2023秋哈工大惯性技术大作业.docx

Assignment ofInertial Technology惯性技术作业（2023 秋）My Chinese Name张My Student No.16S0040Autumn 2023(5) Frequency =50HZFigure 1-23 response of inner ring angle with 50Hz sinusoidal input x 10-5X Y Plot2 1.510.5 s w 0->-0 5-1-1.5-2-1012XAxisx 10-7Figure 1-24 Trajectory of 2-DOF gyro* s response to 50 Hz sinusoidal input analysis of the result:As shown in the figures above, we could learn that the higher the frequency is, the less the vibration of the outer axis is, and the smaller the trajectory of the spinning axis is.1.3. Conclusion of the taskThe previous discussion of the response of a 2-DOF gyro to the three types of torques shows that they all affect the precision of a gyro, but to different degrees.the constant torque has the largest influence, because it makes the gyro process at a constant rate, with its drifting angle increasing with time. In contrast, the impulse torque has the least effect, since it only causes nutation whose magnitude is quite small and, in practice, quickly decays due to damping. In between lies the effect of sinusoidal torque which causes an error (hat swings within a bounded range. However, the lower is the frequency of the sinusoidal torque, the more is it like a constant one, and the larger the range of its swing become.Assignment 2: Single-axis INS simulation2.1. Description of the taskIn a fictitious test of a magnetic levitation train along a track running north-south, it first accelerates and then cruises at a constant speed. Onboard is a single-axis platform INS, working in the way described by the courseware of Unit 5: Basic problems of INS. The motion information and Earth parameters are shown in table 2-1, and the possible error sources are listed in Table 2-2.You are asked to simulate the operation of the INS within 9,900 seconds, and investigate, first one by one and then altogether, the impact of these error sources on (he performance of the INS.Note that the block diagram in the lecture notes (figure 3.6 of both 2023 and 2023 versions) or the old versions of courseware has to be slightly modified before you can obtain reasonable results.Table 2-1 Motion iiifbnnation and Earth paiametersMotion iiifbnnationvaluesunitsEaith parametersvaluesunitsInitial velocity, northward5nVsAcceleration of gravity9.8ni/s2Initial position0mRadius of the Earth6371kmAcceleration, fioni start21 2 nvsDiu ation of acceleration80sTable 2-2 Possible enor soiucesTypesvaluesunitsTypesvaluesunitsInitial position enor20inAccelerometer scale factor error0.00051Initial velocity enor0.05nvsGyroscope scale factor error0.00051Initial plattbrni niisaligninent enor1ItGyioscope ch ifting enor0.01°/liAccelerometer bias enor0.00002iws22.2. Solution of the taskFigure 2-1 Block diagram of the Single-axis INS, The parameters:Ay。= 20m is the Initial position errorVo = 0.05zn/s is the Initial velocity error a0 = 1 is the Initial platform misalignment error Aw = 0.0002m/s2 is the bias error of the accelerometer K。= 0.0005 is the scale factor error of the accelerometer Kg = 0.0005 is the scale factor error of the gyroscope drift = 0.01°/i is the Gyroscope drifting error2.2.1 no error impactFigure 2-2 S i mu I ink model of north-southward single ax i s INS without errorFigure 2-4 reaI velocity without errorFigure 2-5 reaI dispIacement output without errorFigure 2-6 reaI position error without error2.2.2 impact of initial position error Ay0 = 20mFigure 2-7 Si mu I ink model of north-southward single axis INS v/ith Initial position error20mFigure 2-8 reaI acceleration with Initial position error 20 mFigure 2-9 reaI veIocity with Initial position error 20 m analysis of the result:Compared with the no error situation, (he acceleration and velocity of (he train is the same, but the displacement output is 20m higher than that with no error situation, so that the position error shows 20m.2.2.3 impact of initial velocity error AV0 = 0.05m/sFigure 2-14 reaI veIocity with Initial velocity error 0.05 m/sFigure 2-15 reaI dispIacement output v/ith Initial velocity error 0. 