# Incorrect dimensions for matrix multiplication

7 ビュー (過去 30 日間)
Gautami Golani 2019 年 3 月 7 日
コメント済み: Adarsh Ghimire 2019 年 3 月 9 日
%Extracting first 500 points
z = [y(1:500) u(1:500)];
%Plotting the data
figure(1)
idplot(z);
%Using sampling interval of 80ms
figure(1)
idplot(z, 1:500, 0.08);
%Reserving the remaining data
zr = [y(501:1000) u(501:1000)];
%Removing constant levels and making data zero mean
z = dtrend(z);
zr = dtrend(zr);
%Loss function calaculation
V = arxstruc(z, zr, struc(2,2, 1:10));
value = V(1,:);
nk_value = min(value);
V = arxstruc(z, zr, struc(1:5, 1:5, 3));
%Searching for settled-in point
nns = selstruc(V);
i=1;
%Model structure
for na = 1:5
for nb = 1:5
val(i,:) = na+nb;
th = arx(z,[na, nb, 3]);
mse(i,:) = [th.Report.Fit.MSE];
a = th.a;
b = th.b;
ysim = idsim(zr(:, 2), th);
i=i+1;
matx = [val mse];
A = matx;
[A1u,~,idx] = unique(A(:,1));
MinVals = accumarray(idx, A(:,2), [], @(x)min(x));
Result = [A1u, MinVals];
m_one = Result(:,1);
m_two = Result(:,2);
figure(2)
plot(m_one, m_two)
title('Complexity vs Loss Function');
xlabel('Complexity');
ylabel('Loss Function');
end
end
%A-parameter and B-parameter
th = arx(z,[4 4 3]);
th = sett(th, 0.08);
present(th);
A=th.a
B=th.b
%Least Square Estimation
i = 1;
for k=6:length(z)
phi(i,1:5) = [-z(k-1,1) -z(k-2,1) z(k-3,2) z(k-4,2) z(k-5,2)];
i = i+1;
end
Y = z(6:end,1);
L = inv(phi'*phi)*phi*Y;
Getting an error that says 'Incorrect dimensions for matrix multiplication.'
Can anyone explain why this error comes up?
##### 1 件のコメント表示非表示 なし
Walter Roberson 2019 年 3 月 7 日
This code requires idplot(), which existing in the MATLAB 3 (1987) to MATLAB 5 (1997) timeframe and which was removed from MATLAB before R13. A copy of the source appears to be at http://research.jisao.washington.edu/vimont_matlab/System/idplot.html

サインインしてコメントする。

### 回答 (1 件)

Adarsh Ghimire 2019 年 3 月 7 日

L = inv(phi ' * phi ) * phi' * Y;
your phi matrix dimension has to transposed to multiply with y. As i went through your code, I found that
dimensions:
phi = (z-6) x 6
and y = (z-6)x1
so transposing your phi and multiply with y will get you 6x1 matrix which can multiply with inverse result of 6x6 so you get 6x1 matrix as your L.
##### 6 件のコメント表示非表示 5 件の古いコメント
Adarsh Ghimire 2019 年 3 月 9 日
then you manage the dimension

サインインしてコメントする。

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!