惯性导航作业
导航系统大作业
导航系统1．简述捷联惯性系统中地理系到机体系的姿态阵bg C 其含义及其功能。
答:含义:导航坐标系g g g O x y z -到机体坐标系b b b O x y z -的一组欧拉角为,,θγψ，导航坐标系经过3次转动到机体坐标系。
g g g x y z 依次沿g O z -、'b O x -、''b O y -旋转角度-ψ、θ、γ后到b b b x y z 。
姿态矩阵中包含了机体的姿态角方位角ψ、俯仰角θ和横滚角γ。
功能：机体陀螺仪输出的角速度信息经过补偿后,积分得到机体坐标系与导航坐标系的姿态信息和姿态转移矩阵。
捷联惯导系统中,加速度计与载体固连,利用姿态阵完成加速度计输出信息从机体坐标到导航坐标的转换。
转换后的加速度计信息经过积分可得到机体在导航坐标系下的速度和位置。
2．画出并用式表达速度三角形（地速、控速、风速)及航迹角、航向角与偏流角之间的关系.答：风速：空气相对于地面的运动速度;空速:飞机相对于空气运动的速度；地速：飞机相对于地面的运动速度。
=+v v v 风地空航向角：机头在水平面投影与真北方向的夹角ϕ;偏流角:空速矢量和地速矢量之间的夹角,用δ表示；航迹角:飞机速度矢量在水平面投影与真北方向的夹角。
航向角ϕ加上偏流角δ等于地速v 地的方位角α。
v 地v 空v 风3．简述惯性导航系统、卫星导航系统、多普勒导航、塔康、VOR/DME 、天文导航其各自的基本工作原理、特点及误差特性。
答:一、惯性导航系统（1）工作原理以牛顿力学定律为基础，以陀螺仪和加速度计为敏感器件进行导航参数解算。
系统根据陀螺仪的输出建立导航坐标系，根据加速度计输出解算出运载体的速度和位置，从而实现姿态和航向解算. （2）特点惯性导航系统不需要任何外来信息，也不会向外辐射任何信息，仅依靠惯性器件就能全天候，全球性的自主三维定位和三维定向,同时具备自主性、隐蔽性和信息的完备性。
（3）误差特性误差随时间积累，短时间导航精度较高。
哈工大惯性技术(导航原理)大作业讲解
Assignment of Inertial Technology 《惯性技术》作业Autumn 2015Assignment 1: 2-DOF response simulation A 2-DOF gyro is subjected to a sinusoidal torque with amplitude of 4 g.cmand 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 H s M s M s J J s H s J J s H α=-++ 12222()()()()x y x x y x y J H s 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:02222000200222200()sin sin ()()()cos cos ()()ox a ox a 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 ()tα. Fig1.3 is the output of inner ring ()tβ. 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 ()tβFig 1.3 The output of outer ring ()tαI believe the sawtooth wave is caused by the nutation. For the frequency of the nutation isobtained as100002500/3974eHrad s HzJω===≈, which is far higher than the frequencyof the applied sinusoidal torque , namelyaωω.Fig 1.4 Trajectory of 2-DOF gyro’s response to sinusoidal input The 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 isoxMHω, whereas the minor axis is in the direction of the torsion spring effects, with amplitude oxaMHω.The nutation