Figure 13.46 shows the 1 – and 2-axis control torques, M_{1} and M_{2}, respectively. Each control switching corresponds to a sign change of the appropriate spin component. Note that the magnitude of torque M_{1}= F z=125 \mathrm{~N}-m (one jet), whereas the 2-axis torque magnitude is M_{2}=2 F z=250 \mathrm{~N}-m (two jets). Both control torques go to zero for t>15 \mathrm{~s} because each spin component is within the dead zone and the nutation has been removed.

Finally, Figure 13.47 shows the time histories of the nutation angle \theta and “tilt” angle \gamma. The initial nutation angle is

\theta=\tan ^{-1}\left(\frac{I_{1} \omega_{12}}{I_{3} \omega_{3}}\right)=12.23^{\circ}

Figure 13.47 shows that angles \theta and \gamma steadily decrease as the nutation control torques remove the off-axis spin components \omega_{1} and \omega_{2}. At t=15 \mathrm{~s}, both angles have been driven to zero, and hence the 3 axis is aligned with the angular velocity vector \boldsymbol{\omega} and angular momentum vector \mathbf{H}.

As a final note, we can determine the change in angular momentum for the nutation control. At t=0, the initial angular momentum is

\mathbf{H}_{0}=\mathbf{I} \omega_{0}=\left[\begin{array}{ccc} 4,317 & 0 & 0 \\ 0 & 4,317 & 0 \\ 0 & 0 & 2,800 \end{array}\right]\left[\begin{array}{l} 0.7232 \\ 0.4175 \\ 5.9416 \end{array}\right]=\left[\begin{array}{c} 3,122 \\ 1,802 \\ 16,637 \end{array}\right] \mathrm{kg}-\mathrm{m}^{2} / \mathrm{s}

The magnitude is H_{0}=17,023 \mathrm{~kg}-\mathrm{m}^{2} / \mathrm{s}. For t>15 \mathrm{~s}, the satellite is essentially in a pure spin about its 3 axis, and therefore the final angular momentum is H_{f}=I_{3} \omega_{3}=16,637 \mathrm{kg}-\mathrm{m}^{2} / \mathrm{s}. The external torque from firing reaction jets has reduced the magnitude of the angular momentum by about 2 \%.