need help with " Error using vertcat Dimensions of matrices being concatenated are not consistent"
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Can anybody help me, cant understand why do i get this error when i run it?
i have attached my code.
thanks
5 件のコメント
ABDO
2024 年 4 月 26 日
編集済み: Voss
2024 年 4 月 26 日
h= 0.3; % Rayon de lissage
dx=0.3;
N =floor(1/dx);
x =0:dx:(pi/sqrt(3)); % Positions des particules
dt = 1/4;
t = 0:dt:1;
M = length(t) - 1;
% Initialisation de la solution approchée et de la matrice A
A = zeros(N, N);
b=zeros(N,1);
uapp=zeros(N,1);
%sol et cond exacte
u_exact = @(x) cos(sqrt(3)*x);
% Construction de la matrice A et du vecteur b
for i = 1:N
for j =N
r_ij = abs(x(i) - x(j));
if r_ij <= h
d2W_dx2 = second_derivative_monaghan_kernel(r_ij, h);
W_value = monaghan_kernel(r_ij, h);
A(i, j) = (u_exact(x(j))- u_exact(x(i))) * d2W_dx2;
b(i) = -3 * u_exact(x(j)) * W_value ;
end
end
end
u_app(1)=1;
u_app(N)=u_exact(pi/sqrt(3));
uapp = b\A;
uapp = [0; uapp];
% Affichage des résultats
plot(x , uapp, 'r');
hold on;
plot(x, u_exact(x), 'b--');
xlabel('x');
ylabel('u');
legend('Solution approchée (SPH)', 'Solution exacte');
title('Solution approchée et exacte par SPH');
grid on;
% Fonctions du noyau de Monaghan
function W_value =monaghan_kernel(r_ij, h)
C= (1/ h);
if r_ij < h
W_value =C*((2 / 3) - (r_ij / h)^2 +(1/2 *(r_ij / h)^3));
elseif r_ij<2*h
W_value =C*((1/6)*(2-(r_ij / h))^3);
else
W_value=0;
end
end
function d2W_dx2 = second_derivative_monaghan_kernel(r_ij, h)
C = (1 / h); % Corrected the denominator
if r_ij < h
d2W_dx2 = -2 * C * (1 - (3 * (r_ij / h)));
elseif r_ij < 2*h
d2W_dx2 = 2 * C * (2 - (r_ij / h)); % Removed extra characters
else
d2W_dx2 = 0;
end
end
Ihave a problem simultaning a numerical solution
I need some
Voss
2024 年 4 月 26 日
@ABDO: Post a new question. Use this link: https://www.mathworks.com/matlabcentral/answers/questions/new/
回答 (5 件)
Thorsten
2015 年 5 月 18 日
When you vertically connect matrices using ; the matrices must have the same number of columns.
A = ones(10, 3);
B = zeros(20, 4);
C = [A; B]
You get
Error using vertcat
CAT arguments dimensions are not consistent.
Instead you can write
B = zeros(20, 3);
C = [A; B]; % works, A and B have same number of rows
5 件のコメント
Jan
2015 年 5 月 19 日
編集済み: Jan
2015 年 5 月 19 日
I guess that this fails:
t = 0:0.1:10;
...
Qin_doot = [4; 2*t; 10];
the sizes of the scalars 4 and 10 do not match with the size of t.
The description as text "ive get a 3X1 vector by multiplying ..." is less useful than posting the line of code you are talking of.
3 件のコメント
John Amakali
2018 年 5 月 9 日
編集済み: Walter Roberson
2018 年 5 月 9 日
hi guys , i have the same problem... i have attached my code. someone help please.
Warning: Unsuccessful read: A timeout occurred before the Terminator was reached..
Error using vertcat
Dimensions of matrices being concatenated are not consistent.
Error in GUI>pushbutton1_Callback (line 144)
set(M,'Zdata',[Tdata; zeroIndex],'Cdata',[Tdata; Tdata]);
Error in gui_mainfcn (line 95)
feval(varargin{:});
Walter Roberson
2018 年 5 月 9 日
We do not know what size(Tdata) is or size(zeroIndex) .