components are much smaller than that of the forced vibration, which can be eliminated to get the clear static response.2200222()sin sin()cos(1cos)()ox oxa ax aox ox oxa aa a a aM Mt t tJ HM M Mt t tH H Hαωωωωωωβωωωωωωω≈≈-≈-=--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 information and 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 the INS.The block diagram in the courseware might be of some help. However, there lurks an inconspicuous 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 ay A K y ga=∆++-Firstly, the input signal is accelerometer of the platform, and the velocity of the platform is the integration of the acceleration./py y ydty Rω=+=-⎰The acceleration along Yp may contains two parts:cos gsin gypf y yααα=-≈-When accelerometer errors are concerned, the output of accelerometer will be:(1)N a yp Na K f A=++When gyro errors concerned:'(1)p g pKωωε=++Only αis unknown:[(1)]tg pttptK dtdtααωεϕϕω=+++-∆∆=⎰⎰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/y m s ∆=Fig2.9 position bias output when only initial position error exists 010y m ∆=Fig2.10 position bias output bias when only gyro bias errorexists 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/y m 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 the work you have done or understood.。
实验三惯性导航实验
实验三惯性导航实验小组成员：杨曦陈魁吴航杨少帅一、 实验目的1、了解惯性导航设备;2、掌握惯性导航设备的物理连接；3、掌握惯性导航信息的处理方法；4、掌握惯性导航方法并学会用编程实现惯性导航算法。
二、 实验器材YH-5000AHRS ；工业控制计算机；数据采集软件； 稳压电源；串口连接线；三、 实验原理（1） 姿态解算基于四元数法解算姿态矩阵。
p j p i p l Q +++=21 （1）⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--++----+++---+=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡b b b p b b b b p p p z y x C z y x p p p l lp p p lp p p lp p p p p p l lp p p lp p p lp p p p p p l z y x 222123213223113223212223212313212322212)(2)(2)(2)(2)(2)(2 （2） b pbQw Q 21= （3） 上述微分方程表示成矩阵形式：⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡------=⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡321321000021p