精选问答
会用数学软件进行数值分析与拟合的进!解决问题后保证有加分使用压电加速度传感器在实际测量中发现,测试数据常常伴有或大或小的零点漂移问题,以致无法判断数据的实际值,更不用说对数据进行积分求速度和位移了。所谓零点漂移,就是由于受到各种噪声的影响,在所测量的数据中,加入了一个缓慢变化的,但逐渐远离零点的随机误差。为了简化计算,本实验假设加入了一个均值为零,方差为 ( 是常数)的符合正态分布的随机量 在某次高冲击过载加速度试验中,我们选择等间隔的时间 ,对某个试件进行冲击试验,通过压电加速度传感器,获得了256个测

2019-05-23

会用数学软件进行数值分析与拟合的进!解决问题后保证有加分使用压电加速度传感器在实际测量中发现,测试数据常常伴有或大或小的零点漂移问题,以致无法判断数据的实际值,更不用说对数据进行积分求速度和位移了。所谓零点漂移,就是由于受到各种噪声的影响,在所测量的数据中,加入了一个缓慢变化的,但逐渐远离零点的随机误差。为了简化计算,本实验假设加入了一个均值为零,方差为 ( 是常数)的符合正态分布的随机量 在某次高冲击过载加速度试验中,我们选择等间隔的时间 ,对某个试件进行冲击试验,通过压电加速度传感器,获得了256个测量数据,以下是这些数据点在间隔 下的图形。(受阻尼的正弦振动 ) 其中横轴表示时间,每一个相邻的点之间代表一个时间间隔,纵轴表示各时间点所测量的加速度值,在时间间隔 下,这些测量值依次为: {0.266,0.379,0.336,0.106,0.689,0.118,0.587,0.715,0.509,0.814,1.902,1.495,1.389,1.377,1.263,1.4,1.62,1.902,2.05,1.943,1.502,1.886,1.454,2.366,2.304,2.665,2.792,2.406,2.239,2.991,2.601,2.884,2.719,3.587,3.724,3.254,3.187,3.623,3.642,2.92,4.124,3.595,3.203,4.223,4.249,4.683,3.942,3.863,3.982,3.148,4.116,4.454,3.853,4.139,4.835,3.934,4.99,3.698,3.971,3.777,4.71,4.286,3.946,4.545,4.153,4.461,3.94,3.168,3.966,3.996,4.247,3.482,4.44,3.443,3.051,3.518,3.421,2.93,3.134,2.777,1.94,1.992,2.265,2.543,1.851,2.089,1.455,1.43,1.677,0.874,0.951,0.516,0.43,1.058,1.042,0.231,-0.163,0.224,-0.043,-1.06,0.015,-0.747,-1.77,-1.696,-0.995,-1.998,-1.14,-2.463,-1.963,-2.152,-2.075,-1.65,-1.692,-1.872,-2.715,-2.87,-2.626,-2.385,-2.819,-1.603,-2.425,-2.863,-1.378,-2.204,-1.629,-2.291,-1.658,-1.267,-1.454,-1.129,-0.903,-0.497,-0.52,-0.756,-0.434,0.144,-0.017,0.321,0.246,0.687,0.594,0.826,0.667,0.764,1.452,1.553,1.965,1.569,1.705,2.16,2.351,1.997,1.66,1.579,1.829,1.605,1.911,1.261,1.939,1.199,0.21,1.25,0.424,0.646,0.515,0.15,-0.131,0.328,-0.439,-1.086,-1.006,-1.221,-0.668,-1.277,-1.58,-0.935,-1.588,-2.43,-1.087,-1.776,-1.343,-0.981,-2.258,-1.614,-1.592,-0.993,-0.808,-0.608,-0.633,-0.67,-0.514,0.178,0.179,-0.251,0.241,0.865,1.049,0.573,0.839,0.932,1.809,1.565,1.238,1.834,1.08,0.883,2.225,1.411,1.019,1.057,0.28,-0.023,-0.173,0.035,-0.154,-0.663,-0.054,-0.435,-0.531,-0.497,-1.165,-0.999,-0.533,-1.38,-1.737,0.026,-0.882,-1.392,-0.757,-0.857,-1.212,-0.247,-0.284,0.592,0.26,-0.338,-0.602,-0.004,0.065,1.068,0.906,0.975,1.142,0.836,1.854,0.812,1.247,1.082,0.662,0.862,0.887,0.966,1.431,0.482,0.475,-0.538} } 关于本次数学实验试题的内容要求 试题的题目是船舶试件高冲击过载加速度试验的数据处理 (1)从图中可以看出,以上所给的数据 含有随机误差 ,假设此误差 符合正态分布 ,并且进一步假设实际值 是 的光滑函数,试寻找一种算法,从数据 中去除随机误差 ,以获得实际值 。 (2)寻找一种方法,对去除随机误差 后的实际值 进行积分,以得到试件高冲击后,速度的变化情况,进而对速度量进行积分,得到位移的变化情况。 (3)你能否对上面(1)和(2)中的算法,给出其理论分析,或者提供理论上的依据。
