哈工大导航原理作业
哈工大自动控制原理大作业
自动控制原理大作业1.题目在通常情况下，自动导航小车（AGV ）是一种用来搬运物品的自动化设备。
大多数AGV 都需要有某种形式的导轨，但迄今为止，还没有完全解决导航系统的驾驶稳定性问题。
因此，自动导航小车在行驶过程中有时会出现轻微的“蛇行”现象，这表明导航系统还不稳定。
大多数的AGV 在说明书中都声明其最大行驶速度可以达到1m/s ，但实际速度通常只有0.5m/s ,只有在干扰较小的实验室中，才能达到最高速度。
随着速度的增加，要保证小车得稳定和平稳运行将变得越来越困难。
AGV 的导航系统框图如图9所示，其中12=40ms =21ms ττ， 。
为使系统响应斜坡输入的稳态误差仅为1%，要求系统的稳态速度误差系数为100。
试设计合适的滞后校正网络，试系统的相位裕度达到50 ，并估计校正后系统的超调量及峰值时间。
()R s ()Y s2.分析与校正主要过程2.1确定开环放大倍数K100)1021.0)(104.0(lim )(lim =++==s s s sK s sG K v （s →0） 解得K=100)1021.0)(104.0(100++=s s s G s 2.2分析未校正系统的频域特性根据Bode 图：穿越频率s rad c /2.49=ω相位裕度︒---=⨯-⨯--=99.18)2.49021.0(arctan )2.4904.0(arctan 9018011γ 未校正系统频率特性曲线由图可知实际穿越频率为s rad c /5.34=ω2.3根据相角裕度的要求选择校正后的穿越频率1c ω现在进行计算：︒︒︒--=+=---55550)021.0(arctan )04.0(arctan 901801111c c ωω则取s rad c /101=ω可满足要求2.4确定滞后校正网络的校正函数 由于11201~101c ωω）（=因此取s rad c /110111==ωω）（，则由Bode 图可以列出 40)1lg(20)1lg(40)110lg(2022+=+ωω 解得s rad /1.02=ω于是1.0=β 则滞后网络传递函数为1101)(++=s s s G c ，10=T 2.5验证已校正系统的相位裕度已校正系统的开环传递函数为：)110)(1021.0)(104.0()1(100)()(++++=s s s s s s G s G c 相位裕度︒----=-⨯-⨯-+-=2.51)100(arctan )10021.0(arctan )1004.0(arctan )10(arctan 901801111γ校正后的相位裕度大于50°，满足设计要求。
卫星导航系统_哈尔滨工程大学中国大学mooc课后章节答案期末考试题库2023年
卫星导航系统_哈尔滨工程大学中国大学mooc课后章节答案期末考试题库2023年1.在卫星导航系统的结构中，向用户发射导航信号，为用户提供导航服务的是：答案:空间段2.北斗三号卫星导航系统中，MEO、IGSO和GEO距离地面的高度分别为:答案:21528km、35786km、35786km3.卫星导航时间系统的表现形式为周和周内秒，其中，周内秒表示不足一周的秒数，它的最大值是：答案:604800秒4.北斗D1导航电文一个子帧含有多少比特的信息：答案:3005.播发D1导航电文的北斗卫星是：答案:MEO/IGSO卫星6.信号跟踪时哪个支路会输出噪声:答案:Q支路7.GNSS观测值是以接收机（）位置为准的，所以天线的相位中心应该与其几何中心保持一致。
答案:天线的相位中心8.接收信号中存在多普勒频移的原因是：答案:接收机与卫星之间存在相对运动9.下列接收机的哪种启动模式耗时最长:答案:冷启动10.（）指出卫星绕地球做椭圆运动，地球是椭圆的一个焦点。
答案:开普勒第一定律11.GPS和北斗导航电文中播发的电离层延迟模型改正参数对应的经验模型是：答案:Klobuchar模型12.信标差分对的覆盖范围为:答案:300 km13.不同卫星间做差可以消除（）的影响。
答案:接收机钟差14.在无中断情况下，导航系统在整个飞行阶段满足精度和完好性需求的能力，它对应的航空所需导航性能的指标为:答案:连续性15.在非精密进近阶段，基于性能的导航手段可以采用（）来支持。
答案:RAIM16.下列为美国发展的陆基增强系统是：答案:LAAS17.分布式组合法一般分几步处理子系统的数据:答案:两步18.以下项目中，BDS系统与GPS系统的相同之处为：答案:以上均不是19.卫星定位按照接收机位置是否变化可分为静态定位和（）两类答案:动态定位20.BDS导航电文不包含：答案:距离观测值21.相比于第一代卫星导航系统，第二代卫星导航系统的特点有：答案:信号载波频率提升卫星数量增加采用测时-测距体制提升卫星轨道高度22.以下属于GPS现代化内容的是：答案:先进的原子钟新的信号体制23.以下卫星导航系统采用码分多址的是：答案:北斗GalileoGPS24.下面关于伪随机噪声码的说法正确的是：答案:伪随机噪声码可以作为扩频通信中的扩频码伪随机噪声码可以人为控制伪随机噪声声可以作为卫星导航信号中的测距码伪随机噪声码具有白噪声统计特性25.评价信号捕获性能的指标通常有:答案:虚警概率捕获概率漏检概率捕获灵敏度26.对于一颗在无摄运动下的卫星，其（）参数为常数。
L1-导航原理(哈工大导航原理、惯性技术课件)讲解学习
陶瓷 壳体
球形 转子
球形电极 自转轴
钛离子泵
缺陷：结构复杂、昂贵
Lecture 1 -- Introduction
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5.2 低成本、小型化
环形激光陀螺 (Ring laser gyro -- RLG) 1960s 早期开始研制, 1970s 后期进入实用
光纤陀螺 (Fiber Optical Gyro – FOG) 1970s 开始研制, 1980s 早期进入实用
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4.2 历史: 陀螺罗经
陀螺仪被寄予希望, 但面临着自动寻北 的挑战
1908年, Anschutz (德国) 发明了陀螺罗经 (gyro compass)
1909年, Sperry (美国) 也独 立研制出陀螺罗经.
—— 陀螺罗经的出现标志着陀螺仪技术 的现代应用的发端
Lecture 1 -- Introduction
Lecture 1 -- Introduction
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4.2 历史: 在航海的应用
磁罗盘 (Magnetic compass) 被用于早期的航海
19世纪后期，大量的木质 轮船被钢铁材质的轮船取代， 使磁罗盘的效能受到影响.
