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Why can't you just take the DFT itself? And how accurate is this method in spectral density estimation? Are there cases in which it is very accurate?

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  • $\begingroup$ Because that's how it's defined. Also, how would you visualize a complex-valued 2D object? "How accurate" and "when is it very accurate" would require us to first define accuracy, and for that define signal classes, possibly introduce you to the theory of stochastic signals, what an autocorrelation function is and so on... too broad, I'm afraid! $\endgroup$
    – Marcus Müller
    Commented May 23, 2018 at 17:28

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Many applications just want a strictly real value of some sort (for a simple line graph, or peak magnitude frequency estimate, etc.). And the modulus squared is a real value; whereas the result of a DFT is a complex valued vector. (except, depending on definitions, for strictly even functions/waveforms)

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