Why this code the given error?

Question

Sadiq Akbar 2021-9-8

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1449444-why-this-code-the-given-error

评论： Sadiq Akbar 2021-9-13

close all; clear all; clc;
vec = @(MAT) MAT(:);
vecH = @(MAT) MAT(:).';
% M = 5;
M = 10;
N = 4;
% N = 10;
d = 0.5;
K = 3;
SNR_dB = 20;
steerVecT = @(ang) exp(1j*2*pi*d*[0:M-1].'*sin(vecH(ang))); 
steerVecR = @(ang) exp(1j*2*pi*d*[0:N-1].'*sin(vecH(ang))); 
angleGrid = [-80:1e-2:80].';
steerGrid = steerMatTR(deg2rad(angleGrid), steerVecT, steerVecR);
rng(222);
targetAngle = [-30, 0, 40].'+rand(K, 1);
A = ones(K, 1);
steerM = steerMatTR(deg2rad(targetAngle), steerVecT, steerVecR);
r = steerM*A;
noiseVar = r' * r / length(r) / db2pow(SNR_dB);
w =  sqrt(noiseVar / 2) * (randn(size(r)) + 1j * randn(size(r)));
y = r+w;
lambda = 2*sqrt(M*N*log(M*N))*sqrt(noiseVar);
cvx_begin sdp quiet
    variable x(N*M,1) complex
    variable F(M*N,M*N) hermitian
    minimize(norm(y-x))
    subject to
        [F, x; x', lambda^2]>=0
        for idxN1 = 0 : 1 : N-1
            for idxN2 = 0 : 1 : N-1
                Ftmp = F(idxN1*M+1:idxN1*M+M, idxN2*M+1:idxN2*M+M);
                for delta = -(M-1) : 1 : M-1
                    if ((idxN1==idxN2) && (delta==0))
                        sum(diag(Ftmp, delta)) <= 1/N;
                    else
                        sum(diag(Ftmp, delta)) == 0;
                    end
                end
            end
        end
cvx_end
spectrumANM = abs(x'*steerGrid).^2;
spectrumANM = spectrumANM/max(spectrumANM);
figure; plot(angleGrid, spectrumANM);
% User defined function
function steerM = steerMatTR(targetAngle, steerVecT, steerVecR)
steerA = steerVecT(targetAngle);
steerB = steerVecR(targetAngle);
steerM = zeros(size(steerA, 1)*size(steerB, 1), length(targetAngle));
for idxK = 1 : 1 : length(targetAngle) 
    steerM(:, idxK) = kron(steerB(:, idxK), steerA(:, idxK));
end

After running the above code, it gives the following error:

Undefined function 'cvx_begin' for input arguments of type 'char'.
Error in main (line 32)
cvx_begin sdp quiet

Can anyone help me in this regard?

3 个评论
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Adam Danz 2021-9-8

https://www.mathworks.com/matlabcentral/answers/410534-undefined-function-or-variable-cvx_begin#answer_329024

Sadiq Akbar 2021-9-8

Thanks for the responses of both of you dear DGM and Adam Danz. No, I don't know anything about CVX? what is that and how to install that?

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Answer 1

the cyclist 2021-9-8

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1449444-why-this-code-the-given-error#answer_783619

CVX is a third-party MATLAB toolbox (i.e. not from MathWorks), that you need to install according to their instructions. You need to go to their website to do that.

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Walter Roberson 2021-9-9

在 MATLAB Online 中打开

vec = @(MAT) MAT(:);

That creates an anonymous function that accepts a single input parameter, and reshapes the input parameter to be a column vector. For example

A = [1 2; 3 4]
A = 2×2
     1     2
     3     4
vec(A)
ans = 4×1
     1
     3
     2
     4

The vecH anonymous function does something very similar, but reshapes to a row vector instead.

steerVecT = @(ang) exp(1j*2*pi*d*[0:M-1].'*sin(vecH(ang)));

This is an anonmous function that accepts a single parameter.

1j is the imaginary constant, 1*sqrt(-1). 2*pi are what they look like. d is a scalar constant.

[0:M-1] creates the row vector containing 0, 1, 2, 3, 4, up to M-1 -- up to 9 in this case. Then the [] operator around it causes it to be concatenated with the additional input values, but there are no additional input values, so the [] operator is a waste of time here. The [0:M-1] would have been better written as (0:M-1) .

Notice that [] is an active operation in MATLAB, having to do with building vectors and matrices. It is not used to establish priorities for evaluation (though it does have that side-effect)

The .' after the [0:M-1] takes the row vector and converts it into a column vector.

At this point in the expression, the 1j*2*pi*d has been calculated into a single (complex) scalar value. The [0:M-1].' on the right hand side of the * operation has priority over the * operation so the [0:M-1].' is evaluated before the multiplication is.

Now at this point you have scalar * column vector of integers, 0 to M. The first row of output would be 0*constant, the second row would be 1*constant, and so on up to (M-1)*constant, all as a column vector.

We have now reached the * between the .' and the sin() and have evaluated the left side of that to a column vector. The right side now needs to be evaluated.

The vecH(ang) reshapes the function input into a row vector. sin() of the row vector is calculated.

Now we have column vector * row vector. The * operator in MATLAB is algebraic matrix multiplication, also called inner product. In this particular arrangement of column vector * row vector, the mathematical effect is to take the dot product of the column vector and the row vector . Which is to say, to take the sum of multiplying the corresponding elements: C(1).*R(1) + C(2).*R(2) + C(3).*R(3) ... The result of this is going to be a scalar.

And now we calculate exp() of that scalar.

... but why you calculate that is a different question.

cvx_begin sdp quiet
    variable x(N*M,1) complex
    variable F(M*N,M*N) hermitian
    minimize(norm(y-x))
    subject to
        [F, x; x', lambda^2]>=0

those lines all have to do with the third-party program cvx which is a convex optimization program.

If I read correctly, the lines after that until the cvx_end are there to establish constraints for the optimization that is cvx is being asked to do.

the cyclist 2021-9-12

@Sadiq Akbar, you won't be able to officially accept @Walter Roberson's. The reason is that I answered your original question (so you could accept that), but then you asked additional questions in the ensuing comments. Walter was kind enough to answer those additional questions in his comments, but unfortunately comments cannot be accepted or upvoted.

Sadiq Akbar 2021-9-13

Thank you very much dear the cyclist for your guidance. Ok I am going to accept that. But actually I am thankful to all of you as I have mentioned earlier.

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