磁罗盘的使用在地球两极 附近受到限制 寻找替代的方向指示装置
Lecture 1 -- Introduction
Exam: Close book, close note Contact: 15204694662
Lecture 1 -- Introduction
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此课件下载可自行编辑修改，仅供参考！ 感谢您的支持，我们努力做得更好！谢谢
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惯性导航的特点: 自主 (Autonomous, self-contained) 只依赖于对载体的惯性测量 (借助加速度计、陀螺仪) 连续稳定的输出
“导航原理”实验教学课程思政探索与实践
2024.1黑龙江教育·理论与实践一、引言科技发展急需人才，人才培养取决于教育。
当前，随着我国社会改革的不断深化，社会思潮多样并存，各种思想交相融合，多元文化冲突更加频繁。
而学生正处于知识体系、思维方式和价值观念的形成时期，极易受到各种现象、观点、言论的影响。
极端个人主义、拜金主义、享乐主义等不良思潮给学生带来了消极影响，导致部分学生理想信念迷失、道德行为欠缺[1]。
如何培养富有社会责任感的创新人才，是近年来高等教育特别关注和不断探讨的课题。
专业课程教学融入思政元素，正是解决上述问题的一种尝试，也是加强和改进学生素质教育的一种探索[2]。
这种方式符合高等教育与时俱进的发展需要，不仅能够克服人才培养中的种种弊端，也极大地推动了高等教育的改革和创新，具有十分重要的意义[3]。
然而，课程思政在我国高校实施的时间并不长，还有待进一步健全和完善，尚需在理论上深入研究，在实践中总结经验。
文章结合“导航原理”实验教学的特点，将课程思政融入实验教学中，探索与实践该种教学方式对当代学生素质教育的提升效果。
二、“导航原理”实验教学特点“导航原理”作为高等工科院校控制科学与工程学科或航空宇航科学与技术学科的一门专业课程，是学习后续专业课程，如“飞行器控制与制导”“航天器控制”“无人机控制”“最优导航与滤波”等的基础。
“导航原理”实验教学不仅要帮助学生建立起惯性空间的概念，使学生加深对惯性器件结构特点、工作原理和基本特性的了解，实现对理论知识的验证，更重要的是通过实验使学生领悟惯性导航原理的应用规律，提高学生的动手能力、工程实践能力、设计能力和创新能力。
惯性导航系统是导弹和火箭的“眼睛”和“大脑”，提高惯性导航系统的精度是精确打击的关键。
惯性导航器件和系统的设计与制造需要精益求精的工匠精神和创新精神。
“导航原理”实验教学以培育学生科研实践能力和创新精神为目标，融合了国家战略、人才培养内涵式建设、学生个性化发展等多方面的内容[4]，具有深厚的课程思政资源和基础。
卫星定位导航原理知到章节答案智慧树2023年哈尔滨工业大学
卫星定位导航原理知到章节测试答案智慧树2023年最新哈尔滨工业大学第一章测试1.北斗三代系统的组成是（）。
参考答案:24 MEO+3 GEO+3 IGSO2.下列不属于GPS的功能的是（）参考答案:短报文通信3.（）导航系统覆盖范围在2000km~3000km之间。
参考答案:远程导航系统4.对于GPS系统，地面观测者相同方位每天提前（）见到同一颗GPS卫星。
参考答案:4min5.北斗系统包含哪些部分（）。
参考答案:其余都是6.下图星座是（）参考答案:北斗7.导航系统的技术指标主要有（）。
参考答案:定位速率;系统可靠性;覆盖范围;定位精度8.下列属于GPS的功能的是（）参考答案:授时;测速;定位导航9.子午系统的地面控制部分包括（）。
参考答案:控制中心;注入站;跟踪站10.子午星系统有8颗卫星。
（）参考答案:错11.北斗二代系统是被动定位系统。
（）参考答案:对12.以下星座不可以达到全球全天候不间断覆盖的效果。
（）参考答案:错第二章测试1.开普勒方程是描述偏近点角和（）的关系。
参考答案:平近点角2.由于地球自转轴收到地球内部质量不均匀影响，地极点在地球表面上位置随时间而变化的现象称为（）。
参考答案:极移现象3.确定卫星在轨道上的顺势为之的开普勒轨道参数是（）。
参考答案:真近点角4.关于以下电文帧结构，说法不正确的是（）。
参考答案:该电文帧为北斗电文帧5.下式中，旋转哪个参量是使得轨道平面与赤道平面相重合（）。
参考答案:6.开普勒第三定律与（）有关。
参考答案:轨道长半轴;卫星运行周期7.以下关于GPS卫星导航电文的描述正确的是（）。
参考答案:基本单位长度为1500bit;一个主帧包括5个子帧8.星历信息中不包括（）参考答案:卫星时钟校正量;卫星识别号和健康状态;全部卫星的粗略星历9.协议天球坐标系需要扣除（）参考答案:章动现象;岁差现象10.协议地球坐标系最终要转化为协议天球坐标系。
（）参考答案:错11.如果考虑接收机钟差，4颗星即可定位。
1.导航原理第一章2
导航原理（principle of navigation）i) 使用教材：无（主要是没有合适的教材，要自己编）。
ii）参考书：1.惯性导航原理，邓正隆，哈尔滨工业大学出版社，1994；2.GPS卫星导航定位原理与方法，刘基余，科学出版社，2003；3.Elliott D. Kaplan. Understanding GPS：principles andapplications（second edition）.中译本：1）GPS原理与应用（第一版），邱致和（20所），电子工业出版社；2）GPS原理与应用（第二版），寇艳红（北航），电子工业出版社，2007。
4）Pratap Misra，Per Enge. Global Positioning System: Signals, Measurements and Performance(second Edition).中译本： GPS 信号，测量与性能（第二版），罗鸣等，电子工业出版社；iii）课程考核方式：课堂大作业形式。
iv）课程的主要内容：惯性导航部分；北斗部分；GPS部分；天文导航部分；组合导航部分；新增部分：量子导航Simulation-based（粒子滤波）。
瑞典林雪平大学(LinkOping University)的Rickard Karlsson提出一种无需GPS即可定位并导航的新技术。
第一章导航及其发展§1.1 导航的基本概念1、导航的定义在各种复杂的气象条件下，采用最有效的方法并以规定的所需导航性能，引导运载体航行的过程（引导运载体按一定航线从一个地点（出发点）到另一个地点（目的地）的过程）。
2、导航参数导航过程中需要用来完成导航任务的参数。
载体的位置、速度、姿态（角度）等，其中最重要的参数是确定载体的位置，即定位。
所以，导航的核心就是定位。
3、导航的任务1）引导运载体进入并沿预定航线航行；2）导引运载体在夜间和各种气象条件下安全着陆或进港。
导航原理 大作业 哈工大
2. 程序设计说明及代码
2.1 仿真需要的两个子程序 （1）四元数求逆子函数 %四元数求逆函数 function [ qni ] = qiuni( q ) q(1)=q(1); q(2)=-q(2); q(3)=-q(3); q(4)=-q(4); qni=q; end （2）四元数相乘子程序 %四元数相乘计算函数 function [q]=quml(q1,q2); lm=q1(1);p1=q1(2);p2=q1(3);p3=q1(4); q=[lm -p1 -p2 -p3;p1 lm -p3 p2;p2 p3 lm -p1;p3 -p2 p1 lm]*q2; end 2.2 第一种情形：正对导弹进行地面静态测试（导弹质心相对地面静止） （1）用方向余弦矩阵计算，MATLAB 程序如下： function dcm g0=[0;0;9.8];%重力加速度在地里坐标系中的分量表示 wx=15/180*pi; wy=20/180*pi; wz=-10/180*pi; w=sqrt(wx^2+wy^2+wz^2); %四阶近似 I=eye(3); S=1-w^2/6; C=1/2-w^2/24; W=[0,-wz,wy;wz,0,-wx;-wy,wx,0]; c=I+S*W+C*W^2;%载体坐标系到初始坐标系的方向余弦阵 c=inv(c)%初始坐标系到载体坐标系的方向余弦阵 g=c*g0%重力加速度在载体坐标系中的分量 end 在MATLAB命令窗口输入dcm，即得到如下结果： c = 0.9253 0.2129 0.3139 -0.1233 0.9515 -0.2821 -0.3587 0.2223 0.9067
g = -3.5160 2.1791 8.8842 2.3 第二种情形：导弹正在飞行中 MATLAB 程序如下： %主程序
哈工大卫星定位导航原理实验报告
卫星定位导航原理实验专业：班级：学号：姓名：日期：实验一实时卫星位置解算及结果分析一、实验原理实时卫星位置解算在整个GPS接收机导航解算过程中占有重要的位置。
卫星位置的解算是接收机导航解算（即解出本地接收机的纬度、经度、高度的三维位置）的基础。
需要同时解算出至少四颗卫星的实时位置，才能最终确定接收机的三维位置。
对某一颗卫星进行实时位置的解算需要已知这颗卫星的星历和GPS时间。
而星历和GPS时间包含在速率为50比特/秒的导航电文中。
导航电文与测距码（C/A码）共同调制L1载频后，由卫星发出。
本地接收机相关接收到卫星发送的数据后，将导航电文解码得到导航数据。
后续导航解算单元根据导航数据中提供的相应参数进行卫星位置解算、各种实时误差的消除、本地接收机位置解算以及定位精度因子（DOP）的计算等工作。
关于各种实时误差的消除、本地接收机位置解算以及定位精度因子（DOP）的计算将在后续实验中陆续接触，这里不再赘述。
卫星的额定轨道周期是半个恒星日，或者说11小时58分钟2.05秒；各轨道接近于圆形，轨道半径（即从地球质心到卫星的额定距离）大约为26560km。
由此可得卫星的平均角速度ω和平均的切向速度v s为：ω=2π/(11*3600+58*60+2.05)≈0.0001458rad/s (1.1)v s=rs*ω≈26560km*0.0001458≈3874m/s (1.2) 因此，卫星是在高速运动中的，根据GPS时间的不同以及卫星星历的不同（每颗卫星的星历两小时更新一次）可以解算出卫星的实时位置。
本实验同时给出了根据当前星历推算出的卫星在11小时58分钟后的预测位置，以此来验证卫星的额定轨道周期。
本实验另一个重要的实验内容是对卫星进行相隔时间为1s的多点测量（本实验给出了三点），根据多个点的测量值，可以估计Doppler频移。
由于卫星与接收机有相对的径向运动，因此会产生Doppler效应，而出现频率偏移。
哈工大惯性技术(导航原理)大作业
