为什么会出现vertcat错误？ （Matlab）

Question

I am trying to plot the first row of my matrix against time t but I cannot figure out why my matrix yields the error: "vertcat: Dimensions of matrices being concatenated are not consistent."

t = linspace(0,100);

y_mat = (1./t).*([1, t+(1/2)*exp(-3*t)-(1/2)*exp(-t); 
(3/2)*(exp(-t)-exp(-3*t)), 1-(3/2)*exp(-3*t)+ 
(1/2)*exp(-t)] * [(t-4)/3;1]);

plot(t,y_mat(1,:))

Answer 1

You are thinking in the term of symbolic notation but implementing in matrix notation. When you do t = linspace(0,100); it creates a 1x100 matrix (array). So when later on it is used in the definition of y_mat , each expression used in the definition evaluates to 1x100 matrix. So your y_mat definition is tying to do this : [1x100] * [1 1x100 ; 1x100 1x100] * [1x100 ; 1] [1x100] * [1 1x100 ; 1x100 1x100] * [1x100 ; 1] [1x100] * [1 1x100 ; 1x100 1x100] * [1x100 ; 1] which obviously fails.

You have two options: Do all computations in the matrix notation by first computing the matrix multiplication separately and restructuring the matrices to represent the actual multiplication (ensure the 1 s are appropriately replicated).

OR

use Matlabs's symbolic variables and expressions probably like this :

syms t  % creating symbolic variable
% creating symbolic expressions
f0 = 1/t  
f1 = t+(1/2)*exp(-3*t)-(1/2)*exp(-t);
f2 = (3/2)*(exp(-t)-exp(-3*t));
f3 = 1-(3/2)*exp(-3*t)+(1/2)*exp(-t);
f4 = (t-4)/3;
% defining y_mat
y_mat = f0 * [1 f1; f2 f3] * [f4 ; 1]

% putting value in symbolic variable
t = linspace(eps,100); % eps to avoid division by 0 error

% substitute values and evaluate y_mat
y_mat_vals = eval(subs(y_mat));

This gives y_mat_vals a 2x100 matrix, as the answer.

Answer 2

YOu have messed up your code..you need to be careful when typing such functions. To make it simple, I have used a loop.

t = linspace(0,100);

nt = length(t) ;
y_mat = zeros(2,nt) ;

for i = 1:nt
y_mat(:,i) = (1/t(i))*([1           t(i)+(1/2)*exp(-3*t(i))-(1/2)*exp(-t(i));
    (3/2)*(exp(-t(i))-exp(-3*t(i)))   1-(3/2)*exp(-3*t(i))+(1/2)*exp(-t(i))])*[(t(i)-4)/3;1];
end
plot(t,y_mat)

Answer 3

You can also write it out more explicitly. The equation reads:

[ 1,pt2 ; pt3,pt4 ] * [ pt5 ; 1 ] = [ pt5 + pt2 ; pt3.*pt5 + pt4 ]

Since each of those terms is a scalar, you can compute them for all t at the same time using element-wise multiplication:

t = linspace(0,100);

pt2 = t+(1/2)*exp(-3*t)-(1/2)*exp(-t);
pt3 = (3/2)*(exp(-t)-exp(-3*t));
pt4 = 1-(3/2)*exp(-3*t)+(1/2)*exp(-t);

pt5 = (t-4)/3;

y_mat = (1./t) .* [ pt5 + pt2 ; pt3.*pt5 + pt4 ];

plot(t,y_mat)

This might be a bit more verbose, but I don't think it's any less readable than other solutions. And it is much more efficient: 0.0571 ms, versus 483.3 ms ( syms solution ) and 0.681 ms ( loop solution) , for a t with 500 elements.

(Note that multiplying by 1./t uses implicit singleton expansion. This works in MATLAB R2016b and newer. For older versions of MATLAB, use bsxfun .)