A space telescope is to be launched to carry out...

A space telescope is to be launched to carry out astronomical experiments [9]. The pointing control system is desired to achieve 0.01 minute of arc and track solar objects with apparent motion up to 0.21 arc minute per second. The system is illustrated in Figure P5.5(a). The control system is shown in Figure P5.5(b). Assume that $\tau_1=1$ second and $\tau_2=0$ (an approximation). (a) Determine the gain $K=K_1 K_2$ required so that the response to a step command is as rapid as reasonable with an overshoot of less than $5 \%$. (b) Determine the steadystate error of the system for a step and a ramp input. (c) Determine the value of $K_1 K_2$ for an ITAE optimal system for (1) a step input and (2) a ramp input.

Key Concepts

Feedback Control Systems

This concept refers to systems where the output is measured and fed back to be compared with the input, ensuring that the system adheres to desired performance criteria. It is central to understanding how the system’s dynamics can be manipulated via controller elements to achieve stability and accuracy in tracking commands or disturbances.

Transient Response Specifications

Transient response characteristics, including parameters such as overshoot, settling time, and rise time, are crucial for designing a control system that responds rapidly and accurately. In the context of the problem, managing an overshoot of less than 5% while also achieving a rapid step response ties directly to tuning these dynamics.

Steady-State Error Analysis

This involves evaluating how accurately a control system tracks input signals in the long term. Common input types such as step and ramp require different error performance metrics, and analyzing steady-state error is essential to ensure that the system meets precision requirements over time.

System Gain Tuning

Determining the appropriate gain, which is often a product of several controller parameters, is critical for meeting both transient and steady-state specifications. Gain tuning directly affects the stability and responsiveness of the control system by adjusting how aggressively the system corrects errors.

ITAE Optimization

The Integral of Time-weighted Absolute Error (ITAE) is a performance criterion used to design and compare control systems. ITAE optimization typically leads to controllers that minimize the time-weighted error, thereby improving the overall performance in terms of both speed and overshoot, making it particularly useful when designing for both step and ramp inputs.

Transcript

00:01 Hollow plant here it is given a space probe is to be placed in a circular orbit of radius 5600 miles about the venus relics in the specified place as the probe reaches the point of the original trajectory close to the venus it is inserted in the first adaptive transfer orbit by reducing its speed by gila.

00:34 This orbit rings it to a point b with a much reduced velocity the scope is inserted in the second transfer orbit located in the specified plane by changing the direction of its velocity and further reducing its speed by data b this is the speed reduced in the second transfer and finally as the foe approaches to the point b it is inserted in the design and circular orbit while reducing its speed by databases.

01:19 Its mass of the venus is 0 .82 times that of the earth.

01:23 Mark the r -a -r -b are given 9 .3 into 10 to the power 3 am i and rv is 119 to 10 to the power 3 mi we have to calculate reduce velocity as a, b and c given from approaching a along the parabolic trajectory, let us start solving it for earth.

02:10 Radial is 3690 miles, that is 23 .90, double 8, 10 to g power 6.

02:36 G is 32 .g is 32 .2.

02:40 5 per second square the mass of the earth are for venus planet the mass of the planet it is 0 .82 times of the mass of the earth it is 208 2 times so g into m for venus planet becomes 11 .5 543 into 10 to the power 15 6th of 2 per second for parabolic projectory, the radius is given 9 .3 into 10 to the power 3 by that is 49 .104, 10 to the power 6 square.

05:00 So the velocity of approach at 8, you can find that between 2 gm upon ra 2 into 11 .543 into 10 to the power 50.

05:33 Upon radius at 8 that is 49 5104 into 10 to the power 6.

05:47 This better than 3 2 1 .6 .8 .3 into 10 to the power 3 6 per 6.

06:01 1 .14 orbit 1 upon r.

06:23 6 plus gm upon l square abe plus p.

06:34 Of 1x that is gm upon minus a and point b where theta is 0 1 upon r b is equal to g n upon s square a v plus c cause 0 so it can be written as gm as gm s square a v plus it adding equation 1 and 2 we will get 1 upon r a plus 1 upon r b plus equal to twice of gm upon s square ab so from this equation you can find value of h so from here value of s ab you will get twice of gm r a into r b over r b plus r a plus r a now substitute the value.

08:36 The value of angular momentum 12 minute mass in the orbit first class for orbit.

08:44 This will be 2 into g and g into m having the value 11 .453, 10 to the power 50, 49 .10144 into 10204 which r -b plus r -a.

09:27 This is plus -time.

09:43 So on solving it, this value will be.

10:16 So from here, we can find the velocity at a as a session transfer.

10:22 So this can be written as b -a -2 into r -a -v upon r -a.

10:48 R -a -a -a - is 149 .104.

11:14 And velocity at b -1, v -a -2 upon.

11:40 Or directly memorize hav upon rv hab angular momentum per minute mass there is hot -sync throughout the motion it is conserved radius at v is 1 -0 -03 .2 into 10 to the power 6 so on solving it is between second transfer orbit bc at point 3 where theta is 0 1 upon r3 r v is equal to g m s square bc plus c c c c c so it can be written as x point c here beta 6 1.

13:35 R c is 1 upon r c is equal to gm h bc square plus c c c c c c c c adding equation 5 and 6 1 upon r b plus 1 upon r c is called to 2m .m upon s square bc.

14:37 So from here h bc angular momentum per unit marks will be 2 gm r b rc upon r b r c upon r b plus r c substituting the value g into m for venus is 11 .543 into 10 to the power of 15 radius r b 1003 .2 into 10 to 3 power 6 fifth rc 29 .568 into 10 to the power 6 fifth divided by 1003 .2 into 10 to power 6 plus 29 .568 10 to the power 6...

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