Assignment ofInertial Technology《惯性技术》作业My Chinese NameMy Student ID 15S004001Autumn 2015Assignment 1: 2-DOF response simulationA 2-DOF gyro is subjected to a sinusoidal torque with amplitude of 4 g.cm and frequency of 10 Hz along its outer ring axis. The angular moment of its rotor is 10000 g.cm/s , and the angular inertias in its equatorial plane are both 4 g.cm/s 2. Please simulate the response of the gyro within 1 second, and present whatever you can discover or confirm from the result.In this simulation, we are going to discuss the response of a 2-DOF gyro to sinusoidal torque input. According to the transfer function of the 2-DOF gyro, the outputs can be expressed as:12222()()()()yx y x y x y J Hs M s M s J J s H s J J s H α=-++12222()()()()x yx x y x y J Hs M s M s J J s H s J J s H β=+++ The original system transfer function is a 2-input, 2-output coupling system. But the given input only exists one input, we can treat the system as 2 separate FIFO systemAs a consequence, we can establish the block diagram of the system in simulink in Fig 1.1. Substitute the parameters into the system and input, then we have the input signals as follow: 0,4sin(20)y ox M M t π==Then the inverse Laplace transform of the output equals the response of the gyro in timedomain as follows:0222200020222200()sin sin ()()()cos cos ()()ox a oxa x a x a ox ox a ox a a a a a M M t t t J J M M M t t t H H H ωαωωωωωωωωωβωωωωωωωω=-+--=+---Fig 1.1 The block diagram of the system in simulinkAnd the simulation results in time domain within 1 second are shown in the follwing pictures. Fig1.2 is the the output of outer ring ()t α. Fig1.3 is the output of inner ring ()t β. Fig1.4 is the trajectory of 2-DOF gyro ’s response to sinusoidal input. As we can see from the Fig1.3,there are obvious sawtooth wave in the output of the inner ring. It ’s a unexpected phenomenon in my original theoretical analysis.Fig1.2 The output of inner ring ()t βFig 1.3 The output of outer ring ()t αI believe the sawtooth wave is caused by the nutation. For the frequency of the nutation is obtained as010*******/3974e H rad s Hz J ω===≈, which is far higher than the frequencyof the applied sinusoidal torque , namely0a ωω .Fig 1.4 Trajectory of 2-DOF gyro ’s response to sinusoidal inputThe trajectory of 2-DOF gyro ’s response to sinusoidal input are shown in Fig1.4. As we can see, it ’s 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.Note that the major axis of ellipse is in the direction of the forced procession, amplitude of which is 0ox M H ω, whereas the minor axis is in the direction of the torsion spring effects,with amplitude ox aM H ω.The nutation components are much smaller than that of the forced vibration, which can be eliminated to get the clear static response.22002220()sin sin ()cos (1cos )()ox oxaa x a ox ox oxa a a a a aM M t t t J H M M M t t t H H H αωωωωωωβωωωωωωω≈≈-≈-=--To prove it, we eliminate the effects of the nutation namely the quadratic term in the denominator and get Fig 1.5, which is a perfect ellipse. We 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.Fig 1.5 Trajectory of the gyro’s response without nutationAssignment 2: Single-axis INS simulation2.1 description of the problemA magnetic levitation train is being tested 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 informationand Earth parameters are shown in table 2-1, and the possible error sources are shown in Table2-2.Fig2.1 The sketch map of the single-axis INS problemYou are asked to simulate the operation of the INS within 10,000 seconds, and investigate,first one by one and then altogether, the impact of these error sources on the performance of theINS.The block diagram in the courseware might be of some help. However, there lurks aninconspicuous error, which you have to correct before you can obtain reasonable results.There are one core relevant formula, to get the specific form of its solution, we should substitute the unknown parameters.(1)()c N a y A K yga =∆++-Firstly, the input signal is accelerometer of the platform, and the velocity of the platform is the integration of the acceleration.0/p y y ydt yR ω=+=-⎰The acceleration along Yp may contains two parts:cos gsin g yp f y yααα=-≈- When accelerometer errors are concerned, the output of accelerometer will be:(1)N a yp N a K f A =++When gyro errors concerned:'(1)p g p K ωωε=++Onlyαis unknown:000[(1)]tg p t tp t K dt dtααωεϕϕω=+++-∆∆=⎰⎰And the reference block diagram and simulink block diagram are as follows in Fig2.2, Fig2.3. 