why showing 'Array indices must be positive integers or logical values.'

Question

Tazen Moushumi 2020-1-28

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/502306-why-showing-array-indices-must-be-positive-integers-or-logical-values

评论： Walter Roberson 2020-1-28

m2 = zeros(N,1);                 %2Matrix of second moment matrix
    m2_pul = zeros(N,1);  %x(n)x(n+t)Matrix frame
    fs = 50;
    for  t1 = (0:N-1)/fs
          n = 1:N;
            m2_pul(n) = sig(n) .* sig(n+t1);       %x(n)x(n+t)
        
        m2(t1+1) = mean(m2_pul);                        %E[x(n)x(n+t)]
        m2_pul = zeros(N,1);
    end

Array indices must be positive integers or logical values.

Error in bispectrum (line 12)

m2_pul(n) = sig(n) .* sig(n+t1); %x(n)x(n+t)

Error in bi_estimation (line 102)

[c3_w]=bispectrum(sig,sig2,N);

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Answer 1

KSSV 2020-1-28

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https://ww2.mathworks.cn/matlabcentral/answers/502306-why-showing-array-indices-must-be-positive-integers-or-logical-values#answer_412352

编辑：KSSV 2020-1-28

在 MATLAB Online 中打开

m2 = zeros(N,1);                 %2Matrix of second moment matrix
    m2_pul = zeros(N,1);  %x(n)x(n+t)Matrix frame
    fs = 50;
    f = (0:N-1)/fs ; 
    for  t1 = 1:length(f)
          n = 1:N;
            m2_pul(n) = sig(n) .* sig(n+t1);       %x(n)x(n+t)
        
        m2(t1+1) = mean(m2_pul);                        %E[x(n)x(n+t)]
        m2_pul = zeros(N,1);
    end

The indices of array in MATLAB cannot be zero, negative and fractions. When you take:

t1 = (0:N-1)/fs

t1 has fractions, you cannot use them as indices; so error.

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Tazen Moushumi 2020-1-28

sig : noise-less input signal and used as an array.

Walter Roberson 2020-1-28

在 MATLAB Online 中打开

n = 1:N;

Okay, everything in n is going to be a positive integer. It will have entries like [1, 2, 3, 4, 5, ...]

for t1 = (0:N-1)/fs

That would be, for example, [0/8000, 1/8000, 2/8000, ....] which is going to be mostly non-integer. It will be integer at 0 and at each exact integer multiple of fs .

m2_pul(n) = sig(n) .* sig(n+t1); %x(n)x(n+t)

The first sub-expression of that is sig(n) . With n being all positive integers and sig being an array, then sig(n) is valid indexing provided that the maximum value in n does not exceed the number of elements in sig .

The second sub-expression is sig(n+t1) . n has entries like [1, 2, 3, 4, 5, ...] and t1 has entries like [0/8000, 1/8000, 2/8000, ...] and so on, so n+t1 is like [1+0/8000, 2+1/8000, 3+2/8000, 4+3/8000], and so on. The first of those will be an integer, and at one place further than exact integer multiplies of fs, there will be integers (so like 1, 8001, 160001) but most of n+t1 will be non-integer. Those are invalid indices.

But KSSV's suggestion of

    f = (0:N-1)/fs ; 
    for  t1 = 1:length(f)
      n = 1:N;
      m2_pul(n) = sig(n) .* sig(n+t1);       %x(n)x(n+t)

is successful in avoiding using the n+((0:N-1)/fs) as **indices* . It also happens not to use the content of (0:N-1)/fs which might or might not be a problem.

I cannot tell what the original code is intended to do, so I am not sure what to suggest.

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