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I am trying to calculate the SCC of a cyclostationary signal. I have estimated the Spectral Correlation Function (scf) of the signal using FFT Accumulation Method (FAM). I implemented the following pseudocode in matlab:

function [Sx, alphao, f] = scf(x, fs, d_f, d_alpha)
    Compute N' = pow2(nextpow2(fs/df))
            L = N'/4
            P=pow2(nextpow2(fs/dalpha/L));
    N = P*L;    %total no of datat points to process
    x = x(1:N)
    Divide x into chunks of length N' with L overlapping points
    Apply Hamming window across each chunk
    Perform FFT
    Perform the downshift to baseband
    Multiply the baseband with it's complex conjugate
    Execute smoothing by means of P-points FFT.
end

where fs is sampling frequency, d_f is frequency resolution and d_alpha is cyclic frequency resolution

Now I have three matrices : $S_x^\alpha$, which is the SCF, $alphao$, which is a 1D vector the cyclic frequencies and $f$, which is a 1D vector the spectral frequencies. I need to calculate the Spectral Coherence Coefficient which is given by the following equation: $C_x^\alpha(f) = \frac{S_x^\alpha(f)}{\sqrt{S_x^0(f-\alpha/2)S_x^0(f+\alpha/2)}} \forall f$

My questions are:

  1. If I put $S_x^0(f-\alpha/2)$ in the denominator then won't the denominator essentially becomes $S_x^0(f)$ since $\alpha = 0$?

  2. Will $C_x^\alpha$ be of the same size as $S^\alpha_x$?

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I am just working through this information myself, but I believe I understand your questions.

  1. If I put $S_x^0(f-\alpha/2)$ in the denominator then won't the denominator essentially become $S_x^0(f)$, since $\alpha = 0$?

It will become $|S_x^0(f)|$ at $\alpha = 0$, but at other values of $\alpha$, you would need the conjugate multiply at offsets of $f$.

  1. Will $C_x^\alpha$ be of the same size as $S^\alpha_x$?

Yes, one would think. The coherence SOF function, is just supposed to be some normalized version of the SCF.

I am struggling with the FAM, it appears to output data on a tilted spectral plane, where $\alpha=0$ lies on the matrix diagonal.

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