p p l w w w w w w w w w w w w p p p l b pbxb pbyb pbz b pbx b pbz b pbyb pby b pbz b pbxb pbz b pbyb pbx（4） 初始四元数的确定计算如下：⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡--++=⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡2c o s 2c o s 2s i n 2s i n 2s i n 2c o s 2c o s 2s i n 2s i n 2s i n 2c o s 2c o s 2s i n 2c o s 2s i n 2c o s 2s i n 2c o s 2s i n 2s i n 2s i n 2c o s 2c o s 2co s )0(3)0(2)0()0(0000000000000000000000001γθϕγθϕγθϕγθϕγθϕγθϕγθϕγθϕG G G G G G G G p p p l （5） 用四阶龙格库塔法解（4）的微分方程；⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=333231232221131211T T T T T T T T T C p b 由p b C 中提取γϕλ，，G231sin T -=主λ 22211tan T T G -=主ϕ)(tan 33131T T --=主γ 从而可得到：主λλ=⎪⎩⎪⎨⎧<>+>><+=0,020,002122212222T T T T T GG G G πϕϕπϕϕ主主主⎩⎨⎧<>=0,)(-0331333T T sign T πγγγ （2） 速率位置解算将加速度测量的沿坐标系轴向的比力bib a 转换成沿着导航坐标系轴向的比力p ib a ,则速度方程为:p p epp ep p ib p ep g V w a V +⨯+Ω-=)2( 展开得到：⎪⎩⎪⎨⎧-+Ω-+Ω+=+Ω+Ω-=+Ω-Ω+=gV w V w a V V w V a V V w V a V p epy p epx p x p epx p epy p y p ibz pepz p epzp epx p x p epx p z p iby p epy p epzp epy p y p epy p z p ibx p epx )2()2()2(2)2(2 由于Ω,pep w 都很小，故而速度方程简化为：⎪⎩⎪⎨⎧-===ga V a V a V p ibz pepz piby p epy p ibxp epx用一阶欧拉法解，则：⎪⎩⎪⎨⎧+-=++=++=+)(*)()()(*)()(*)(t V T g a T t V t V T a T t V t V T a T t V p epz p ibz p epzpepy p iby p epy pepx p ibx p epx 其中T 为采样时间。
导航原理大作业
导航原理作业（惯性导航部分）一、题目内容一枚导弹采用捷联惯性导航系统，三个速率陀螺仪Gx, Gy, Gz和三个加速度计Ax, Ay, Az的敏感轴分别沿着着弹体坐标系的Xb, Yb, Zb轴。
初始时刻该导弹处在北纬45.75度，东经126.63度。
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）。
初始时刻弹体坐标系和地理坐标系重合，如图所示，弹体的Xb轴指东，Yb轴指北，Zb轴指天。
此后弹体坐标系Xb-Yb-Zb相对地理坐标系的转动如下：首先，弹体绕Zb（方位轴）转过-10 度；接着，弹体绕Xb（俯仰轴）转过15 度；然后，弹体绕Yb（滚动轴）转过20 度；最后弹体相对地面停止旋转。
请分别用方向余弦矩阵和四元数两种方法计算：弹体经过三次旋转并停止之后，弹体上三个加速度计Ax, Ay, Az的输出。
取重力加速度的大小g = 9.8m/s2。
第二种情形：导弹正在飞行中。
初始时刻弹体坐标系仍和地理坐标系重合；且导弹初始高度200m，初始北向速度1800 m/s，初始东向速度和垂直速度都为零。
陀螺仪和加速度计的输出都为脉冲数形式，陀螺输出的每个脉冲代表0.00001弧度的角增量。
加速度计输出的每个脉冲代表1μg，1g = 9.8m/s2。
