I am studying Fourier analysis and I am still new to this topic. If I understand that the maximum frequency that can be used in a DFT is given by $N/2$, where $N$ is the number of samples in our discrete signal. This always assumes that the sinusoid is an harmonic of $f$, where $f = 1$ with $\sin(2 \pi f t)$. In other words, we assume the first harmonic is a full cycle. How can this be justified? I am trying to make sense of this. Why don't we use a half cycle for example as in the case of DCTs? Why does the 1st harmonic have to be a full cycle?

EDIT: asking the question made me think;-). I would actually say that this is because the discrete signal is considered to be periodic and thus the harmonics have to be periodic as well thus $f = 1$. But could someone confirm please that this is the right reasoning?

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    $\begingroup$ I find it very hard to figure out what you may be asking about. If $\sin(2\pi ft)$ is periodic in $t$ with period $1$, it's clear that $f$ must be an integer. Moreover, if $f$ is changed by a multiple of $N$, $\sin(2\pi ft)$ doesn't change for $t$ equal to a multiple of $1/N$ at all because the argument changes by a multiple of $2\pi$. That's everything that one needs to know and the answer to all conceivable meaningful questions about the periodicity and quantization are included. $\endgroup$ Commented Oct 12, 2013 at 13:14
  • $\begingroup$ "I find it very hard to figure out what you may be asking about." that bit doesn't contribute to the answer but I appreciate the rest of it. Thank you. $\endgroup$
    – user18490
    Commented Oct 12, 2013 at 13:29
  • $\begingroup$ Just for reference though. I understand that you can by convention assume that you N samples in 1 unit time. However the standard way of looking at the problem is to consider that if a signal as a period of N samples then the frequency of the 1st harmonic is $f = {1 \over N}$. And this clearly not an integer. What you refer to in the fourier eq. would be the factor k which corresponds to the harmonic index. Because you consider a period of 1 this works, however not every author look at this problem that way. $\endgroup$
    – user18490
    Commented Oct 12, 2013 at 16:53


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