Unable to perform assignment because the left and right sides have a different number of elements.
2 views (last 30 days)
Show older comments
I got "Unable to perform assignment because the left and right sides have a different number of elements." error and I don't understand how to fix it. Here is my code:
%% Parameters
rho = 0.95; % Correlation coefficient
N = 2048; % Sequence length
sigma2 = 1; % Noise variance
%% Generate AR-1 sequence
x = zeros(N,1);
x(1) = randn; % Initial value
for n = 2:N
x(n) = rho*x(n-1) + sqrt(sigma2)*randn; % AR-1 model
end
%% Compute theoretical prediction filters
R = autocorr(x,3); % Autocorrelation function up to lag 3
r = R(2:end); % Cross-correlation function up to lag 2
h = [rho; zeros(2,1)]; % Initialize filter coefficients
for k = 1:2
h(k+1) = -r(1:k)'*inv(toeplitz(R(1:k))); % Levinson-Durbin recursion
end
%% Apply theoretical prediction filter
xp = filter(h,1,x); % Predicted sequence
e = x - xp; % Prediction error
Pe_theory = e'*e/N; % Prediction error energy
%% Estimate autocorrelation function
R_est = xcorr(x,'biased'); % Autocorrelation function estimation
R_est = R_est(N:2*N-1); % Keep lags up to 3
r_est = R_est(2:end); % Cross-correlation function estimation up to lag 2
%% Compute optimal prediction filter from estimated autocorrelation function
h_est = [1; -r_est(1)/R_est(2); -r_est(2)/R_est(3)]; % Optimal filter coefficients
%% Apply estimated prediction filter
xp_est = filter(h_est,1,x); % Predicted sequence using estimated filter
e_est = x - xp_est; % Prediction error using estimated filter
Pe_est = e_est'*e_est/N; % Prediction error energy using estimated filter
%% Display results
fprintf('Theoretical optimal filter coefficients: h = [%f, %f, %f]\n', h)
fprintf('Theoretical prediction error energy: Pe = %f\n', Pe_theory)
fprintf('Estimated optimal filter coefficients: h_est = [%f, %f, %f]\n', h_est)
fprintf('Estimated prediction error energy: Pe_est = %f\n', Pe_est)
0 Comments
Answers (2)
Torsten
on 8 Apr 2023
Edited: Torsten
on 8 Apr 2023
Execute the code after changing
for k = 1:2
h(k+1) = -r(1:k)'*inv(toeplitz(R(1:k))); % Levinson-Durbin recursion
end
to
for k = 1:2
disp(size(-r(1:k)'*inv(toeplitz(R(1:k)))))
h(k+1) = -r(1:k)'*inv(toeplitz(R(1:k))); % Levinson-Durbin recursion
end
and you will see that for k=2, -r(1:k)'*inv(toeplitz(R(1:k))) has size 1x2 which cannot be saved in the array element h(k+1) which has size 1x1.
0 Comments
Walter Roberson
on 8 Apr 2023
h(k+1) = -r(1:k)'*inv(toeplitz(R(1:k))); % Levinson-Durbin recursion
When k becomes 2, toeplitz(R(1:k)) becomes 2 x 2, and inv() of that becomes 2 x 2.
When you do matrix multiplication of an M x 2 matrix and a 2 x 2 matrix, you are going to get out an M x 2 matrix. If M is 1 (which it is in this particular case) then you get out a 1 x 2 matrix. And there is no way to store that in a single location h(k+1)
This cannot be solved by using a different size of matrix on the left side of the * operator: no matter what you multiply by, the result is going to have at least two elements (well, unless it is 0 x 2, in which case the * would give a 0 x 2 result, which you cannot store in a scalar location anyhow.)
Therefore there is no way to solve this problem if inv(toeplitz(R(1:k))) is a correct term -- not unless you store the result into more than one output location.
0 Comments
See Also
Categories
Find more on Global or Multiple Starting Point Search in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!