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.The results of the simulation are shown in Fig 2.4 to Fig 2.13.Fig2.2 The reference block diagram in the courseware(unrectified)Fig2.3 The simulink block diagram for Assignment 22.2 results and analysis of the problemFig.2.4 real acceleration,velocity and displacement output without error sourcesKFig2.5 position bias when only accelerometer scale factor error exists0.0001aFig2.6 position bias output when only gyro scale factor error exists 0.0001g K =Fig2.7 position bias when only acceleromete bias error exists 20.0002/N A m s ∆=Fig2.8 position bias output when only initial velocity error exists 200.01/ym s ∆=Fig2.9 position bias output when only initial position error exists 010y m ∆=Fig2.10 position bias output bias when only gyro bias error exists 0.000000024240.00681/3/h rad s ε=︒=Fig2.11 position bias output when only initial platform misalignment angle exists0.000012120''345radα==Fig2.12 output considering all the error sourcesFig2.13 position bias output considering all error sourcesAs we can see in the above simulation results, if there is no error we can navigate the train ’s motion correctly, which comes from north to the south as shown in Fig2.4, beginning with an constant acceleration within 60 seconds then cruises at a constant speed, approximately 65 m/s. However, the situation will change a lot when different errors put into the simulation. The initial position error 010y m ∆= effects least as Fig.2.8, for this error doesn ’t enter into the closed loop and it won ’t influence the iterative process. The position bias is constant and can be negligible.In the second case, when the accelerometer scale factor error exists, 0.0001a K =, as shown in Fig2.5, the result are stable and almost accurate, the position bias is a sinusoidal output. So it is with the accelerometer bias error situation, 20.0002/N A m s ∆=, in Fig2.7, the initialvelocity error, 200.01/ym s ∆= in Fig2.9, and the initial platform misalignment angle, 05''α=, in Fig2.11. However, the influence degrees of the different factors are not in the samemagnitude. The accelerometer scale factor influences the least with magnitude of 5, then the initial velocity larger magnitude of 8, and the accelerometer bias magnitude of 25. The influence of the initial platform misalignment angle is much more significant with a magnitude of 150. All the navigation bias in the second kind case is sinusoidal, which means they ’re limited and negligible as time passes by.In the third case, such as the gyro scale factor error situation, 0.0001g K =, in Fig2.6, and the gyro bias error,0.01/h ε=︒, results in Fig2.10, effects the most significant, the trajectory of the navigation disvergence 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 bias errors and the gyroscope scale factor error, results considering all the error sources disverge, and the navigationposition of the motion will be away from the real motion after a enough long time, as shown in Fig2.10. The gyro bias error is the most significant effect factor of all errors. By the time of 10000s, it has reaches 1600m, and it’s nearly the quantity of the position bias considering all error sources. Through contrasting all the results, We can conclude that the gyro bias error is the main component of the whole position bias.Impression of the Whole simulation experimentThrough contrasting all the results, We can conclude that the gyro bias error is the main component of the whole position bias, and the 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 3: SINS simulation3.1 Task descriptionA missile equipped with SINS is initially at the position of 46o NL and 123 o EL, stationary on a launch pad. Three gyros, GX, GY, GZ, and three accelerometers, AX, AY, AZ, are installed along the axes Xb, Yb, Zb of its body frame respectively.Case 1: stationary testThe body frame of the missile initially coincides with the geographical frame, as shown in the figure, with its pitching axis Xb pointing to the east, rolling axis Yb to the north, and azimuth axis Zb upward. Then the body of the missile is made to rotate in 3 steps:(1) -22o around Xb(2) 78o around Yb(3) –16o around ZbFig 3.1 Introduction to assignment 3After that, the body of the missile stops rotating. You are required to compute the final outputs of the three accelerometers on the missile, using quaternion and ignoring the device errors. It is known that the magnitude of gravity acceleration is g = 9.8m/s2.Case 2: flight tes tInitially, the missile is stationary on the launch pad, 400m above the sea level. Its rolling axis is vertical up,and its pitching axis is to the east. Then the missile is fired up. The outputs of the gyros and accelerometers are both pulse numbers. Each gyro pulse is an angular increment of 0.01arc-sec, and each accelerometer pulse is 1e-7g, with g =9.8m/s2. The gyro output frequency is 200Hz, and the accelerometer’s is 10Hz. The outputs of the gyros and accelerometers within 1315s are stored in a MA TLAB data file named imu.mat, containing matrices gm of 263000× 3 and am of 13150× 3 respectively. The format of the data is as shown in the tables, with 10 rows of each matrix selected. Each row represents the outputs of the type of sensors at each sampling time. The Earth can be seen as an ideal sphere, with radius 6371.00km and spinning rate 7.292× 10-5 rad/s, The errors of the sensors are ignored, so is the effect of height on the magnitude of gravity. The outputs of the gyros are to be integrated every 0.005s. The rotation of the geographical frame is to be updated every 0.1s, so are the velocities and positions of the missile.You are required to:(1) compute the final attitude quaternion, longitude, latitude, height, and east, north, vertical velocities of the missile.