05 m/sFigure 2-16 position error with Initial velocity error 0.05 m/s analysis of the result:Compared with the no error situation, the acceleration of the train is (he same, but ihe velocity is 0.05m/s faster than that with no error situation, so that the displacement output and position error changes as the figures shows.2.2.4 impact of initial platform misalignment error a0 = 1F i gure 2-18 reaI acceI erat i on with Initial platform mi saIi gnment error 1"Figure 2-20 reaI d i spIacement output with Initial platform mi salignment error 1"Figure 2-21 reaI position error with Initial pIatform misaIignment error 1”2.2.5 impact of accelerometer bias error AN = 0.0002m/s2Figure 2-22 si mu I ink model of north-southward single axis INS with accelerometer bias error 0.0002m/s2Figure 2-23 reaI acceleration with accelerometer bias error 0.0002m/s2Figure 2-24 real velocity with accelerometer bias error 0.00027n/s2Figure 2-25 reaI dispIacement output with accelerometer bias error 0.0002?n/s22.2.6 impact of the scale factor error of accelerometer Ka = 0.0005Figure 2-27 si mu I ink model of north-southward single axis INS v/i th acceIerometer sea Ie factor error 0. 0005Figure 2-28 reaI acceleration with accelerometer scale factor error 0. 0005Figure 2-29 reaI velocity with acceIerometer scale factor error 0. 0005Figure 2-30 reaI displacement output with accelerometer scale factor error 0. 0005The report is to contain:1. Description of the tasks - contents of the next two pages and the previous assignments.2. The code of your programs, and (heir explanation.3. The results of your computation or simulation (as listed by the requirement).4. Your analysis of the result, and your reflection on the programming or simulation5. Originality statements or reference/assistance acknowledgements.English is expected in writing, though Chinese is also accepted.Figure 2-31 position error with accelerometer scale factor error 0.00052.2.7 impact of the scale factor error of gyroscope Kg = 0.0005Figure 2-32 si mu I ink model of north-southward single axis INS withFigure 2-34 reaI velocity with Gyroscope sea Ie factor error 0. 0005Figure 2-36 position error with Gyroscope scale factor error 0. 00052.2.8 impact of the Gyroscope drifting error drift = 0.01°/hFigure 2-37 si mu I ink mode I of north-southward single axis INS withFigure 2-38 reaI acceleration with Gyroscope drifting error 0.01 */hFigure 2-40 dispIacement output with Gyroscope dr ift ing error 0.01 */hFigure 2-41 position error v/ith Gyroscope drifting error 0. 