哈工大慣性技術(shù)(導(dǎo)航原理)大作業(yè)2016

1、Assignment ofInertial Technology慣性技術(shù)作業(yè)（2016 秋）My Chinese NameMy Student No.16S104The 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 their explanation.3. The results of your computation or simulation (as l

2、isted 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.Assignment 1: 2-DOF response simulation1.Description of the

3、 tasksA 2-DOF gyro has a rotor with angular moment 10000 g × cm × s . Its equatorial angular inertias are both 4 g × cm × s 2 .Please investigate the response of the gyro to the following types of torques as listed in the table, and present whatever you can discover or confirm fr

4、om the results.Table 1-1 Types of input torquesTorque descriptionDirectionSuggested simulation timeImpulse, magnitude 40000 g × cm , duration 1e-5sInner ring axis0.003 0.02sConstant, magnitude 1 g × cm .Inner ring axis0.005 0.02sSinusoidal, amplitude 1 g × cm , frequency 2HzOuter ring

5、 axis0.5 2sSinusoidal, amplitude 1 g × cm , frequency 5HzOuter ring axis0.2 0.8sSinusoidal, amplitude 1 g × cm , frequency 10HzOuter ring axis0.1 0.4sSinusoidal, amplitude 1 g × cm , frequency 20HzOuter ring axis0.05 0.4sSinusoidal, amplitude 1 g × cm , frequency 50HzOuter ring a

6、xis0.02 0.2sNote that the default configurations of the simulink parameters, such as those of step size, might not be adequate to bring about the nuances in the simulation result.2.Simulation and AnalysisAccording to the transfer function of the 2-DOF gyro, the outputs can be expressed as:Given this

7、 function, we can establish the block diagram of the system in Simulink. 2.1 Input1: Impulse, magnitude 40000 g.cm , duration 1e-5s, on inner ring axisIn this case:The frequency of the nutation is obtained as:Radius of nutation is obtained as:The block diagram of the system in Simulink is showed in

8、Fig 2.1.Fig 2.1 The block diagram of input1 in SimulinkIn the block diagram, we can add a XY Graph to show the relationship between and . However, the graph of their relationship cannot be edited. So I log the data of Scope and Scope1into workspace of Matlab and name them separately as Alpha and Bet

9、a.Firstly, I set the simulation stop time as 0.003s, then run the system. In workspace, I draw the graph of the two response into one axes and get the following Fig 2.2. Then I also draw the trajectory of 2-DOF gyros response to this impulse input in Fig 2.3.Fig 2.2 2-DOF gyros two responses to impu

10、lse input(0.003s)Fig 2.3 Trajectory of 2-DOF gyros response to impulse input(0.003s)From Fig2.3 we can obviously get that the trajectory of 2-DOF gyros response to impulse input is a circle, whose center is, and radius is .In order to compare the data with different simulation time, I change the sim

11、ulation stop time to 0.02s, and get the trajectory in Fig2.4.Fig 2.4 Trajectory of 2-DOF gyros response to impulse input(0.02s)From Fig2.4 we can see that the trajectory is not smooth, why? For one thing we can confirm is that it is related to the simulation time. Because the only difference between

12、 Fig2.4 and Fig2.3 is the simulation time. At the beginning, we have figured out that Which means that the period isSo 0.003s is about one period and 0.02s is about nine periods. Owing to the simulation step of Simulink is fixed (or maybe I dont know how to change), the result will have bias when si

13、mulation time is long, which may explain the sawtooth around the edge.2.2 Input2: Constant, magnitude 1 g.cm, on inner ring axisIn this case:The frequency of the nutation is obtained as:The drift rate is obtained as:The radius of nutation is obtained as:The block diagram of the system in Simulink is

14、 showed in Fig 2.5. The two responses to the constant input within 0.01s is showed in Fig2.6. The trajectory of 2-DOF gyros response to this constant input is showed in Fig 2.7.Fig 2.5 The block diagram of input2 in SimulinkFig 2.6 2-DOF gyros two responses to constant input(0.01s)Fig 2.7 Trajectory

15、 of 2-DOF gyros response to constant input(0.01s)From Fig2.7 we can obviously get that the trajectory of 2-DOF gyros response to constant input is a cycloid.2.3 Input3: Sinusoidal, amplitude 1 g.cm, frequency 2Hz, on outer ring axisIn this case:The frequency of the nutation is obtained as:The block

16、diagram of the system in Simulink is showed in Fig 2.8. The simulation results in time domain are showed in the following figures. Fig2.9 is the output of 2-DOF gyros two responses within 2s. Fig2.10 is the output of outer ring within 2s. Fig2.11 is the trajectory of 2-DOF gyros response to sinusoid

17、al input within 0.5s. As we can see from the Fig2.10, there are obvious sawtooth wave in the output of the inner ring. Its an unexpected phenomenon in my original theoretical analysis.Fig 2.8 The block diagram of input3 in SimulinkFig 2.9 2-DOF gyros two responses to sinusoidal input(2s)Fig2.10 The