The message about timeout suggests that you might be doing a serial read or read from arduino or raspberry, and that it is not returning as much data as you expect. If your Tdata is the actual received data and zeroIndex is built under the assumption that all of the requested data was received, then they could easily end up with incompatible sizes.
A N
2018 年 2 月 10 日
編集済み: Stephen23
2018 年 2 月 10 日
I've got another problem with using vertcat Dimensions of matrices being concatenated are not consistent.
I'm trying to write a code to make a transmission characteristic of the Gaussian filter from this
My code is:
lambdac=0.8; %mm
dx=0.001; %mm
alpha=sqrt(log(2)/pi);
disp(alpha);
x=(-lambdac:dx:lambdac);
S=(1/alpha*lambdac).*exp(-pi*(x/alpha*lambdac).^2);
S=S/sum(S);
figure(1)
plot(x,S)
xlabel('Odległość [mm]')
ylabel('Wzmocnienie')
xlim([-1 1])
%Przykład 5.2
m=size(S,1); %długość filtru gaussowskiego
l=8; %długość punktów pomiarowych na profilu
n=1/dx; %liczba punktów pomiarowych na profilu
S=[zeros(n/2-floor(m/2),1); S; zeros(n/ 2-floor(m/2)-1,1)]; %copied from %previous link
Sf=fft(S); %transformata Fouriera S
j=(2:1:floor (n/2)+1); %generuje falę o długości dla danych X
wave=n*dx./(j-1);
figure (2)
semilogx (wave, 100*abs(Sf(2:floor (n/2)+1,1)));
xlabel('Długość fali [mm]')
ylabel('Amplituda [%]')
hold off
I've tried also with
S=[zeros(n/2-floor(m/2)-1,1); S; zeros(n/ 2-floor(m/2)+1,1)];
but no result. After n=1/dx I've found:
a=zeros(n/2-floor(m/2),1);
size(a);
disp(size(a));
S;
size(S);
disp(size(S));
c=zeros(n/2-floor(m/2)+1,1);
size(c);
disp(size(c));
and result:
500 1
1 1601
501 1
Any idea how to solve it?
1 件のコメント
Stephen23
2018 年 2 月 10 日
編集済み: Stephen23
2018 年 2 月 10 日
So S has 1601 columns and you are trying to vertically concatenate it with two arrays, each of which has just one column. You need to ensure that the number of columns or rows is the same, e.g.:
R = n/2-floor(m/2);
S = [zeros(1,R-1), S, zeros(1,R+1)];
A N
2018 年 2 月 10 日
編集済み: Walter Roberson
2018 年 2 月 10 日
Great! Works!
After it I would like to do a FFT of S, but
My code:
Sf=fft(S); %transformata Fouriera
j=(2:1:floor(n/2)+1); %faa o danej długości
size(j);
disp(size(j));
wave=n*dx./(j-1);
size(wave);
disp(size(wave));
figure(2)
semilogx(wave,100*abs(Sf(2:floor(n/2)+1,1)))
xlabel('Długość fali [mm]')
ylabel('Amplituda [%]')
Command Windows shows as a result of those displays:
1 500
1 500
Index exceeds matrix dimensions.
Error in proba (line 61)
semilogx(wave,100*abs(Sf(2:floor(n/2)+1,1)));
Error in proba (line 65)
semilogx(wave,100*abs (2:floor(n/2)));"
Do you know how to solve it? Maybe my error is easy but I've been doing only spectral FFTs and would like to know more about examples like this (no Fs and spectral). In Help I found only spectral FFT. Line with semilog I copied from this book, and really hard to find out - is this wrong?
Edit:
Writing:
semilogx(wave,100*abs(Sf(2:floor(n/2)+1)));
impacts of this plot:
7 件のコメント
MANYINGA MARC
2023 年 3 月 24 日
Bonjour la communauté. J'ai un code de calcul, Dan lequel je voudrais faire varier les éléments d'une matrice À, en fonction d'une variable temps t. Par exemple
t=0:10:60 A=[1 2*t 3 3*t 5 8 5 6 3] Affiche error using vertcat Besoin d'aide
1 件のコメント
Walter Roberson
2023 年 3 月 24 日
A=[1 2*t 3 3*t 5 8 5 6 3] would be an error if t is not a row vector.
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