01 */h2.2.9 impact of all the errorsIntegrator2Figure 2-42 BIock d i agram of north-southward single axis INS with error a I together.Figure 2-43 real acceleration with error altogether.Figure 2-44 reaI velocity with error altogether.Figure 2-45 real displacement output with error a I together.Figure 2-46 position error with error altogether.2.3. Conclusion of the taskThrough contrasting all the results, we can conclude that the Initial platform misalignment error and gyroscope drifting error is the main component of the whole position bias, and they do most harm to our navigation. So it is a must for us to weaken or eliminate it anyway. In spite of all the disadvantages discussed above, the INS still shows us a relatively accurate results of single-axis navigation.Assignment 3: SINS simulation3.1. Description of the taskIn an fictitious mission, a spaceship is to be lifted from a launching site located at 19°37' NL and 110°57'EL, into a circular orbit 400 kilometers high along the equator. The spaceship is equipped with a strapdown INS whose three gyros, GX, GY, GZ, and three accclcromctcrs, AX, AY, AZ, arc installed respectively along the axes Xb. Yb, Zb of the body frame.Case 1： Stationary testThe body frame of the spaceship initially coincides with (he geographical frame, as shown in the figure, with its pitching axis Xb pointing to the east, rolling axis Yb to the north, and heading axis Zb upward. Then the body of the missile is made to rotate in 3 steps:(1) 80° around Xb(2) 90° around Yb(3) 170° around ZbAfter that, the body of the spaceship stops rotating. You arc required to compute the final outputs of the three accelerometers in (he spaceship, using quaternion and ignoring the device errors. It is known that the magnitude of gravity acceleration is gO = 9.79m/s2.Case 2: The launching processThe spaceship is installed on the top of an vertically erected rocket. Its initial heading, pitching and rolling angles with respect to the local geographical frame arc -90,90 and 0 degrees respectively. The default rotation sequence is heading - pitching -* rolling. The top of the rocket is initially 100m above the sea level. Then the rocket is fired up. The outputs of the gyros and accclcromctcrs in the spaceship are both pulse numbers. Each gyro pulse is an angular increment of 0.01 arcsec, and each accelerometer pulse is le-7g0, with gO = 9.79m/s2. The gyro output frequency is 100Hz, and the accelerometer's is 5Hz. The outputs of the gyros and accelerometers within 1800s are stored in a MATLAB data file named mission.mat, containing matrices GGM of 180000x3 from gyros and AAM of 9(M)0x3 from accelerometers respectively. The format of the data in the two matrices is as shown in (he tables, with 10 rows of each matrix selected. Each row represents the outputs of the type of sensors at a sampling time.The Earth can be seen as an ideal sphere, with radius 6371.00km and spinning rate 7.292x 10-5 rad/s, The errors of the sensors arc ignored, so is the effect of height on the magnitude of gravity. Besides, the influence of height on the angular rates of the geographical frame and the changing rates of latitude and longitude should also be considered. Velocity, position and the geographical frame can be updated every 0.2s, within which the attitude of the vehicle can be updated multiple times, depending on the chosen algorithm (20 for 1-S, 10 for 2-S. and 5 for 4-S).You are required to:(1) compute the final attitude quaternion, longitude, latitude, height, and east, north, vertical velocities of the spaceship.(2) draw the latitude-versus-longitude trajectory of the spaceship, with horizontal longitude axis.(3) draw the curve of the height of the spaceship, with horizontal time axis.