陀螺仪和加速度计输出的采样频率都为10Hz，在200秒内三个陀螺仪和三个加速度计的输出存在了数据文件gaout.mat中，内含一矩阵变量ga，有2000行，6列。
每一行中的数据代表每个采样时刻三个陀螺Gx, Gy, Gz将地球视为理想的球体，半径6371.00公里，且不考虑仪表误差，也不考虑弹体高度对重力加速度的影响。
选取弹体的姿态计算周期为0.1秒，速度和位置的计算周期为1秒。
（1） 请计算200秒后弹体到达的经纬度和高度，东向和北向速度； （2） 请计算200秒后弹体相对当地地理坐标系的姿态四元数；（3） 请绘制出200秒内导弹的经、纬度变化曲线（以经度为横轴，纬度为纵轴）； （4） 请绘制出200秒内导弹的高度变化曲线（以时间为横轴，高度为纵轴）。
11惯性导航与组合导航作业(无水印)
惯导作业一、填空题1.惯性导航系统是一种不依赖任何外部信息、也不向外部辐射能量的______导航系统。
答案：自主式2.不依赖外界信息，只靠对载体本身的______、来完成导航任务的技术称做惯性导航，也称为自主式导航答案：惯性测量3. 加速度计其输出一般是______、，但在积分加速度计的情况下则输出为______、。
答案：速度、加速度4. 惯性器件就是测量载体______、和______、参数的传感器。
答案：线运动、角运动5. 加速度经过一次积分可以得到______，经过二次积分得到______。
答案：运动速度、运动距离6. 描述角运动的参数有______、______。
答案：姿态角、姿态角速度7. 描述线运动的参数有______、______、______。
答案：位移、速度、加速度8. 高速旋转的自由陀螺仪，当不受外力矩作用时，其主轴将保持它在空间的______方向不变。
答案：初始9. 由表观运动所引起的陀螺______偏离当地地垂线的误差，称之为陀螺仪的“表观误差”。
答案：自转轴二、单选题1．陀螺自转轴方向相对惯性空间保持不变，以地球作为参考基础，陀螺自转轴相对地球表面的转动，为（）。
A．表观运动B．自转运动C．定轴运动D．进动运动答案：A三、多选题（每题1分）1．惯性导航系统的核心有（）A．加速度计、B．陀螺仪C．导航计算机D．GPS答案：ABC2．惯性导航系统的基本组成（）A．加速度计B．模拟某一坐标系的惯性平台C．导航计算机D．控制显示器答案：ABCD3．激光陀螺特点有哪些（）。
A．抗干扰能力弱B．启动快C．动态特性较宽D．稳定性好答案：BCD4.关于组合导航系统，下列说法正确的是（）。
A．提高导航系统的精度B．提高导航系统的可靠性C．提高导航系统的安全性D．启动快答案：ABC四、判断题1. 一个沿直线运动的载体，只要借助于加速度计测出它的加速度，那么，载体在任何时刻的速度和相对出发点的距离就可以实时地计算出来。
北航惯导第一次大作业
《惯性导航原理》第一次大作业一、 原理分析惯导系统为指北方位的平台系统，则利用比力方程以及陀螺提供的东、北、天三个比力数据，即可计算得到在每个数据采集点的平台即时速度，再通过经纬度的计算公式，就可以得到每个数据采集点平台的即时经纬度，以每个数据采集点为下一个采集点的起点，即可对速度和经纬度进行累计计算，从而得到平台在运动过程中任意时刻的速度和位置情况。
运动过程中任意时刻的速度和位置情况。
1.模型公式的推导载体相对地球运动时，载体相对地球运动时，加速度计测得的比力表达式，加速度计测得的比力表达式，加速度计测得的比力表达式，称为比力方程，称为比力方程，称为比力方程，方程如方程如下：下：g V V f epep ieep-´++=)2(vv （1）在指北方案中，平台模拟地理坐标系，将上式中平台坐标系用地理坐标系代入得：入得：t tt ett iettgV f V+´+-=)2(v v （2）系统中测量的是比力分量，将上式写成分量形式系统中测量的是比力分量，将上式写成分量形式=-+ （3） 又因为地球的自转角速率为：又因为地球的自转角速率为：(4)地理坐标系相对于地球坐标系的角速率为：地理坐标系相对于地球坐标系的角速率为：= (5)将（将（44）（5）两个式子带入（）两个式子带入（33）式，即可得到如下方程组：）式，即可得到如下方程组：（6）2.速度计算作业要求只考虑水平通道，作业要求只考虑水平通道，因此只需要计算正东、因此只需要计算正东、因此只需要计算正东、正北两个方向的速度即可。
正北两个方向的速度即可。
正北两个方向的速度即可。