(2) compute the total horizontal distance traveled by the missile.(3) draw the latitude-versus-longitude trajectory of the missile, with horizontal longitude axis.(4) draw the curve of the height of the missile, with horizontal time axis.Fig 3.2 simplified navigation algorithm for SINS 3.2Procedure code3.2.1 Sub function code：quaternion multiply code:function [q1]=quml(q1,q2);lm=q1(1);p1=q1(2);p2=q1(3);p3=q1(4);q1=[lm -p1 -p2 -p3;p1 lm -p3 p2;p2 p3 lm -p1;p3 -p2 p1 lm]*q2;endquaternion inversion code:function [qni] =qinv(q)q(1)=q(1);q(2)=-q(2);q(3)=-q(3);q(4)=-q(4);qni=q;end3.2.2case1 DCM algorithm：function ans11cz=[cos(-22/180*pi) sin(-22/180*pi) 0 ;-sin(-22/180*pi) cos(-22/180*pi) 0;0 0 1]; %The third rotation DCMcx=[1 0 0 ;0 cos(78/180*pi) sin(78/180*pi);0 -sin(78/180*pi) cos(78/180*pi)]; %The first rotation DCMcy=[cos(-16/180*pi) 0 -sin(-16/180*pi);0 1 0;sin(-16/180*pi) 0 cos(-16/180*pi)]; %The second rotation DCM A=cz*cy*cx*[0;0;-9.8]end3.2.3case1 quaternion algorithm：function ans12g=[0;0;0;-9.8];q1=[cos(-11/180*pi);sin(-11/180*pi);0;0]; %The first rotation quaternionq2=[cos(39/180*pi);0;sin(39/180*pi);0]; %The second rotation quaternionq3=[cos(-8/180*pi);0;0;-sin(-8/180*pi)]; %The third rotation quaternionr=quml(q1,q2); %call the quaternion multiplication subfunction q=quml(r,q3);P1=[q(1) q(2) q(3) q(4);-q(2) q(1) q(4) -q(3);-q(3) -q(4) q(1) q(2);-q(4) q(3) -q(2) q(1)];P2=[q(1) -q(2) -q(3) -q(4);q(2) q(1) q(4) -q(3);q(3) -q(4) q(1) q(2);q(4) q(3) -q(2) q(1)];P=P1*P2;gn=P*g;gn=gn(2:4)end3.2.4 case2 SINS quaternion algorithm code：function ans2clc;clear;%parameters initializing：T=0.005;K=13150;R=6371000; %radius of earthwE=7.292*10^(-5); %spinning rate of earthQ=zeros(4, 263001) ;%quaternion matrix initializinglongitude=zeros(1,13151);latitude=zeros(1,13151);H=zeros(1,13151); %altitude matrixQ(:,1)=[cos(45/180*pi);-sin(45/180*pi);0;0]; % initial quaternion longitude(1)=123; %initial longitudelatitude(1)=46; % initial latitudeH(1)=400; %initial altitudelength=0;g=9.8;vE = zeros(1,13151); %eastern velocityvN = zeros(1,13151); %northern velocityvH = zeros(1,13151); %upward velocityvE(1)=0;vN(1)=0;vH(1)=0;load imu.mat %data loading%main calculation sectionfor N=1:Kq1=zeros(4,11);q1(:,1)=Q(:,N);for n=1:20 % Attitude iteration wx=0.01/(3600*180)*pi*gm((N-1)*10+n,1); % Angle incrementwy=0.01/(3600*180)*pi*gm((N-1)*10+n,2);wz=0.01/(3600*180)*pi*gm((N-1)*10+n,3);w=[wx,wy,wz]';normw=norm(w); % Norm calculationW=[0,-w(1),-w(2),-w(3);w(1),0,w(3),-w(2);w(2),-w(3),0,w(1);w(3),w(2),-w(1),0];I=eye(4);S=1/2-normw^2/48;C=1-normw^2/8;q1(:,n+1)=(C*I+S*W)*q1(:,n);Q(:,N+1)=q1(:,n+1);endWE=-vN(N)/R; % rotational angular velocity component of a geographic coordinate systemWN=vE(N)/R+wE*cos(latitude(N)/180*pi);WH=vE(N)/R*tan(latitude(N)/180*pi)+wE*sin(latitude(N)/180*pi);attitude=[WE,WN,WH]'*T; %correction of the quaternion by updating the rotation of geographic coordinatenormattitude=norm(attitude);e=attitude/normattitude;QG=[cos(normattitude/2);sin(normattitude/2)*e];Q(:,N+1)=quml(qinv(QG),Q(:,N+1));fx=1e-7*9.8*am(N,1); %specific force measured by accelerometerfy=1e-7*9.8*am(N,2);fz=1e-7*9.8*am(N,3);Fb=[fx fy fz]';F=quml(Q(:,N+1),quml([0;Fb],qinv(Q(:,N+1)))); %The specific force is decomposed into geographic coordinate system.FE(N)=F(2);FN(N)=F(3);FU(N)=F(4);%calculate the velocity of the vehicle:VED(N)=FE(N)+vE(N)*vN(N)/R*tan(latitude(N)/180*pi)-(vE(N)/R+2*wE*cos(latitude(N)/ 180*pi))*vH(N)+2*vN(N)*wE*sin(latitude(N)/180*pi);VND(N)=FN(N)-2*vE(N)*wE*sin(latitude(N)/180*pi)-vE(N)*vE(N)/R*tan(latitude(N)/180 *pi)-vN(N)*vH(N)/R;VUD(N)=FU(N)+2*vE(N)*wE*cos(latitude(N)/180*pi)+(vE(N)^2+vN(N)^2)/R-g;%integration and get the relative velocity of vehicle:vE(N+1)=VED(N)*T+vE(N);vN(N+1)=VND(N)*T+vN(N);vH(N+1)=VUD(N)*T+vH(N);% integration and get the position of vehicle:longitude(N+1)=vE(N)/(R*cos(latitude(N)/180*pi))*T/pi*180+longitude(N);latitude(N+1)=vN(N)/R*T/pi*180+latitude(N);H(N+1)=vH(N)*T+H(N);length=sqrt((vE(N))^2+(vN(N))^2)+length;endfigure(1) %picture the longitude-latitude curve of the motion within 1315 secondstitle(' longitude-latitude ');hold ongrid onplot(longitude,latitude);figure(2) % picture the altitude curve of the motion within 1315 secondstitle('altitude');hold ongrid onplot(0:13150,H);3.3Simulation outputs and results analysisIn case 