Acceleration to displacement data based on accelerometer

Question

MARLIANA HAFIZA 2023-4-12

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链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1945624-acceleration-to-displacement-data-based-on-accelerometer

编辑： Chunru 2023-4-12

Book5.csv

Hi,

Can anyone help to guide me on how to find the displacement value based on the acceleration result from accelerometer ? I already try to use several method that have been found but it seems not provide the good result and not understand much the given method. The attach excel file is acceleration output value from a wheel that have rotation of 400 rpm and my previous code that I used.

Thanks in advance !

num=csvread("Book5.csv");                                                                            
Fn = Fs/2;                                                     
Tr = linspace(num(1,1), num(1,end), size(num,1));             
Dr = resample(num(:,2), Tr);                               
Dr_mc  = Dr - mean(Dr,1);                                       
FDr_mc = fft(Dr_mc, [], 1);                                     
Fv = linspace(0, 1, fix(size(FDr_mc,1)/2)+1)*Fn;                
Iv = 1:numel(Fv);                                              
figure
plot(Fv, abs(FDr_mc(Iv,:))*2)
grid
xlabel('Frequency (Hz)')
ylabel('Amplitude')

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Answer 1

Chunru 2023-4-12

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1945624-acceleration-to-displacement-data-based-on-accelerometer#answer_1214044

编辑：Chunru 2023-4-12

在 MATLAB Online 中打开

num=readtable("https://www.mathworks.com/matlabcentral/answers/uploaded_files/1352749/Book5.csv");

head(num);

Var1 Var2 Var3 Var4 ______ _______ _______ _______ 0 -1.7658 -0.2943 -0.5886 0.0238 -1.8639 -0.2943 -0.4905 0.0476 0 -0.1962 -0.4905 0.0714 -1.2753 -0.1962 -0.4905 0.0952 -1.3734 -0.1962 -0.4905 0.119 -1.2753 -0.1962 -0.4905 0.1429 -1.3734 -0.1962 -0.4905 0.1667 -1.3734 -0.1962 -0.4905

% initialize the velocity and position

v0 = [0 0 0];

s0 = [0 0 0]; % 3d

vt = v0 + cumtrapz(num{:, 1}, num{:, 2:end});

st = s0 + cumtrapz(num{:, 1}, vt);

whos

Name Size Bytes Class Attributes cmdout 1x33 66 char num 612x4 21215 table s0 1x3 24 double st 612x3 14688 double v0 1x3 24 double vt 612x3 14688 double

plot3(st(:,1), st(:,2), st(:,3), 'r.')

box on; grid on;

xlabel("x"); ylabel("y"); zlabel("z")

figure

plot(num{:, 1}, num{:, 2:end})

figure

plot(num{:, 1}, st);

figure;

plot(num{:, 1}, vt);

mean(num{:, 2:end}, 'omitnan')

ans = 1×3

-1.2512 -0.2125 -0.5284

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MARLIANA HAFIZA 2023-4-12

Thank you for your help. But, I quite confuse regarding the displacement value. That means based on the output given, the displacement keeps decreasing downwards even the acceleration result got upside down waveform ?

Chunru 2023-4-12

编辑：Chunru 2023-4-12

See the update above. The general trend is determined by the mean value of the acceleration data, which are all negative as shown above. You can see the velocity is negative though fluctuating (due to the acceleration up and down)

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