理理论上计算得到t x V 、t y V 后，再积分一次可得到速度值，即后，再积分一次可得到速度值，即ïîïíì+=+=òòt t y t y t ytt x t x tx V dt V V V dt V V 000但在本次计算过程中，三个方向的速度均是从零开始在各时间节点上的累加，并不是t的函数，因此速度计算可以由以下方程组实现：（7）此方程组表示了从第i 个采集点到第（个采集点到第（i+1i+1i+1）个采集点的速度递推公式。
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.。
惯性导航课程实验报告问问
一.实验目的1．认识三轴惯性平台的各个组成器件2.讨论验证三轴平台的工作原理，并对其稳定回路及工作过程做出分析二．实验原理一个双自由度陀螺有两个测量轴，可为平台提供两个轴的稳定基准，而三轴平台要求陀螺为平台提供三个轴的稳定基准，所以三轴平台需要两个双自由度陀螺。
设两个陀螺的外环轴均平行于平台的方位轴安装，则内环轴自然平行于平台平面。
在正常工作状态下，两个陀螺的自转轴也平行于平台台面，且相互之间保持垂直关系，即两个陀螺的内环轴之间也保持垂直关系。
两个陀螺的内环轴作为平台绕两个水平轴稳定的基准，而两个陀螺的外环轴之一，作为平台绕方位轴稳定的基准。
三.实验内容1.方位稳定轴的空间积分状态在双自由度陀螺构建的三轴惯性平台中，平台的方位稳定回路陀螺2外环轴上的信号器，放大器，平台方位轴上的稳定电机等组成。
当干扰力矩作用在平台的方位轴上时，平台绕方位轴转动偏离原有的方位，而平台上的陀螺却具有稳定性。
这样，平台相对陀螺外环出现了偏转角，陀螺2外环轴上的信号器必有信号输出，经放大器放大后送至平台方位轴上的稳定电机，方位稳定电机输出稳定力矩作用到平台方位轴上，从而平衡作用在平台方位轴上的干扰力矩，使平台绕方位轴保持稳定。
同样，给陀螺2内环轴上的力矩器输入与指令角速度大小成比例的电流，可实现方位稳定轴的空间积分要求2.水平稳定回路的工作如下图所示由三个单轴平台直接叠加的三轴平台在航向变化时，平台上的陀螺与稳定电机之间的相对位置关系．图(a)表示航向为零，即方位环环对俯仰环没有转角时陀螺与稳定电机之间的相对位置关系，此时的陀螺Ⅱ感受沿横滚轴（纵向）方向作用到平台上的干扰力矩，信号器输出的信号经横滚放大器A．放大后给横滚轴稳定电机，产生纵向稳定力矩，使平台沿纵向（x.轴）保持稳定，陀螺I感受沿俯仰轴（横向）方向作用到平台上的干扰力矩。
经信号器．放大器和俯仰轴稳定电机，产生沿横向的稳定力矩．使平台沿横向保持稳定。
同样，若给两个陀螺的力矩器输入与指令角速度成比例的电流，平台也可正常工作在空间积分状态。
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惯性导航作业一、数据说明：1：惯导系统为指北方位的捷连系统。
初始经度为116.344695283度、纬度为39.975172度,高度h为30米。
初速度v0=[-9.993908270;0.000000000;0.348994967]。
2：jlfw中为600秒的数据，陀螺仪和加速度计采样周期分别为为1/100秒和1/100秒。
3：初始姿态角为[2 1 90]（俯仰，横滚，航向,单位为度）,jlfw.mat中保存的为比力信息f_INSc(单位m/s^2)、陀螺仪角速率信息wib_INSc(单位rad/s)，排列顺序为一~三行分别为X、Y、Z向信息.4: 航向角以逆时针为正。
5:地球椭球长半径re=6378245;地球自转角速度wie=7.292115147e-5;重力加速度g=g0*(1+gk1*c33^2)*(1-2*h/re)/sqrt(1-gk2*c33^2)；g0=9.7803267714;gk1=0.00193185138639;gk2=0.00669437999013;c33=sin(lat纬度);二、作业要求：1：可使用MATLAB语言编程，用MATLAB编程时可使用如下形式的语句读取数据：load D:\...文件路径...\jlfw,便可得到比力信息和陀螺仪角速率信息。
用角增量法。
2：(1) 以系统经度为横轴，纬度为纵轴（单位均要转换为：度）做出系统位置曲线图；(2) 做出系统东向速度和北向速度随时间变化曲线图（速度单位：m/s，时间单位：s）；(3) 分别做出系统姿态角随时间变化曲线图（俯仰，横滚，航向,单位转换为：度，时间单位：s）；以上结果均要附在作业报告中。
3：在作业报告中要写出“程序流程图、现阶段学习小结”，写明联系方式。
