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时间:2020-10-28 09:52 来源: Essay代写

摘要:卫星研究report范文 The Research of Satellite 这份report是关于研究卫星的一些简单功能,介绍卫星主要从哪些方面可以进行数据的精确传递和输送。另外还说明了卫星的定位功能,同时也涉及到了牛......

卫星研究report范文 The Research of Satellite
这份report是关于研究卫星的一些简单功能,介绍卫星主要从哪些方面可以进行数据的精确传递和输送。另外还说明了卫星的定位功能,同时也涉及到了牛顿定律等重要定理的运用。
Mathematical Approach-  数学方法
Satellite 卫星
  Beginning with the approach towards our satellite, conversion programs were written in order to convert angles from degrees, minutes, and seconds to radians, and vice versa. Next, we used the formula derived from exercise #4 from equations; (19), (18), (17), (14), and (13) in the project homework assignment. This gives us a formula from converting position and general time t as given in [ tV  Ψd  Ψm  Ψs  NS  λd  λm  λs  EW  h (10)] into Cartesian coordinates. Since the main purpose for each of our programs is to take in standard input, and out standard output, we used the information gained in exercise 6 to convert the data from Cartesian coordinates into form (10). In order to complete this task we used and added some additional parameters to (20). 
  We were then compelled to find a condition set to assure us the satellites used for our data were above the horizon. Given in Cartesian coordinates of the following equation (31), we figure x being a point on earth and s, in space, this tells us whether s viewed from x is in fact above the horizon.
从接近我们的卫星开始,我们编写了转换程序,以便将角度、分、秒转换为弧度,反之亦然。接下来,我们使用练习#4从方程式中导出的公式;(19)、(18)、(17)、(14)、(13)。这给了我们一个公式从转换位置和时间t在[电视ΨdΨmΨm d NSλλλ年代电子战h(10)]到笛卡尔坐标系。由于我们每个程序的主要目的是接收标准输入,输出标准输出,因此我们使用练习6中获得的信息将数据从笛卡尔坐标转换为形式(10)。为了完成这个任务,我们使用并向(20)添加了一些额外的参数。
然后,我们被迫找到一个条件,以确保我们用于数据的卫星在地平线之上。在笛卡尔坐标系下,给出方程(31),我们认为x是地球上的一点,而s,在空间中,这告诉我们,从x上看,s是否在视界之上。
t0 = tV - ǁxs(tV)-xVǁ2/c
Our final part in solving the satellite, is to compute our ts and xs, being time and position of the satellite respectively. We continued following the iteration from exercise 9 and solving t0 in the form (40). 
我们求解卫星的最后一部分,是计算我们的ts和xs,分别是卫星的时间和位置。我们继续遵循练习9中的迭代,并以(40)的形式求解t0。
Receiver
  As mentioned in the project assignment, we have 4 unknowns with coordinates of location and time, which leaves us a range within c(tS – tV), where we do not know tV. For our mathematical approach in solving our system of 4 nonlinear equations, we chose to follow exercise 14, and solving thru Newton’s method, the nonlinear system obtained by the Least Squares approach. Due to the possibility of failure in solving this non-linear system, we chose our initial points as such: for the receiver, denote geo0 and geo1 be last two geographic coordinates of vehicle's solution. Let d_geo be the difference of geo1 and geo0 and geo = d_geo + geo1.  Then init_t = t_s[1]+.08 and init_x = geography2cartesian(init_t, geo). We used .08s for the addition to t_s[1], because it is experimentally the average time that the signal takes to go to the vehicle. Also, let it be known that the init_x is approximated by assuming the velocity of the vehicle, constant over a short period.
接收机
正如在项目作业中提到的,我们有4个未知数,位置和时间的坐标,这给我们留下了一个c(tS - tV)的范围,我们不知道电视。对于我们求解4个非线性方程组的数学方法,我们选择了练习14,并通过牛顿法求解,通过最小二乘法得到的非线性方程组。由于求解该非线性系统存在失败的可能性,我们的初始点如下:对于接收器,表示geo0, geo1为车辆解的最后两个地理坐标。设d_geo为geo1和geo0的差值,geo = d_geo + geo1。然后init_t = t_s[1]+。和init_x = geography2cartesian(init_t, geo)我们用0。08来表示t_s[1]的加法,因为从实验上看,它是信号到达飞行器的平均时间。同样,让我们知道,init_x是通过假设飞行器的速度来近似的,在短时间内恒定。
Directions/How to use- 方向/如何使用
    Our programming language consisted of using C++.
  1. Compile: g++ -o satellite satellite.cpp g++ -o receiver receiver.cpp
  2. ./satellite , ./ receiver
Satellite takes geographic data from standard input [ tV  Ψd  Ψm  Ψs  NS  λd  λm  λs  EW  h (10)] and generates data from satellites above the horizon in the form [iS tS xS] (26). The receiver takes data in form (26), and prints to standard output in (10). Both the satellite and receiver use data.dat parameters. These programs can be used by giving a command like:
cat bm.dat | java vehicle | ./satellite | ./receiver
Lesson Learned-
  We noticed in derivatives that there could have been problems in diving the Ni = ǁxS - xǁ2, since Ni is large, we avoided dividing by zero. From this, we learned that we should divide by larger numbers, but if in another project Ni was small, we would know to modify the problem or equation to divide by a reasonably large number to avoid having to divide by zero.
Other worth mentioning-
  The project required that the accuracy of the GPS be within .01m. But to make it safe, the receiver was programmed to have .0001m as the condition of convergence. Also, the accuracy of time is c times of length accuracy, thus making both items at a higher degree of accuracy.
Additionally, we have also tried to solve satellites by using Newton’s method (form 38) and solve receiver by taking first 4 satellites and applying Newton’s method (exercise 13). It turns out that least square method for receiver and Newton’s method for satellite requires more actual initial point.
我们的编程语言由使用c++组成。
1. 编译:g++ -o卫星。cpp接收机。cpp
. ./卫星,. ./接收器
卫星将地理数据从标准输入(电视ΨdΨmΨm d NSλλλ年代电子战h(10)]和生成数据从卫星在地平线形式是tS xS(26)。接收器以(26)形式接收数据,并以(10)形式打印到标准输出。卫星和接收器都使用数据。dat参数。这些程序可以使用的命令,如:
cat bm.dat | java车辆| ./卫星| ./接收机
教训,
我们注意到在衍生品可能是问题在潜水倪=ǁxS - xǁ2,因为倪很大,我们避免了除以零。从这里,我们知道我们应该除以更大的数,但如果在另一个项目中,Ni很小,我们就会知道修改问题或方程,除以一个相当大的数,以避免必须除以零。
其他值得一提的
该项目要求GPS的精度在0.01米以内。但为了安全起见,接收机的收敛条件设置为。0001m。同时,时间的精度是长度精度的c倍,使得两个项目的精度都更高。
此外,我们还尝试用牛顿法(form 38)求解卫星,用前4颗卫星用牛顿法求解接收器(练习13)。结果表明,接收机的最小二乘法和卫星的牛顿法需要更实际的初始点。

report写作范文卫星研究http://www.0592w.com/Reportfanwen/fanwen548.html

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