While learning Fourier series I read the definitions of representation for a continuous time signal $x(t)$ as:
$$x(t)=A_0 + 2 \sum_{k=1}^{\infty} A_k \cos(k \omega_0 t) - B_k \sin(k \omega_0 t) \tag{1}$$
where $A_k$ and $B_k$ are real.
Another definition was: $$x(t)=a_0 + \sum_{k=1}^{\infty} a_k \cos(k \omega_0 t) + b_k \sin(k \omega_0 t) \tag{2}$$
Why two different definitions? I mean in $(1)$ there is a coefficient of $2$ for sinusoids and sine coefficients are negated.
alright, let's review a little bit of Euler before we get to the Fourier.
$$ e^{j \theta} \ = \ \cos(\theta) \ + \ j \sin(\theta) $$
from that you can get
$$ \cos(\theta) = \frac{e^{j \theta} + e^{-j \theta}}{2} \quad\quad\quad \sin(\theta) = \frac{e^{j \theta} - e^{-j \theta}}{2j} $$
so now let's look at Eq (1)
$$ \begin{align} x(t) \ &= \ A_0 \ + \ 2 \sum_{k=1}^{\infty} A_k \cos(k \omega_0 t) - B_k \sin(k \omega_0 t) \\ &= \ A_0 \ + \ 2 \sum_{k=1}^{\infty} A_k \frac{e^{j k \omega_0 t} + e^{-j k \omega_0 t}}{2} \ - \ B_k \frac{e^{j k \omega_0 t} - e^{-j k \omega_0 t}}{2j} \\ &= \ A_0 \ + \ \sum_{k=1}^{\infty} (A_k + jB_k) e^{j k \omega_0 t} \ + \ (A_k - jB_k) e^{-j k \omega_0 t} \\ &= \ \sum_{k=-\infty}^{\infty} c_k \ e^{j k \omega_0 t} \\ \end{align} $$
where
$$ c_k = \begin{cases} A_{-k} - jB_{-k}, & \quad \text{for } k < 0 \\ A_0, & \quad\quad k = 0 \\ A_k + jB_k, & \quad\quad k > 0 \end{cases} $$
and, going in the other direction,
$$ \begin{array}{lcl} A_0 & = & c_0 \\ A_k & = & \Re\{c_k \} \quad\quad \text{for } k>0\\ B_k & = & \Im\{c_k \} \quad\quad\quad k>0 \end{array} $$
note that for real $A_k$ and real $B_k$, then
$$ c_{-k} = c_k^* = \text{"complex conjugate of } c_k \text{ "}$$
so, for Eq (1), the simplicity of having the real and imaginary parts of the $c_k$ coefficients be simply $A_k$ and $B_k$ (for $k \ge 0$) is the motivation for the convention of the leading "2" before the summation and for the minus sign. it's just a convention. this convention is useful because it's much easier to derive the coefficients $c_k$ than it is to derive the coefficients $A_k$ and $B_k$.
$$ c_k \ = \ \frac{\omega_0}{2 \pi}\int_{t_0 - \pi/\omega_0}^{t_0 + \pi/\omega_0} x(t) \ e^{-j k \omega_0 t} \ dt $$
where $ -\infty < t_0 < +\infty $ can be any convenient real value.
the convention for Eq. (2) comes about from looking at Fourier series first as a real analysis problem without the use of complex variables. then this is a simple first statement:
$$x(t) \ = \ a_0 \ + \ \sum_{k=1}^{\infty} a_k \cos(k \omega_0 t) + b_k \sin(k \omega_0 t)$$
as you can see $a_k$ and $b_k$ are related to $A_k$ and $B_k$ in a very simple and straight forward manner. you can derive that simple relationship. but the formulae for getting $a_k$ and $b_k$ from $x(t)$ (and $\omega_0$) is less straight forward to derive and to express. if fact, it's two equations, not one simpler equation.
The first represented is single sided (only positive frequency) that usually is introduced with reference to the double ended complex form of the Fourier series. The coefficients B and C are real because of how they relate to the generally complex coefficients from the negative and positive frequency components in the complex form.
Going back to the first form, the complex representation uses a sum from $-\infty$ to $+\infty$. Typically a complex exponential is used in the sum, but $\cos + i\sin$ works just fine. Say the sum is defined with COMPLEX coefficients using the letter $a$.
There is a relationship between the two sets of coefficients:
$B_k = 2\operatorname{Re}(a_k)$
$C_k = -2\operatorname{Im}(a_k)$
Hence the negative sign on $C$.
The 2nd representation is a valid representation but because the series is defined only in terms of positive frequency and real functions, the coefficients are calculated differently.
This answer is correct. For anyone that doesn't understand what a singled ended representation of a Fourier series is, please read the following:
i.e. robert bristow-johnson
You can also consult the standard bearer on LTI systems, Athanasios Papoulis, "Circuits and Systems" pp 336.