18、output of outer ring from sinusoidal input (2s)Fig 2.11 Trajectory of 2-DOF gyros response to sinusoidal input(0.5s)I believe the sawtooth wave is caused by the nutation. For the frequency of the nutation is obtained as:which is far higher than the frequency of the applied sinusoidal torque, namely:

19、 wa <<w0.The trajectory of 2-DOF gyros response to sinusoidal input are shown in Fig2.11. As we can see, its coupling of X and Y channel scope output. The overall shape is an ellipse, which is not perfect for there are so many sawteeth on the top of it.By the way, the inverse Laplace transform

20、 of the output equals the response of the gyro in time domain as follows:Note that the major axis of ellipse is in the direction of the forced procession, amplitude of which is , whereas the minor axis is in the direction of the torsion spring effects, with amplitude . The nutation components are mu

21、ch smaller than that of the forced vibration, which can be eliminated to get the clear static response.To prove it, we eliminate the effects of the nutation namely the quadratic term in the denominator and get the following equations. Then we get Fig 2.12, which is a perfect ellipse.Fig 2.12 Traject

22、ory of the gyros response without nutationWe can conclude that when input to the 2-DOF gyro is sinusoidal torque, the gyro will do an ellipse conical pendulum as a static response, including procession and the torsion spring effects, together with a high-frequency vibration as the dynamic response.2

23、.4 Input4: Sinusoidal, amplitude 1 g.cm, frequency 5Hz, on outer ring axisIn this case:Due to input4 is similar to input3, only the frequency is different, so here we just analyze the output of outer ring and the trajectory of the gyros response.From Fig 2.8 we can get the output of outer ring in ti

24、me domain as showed in Fig 2.13.Fig2.13 The output of outer ring from Fig 2.14 Trajectory of 2-DOF gyrossinusoidal input (0.2s) response to sinusoidal input(0.2s)2.5 Input5: Sinusoidal, amplitude 1 g.cm, frequency 10Hz, on outer ring axisIn this case:Due to input5 is similar to input3, only the freq

25、uency is different, so here we just analyze the output of outer ring and the trajectory of the gyros response.From Fig 2.8 we can get the output of outer ring in time domain as showed in Fig 2.15.Fig2.15 The output of outer ring from Fig 2.16 Trajectory of 2-DOF gyrossinusoidal input (0.1s) response

26、 to sinusoidal input(0.1s)2.6 Input6: Sinusoidal, amplitude 1 g.cm, frequency 20Hz, on outer ring axisIn this case:Due to input5 is similar to input3, only the frequency is different, so here we just analyze the output of outer ring and the trajectory of the gyros response.From Fig 2.8 we can get th

27、e output of outer ring in time domain as showed in Fig 2.17.Fig2.17 The output of outer ring from Fig 2.18 Trajectory of 2-DOF gyrossinusoidal input (0.05s) response to sinusoidal input(0.05s)2.7 Input7: Sinusoidal, amplitude 1 g.cm, frequency 50Hz, on outer ring axisIn this case:Due to input5 is si

28、milar to input3, only the frequency is different, so here we just analyze the output of outer ring and the trajectory of the gyros response.From Fig 2.8 we can get the output of outer ring in time domain as showed in Fig 2.19.Fig2.19 The output of outer ring from Fig 2.20 Trajectory of 2-DOF gyrossi

29、nusoidal input (0.02s) response to sinusoidal input(0.02s)3.ConclusionAfter comparing the simulation results, its clear that the trajectory of 2-DOF gyros response is a precise circle with an impulse input torque, and cycloid with a constant input torque. However, the circle becomes ellipse and with

30、 sawtooth when the input torque becomes a sinusoidal signal. Besides, when frequency of signal increases, the trajectory becomes more irregular, that maybe because the outer rings vibration includes more noise.Assignment 2: Single-axis INS simulation1.Description of the tasksIn a fictitious test of

31、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 t

32、able 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 the performance of the INS.Note that the block diagram in the lecture

33、notes (figure 3.6 of both 2015 and 2016 versions) or the old ver- sions of courseware has to be slightly modified before you can obtain reasonable results.Table 2-1 Motion information and Earth parametersMotion informationvaluesunitsEarth parametersvaluesunitsInitial velocity, northward5m/sAccelerat

34、ion of gravity9.8m/s2Initial position0mRadius of the Earth6371kmAcceleration, from start2m/s2Duration of acceleration80sTable 2-2 Possible error sourcesTypesvaluesunitsTypesvaluesunitsInitial position error20mAccelerometer scale factor error0.00051Initial velocity error0.05m/sGyroscope scale factor

35、error0.00051Initial platform misalignment error1"Gyroscope drifting error0.01o/hAccelerometer bias error0.00002m/s22.Simulation and AnalysisThere is one core relevant formula, to get the specific form of its solution, we should substitute the unknown parameters. Firstly, the input signal is acc

36、elerometer of the platform, and the velocity of the platform is the integration of the acceleration. The acceleration along Yp may contains two parts:When accelerometer errors are concerned, the output of accelerometer will be:When gyro errors concerned:Only is unknown:Fig 2.1 The reference block di