1, the DCM algorithm and quaternion results are the same as follow:A =7.53165.9788-1.8892And the results suggest the solution of DCM and quaternion is equivalence in this problem.In case 2, the latitude-versus-longitude trajectory of the missile(with horizontal longitude axis) and the altitude curve of the missile are shown as follows in Fig 3.3 and Fig 3.4.Fig3.3 the latitude-versus-longitude trajectory of the missile(with horizontal longitude axis)Fig3.4the altitude curve of the missileImpression of the Assignment 3In this assignment, we simulate the missile navigation situation in a real problem, as the problem we have done before. I did this based on my own original code, but to my surprise, it’s harder than I have expected. For the detailed thoughts for my program I have forgotten. I have to recheck the code line by line, But I am still troubled in correcting the initial parameters for many times. It has taught me a lesson, which is never to be egotistical for your ability to memory and thework you have done or understood.。
L1-导航原理(哈工大导航原理、惯性技术)
05
前沿科技与未来发展趋 势
量子惯性传感器研究进展及挑战
量子惯性传感器原理
利用量子力学原理，通过测量微观粒 子（如原子、光子）的状态变化来感 知物体的运动状态。
研究进展
近年来，量子惯性传感器在精度、稳 定性和可靠性方面取得了显著进展， 但仍面临一些技术挑战，如量子态的 制备、操控和测量等。
挑战与前景
挑战与前景
人工智能在导航领域的应用需要解决数据获取、算法优化和实时性等问题，未来有望与 量子计算、生物计算和光计算等前沿技术相结合，推动导航技术的创新发展。
多源信息融合技术在导航中的应用探讨
多源信息融合原理
将来自不同传感器的信息进行融合处 理，提取出更准确、全面的导航信息 。
应用实例
多源信息融合技术已广泛应用于组合 导航、室内外无缝定位等领域，提高 了导航系统的性能和可靠性。
可靠性评估
可靠性评估旨在评估组合导航系统在长时间运行或复杂环境下的性能表现。通过对系统在不同条件下的 故障率、恢复时间等进行分析，可以评估系统的可靠性。
惯性/卫星组合导航系统优化策略
卡尔曼滤波算法
粒子滤波算法
深度学习算法
卡尔曼滤波是一种高效的递归 滤波器，适用于线性系统。在 惯性/卫星组合导航系统中，可 以利用卡尔曼滤波算法对IMU 和卫星信号进行融合处理，提 高系统的定位精度和稳定性。
螺等。
加速度计类型
包括压电式、压阻式、电容式等。
惯性器件特点
不同类型惯性器件具有不同的测量 精度、动态范围、稳定性等特点。
惯性系统误差来源与补偿方法
误差来源
包括初始对准误差、器件误差（如刻度因数误差、零偏误差 等）、计算误差等。
补偿方法
通过误差建模、滤波算法（如卡尔曼滤波、粒子滤波等）对 误差进行估计和补偿，提高导航精度。同时，可采用组合导 航技术，融合其他传感器信息（如GPS、里程计等），进一 步提高导航系统的性能和可靠性。
哈尔滨工业大学导航原理大作业报告
Assignment of Principles of Navigation 《导航原理》作业(惯性导航部分2016 秋)My Chinese NameMy Class No.My Student No.Task descriptionIn an fictitious mission, a spaceship is to be lifted from a launching site located at 19o37' NL and 110o57' EL, into a circular orbit 400 kilometers high along the equator. The spaceship is equipped with a strapdown INS whose three gyros, G X , G Y , G Z , and three accelerometers, A X , A Y , A Z , are installed respectively along the axes X b , Y b , Z b 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 X b pointing to the east, rolling axis Y b to the north, and heading axis Z b upwards. Then the body of the spaceship is made to rotate in 3 steps: (1) 80oaround X b(2) 90oaround Y b (3) 170oaround Z bAfter that, the body of the spaceship stops rotating. You are required to compute the final outputs of the three accelerometers in the spaceship, using quaternion and ignoring the device errors. It is assumed that the magnitude of gravity acceleration at the ground level is g 0 = 9.79m/s 2.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 are -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 accelerometers in the spaceship are both pulse numbers. Each gyro pulse is an angular increment of 0.01 arcsec, and each accelerometer pulse is 1e -7g 0, with g 0 = 9.79m/s 2. The gyro output fre- quency is 100Hz, and the accelerometer’s is 5Hz.The outputs of the gyros and accelerometers within 1460s are stored in a MATLAB data file named mission.mat, con-taining matrices GGM of 146000×3 from gyros and AAM of 7300×3 from accelerometers respectively. The format of the data in the two matrices is as shown in the 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 R = 6371.00km and spinning rate is 7.292×10-5rad/s, The errors of the gyros and ac- celerometers can be ignored, but the effect ofheight on the magnitude of gravity has to be taken into account. The gravity acceleration at the sea level of the near equator region can be chosen as g 0=9.79 m/s 2, and the magnitude ofgravity acceleration at a height of h can be computed asBesides, the influence of height on the angular rates of the geographical frame and the changing rates oflatitude and longitude should also be considered.The outputs of the gyros can be integrated every 0.01s. The rotation of the geographical frame is to be up- dated every 0.2s, so are the velocities and positions of the spaceship. 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.g h =g 0R 2(R +h )2一 任务一1 利用四元数计算飞船上加速度计的最终输出初始时刻三个加速度计的输出A=[0,0,9.79]。
哈工大-导航原理作业