优质解答
x=0:1:255;y=[0.266,0.379,0.336,0.106,0.689,0.118,0.587,0.715,0.509,0.814,1.902,1.495,1.389,1.377,1.263,1.4,1.62,1.902,2.05,1.943,1.502,1.886,1.454,2.366,2.304,2.665,2.792,2.406,2.239,2.991,2.601,2.884,2.719,3.587,3.724,3.254,3.187,3.623,3.642,2.92,4.124,3.595,3.203,4.223,4.249,4.683,3.942,3.863,3.982,3.148,4.116,4.454,3.853,4.139,4.835,3.934,4.99,3.698,3.971,3.777,4.71,4.286,3.946,4.545,4.153,4.461,3.94,3.168,3.966,3.996,4.247,3.482,4.44,3.443,3.051,3.518,3.421,2.93,3.134,2.777,1.94,1.992,2.265,2.543,1.851,2.089,1.455,1.43,1.677,0.874,0.951,0.516,0.43,1.058,1.042,0.231,-0.163,0.224,-0.043,-1.06,0.015,-0.747,-1.77,-1.696,-0.995,-1.998,-1.14,-2.463,-1.963,-2.152,-2.075,-1.65,-1.692,-1.872,-2.715,-2.87,-2.626,-2.385,-2.819,-1.603,-2.425,-2.863,-1.378,-2.204,-1.629,-2.291,-1.658,-1.267,-1.454,-1.129,-0.903,-0.497,-0.52,-0.756,-0.434,0.144,-0.017,0.321,0.246,0.687,0.594,0.826,0.667,0.764,1.452,1.553,1.965,1.569,1.705,2.16,2.351,1.997,1.66,1.579,1.829,1.605,1.911,1.261,1.939,1.199,0.21,1.25,0.424,0.646,0.515,0.15,-0.131,0.328,-0.439,-1.086,-1.006,-1.221,-0.668,-1.277,-1.58,-0.935,-1.588,-2.43,-1.087,-1.776,-1.343,-0.981,-2.258,-1.614,-1.592,-0.993,-0.808,-0.608,-0.633,-0.67,-0.514,0.178,0.179,-0.251,0.241,0.865,1.049,0.573,0.839,0.932,1.809,1.565,1.238,1.834,1.08,0.883,2.225,1.411,1.019,1.057,0.28,-0.023,-0.173,0.035,-0.154,-0.663,-0.054,-0.435,-0.531,-0.497,-1.165,-0.999,-0.533,-1.38,-1.737,0.026,-0.882,-1.392,-0.757,-0.857,-1.212,-0.247,-0.284,0.592,0.26,-0.338,-0.602,-0.004,0.065,1.068,0.906,0.975,1.142,0.836,1.854,0.812,1.247,1.082,0.662,0.862,0.887,0.966,1.431,0.482,0.475,-0.538]; X=fft(y); threshold=max(abs(X))/10; Y=X.*(abs(X)> threshold); yy=ifft(Y); plot(x,y,x,yy,'r') %YY就是速度 %yy=delta(YY)/delta(XX) %delta(YY)=yy*delta(XX) %YY2-YY1=yy(1)*(XX2-XX1) %YY2=YY1+yy(1)*(XX2-XX1)=YY1+yy(1) YY(1)=yy(1); for k=1:length(yy)-1 YY(k+1)=YY(k)+yy(k) end figure plotyy(x,yy,x,YY) test=diff(YY)-yy(1:end-1) %diff(YY)求导数,得到加速度,与yy相减,差值为1.0e-013*,说明方法正确。 