(注意程序流程图不是课本上的惯导解算流程，而是你程序分为哪几个模块、是怎样一步步执行的,什么位置循环等,让别人根据该流程图能够编出相应程序) (学习小结按条写，不用写套话) 4：作业以纸质报告形式提交，附源程序。
三、基本原理和公式1、初始姿态矩阵的确定：根据初始姿态角求四元数：0123cos cos cos sin sin sin 222222cossin cos sin cos sin 222222cos cos sin sin sin cos 222222cossin sin sin cos cos 222222abab abab ab ab abab q q q q ψψθγθγψψθγθγψψθγθγψψθγθγ=-=-=+=+再根据四元数求方向余弦矩阵的初始矩阵：()()()()()()222201231203130222221203012323012222130223010123222222b t q q q q q q q q q q q q C q q q q q q q q q q q q q q q q q q q q q q q q ⎡⎤+--+-⎢⎥=--+-+⎢⎥⎢⎥+---+⎣⎦2、指北方位系统的运动解算：“平台”指令角速度为：()()cos sin tan()ty yt tt x it ie xt tx ie xt V R V L R V L L R ωωω⎡⎤-⎢⎥⎢⎥⎢⎥⎢⎥=+⎢⎥⎢⎥⎢⎥+⎢⎥⎣⎦加速度计获得的比力信息bib f 为载体坐标系中各个轴向的比力，而我们需要的比力t it f 为地理坐标系中各个轴向的比力，它们之间应用矩阵t b C 做变换：()()1t t bit b ibT t b b bttf C f C CC -=⨯==根据比力信息可以求出各个方向上的加速度：()()()()(2sin tan())(2cos )(2sin tan())(2cos )t t t t tt x x x ibxie y ie zxt xt t t y tt tt x y ibyie x zxt ytt ty t tt t x z ibz ie x y xt ytV V V fL L V L V R R V V V fL L V V R R V V V f L V V gR R ωωωω•••=++-+=-++=+++-因此可以求得速度为：t tt t xx x t t t t y y y V V dt V V V dt V ••=+=+⎰⎰载体所在位置的地理纬度L 、经度λ可由下列方程求得：000)sec(λλ+=+=⎰⎰dt L R V L dt R V L txtxttytyt3、四元数姿态矩阵的更新：b b b t tb ib t it C ωωω=-式中，bib ω为陀螺所测得的角速度。
用毕卡逼近法更新b t C 的值，T 为采样时间bib T θω∆=⨯[]0000x y z xz y y z x z yxθθθθθθθθθθθθθ-∆-∆∆⎡⎤⎢⎥∆∆-∆⎢⎥∆=⎢⎥∆-∆∆⎢⎥∆∆-∆⎢⎥⎣⎦ 22220x y z θθθθ∆=∆+∆+∆()()()()[]()2420001118384248q n I q n θθθθ⎧⎫⎛⎫⎛⎫∆∆∆⎪⎪+=-++-∆ ⎪ ⎪⎨⎬ ⎪ ⎪⎪⎪⎝⎭⎝⎭⎩⎭4、姿态角的求解：姿态角与姿态矩阵的关系：cos cos sin sin sin cos sin sin sin cos sin cos cos sin cos cos sin sin cos cos sin sin sin sin cos sin cos cos cos ab abab abb t ababab abab abC γψγθψγψγθψγθθψθψθγψγθψγψγθψγθ-+-⎡⎤⎢⎥=-⎢⎥⎢⎥+--⎣⎦式中θ,γ,ab ψ分别为俯仰角，滚转角和偏航角，以逆时针为正方向，而课本上是以顺时针为正，因此需要对课本上的公式进行修改，将ab ψ-代入原公式可得现公式。
如果记111213212223313233b t T T T C T T T T T T ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦则由以上两式即可解算出姿态角：()1231133312122sin tan tan T T T T T θγϕ---=⎛⎫=- ⎪⎝⎭⎛⎫=⎪⎝⎭四、程序流程图五、结果六、小结这次作业是捷联惯导的解算，是利用上次作业的结果对数据进行处理。