Assignments of Inertial Navigation 《惯性导航》作业Autumn 2017Assignment 1: Coordinate transformation for 3D animationfigure 1.1 Interface for rotating animation control of missileAttached is a group of MATLAB programs for 3D animation control of a missile. Initially, the body frame of the missile coincides with the local geographical frame, with its pitching axis X pointing to the east, rolling axis Y to the north, and heading axis Z to the sky, as shown in figure 1.1 which is the controlling in-terface produced by running the program main.m.The three coordinates of each vertex of the 3D model occupies a row in the matrix VTX, The color property and surface information of the model are defined by matrices VTXcolor and faceM respectively. These matrices are loaded from the data file missiledata.mat.Input an angle (in degree) in the "Rotation Angle" box, and click one of the rotation buttons ("Heading Ro-tation", "Pitching Rotation", or "Rolling Rotation"), then it is expected that the missile will rotate from its current attitude for the input angle around the chosen axis of its current body.The animation is to be achieved by the program redraw.m which redraws the missile every once it rotates for one degree, using patch command. During each step of the animation, the current rotation angle of the missile relative to its pre-animation attitude has been generated and stored in one of the variables head, pitch and roll, with only one of them being nonzero. Before each redrawing, you need to re-calculate the coordinates of the missile in VTX resolved in the geographical frame, instead of keeping them the same as their values at the very beginning (VTX0). That is, in the program redraw.m, you need to replace the command line VTX=VTX0with your own codes for re-calculating VTX. Elsewhere, additional codes also might be required to make it work.Please rewrite the program redraw.m, or others if necessary, so that successive rotating animations can be achieved, and explain the rationale behind your rewriting.一、 任务分析本作业需要实现的功能为实现火箭的连续旋转。
哈工大导航原理大作业
《导航原理》作业(惯性导航部分)一、题目要求A fighter equipped with SINS is initially at the position of ︒35 NL and︒122 EL,stationary on a motionless carrier. Threegyros X G ,Y G ,Z G ,and three accelerometers, X A ,Y A ,Z A are installed along the axes b X ,b Y ,b Z of the body frame respectively. Case 1:stationary onboard testThe body frame of the fighter initially coincides with the geographical frame, as shown in the figure, with its pitching axis b X pointing to the east,rolling axis b Y to the north, and azimuth axis b Z upward. Then the body of the fighter is made to rotate step by step relative to the geographical frame.(1) ︒10around b X (2) ︒30around b Y (3) ︒50-around b ZAfter that, the body of the fighter stops rotating. You are required to compute the final output of the three accelerometers on the fighter, using both DCM and quaternion respectively,and ignoring the device errors. It is known that the magnitude of gravity acceleration is 2/8.9g s m =. Case 2:flight navigationInitially, the fighter is stationary on the motionless carrier with its board 25m above the sea level. Its pitching and rolling axes are both in the local horizon, and its rolling axis is ︒45on the north by east, parallel with the runway onboard. Then the fighter accelerate along therunway and take off from the carrier.The output of the gyros and accelerometers are both pulse numbers,Each gyro pulse is an angular increment of sec arc 1.0-,and each accelerometer pulse is g 6e 1-,with 2/8.9g s m =.The gyro output frequency is 10 Hz,and the accelerometer ’s is 1Hz. The output of gyros and accelerometers within 5400s are stored in MATLAB data files named and , containing matrices gm of 35400⨯ and am of 35400⨯ respectively. The format of data as shown in the tables, with 10 rows of each matrix selected. Each row represents the out of the type of sensors at each sample time.The Earth can be seen as an ideal sphere, with radius and spinning raterad/s 10292.75-⨯, Theerrors of sensors are ignored, so is theeffect of height on the magnitude of gravity. The output of the gyros are to integrated every . The rotation of geographical frame is to be updated every 1s, so are the velocities and position of the figure. You are required to:(1)Compute the final attitude quaternion, longitude, latitude, height, and east, north, vertical velocities of the fighter.(2)Compute the total horizontal distance traveled by the fighter. (3)Draw the latitude-versus-longitude trajectory of the fighter, with horizontal longitude axis.(4)Draw the curve of the height of fighter, with horizontal time axis.二、Case1解答方向余弦阵法（1） 绕Xb 轴转过ψ︒=10ϕ⎪⎪⎪⎭⎫ ⎝⎛︒︒-︒︒=⎪⎪⎪⎭⎫ ⎝⎛-=10cos 10sin 010sin 10cos 0001cos sin 0sin cos 0001ϕϕϕϕϕC（2） 绕Yb 轴转过︒=30θ⎪⎪⎪⎭⎫ ⎝⎛︒︒︒-︒=⎪⎪⎪⎭⎫ ⎝⎛-=30cos 030sin 01030sin 030cos cos 0sin 010sin 0cos θθθθθC（3） 绕Zb 轴转过︒-=50ψ⎪⎪⎪⎭⎫ ⎝⎛︒-︒-︒-︒-=⎪⎪⎪⎭⎫ ⎝⎛=1000）50（cos ）50（sin -0）50（sin ）50（cos 1000cos sin -0sin cos ψψψψψC 所以变换后的坐标X E Y C C C N Z ϕθψζ⎛⎫⎛⎫ ⎪⎪= ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭由于初始时刻有009.8E N ζ⎛⎫⎛⎫ ⎪ ⎪= ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭所以计算得三个加速度计的输出分别是 ,/4504.42x s m A -=,/6027.22y s m A -=s m A /3581.8z =2计算程序见附录一四元数法（1）绕Xb 轴转过ψ︒=10ϕ i 2sin2cos q ϕϕϕ+=（2）绕Yb 轴转过︒=30θj 2sin2cosq θθθ+=（3）绕Zb 轴转过︒-=50ψk 2sin2cosq ψψψ+=则合成四元数)2sin 2(cos )2sin 2(cos )2sin 2(cosk j i q q q q ⋅+⋅+⋅+==ψψθθϕϕψθϕ合成四元数的逆1123q p i p j p k λ-=---由公式 q N E qZ YX ⎪⎪⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫⎝⎛-ξ1计算得三个加速度计的输出分别是 ,/4504.42x s m A -=,/6027.22y s m A -=s m A /3581.8z =2由两种计算方法的计算结果可以看出，方向余弦阵法和四元数法的计算结果是一致的。
