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This question already has an answer here:

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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marked as duplicate by Marcus Müller, MBaz, Stanley Pawlukiewicz, A_A, lennon310 May 24 '18 at 3:59

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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 May 23 '18 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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