和上次不同，这次遇到了较多的问题。
首先，对捷联惯导的基本原理理解的不够深刻，比如坐标系的转换，四元数微分方程的求解。
其次，由于课本的姿态角是以顺时针为正，而原始数据是以逆时针为正，因此需要对书上的公式进行修改，在这个过程中就出现了许多问题，比如正负号问题。
总之，这次作业弥补了学习上的不足，使我对基本原理理解更为深刻，也初步了解惯导的基本操作。
七、程序clccleara=load('C:\Users\Administrator\Documents\MATLAB/jlfw.dat');Wib_INSc=a(:,2:4)';f_INSc=a(:,5:7)';%第一列：数据包序号第二至四列：分别为东、北、天向陀螺仪角速率信息wib_INSc（单位：rad/s）%第五至七列：分别为东、北、天向比力信息f_INSc（单位：m/s^2）.L(1,:)=zeros(1,60001);Lambda(1,:)=zeros(1,60001);Vx(1,:)=zeros(1,60001);Vy(1,:)=zeros(1,60001);Vz(1,:)=zeros(1,60001);Rx(1,:)=zeros(1,60001);%定义存放卯酉圈曲率半径数据的矩阵Ry(1,:)=zeros(1,60001);%定义存放子午圈曲率半径数据的矩阵psi(1,:)=zeros(1,60001);%定义存放偏航角数据的矩阵theta(1,:)=zeros(1,60001);%定义存放俯仰角数据的矩阵gamma(1,:)=zeros(1,60001);%定义存放滚转角数据的矩阵I=eye(4); %定义四阶矩阵L(1,1)=39.975172/180*pi;%纬度初始值单位：弧度Lambda(1,1)=116.344695283/180*pi;%经度初始值单位：弧度Vx(1,1)=-9.993908270;%初始速度x方向分量Vy(1,1)=0;%初始速度y方向分量Vz(1,1)=0.348994967;%初始速度z方向分量Wibx(1,:)=a(:,2); %提取陀螺正东方向角速率并定义Wiby(1,:)=a(:,3); %提取陀螺正北方向角速率并定义Wibz(1,:)=a(:,4); %提取陀螺天向角速率并定义fibbx(1,:)=a(:,5);%x方向的比力数据fibby(1,:)=a(:,6);%y方向的比力数据fibbz(1,:)=a(:,7);%z方向的比力数据g0=9.7803267714;Wie=7.292115147e-5;%地球自转角速度Re=6378245;%长半径e=1/298.3;%椭圆度t=0.01;%采样时间psi(1,1)=90/180*pi;%偏航角初始值单位：弧度theta(1,1)=2/180*pi;%俯仰角初始值单位：弧度gamma(1,1)=1/180*pi;%滚转角初始值单位：弧度gk1=0.00193185138639;gk2=0.00669437999013;H=30;%高度%求解四元数系数q0,q1,q2,q3的初值q(1,1)=cos(psi(1,1)/2)*cos(theta(1,1)/2)*cos(gamma(1,1)/2)-sin(psi(1,1)/2)*sin(theta( 1,1)/2)*sin(gamma(1,1)/2); %q0q(2,1)=cos(psi(1,1)/2)*sin(theta(1,1)/2)*cos(gamma(1,1)/2)-sin(psi(1,1)/2)*cos(theta( 1,1)/2)*sin(gamma(1,1)/2); %q1q(3,1)=cos(psi(1,1)/2)*cos(theta(1,1)/2)*sin(gamma(1,1)/2)+sin(psi(1,1)/2)*sin(theta( 1,1)/2)*cos(gamma(1,1)/2); %q2q(4,1)=cos(psi(1,1)/2)*sin(theta(1,1)/2)*sin(gamma(1,1)/2)+sin(psi(1,1)/2)*cos(theta( 1,1)/2)*cos(gamma(1,1)/2); %q3for