x=0:1:255;y=[0.266,0.379,0.336,0.106,0.689,0.118,0.587,0.715,0.509,0.814,1.902,1.495,1.389,1.377,1.263,1.4,1.62,1.902,2.05,1.943,1.502,1.886,1.454,2.366,2.304,2.665,2.792,2.406,2.239,2.991,2.601,2.884,2.719,3.587,3.724,3.254,3.187,3.623,3.642,2.92,4.124,3.595,3.203,4.223,4.249,4.683,3.942,3.863,3.982,3.148,4.116,4.454,3.853,4.139,4.835,3.934,4.99,3.698,3.971,3.777,4.71,4.286,3.946,4.545,4.153,4.461,3.94,3.168,3.966,3.996,4.247,3.482,4.44,3.443,3.051,3.518,3.421,2.93,3.134,2.777,1.94,1.992,2.265,2.543,1.851,2.089,1.455,1.43,1.677,0.874,0.951,0.516,0.43,1.058,1.042,0.231,-0.163,0.224,-0.043,-1.06,0.015,-0.747,-1.77,-1.696,-0.995,-1.998,-1.14,-2.463,-1.963,-2.152,-2.075,-1.65,-1.692,-1.872,-2.715,-2.87,-2.626,-2.385,-2.819,-1.603,-2.425,-2.863,-1.378,-2.204,-1.629,-2.291,-1.658,-1.267,-1.454,-1.129,-0.903,-0.497,-0.52,-0.756,-0.434,0.144,-0.017,0.321,0.246,0.687,0.594,0.826,0.667,0.764,1.452,1.553,1.965,1.569,1.705,2.16,2.351,1.997,1.66,1.579,1.829,1.605,1.911,1.261,1.939,1.199,0.21,1.25,0.424,0.646,0.515,0.15,-0.131,0.328,-0.439,-1.086,-1.006,-1.221,-0.668,-1.277,-1.58,-0.935,-1.588,-2.43,-1.087,-1.776,-1.343,-0.981,-2.258,-1.614,-1.592,-0.993,-0.808,-0.608,-0.633,-0.67,-0.514,0.178,0.179,-0.251,0.241,0.865,1.049,0.573,0.839,0.932,1.809,1.565,1.238,1.834,1.08,0.883,2.225,1.411,1.019,1.057,0.28,-0.023,-0.173,0.035,-0.154,-0.663,-0.054,-0.435,-0.531,-0.497,-1.165,-0.999,-0.533,-1.38,-1.737,0.026,-0.882,-1.392,-0.757,-0.857,-1.212,-0.247,-0.284,0.592,0.26,-0.338,-0.602,-0.004,0.065,1.068,0.906,0.975,1.142,0.836,1.854,0.812,1.247,1.082,0.662,0.862,0.887,0.966,1.431,0.482,0.475,-0.538]; X=fft(y); threshold=max(abs(X))/10; Y=X.*(abs(X)> threshold); yy=ifft(Y); plot(x,y,x,yy,'r') %YY就是速度 %yy=delta(YY)/delta(XX) %delta(YY)=yy*delta(XX) %YY2-YY1=yy(1)*(XX2-XX1) %YY2=YY1+yy(1)*(XX2-XX1)=YY1+yy(1) YY(1)=yy(1); for k=1:length(yy)-1 YY(k+1)=YY(k)+yy(k) end figure plotyy(x,yy,x,YY) test=diff(YY)-yy(1:end-1) %diff(YY)求导数,得到加速度,与yy相减,差值为1.0e-013*,说明方法正确。
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