Matrix Dimensions Must agree
4 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Tristen Hernandez
el 12 de Dic. de 2019
Comentada: Walter Roberson
el 13 de Dic. de 2019
Hi, currently I'm trying to create a function that with the given inputs will output various graphs of the system, however I'm running into problem with the matrix dimensions on line 49 not working (the S function). I was wondering if a fresh set of eyes could help me see where I'm going wrong with it. Also note this is a function that is to be used in another script with a plotting function. Any and all help very much appreciated.
function [t,S,i_arr,Upeak,Lpeak] =calculation(m,time,tsd,k,eta)
%%%%%%%%%% Write 4-5 lines (comments about what this function does and sends back)
%%%%%%%%%%
%%%%%%%%%%
%%%%%%%%%%
% This function needs to take the following as input values:
% (1) m (mass)
% (2) time
% (3) tsd
% (4) k (stiffness)
% (5) eta
tsd = 0.01;
m = 2.5;
k = 175; %stiffness
time = [0 4];
eta = [0.015 0.09 1]
eta2 = [0.02 1 0.08 2 1.5 0];
t=(0:tsd:4);
A = 5;
w0= sqrt(k/m) ; % Determine w0
% Build three empty arrays
i_arr=[]; % Saves all etas less than 1
Upeak=[]; % Saves all upper peaks
Lpeak=[]; % Saves all lower peaks
%%
for i=1:length(eta2) % Run index 'i' from 1 to the length of 'eta' array
count = 1; % Initialize both counters to 1
count1 = 1; %
if i < 1 %check if the current iteration (i) of eta is less than 1
% Set first row and the 2*i-1 column for both Upeak and Lpeak to 0
Upeak(1,2*i-1)=0;
Lpeak(1,2*i-1)=0;
Upeak(1,2*i)=A;
Lpeak(1,2*i)=-A;
% Add this iteration of eta to i_arr
i_arr=[i_arr i];
end
wd = w0.*(sqrt(1-eta.^2)) ;% Fill in for wd
%%
for j=1:length(t)
S(i,j) = exp((-i).*w0.*t).*(A.*cos(wd.*j)) ;
if i > 1 && j > 3
right = S(i,j)==exp(-i*w0.*t)*A*cos(wd.*j);
middle = S(i,j)==exp(-i*w0.*t)*A*cos(wd.*(j-1));
left =S(i,j)==exp(-i*w0.*t)*A*cos(wd.*(j-2));
%%
if middle > left && middle > right% Middle value is greater than both left AND right
Upeak(count+1,2*i-1)= j; %the current value of time
Upeak(count+1,2*i)=S(i,j-1) ; %S(<current i>, <current j - 1>)
count=count + 1 ; %add 1 to count
end
% Check if this is lower peak value
if middle < left && middle < right % Middle value is less than both left AND right
Lpeak(count+1,2*i-1)=j ; %the current value of time
Lpeak(count+1,2*i)=S(i,j-1) ; %S(<current i>, <current j - 1>)
count1= count1 + 1; %add 1 to count1
end
end
end
end
%%
if i == length(eta) % Check if the length of eta is 3 (for part (a))
% Output a table, with proper headings, of the "Upeak" array
fprintf(' Upeak Data Points \n')
fprintf('%3.3f \n', Upeak(i,j));
else
% Output a table, with proper headings, of the "Lpeak" array
fprintf(' Lpeak Data Points \n')
fprintf('%3.3f \n', Lpeak(i,j));
end
end
0 comentarios
Respuesta aceptada
Walter Roberson
el 12 de Dic. de 2019
t=(0:tsd:4);
t is a vector
for j=1:length(t)
taking length(t) again implies t is a vector.
S(i,j) = exp((-i).*w0.*t).*(A.*cos(wd.*j)) ;
t is a vector and you use all of t, so the right hand side is a vector.
Considering that you loop j to length(t) you should probably be using t(j)
0 comentarios
Más respuestas (1)
Tristen Hernandez
el 12 de Dic. de 2019
1 comentario
Walter Roberson
el 13 de Dic. de 2019
eta = [0.015 0.09 1]
eta is a vector
wd = w0.*(sqrt(1-eta.^2)) ;% Fill in for wd
so wd is a vector the same size as eta. (Note: it is being calculated inside a loop even though w0 and eta are not changed inside the loop.)
S(i,j) = exp((-eta(i)).*w0.*t(j)).*(A.*cos(wd.*t(j))) ;
wd is a vector so the right hand side is a vector.
Since wd is the same size as eta, perhaps you should be indexing wd by the same thing you index eta by.
Ver también
Categorías
Más información sobre Matrix Indexing en Help Center y File Exchange.
Productos
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!