Example 7.1. In this example the control law in Eq. (7.12) is used to perform a reorientation maneuver with large initial errors. The inertia matrix of the spacecraft is given by 10000 0 0 J = 0 9000 0 kg-m 0 0 12000 The initial quaternion is given by q(to) = [0.6853 0.6953 0.1531 0.1531] and the initial angular velocity is given by (to) = [0.5300 0.5300 0.0530] deg/s. The desired quaternion is the identity quaternion. The control gains are set to kp = 50 and ka = 500. A plot of the quaternion errors is shown in Fig. 7.1a. The fourth error-quaternion component approaches 1 while the other three components approach zero. The control torques are shown in Fig. 7.1b. Note the large control torques at the beginning of the maneuver, which are due to the large initial errors. From these plots it is clear that the control law provides a reorientation maneuver to the desired attitude. 2. Compute the control regulation case example 7.1 from the book. The initial quaternion is q = [0.6853,0.6953,0.1531,0.1531] and the initial angular velocity is = [0.53000.53000.0530] = 9000 kg. m, Izz = = degrees per second. For the inertia tensor has the following principal inertia values, Ixx 10000 kg. m, Iyy 12000 kg.m. The control quaternion is the unit quaterion. For the control gains the values kp = 50 and kd = 500 may be used. Outline the single steps to get to a solution and make a plot of the quaternion errors. What are your observations? Also: What is the meaning of the control gain values?