37、agram in the courseware (rectified)Fig2.2 The Simulink block diagram for Assignment 2 with all error sourcesAnd the reference block diagram and Simulink block diagram are as above in Fig2.1, Fig2.2. There is a small fault in the reference block, which is that the sign of the marked add operation sho

38、uld be positive instead of negative. However, the reference block here has been rectified.Following figures are the results of the impacts of these error sources on the performance of the INS, first one by one and then altogether.Fig 2.3 Real acceleration output without error sourcesFig 2.4 Real vel

39、ocity output without error sourcesFig 2.5 Real displacement output without error sourcesFig2.6 Position bias output when only initial position error existsFig2.7 Position bias output when only initial velocity error existsFig2.8 Position bias output when only initial platform misalignment error exis

40、tsFig2.9 Position bias when only accelerometer bias error existsFig2.10 Position bias when only accelerometer scale factor error existsFig2.11 Position bias output when only gyro scale factor error exists Fig2.12 position bias output bias when only gyro drifting errorFig2.13 Position bias output con

41、sidering all error sourcesFig2.14 Position bias output considering no error sourcesAs we can see in the above simulation results, if there is no error we can navigate the trains motion correctly, which comes from north to the south as shown in Fig2.3 to Fig2.5, beginning with a constant acceleration

42、 within 80 seconds then cruises at a constant speed, approximately 165 m/s. However, the situation will change a lot when different errors put into the simulation. The initial position error effects least as Fig.2.6, for this error doesnt enter into the closed loop and it wont influence the iterativ

43、e process. The position bias is constant and can be negligible.In the second case, when the accelerometer scale factor error exists, , as shown in Fig2.10, the result are stable and almost accurate, the position bias is a sinusoidal output. So it is with the accelerometer bias error situation, , in

44、Fig2.9, the initial velocity error, in Fig2.7, and the initial platform misalignment angle, , in Fig2.11. However, the influence degrees of the different factors are not in the same magnitude. The accelerometer scale factor influences the least with magnitude of 62, then the initial velocity smaller

45、 magnitude of 40, and the accelerometer bias magnitude of 25. The influence of the initial platform misalignment angle is much more significant with a magnitude of 62. All the navigation bias in the second kind case is sinusoidal, which means theyre limited and negligible as time passes by. In the t

46、hird case, such as the gyro scale factor error situation, , in Fig2.11, and the gyro drifting error, results in Fig2.12, effects the most significant, the trajectory of the navigation deviation accumulated as time goes. The position bias is a combination of sinusoidal signal and ramp signal. They al

47、so show that the longitudinal and distance errors resulted from gyro drifts are not convergent in time. It means the errors in the gyroscope do most harm to our navigation. And due to the significant influence of the gyro drifting errors and the gyroscope scale factor error, results considering all

48、the error sources, and the navigation position of the motion will be away from the real motion after an enough long time, as shown in Fig2.12. The gyro drifting error is the most significant effect factor of all errors. By the time of 9900s, it has reaches 2250m, and its nearly the quantity of the p

49、osition bias considering all error sources. Through contrasting all the results, we can conclude that the gyro drifting error is the main component of the whole position bias. 3.ConclusionThrough contrasting all the results, we can conclude that the gyro drifting error is the main component of the w

50、hole position bias, and the gyro bias or the drift error 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 31.Descriptio

51、n of the tasksIn an fictitious mission, a spaceship is to be lifted from a launching site located at 19o 37' NL and 110o 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 accelerometers

52、, AX, AY, AZ, are installed respectively along the axes Xb, Yb, Zb of the body frame.Case 1: Stationary testDuring a pre-launching ground test, the body frame of the spaceship initially coincides with the local geographical frame, with its pitching axis Xb pointing to the east, rolling axis Yb to th

53、e north, and heading axis Zb upwards. Then the body of the spaceship is made to rotate in 3 steps:(1) 80o around Xb (2)90o around Yb (3)170o around ZbAfter that, the body of the spaceship stops rotating. You are required to compute the final outputs of the three accelerometers in the spaceship, usin

54、g quaternion and ignoring the device errors. It is assumed that the magnitude of gravity acceleration at the ground level is g0 = 9.79m/s2.Case 2: The launching processThe spaceship is installed on the top of a vertically erected rocket. Its initial heading, pitching and rolling angles with respect

55、to the local geographical frame are-90, 90 and 0 degrees respectively. The default rotation sequence is heading pitching rolling. The top ofthe rocket is initially 100m above the sea level. Then the rocket is fired up.The outputs of the gyros and accelerometers in the spaceship are both pulse number

56、s. Each gyro pulse is an angular increment of 0.01 arcsec, and each accelerometer pulse is 1e-7g0, with g0 = 9.79m/s2. The gyro output fre- quency is 100Hz, and the accelerometers is 5Hz. The outputs of the gyros and accelerometers within 1800s are stored in a MATLAB data file named mission.mat, con-taining matrices GGM of 180000×3 from gyros and AAM of 9000×3 from accelerometers respectively. The format of the data in the two matrices is as shownin the tables, with