Matrix Dimensions Must agree
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Tristen Hernandez
on 12 Dec 2019
Commented: Walter Roberson
on 13 Dec 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 Comments
Accepted Answer
Walter Roberson
on 12 Dec 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)
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More Answers (1)
Tristen Hernandez
on 12 Dec 2019
1 Comment
Walter Roberson
on 13 Dec 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.
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