Question 13.6: Consider again the satellite and its initial conditions pres...

The initial state in the phase plane is attitude error \phi_{e}(0)=\phi_{0}-\phi_{\text {ref }}=-45^{\circ}(=-0.7854 \mathrm{rad}) and \dot{\phi}_{e}(0)=-10  \mathrm{deg} / \mathrm{s}(=-0.1745  \mathrm{rad} / \mathrm{s}). Because the initial state is “below” the switching curve (Figure 13.40), we select the positive torque jets [or, we can show that the initial switching signal determined by Eq. (13.64)

M_c=\left\{\begin{array}{l}+2 F r \text { if } \quad \sigma=-\alpha \phi_e-0.5|\dot{\phi}| \dot{\phi}>0 \\-2 F r \text { if } \quad \sigma=-\alpha \phi_e-0.5|\dot{\phi}| \dot{\phi}<0\end{array}\right.       (13.64)

is positive]. The phase plane profile is determined by Eq. (13.57):

\frac{1}{2} \dot{\phi}_{e}^{2}-u \phi_{e}+C=0       (13.66)

where u=+2 F r / I=0.25  \mathrm{rad} / \mathrm{s}^{2} is the positive torque acceleration. Substituting the initial error state into Eq. (13.66), we find that C=-0.2116(\mathrm{rad} / \mathrm{s})^{2}. The switching point in the phase plane is the intersection of the positive-torque parabolic curve [Eq. (13.66] and the negative-torque switching curve (with C=0; see Figure 13.32). The intersection of these two parabolic curves is

\underbrace{\frac{1}{2} \dot{\phi}_{e}^{2}-0.25 \phi_{e}+C}_{\text {Positive jet-torque }}=\underbrace{\frac{1}{2} \dot{\phi}_{e}^{2}+0.25 \phi_{e}}_{\begin{array}{l} \text { Negative-torque } \\ \text { switching curve } \end{array}}

Using C=-0.2116(\mathrm{rad} / \mathrm{s})^{2}, we find that intersection occurs at the attitude error \phi_{e}= -0.4232  \mathrm{rad}\left(=-24.25^{\circ}\right). The attitude error rate at the switching curve can be determined by setting the switching signal, Eq. (13.63), to zero:

\sigma=-\alpha \phi_{e}-\frac{1}{2}\left|\dot{\phi}_{e}\right| \dot{\phi}_{e}=0      (13.67)

Using \phi_{e}=-0.4232  \mathrm{rad} and \alpha=0.25  \mathrm{rad} / \mathrm{s}^{2} in Eq. (13.67), we find that the attitude rate at the switching curve is \dot{\phi}_{e}=0.4560  \mathrm{rad} / \mathrm{s}. In summary, the switching point is

\phi_{e}\left(t_{1}\right)=-0.4232  \mathrm{rad}, \dot{\phi}_{e}\left(t_{1}\right)=0.4560  \mathrm{rad} / \mathrm{s}

Or,

\phi_{e}\left(t_{1}\right)=-24.25^{\circ}, \dot{\phi}_{e}\left(t_{1}\right)=26.35  \mathrm{deg} / \mathrm{s}

Time t_{1} is the switching time. This switching point matches the numerical simulation results presented in Example 13.5 (see Figure 13.40).

The second part of this problem involves estimating the total maneuver time. We neglect the effect of the relay’s dead-zone and assume “perfect switching” when \sigma=0. Because the torque acceleration is constant, it is easy to integrate the angular acceleration of the attitude error by combining Eqs. (13.51), (13.47), and (13.48):

Attitude error acceleration: \ddot{ϕ}_e =\ddot{ϕ}−\ddot{ϕ}_{\text{ref}}^{\nearrow 0} =\ddot{ϕ}       (13.51)

M_{c}=I{\ddot{\phi}}         (13.47)

{M_{c}=\pm\,2F r}      (13.48)

\ddot{\phi}_{e}=\frac{ \pm 2 F r}{I}= \pm 0.25  \mathrm{rad} / \mathrm{s}^{2}

The integral is

\dot{\phi}_{e}(t)=\dot{\phi}_{e}(0) \pm 0.25 t  \mathrm{rad} / \mathrm{s}     (13.68)

For the initial positive-torque segment, we use positive torque acceleration, +0.25  \mathrm{rad} / \mathrm{s}^{2}. Using the initial attitude rate \left[\dot{\phi}_{e}(0)=-0.1745  \mathrm{rad} / \mathrm{s}\right] and the attitude rate on the switching curve \left[\dot{\phi}_{e}\left(t_{1}\right)=0.4560  \mathrm{rad} / \mathrm{s}\right], we determine the switching time:

t_{1}=\frac{\dot{\phi}_{e}\left(t_{1}\right)-\dot{\phi}_{e}(0)}{0.25}=2.538 \mathrm{~s}

This switching time corresponds to the numerical simulation results from Example 13.5, and can be seen in Figures 13.38, 13.39, and 13.41. The remaining maneuver time to the origin (time on the negative-torque switching curve) can be computed using a modified version of Eq. (13.68):

\Delta t=\frac{\dot{\phi}_{e}\left(t_{f}\right)-\dot{\phi}_{e}\left(t_{1}\right)}{-0.25}=1.840 \mathrm{~s}

where \dot{\phi}_{e}\left(t_{f}\right)=0 is the final attitude rate (the origin of the phase plane). The total maneuver time is t_{f}=t_{1}+\Delta t=4.38 \mathrm{~s}. This calculation shows a good match with the simulation results of Example 13.5 shown in Figures 13.38 and 13.39.