1、Assignment ofInertial Technology惯性技术作业（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
2、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.Assignment 1: 2-DOF response simulation1.Description of th
3、e 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 from the results.Table 1-1 Ty
4、pes of input torquesTorque description Direction Suggested simulation timeImpulse, magnitude 40000 g cm , duration 1e-5s Inner ring axis 0.003 0.02sConstant, magnitude 1 g cm . Inner ring axis 0.005 0.02sSinusoidal, amplitude 1 g cm , frequency 2Hz Outer ring axis 0.5 2sSinusoidal, amplitude 1 g cm
5、, frequency 5Hz Outer ring axis 0.2 0.8sSinusoidal, amplitude 1 g cm , frequency 10Hz Outer ring axis 0.1 0.4sSinusoidal, amplitude 1 g cm , frequency 20Hz Outer ring axis 0.05 0.4sSinusoidal, amplitude 1 g cm , frequency 50Hz Outer ring axis 0.02 0.2sNote that the default configurations of the simu
6、link 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:12 212 2yx yx xyy xxy xyJHsMsMsHJsJGiven this function, we can establi
7、sh 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: 140(t)0yxMA，The frequency of the nutation is obtained as:0125rad/s397HzeHJRadius of nutation is obtained as: 54r0.1rcminyThe block diagram of the system in Simu
8、link is showed in 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 separatel
9、y as Alpha and Beta.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
10、 responses to impulse 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 .5(410,)rad5410radIn order to compare the data with differe
11、nt simulation time, I change the simulation 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.
12、 Because the only difference between Fig2.4 and Fig2.3 is the simulation time. At the beginning, we have figured out that 01025rad/s397Hz4eHJWhich means that the period is =.s.0397TfSo 0.003s is about one period and 0.02s is about nine periods. Owing to the simulation step of Simulink is fixed (or m
13、aybe I dont know how to change), the result will have bias when simulation 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: 1(t)0yxMlA，The frequency of the nutation is obtained as:01025rad/s397Hz4eHJThe drift rate is
14、 obtained as: 4rad/s.3eg/min0.e/hyThe radius of nutation is obtained as: 83210r.61arcseyJMHThe block diagram of the system in Simulink is 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 show
15、ed 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 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 In
16、put3: Sinusoidal, amplitude 1 g.cm, frequency 2Hz, on outer ring axisIn this case: 1sin(4t)0xyMA，The frequency of the nutation is obtained as:0025rad/s397HzeHJThe block diagram of the system in Simulink is showed in Fig 2.8. The simulation results in time domain are showed in the following figures.
17、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 sinusoidal 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 unexp
18、ected 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 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 s
19、awtooth wave is caused by the nutation. For the frequency of the nutation is obtained as:01025rad/s397Hz4eHJwhich is far higher than the frequency of the applied sinusoidal torque, namely: wa w0.The trajectory of 2-DOF gyros response to sinusoidal input are shown in Fig2.11. As we can see, its coupl
20、ing 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 of the output equals the response of the gyro in time domain as follows:0220 0202()sinsin()()cocooxaoxaxx xaaaaMMt ttJJ
21、t t tHHNote 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 . 0/oxMH /oxaMHThe nutation components are much smaller than that of the forced vibration, which c
22、an 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. 20020()sinsin)co(1cos)(oxoxaaaxx xaaaMt ttJHMt ttH Then we get Fig 2.12, which is a perfect ellipse.Fig 2.12 Trajector
23、y 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.2.4
24、 Input4: Sinusoidal, amplitude 1 g.cm, frequency 5Hz, on outer ring axisIn this case: 1sin(0t)0xyMA，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 out
25、er ring in time 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: 1sin(20t)xyMA，Due to input5 is simi
26、lar 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 2.15.Fig2.15 The output of outer ring from Fig 2.16 Trajectory of 2-DOF gyrossinu
27、soidal input (0.1s) response to sinusoidal input(0.1s)2.6 Input6: Sinusoidal, amplitude 1 g.cm, frequency 20Hz, on outer ring axisIn this case: 1sin(40t)xyMA，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 th
28、e gyros response.From Fig 2.8 we can get the 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 oute
29、r ring axisIn this case: 1sin(0t)0x yMA，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 the output of outer ring in time domain as showed in Fig 2.19.Fig2.19 The out
30、put of outer ring from Fig 2.20 Trajectory of 2-DOF gyrossinusoidal 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
31、 input torque. However, the circle becomes ellipse and with 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 s
32、imulation1.Description of the tasksIn 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.
33、 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 the perform
34、ance of the INS.Note that the block diagram in the lecture 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 information values units Earth
35、 parameters values unitsInitial velocity, northward 5 m/s Acceleration of gravity 9.8 m/s2Initial position 0 m Radius of the Earth 6371 kmAcceleration, from start 2 m/s2Duration of acceleration 80 sTable 2-2 Possible error sourcesTypes values units Types values unitsInitial position error 20 m Accel
36、erometer scale factor error 0.0005 1Initial velocity error 0.05 m/s Gyroscope scale factor error 0.0005 1Initial platform misalignment error 1 “ Gyroscope drifting error 0.01 o/hAccelerometer bias error 0.00002 m/s22.Simulation and AnalysisThere is one core relevant formula, to get the specific form
37、 of its solution, we should substitute the unknown parameters. (1)()cNayAKygFirstly, the input signal is accelerometer of the platform, and the velocity of the platform is the integration of the acceleration. 0/pyydtRThe acceleration along Yp may contains two parts:cosgingypf yWhen accelerometer err
38、ors are concerned, the output of accelerometer will be:(1)NaypNKfAWhen gyro errors concerned: pgpOnly is unknown:00(1)tgptpdtdFig 2.1 The reference block diagram in the courseware (rectified)Fig2.2 The Simulink block diagram for Assignment 2 with all error sourcesAnd the reference block diagram and
39、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 should be positive instead of negative. However, the reference block here has been rectified.Following figures are the results of the impacts of th
40、ese 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 velocity output without error sourcesFig 2.5 Real displacement output without error sourcesFig2.6 Position bias output when only initial position e
41、rror existsFig2.7 Position bias output when only initial velocity error existsFig2.8 Position bias output when only initial platform misalignment error existsFig2.9 Position bias when only accelerometer bias error existsFig2.10 Position bias when only accelerometer scale factor error existsFig2.11 P
42、osition bias output when only gyro scale factor error existsFig2.12 position bias output bias when only gyro drifting errorFig2.13 Position bias output considering all error sourcesFig2.14 Position bias output considering no error sourcesAs we can see in the above simulation results, if there is no
43、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 within 80 seconds then cruises at a constant speed, approximately 165 m/s. However, the situation will change a lot when different errors put in
44、to 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 02mythe iterative process. The position bias is constant and can be negligible.In the second case, when the accelerometer scale factor error exists, , as sho
45、wn in Fig2.10, 0.5aKthe result are stable and almost accurate, the position bias is a sinusoidal output. So it is with the accelerometer bias error situation, , in Fig2.9, the initial velocity error, 20./NAmsin Fig2.7, and the initial platform misalignment angle, , in Fig2.11. However, 20.5/sy 01the
46、 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 magnitude of 40, and the accelerometer bias magnitude of 25. The influence of the initial platform misalignment angle
47、 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 third case, such as the gyro scale factor error situation, , in Fig2.11, and the gyro 0.5gKdrifting error, , results in
48、 Fig2.12, effects the most significant, the trajectory of the navigation 0.1/hdeviation accumulated as time goes. The position bias is a combination of sinusoidal signal and ramp signal. They also 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 the error sources, and the navigation position of the mot