哈工大航概实验作业27
姓名：学号：院系：27各种飞行器导航控制执行机构的定义、分类、特点、功能、主要技术性能，举例说明（如导弹舵机、空间飞行器反作用飞轮）制导系统制导系统由控制系统和引导系统构成，引导系统：引导系统通过探测装置确定导弹相对目标或发射点的位置形成引导指令送给控制系统。
由探测设备和导引指令形成装置组成。
控制系统：响应引导系统的引导指令信息，产生作用力，迫使导弹改变飞行轨迹。
使导弹沿要求的弹道飞行；或者稳定导弹的飞行由导弹姿态敏感元件、操纵面位置敏感元件、计算机、作动装置、操纵面和弹体组成。
制导技术的种类按制导原理可分为:1,自主式制导系统2,遥控制导系统3,寻的制导系统4,复合制导系统1, 自主式制导不需要提供目标的直接信息，也不需要弹体以外的设备配合，而仅靠弹体自身装载的测量仪器测量地球的某些物理特征，从而确定弹体的飞行轨道，控制引导弹体命中目标。
特点：自主式制导的弹体的飞行完全自主，因而不易受干扰。
但由于制导程序是预先确定的，所以这种制导方式只适于攻击地面固定目标。
自主式制导又分为相关制导和惯性制导两种。
功用：大多数地地弹道导弹，如美国的“大力神”(Titan)、“民兵”洲际弹道导弹等都采用惯性制导。
随着制导技术的发展，还可采用天文或地形地图匹配的方式来提高制导精度。
大部分地（潜）地导弹采用自主式制导系统。
2,_遥控制导系统由设在导弹以外的制导站导引和控制导弹飞行的制导系统.遥控制导系统可分为有线指令制导,无线电指令制导和波束制导系统.特点：遥控制导的优点是弹上设备简单，在一定射程范围内可获得较高的制导精度。
缺点是射程受跟踪测量系统作用距离的限制，制导精度随射程的增加而降低，并易受干扰。
随着导弹技术的发展，遥控制导一般可与其他制导方式组成复合制导。
只有射程较近的战术导弹才采用全程遥控制导,通常还同时采用多种跟踪测量手段,以提高导弹的战斗性能。
测量和控制误差是影响遥控制导精度的主要因素。
因此，研制新型测量装置和应用现代控制理论设计制导系统，受到普遍重视。
哈尔滨工业大学—第5章智能网联汽车导航定位系统
SCHOOL OF AUTOMOBILE ENGINEERING
规格严格 功夫到家
第3页
（威海） 汽车工程学院
HARBIN INSTITUTE OF TECHNOLOGY
SCHOOL OF AUTOMOBILE ENGINEERING
规格严格 功夫到家
5.1 导航定位的定义与类型—定义
➢ 智能网联汽车或无人驾驶汽车的导航定位通过全球定位系统
期载波相位差分技术）
与IMU（惯性测量单元）融合的方案，多种传感器融合加上一
个误差状态卡尔曼滤波器使得定位精度可以达到5~10cm，且具
备高可靠性和鲁棒性，市区允许最高时速超过60km/h。
2020/6/4
（威海） 汽车工程学院
（威海） 汽车工程学院
HARBIN INSTITUTE OF TECHNOLOGY
SCHOOL OF AUTOMOBILE ENGINEERING
规格严格 功夫到家
第5章 智能网联汽车导航定位系统
➢5.1 导航定位的定义与类型
➢5.2 全球定位系统
➢5.3 北斗卫星导航定位系统
➢5.4 惯性导航系统
2020/6/4
（威海） 汽车工程学院
HARBIN INSTITUTE OF TECHNOLOGY
SCHOOL OF AUTOMOBILE ENGINEERING
蔽GPS信号，而且GPS定位精度低，更新周期长，远远不能满
足自动驾驶的需求
2020/6/4
（威海） 汽车工程学院
HARBIN INSTITUTE OF TECHNOLOGY
SCHOOL OF AUTOMOBILE ENGINEERING
规格严格 功夫到家
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Assignments of Inertial Navigation 《惯性导航》作业1
Autumn 2018
Assignment 1: Aircraft Displaying on the Earth
Attached is a group of MA TLAB programs for displaying an aircraft near the surface of the Earth, as shown in figure 1.1 which is drawn by running the program main.m.
If you input the longitude, latitude, heading, pitching, and rolling angles (all in degree), and click the updating button, then the aircraft are expected to be displayed at a proper position on the Earth, and with proper attitude. The coordinates of each vertex of the aircraft 3D model occupies a row in the matrix vtx.
Before you click the updating button for the first time, the aircraft is hidden at the center of the Earth. Please rewrite the program redraw.m by adding the necessary rotation and shifting operations to the vertex coordinates of the aircraft. The acquired position and attitude information has been stored in the variables lon, lat, head, pitch and roll. With each updating, the coordinates of the aircraft in vtx in the MA TLAB drawing frame (same as the Earth frame) need to be re-calculated, instead of being kept the same as their values in the original 3D model (vtx0). That is, in the program redraw.m, you need to replace the command line vtx=vtx0 with your own codes for re-calculating vtx.
And please explain the rationale behind your rewriting.
作业报告
一．理论基础
a. 旋转变换矩阵
在坐标系中，可以通过旋转角度知道一个点在新旧坐标系的坐标 在这样的旋转变化下存在如下的坐标变换矩阵：
[x ′y ′z ′
]=[100
0cosα
sinα0−sinαcosα][x y z
] 同样的，绕y 轴和z 轴旋转的坐标变换矩阵为：
[x ′
y ′z ′
]=[cosβ0−sinβ0
10sinβ0cosβ][x
y z ] [x ′y ′z ′
]=[cosγsinγ0−sinγ
cosγ000
1][x y z ] 其中α、β、γ分别代表绕轴旋转的角度。
b. 平移变换矩阵
为了将飞机平移到给定经纬度的地球表面，需要在旋转之后的坐标上加上地球半径的分量
在球坐标系和直角坐标系中有如下的转换关系：
{x =Rsinθcosφy =Rsinθsinφz =Rcosθ
二．设计思路
程序的整体思路分为两部分，第一部分在初始位置根据经度、维度、航向角、俯仰角和滚转角计算出坐标旋转矩阵，第二部分根据经度纬度计算出坐标平移矩阵，将转换好的飞机平移到平面。
a.坐标旋转
1.由经纬度计算出地球坐标系到给定经纬度下的地理坐标系的方向余弦矩阵。
rotate_z = [cosd(lon+90),sind(lon+90),0;-sind(lon+90),cosd(lon+90),0; 0,0,1]; %地球坐标系到地理坐标系绕Z 轴旋转
rotate_x = [1,0,0;0,cosd((90-lat)),sind((90-lat));0,-sind((90-lat)),cosd((90-lat))]; %地球坐标系到地理坐标系绕X 轴旋转
rotate_y = [1,0,0;0,1,0;0,0,1]; %地球坐标系到地理坐标系绕Y 轴旋转
rotate=rotate_y*rotate_x*rotate_z; %地球坐标系到地理坐标系方向余弦矩阵
2.由航向角、俯仰角、偏航角计算出当前机体坐标系到旋转后机体坐标系的方向余弦矩阵 pose_z = [cosd(head),sind(head),0;-sind(head),cosd(head),0; 0,0,1]; %地理坐标系到机体坐标系绕Z 轴旋转
%地理坐标系到机体坐标系绕X轴旋转
pose_y = [cosd(roll),0,-sind(roll);0,1,0;sind(roll),0,cosd(roll)];
%地理坐标系到机体坐标系绕Y轴旋转
pose = pose_y*pose_x*pose_z;
%地理坐标系到机体坐标系方向余弦矩阵
3.将之前得到的方向余弦矩阵相乘得到最终的方向余弦矩阵。
由于在两次旋转之后的机体坐标系下，所有点的坐标应与初始机体坐标系下所有点的坐标相同，故可得
rotate =pose*rotate;
%最终的坐标转换矩阵
vtx1 = vtx0';
for i=1:5710
vtx2(:,i) =rotate\vtx1(:,i);
end
vtx3 = vtx2';
b.坐标平移
vtx(:,1) = vtx3(:,1)+re*cosd(lat)*cosd(lon);
vtx(:,2) = vtx3(:,2)+re*cosd(lat)*sind(lon);
vtx(:,3) = vtx3(:,3)+re*sind(lat);
三．结果展示
最终效果图显示如下。