[SOLVED] 代写 R parallel react theory 1. Objective:

1. Objective:
UNIVERSITY COLLEGE DUBLIN
SCHOOL OF ELECTRICAL, ELECTRONIC & COMMUNICATIONS ENG.
EEEN40010 CONTROL THEORY
Experiment 4CT1
LINEAR STATE FEEDBACK CONTROL
To investigate the design of linear state feedback controllers with observer.
2. Background Information:
See earlier handouts.
The command place will be useful. Otherwise consult lecture notes. 3. Problem
In aerospace a known problem, which has caused the failure of several missions, is that of sloshing of liquid fuel. We generally carry liquid fuel because it offers a more flexible form of propellant than solid fuel since it can be throttled. Solid fuel on the other hand is more inclined to provide a fixed, uncontrollable thrust. A commonly-used model for slosh is a simple pendulum. This model is only approximately valid if the liquid surface remains flat and the slosh angle is fairly small.
Figure 1: Vega rocket upper stage

We consider the upper stage where gravity and drag can be neglected. As inertial frame we choose E1 pointing to the left, E2 pointing outward from page and E3 pointing upwards. The main rocket
is modelled as a rigid body which is assumed to rotate in the 1-3 plane only. Accordingly, in the notation of the minor project, the yaw   0 and the roll   0 . The pitch is the angle  shown in Fig. (1). Again from minor project the angular velocity vector of the main rocket is determined to
be Eb 222
in terms of the body-fixed frame, although in this case Eb  E
2-axis only. The fuel is modelled as the pendulum mass. The interaction between the fuel and the
since rotation is about the
main rocket is modelled as the main rocket applying reaction force R
applying equal but opposite force on the main rocket. It is assumed that the reaction force vector is parallel to the line going from the pivot point to the centre of the pendulum mass, i.e. that this reaction force does not apply any moment to the fuel mass.
Show that the total force on the main rocket is:
fEb  FEb  RsinEb  RcosEb 1313
on the fuel and the fuel
com expressed relative to body-fixed frame. This force will equal m d 2 r
where m is the mass of the is the position vector of the centre of mass of the main rocket relative to the
dt2
inertial frame. We assume that the centre of mass of the main rocket and the centre of mass of the
main rocket and r com
fuel both remain in the 1-3 plane, i.e. that:
r r1 Eb0Ebr3 Eb
andr r1 Eb0Ebr3 Eb com com 1 2 com 3 fuel fuel 1 2 fuel 3
.
As stated let the yaw and roll be zero and assume that J, the mass moment of inertia matrix relative to the body-fixed frame, is diagonal. With reference to the minor project show that:
deducingthat: J fdbRsin. 22
Show that the position vector for the centre of the pendulum mass is:
r r bEb asinEb acosEb fuel com 3 1 3
Note that from minor project: d Eb Eb Eb Eb
dt
if i = 3. Hence show that:
0 0 Rsin d Jbb Jb Mb fd0 0 
dt        0  b Rcos
     
. Show that this equals Eb ii2i3
if i = 1,
0
if i = 2 and  E1b

d r  d r bEb acosEb asinEb asinEb acosEb dtfuel dtcom 1 1 3 3 1
d
 r bEb   acosEb   asinEb  
dt com 1 1 3
d2 d2
r  r bEb b2Eb   acosEb   asinEb  
dt2 fuel dt2 com 1 3 1 1
   acosEb    asinEb    acosEb    asinEb 
3331
d2
dt2com 1 3 1 3
and this will equal m
r bEb b2Eb   acosEb   asinEb  

   asinEb    acosEb .  
The force on the fuel mass is: RsinEb  RcosEb 1 3
Hence or otherwise show that the equations of motion are:
r1 fRsin , r3 FRcos com m com m
   fd  bR sin   J22
22
13
d 2rfuel . fuel dt2
f211 mb  acos  asinRsinmm 
  fuel 
F211
b2 asin acosRcos 
m mm
the system where the offset components of the thrust, F
the inputs. Show that the linearised system decouples and determine the state-space
.  fuel 
By employing the third equation eliminate R in the fourth equation. By adding sin   multiplied by the fifth equation to cos multiplied by the fourth equation eliminate R in the fifth equation.
Note that as a result the system decouples.
1. Select operating values 0  0 and 0  0 . Assume that the operating value of thrust, F0 , is non-zero. Find the operating values for all other quantities. Derive a state-space model for
and lateral thrust, f
, are treated as
equations for the subsystem with state variables:  ,  ,  , 
and single input f
.
Following Thompson and O’Connor (2015, “Wave-Based Attitude Control of Spacecraft with Fuel Sloshing Dynamics”, ECCOMAS Thematic Conference on Multibody Dynamics, Barcelona, Catalonia, Spain, pp. 263-275) take:

Parameter
Value
Unit
m
2105
kg
mf
88
kg
J22
1883
kg m2
a
0.53
m
b
-1.43
m
F0
2450
N
The negative value for b is not an error, but reflects the location of the fuel tank. Thompson and O’Connor do not offer a value for parameter d. Take it to be a value which seems sensible to you.
2. Suppose the output is taken to equal the pitch angle . It is desired to drive this angle to 5o. Design a linear state feedback controller with preamplifier and observer so that the resulting pitch response has zero steady-state error, a 2% settling time of no more than 10 sec. and an overshoot of no more than 10%
3. A signal of concern is the slosh angle  Determine the maximum value of the slosh angle as predicted by the linearised, local model for the resulting closed-loop system. Plot the slosh response for this system.
4. Assume that the thrust F is held constant at the operating value. Implement the designed controller on the system described by the global model. Plot the pitch response and the slosh response for the system and make a judgement as to whether the controller should be deemed acceptable.