i=1:60000g=g0*(1+gk1*sin(L(i)^2)*(1-2*H/Re)/sqrt(1-gk2*sin(L(i)^2)));%计算重力加速度Rx(i)=Re/(1-e*(sin(L(i)))^2);%根据纬度计算卯酉圈曲率半径Ry(i)=Re/(1+2*e-3*e*(sin(L(i)))^2);%根据纬度计算子午圈曲率半径%求解四元数姿态矩阵q0=q(1,i);q1=q(2,i);q2=q(3,i);q3=q(4,i);Ctb=[q0^2+q1^2-q2^2-q3^2,2*(q1*q2+q0*q3),2*(q1*q3-q0*q2);2*(q1*q2-q0*q3),q2^2-q3^2+q0^2-q1^2,2*(q2*q3+q0*q1);2*(q1*q3+q0*q2),2*(q2*q3-q0*q1),q3^2-q2^2-q1^2+q0^2;];Cbt=Ctb';fibt=Cbt*[fibbx(i);fibby(i);fibbz(i)];%比力数据fibtx(i)=fibt(1,1);fibty(i)=fibt(2,1);fibtz(i)=fibt(3,1);Vx(1,i+1)=(fibtx(i)+(2*Wie*sin(L(i))+Vx(i)*tan(L(i))/Rx(i))*Vy(i)-(2*Wie*cos(L(i)) +Vx(i)/Rx(i))*Vz(i))*t+Vx(i);%计算速度x方向分量Vy(1,i+1)=(fibty(i)-(2*Wie*sin(L(i))+Vx(i)*tan(L(i))/Rx(i))*Vx(i)+Vy(i)*Vz(i)/Ry(i) )*t+Vy(i);%计算速度y方向分量Vz(1,i+1)=(fibtz(i)+(2*Wie*cos(L(i)+Vx(i))/Rx(i))*Vx(i)+Vy(i)*Vy(i)/Ry(i)-g)*t+Vz (i);%计算速度z方向分量Witt=[-Vy(i)/Ry(i);Wie*cos(L(i))+Vx(i)/Rx(i);Wie*sin(L(i))+Vx(i)*tan(L(i))/Rx(i)];%求出平台指令角速度值Wibb=[Wibx(i);Wiby(i);Wibz(i)];Wtbb=Wibb-Ctb*Witt;%将指令角速度转换到平台坐标系,并求解WtbbL(1,i+1)=t*Vy(i)/Ry(i)+L(i);Lambda(1,i+1)=t*Vx(i)/(Rx(i)*cos(L(i)))+ Lambda(i);x=Wtbb(1,1)*t;y=Wtbb(2,1)*t;z=Wtbb(3,1)*t; %求取迭代矩阵中的各ΔthetaA=[0 -x -y -z;x 0 z -y;y -z 0 x;z y -x 0];%求取迭代矩阵[Δtheta]T=x^2+y^2+z^2; % 计算[Δtheta]^2的q(:,i+1)=((1-T/8+T^2/384)*I+(1/2-T/48)*A)*q(:,i);%求q0theta(i+1)=asin(Ctb(2,3));if(Ctb(2,2)>=0)psi(i+1)=atan(-Ctb(2,1)/Ctb(2,2));elseif(Ctb(2,1)>0)psi(i+1)=atan(-Ctb(2,1)/Ctb(2,2))+pi;elsepsi(i+1)=atan(-Ctb(2,1)/Ctb(2,2))-pi;endif(Ctb(3,3)>0)gamma(i+1)=atan(-Ctb(1,3)/Ctb(3,3));elseif(Ctb(1,3)<0)gamma(i+1)=atan(-Ctb(1,3)/Ctb(3,3))+pi;elsegamma(i+1)=atan(-Ctb(1,3)/Ctb(3,3))-pi;endendfigure(1)plot(L*180/pi,Lambda*180/pi);title('经纬度位置曲线');xlabel('经度/°');ylabel('纬度/°');grid ont=0:0.01:600;figure(2)plot(t,Vx);title('东西方向速度');xlabel('时间/s');ylabel('速度/m/s');grid on figure(3)plot(t,Vy);title('南北方向速度');xlabel('时间/s');ylabel('速度/m/s');grid on figure(4)plot(t,theta*180/pi);title('俯仰角');xlabel('时间/s');ylabel('度');grid on figure(5)plot(t,gamma*180/pi);title('滚转角');xlabel('时间/s');ylabel('度');grid on figure(6)plot(t,psi*180/pi);title('偏航角');xlabel('时间/s');ylabel('度');grid on。