Fourier Series: how to get magnitiude and phase? So, I know how to calculate Fourier Series, Continuous Fourier Transform, and I'm learning how to calculate Discrete Fourier Transform. But there's one thing I don't know for sure.
Any periodic function can be represented as a sum:
$$
f(t) = \sum_{N=-\infty}^{\infty}c_ne^{2\pi i n t}
$$
And $c_n$ are complex numbers, and $c_n = \overline{c_{-n}}$ (only if $f(t)$ is real-valued). Let's write that $c_n = A_ne^{i\phi_n}$
I'm almost sure that $A_n$ is the magnituide of a given frequency, though wikipedia states that $c_n = \frac{A_n}{2}e^{i\phi_n}$. So, is $|c_n|$ only half of the magnitiude (it has something to do with the negative frequencies, I think)?
I'd also guess that $\phi_n$ is the phase of the n-th sinusoid in the spectrum, however, I couldn't find the confirmation of my guess anywhere.
Does it work the same way for continuos and discrete Fourier Transform? Say, if:
$$
F(\mu) = A_\mu e ^{i\phi_{\mu}}
$$
Is the $A_\mu$ the magnitiude and $\phi_\mu$ the phase?
It would be great if you could also point out some resources about it
EDIT: after some more looking, I've found out whether my guess about the amplitiude and the phase was correct, and I've posted the answer here. I still don't know what to do about those negative frequencies though.
 A: Yes, the reason is exactly due to the negative terms. To see this, look at the sum of the terms corresponding to $-n$ and $n$ in the Fourier series. Summing both terms, we obtain
$$ c_n e^{2\pi i n t} + c_{-n} e^{-2\pi i n t} = 2 \mathrm{Re}(c_n e^{2\pi i n t}) = 2 \mathrm{Re}\left(\frac{A_n}{2} e^{i(2\pi n t + \phi_n)}\right) = A_n cos(2\pi n t+\phi_n),  $$
where in the first step we used that the sum of a complex number and its complex conjugate is two times its real value. Therefore, $A_n$ is the amplitude of the frequency $2\pi n$ and $\phi_n$ its phase.
In the continuous case, it is a bit different, however conceptually similar. As $f(t)$ is obtained via the integration over $F(\mu)$, $F(\mu)$ is the amplitude and phase \emph{density} of a certain frequency.
A: So, according to the wikipedia page I've linked:

The Fourier transform of a function of time is a complex-valued function of frequency, whose magnitude (absolute value) represents the amount of that frequency present in the original function, and whose argument is the phase offset of the basic sinusoid in that frequency.

According to the same wikipedia page:
$$
\hat{f}(\xi) = A(\xi)e^{i\phi(\xi)}
$$
And
$$
A(\xi) = |\hat{f}(\xi)|
$$
is the amplitude, and the
$$
\phi(\xi) = Arg(\hat{f}(\xi))
$$
is the phase.
(I still don't know what to do about those negative frequencies)