(4) draw the curves of the attitude angles of the spaceship, with horizontal time axis.GxGyGz-106584145-106487147-106686148-106486147-106687146-106585145-106587146-106585145-106784147-106587147AxAyAz-159846814634887-1279082-161193414636446-1253706-162539314637974-1228346-163885614639472-1202978-165231914640940-1177617-167562014635359-1221229-168684614636804-1196141-169807114638220-1171047-170930114639605-1145960-172054414640958-11208823.2. Solution of the task3.2.1 easel Stationary testCode:% case 1%The first rotation quaternion%The second rotation quaternion%The third rotation quaternion%call the quaternion multiplication subfunctiong=0;0;0;-9.79;ql=cos(40/l80*pi) sin(40/180*pi) 0 OJ; q2=cos(45/180*pi) 0 sin(45/180*pi) 0; q3=lcos(85/l8()*pi) 0 0 sin(85/l80*pi)J; r=quatmultiply(q 1 ,q2);q=quatmultiply(r.q3);PL=q(l) q(2) q(3) q(4);- q(2) q(l)q(4) -q(3);- q(3)-q(4) q(l)q(2);- q(4) q(3) -q(2) q(l);PR=q(l)-q(2) -q(3) -q(4);q(2) q(l)q(4)-q(3);q(3) -q(4) q(l)q(2);q(4) q(3)-qq(l)；P=PL*PR;gn=P*g；gn=gn(2:4)3.2.2 case2 The launching processFlow chart:Figure 3-1 simpIif ied navigation algor ithm for SINSCode:% case 2T=0.2;K=1800/T;R=6371000;%radius of earthwE=7.292*10A(-5);%spinning rate of earthQ=zeros(K+1.4);%quaternion matrix initializinglongitude=zeros( 1 ,K/5+1);latitudc=zcros( 1 ,K/5+1);H=zeros(l.K/5+l);%altitude matrixql=cos(-45/180*pi)00 sin(-45/180*pi);%The first rotationquaternionq2=cos(45/180*pi)sin(45/180*pi)00; %Thc second rotation quaternionq3= 1 0 0 0; %The third rotation quaternionQ( I ,:)=quatmultiply(qua(mulliply(q I ,q2),q3);% initial quaternionEUL( I ,:)=q2eul(Q( 1longitude(l)=l 10+57/60:%initial longitudelalitude( 1)=19+37/60;%initial latitudeH( 1)=100;%initial altitudelength=0;g=9.79;vE = zcros(l,K/5+l); %eastern velocityvN = zcros(l,K/5+l);%northem velocityvH = zeros(l,K/5+l);%upward velocityvE(l)=0;vN(l)=0;vH(l)=0;load miss ion. mat %data loading % ma in calculation sectionfor N=1:Kq 1 =zeros( 1,4,1);ql(l,:)=Q(N,:);forn=l:20-l% Aililude iterationwx=0.01 /(3600* 180)*pi*GGM(N- l)*20+n,l );% Angle incrementwy=0.01/(3600*180)*pi*GGM(N- l)*20+n,2);wz=0.01/(3600* 180)*pi*GGM(N-1 )*20+n,3);w=wx,wy,wz'normw=norm(w);% Norm calculationW=0,-w(l ),-w(2),-w(3);w(1)OM3),-w(2);w(2),-w(3),0,w(l);w ,w(2),w(l),0;I=eye(4);S=l/2-normwA2/48;C= 1 -normwA2/8+normwA4/384;q l(n+1 ,:)=q l(n,:)*(C*I+S*W);Q(N+l,:)=ql(n+l,:);endWE=-vN(N)/(R+H(N); % rotational angular velocity component of a geographic coordinate systemWN=vE(N)/(R+H(N)+wE*cos(latitude(N)/18 0*pi);WH=vE(N)/(R+H(N)*tan(latitude(N)/180*pi )+wE*sin(latitude(N)/l 80*pi);atlitude=WE;WN;WH'*T;%correction of the quaternion by updating the rotation of geographic coordinatenormattitudc=nonn(attitudc);e=a(tilude/nonna(tilude;QG=cos(nonnaltitude/2),sin(nonnattitude/2)* e；Q(N+1 ,:)=quatmultiply(qinv(QG), Q(N+1,:);EUL(N+ l,:)=q2cul(Q(N+1,:); fx=le-7*9.79* AAM(N J);%specific force measured by accelerometer fy=le-7*9.79*AAM(N,2);fz=le-7*9.79*AAM(N,3);Fb=fx fy fz;F=quatmultiply(Q(N+1 ,:),quatmultiply(O,Fb, qinv( Q(N+1,:);%The specific force isdecomposed into geographic coordinate system.FE(N)=F(2);FN(N)=F(3);FU(N)=F(4); %calculate the relative acceleration of the vehicle:VED(N)=FE(N)+vE(N)*vN(N)/(R+H(N)* (an(la(itude(N)Zl 80*pi)- (vE(N)/(R+H(N)+2*wE*cos(latitude(N)/180 *pi)*vH(N)+2*vN(N)*wE*sin(latitude(N)/18 0*pi);VND(N)=FN(N)-2* vE(N)* wE*sin(latitude(N)/180*pi)-vE(N)*vE(N)/(R+H(N)*tan(latitudc(N)/180* pi)-vN(N)*vH(N)/(R+H(N);VUD(N)=FU(N)+2*vE(N)*wE*cos(latitud e(N)/180*pi)+(vE(N)A2+vN(N)A2)/(R+H(N)- g*RA2/(R+H(N)A2;%integration and get the relative velocity of vehicle:vE(N+ l)=VED(N)*T+vE(N);vN(N+1 )=VND(N)*T+vN(N);vH(N+ l)=VUD(N)*T+vH(N);% integration and get the position of vehicle:longitude(N+1 )=vE(N)/(R+H(N)*cos(latitud e(N)/180*pi)*T/pi* 180+longitudc(N);length=sqrt(vE(N)A2+(vN(N)A2)+length;enddisplay(Q(K+1'final attitude quaternion') display(latitude(K+1),'final latitude') disp