Dimensions of arrays being concatenated are not consistent.

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Frane 2021-7-5

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https://ww2.mathworks.cn/matlabcentral/answers/872133-dimensions-of-arrays-being-concatenated-are-not-consistent

评论： Stephan 2021-7-6

epas_steering_system.m

Hello there,

When I run my programm it says the error message written in the tittle. It says that for the 1st row of matrix A (line 79). Any suggestions?

Thanks.

P.S. I've uploaded the file in the message.

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Answer 1

Stephan 2021-7-5

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https://ww2.mathworks.cn/matlabcentral/answers/872133-dimensions-of-arrays-being-concatenated-are-not-consistent#answer_740218

在 MATLAB Online 中打开

clc;
clear all;
close all;
%% Parametri
    %% Inercija; {kg.m2}
    Juc = 2.6e-2;
    Jlc = 1e-3;
    
    %% Krutost; {N.m/rad}  
    kuc = 114.59; 
    klc = 687.55;
    kr = 400000;
    
    %% Prigusenje; {N.m.s/rad}
    duc = 6e-2; 
    dlc = 6;
    dr = 12000;
    
    %% Radijus; {m}
    rpin = 7.78e-3;
    
    %% Sila; {N}
    Fr = 1;
    Flim = 75;
    Flim_column = 0.2;
    Flim_torsion = 0.01;
    
    %% Masa; {kg}
    mr = 10;
    
    %% Moment; {N.m}
    MsAmp = 4.5;
    
    %% Kutna brzina; {rad/s}
    freq = 2;
    
    %%
    Kt = {0.3 , 'N.m/a'}; 
    J = 0.01;
    b = 0.1;
    L = 0.5;
    R = 1;
    K = 0.01;
    kESF = 8e6;
    kESF_column = 7000;
    kESF_torsion = 7000;
    
%% Ulazni podatci
time = 0.01: 0.01: 0.2;
Ms = MsAmp*sin(2*pi*freq*time);
dm = 0;
Jm = 0;
%% Algoritam prema kojem se racuna I iz Ms   
if Ms >=3.5
    I = 21.29*Ms-69.4;
elseif Ms < 3.5 & Ms >= 2
    I = 2.73 * Ms - 4.47;
elseif Ms < 2 & Ms >= 0
    I = 0.5 * Ms;
elseif Ms < 0 & Ms >= (-2)
    I = 0.5 * Ms;
elseif Ms < (-2) & Ms >= (-3.5)
    I = 2.73 * Ms + 4.47;
elseif Ms < (-3.5) 
    I = 21.29 * Ms + 69.4;
end
%% Matrice A, B, C i D
A = [0 1 0 0 0 0 0 0 0;
    -kuc/Juc -duc/Juc kuc/Juc duc/Juc 0 0 0 0 0;
    0 0 0 1 0 0 0 0 0;
    kuc/Jlc duc/Jlc -kuc/Jlc -duc/Jlc -klc/Jlc -dlc/Jlc klc/Jlc*rpin dlc/Jlc*rpin 0;
    0 0 0 0 0 1 0 0 0;
    0 0 0 0 0 0 0 1 0;
    0 0 0 0 -klc/mr*rpin -dlc/mr*rpin klc/mr*rpin dlc/mr*rpin 0;
    0 0 0 0 0 0 0 0 1;
    0 0 0 0 0 0 0 0 -dm/Jm];
B = [0 0 0;
    1 -K 0;
    0 0 0;
    0 K/Jlc 0;
    0 0 0;
    0 0 0;
    0 0 1;
    0 0 0;
    0 K/Jm 0];
C = B.';
D = 0;

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Frane 2021-7-5

编辑：Frane 2021-7-5

It works. Can you tell me what did you change so I know in the future?

Stephan 2021-7-6

在 MATLAB Online 中打开

A = [0 1 0 0 0 0 0 0 0;
    -kuc/Juc -duc/Juc kuc/Juc duc/Juc 0 0 0 0 0;
    0 0 0 1 0 0 0 0 0;
    kuc/Jlc duc/Jlc -kuc/Jlc -duc/Jlc -klc/Jlc - dlc/Jlc klc/Jlc*rpin dlc/Jlc*rpin 0;
    %                                           ^
    %                                           |
    %                                           ---- change this
    
A = [0 1 0 0 0 0 0 0 0;
    -kuc/Juc -duc/Juc kuc/Juc duc/Juc 0 0 0 0 0;
    0 0 0 1 0 0 0 0 0;
    kuc/Jlc duc/Jlc -kuc/Jlc -duc/Jlc -klc/Jlc -dlc/Jlc klc/Jlc*rpin dlc/Jlc*rpin 0;
    %                                           ^
    %                                           |
    %                                           ---- to this. Otherwise MATLB
    %                                                interprets:
    %                                                -klc/Jlc - dlc/Jlc 
    %                                                as one expression, which leads
    %                                                to having one column
    %                                                missing. Better: use
    %                                                comma to seperate
    %                                                entries in matrices
    %                                                and vectors