$$

\[ \newcommand{\d}{\text{d}} \newcommand{\dt}{\d t} \newcommand{\euler}{\mathrm{e}} \]

Fouries Series Representation of Continuous Signal

Given a signal $x(t)$ that can be presented as the sum of harmonically related signals:

\[ x(t) = \sum_{k=-\infty}^{\infty}a_ke^{jkw_0t} \]

where $w_0$ is the base frequency. The parameter $a_k$ can be computed as

\[ a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0t}\dt \]

Proof. Multiple $e^{-jn\omega_0t}$ on both sides of the series expression of $x(t)$ and integrate from $0$ to $T$ respect to $t$:

\[ \begin{align} \int_0^T x(t)e^{-jn\omega_0t} \dt &= \int_0^T \sum_{k=-\infty}^{\infty}a_ke^{j(k-n)w_0t}\dt \\ &= \sum_{k=-\infty}^{\infty}a_k \int_0^T e^{j(k-n)w_0t}\dt \\ \end{align} \]

Notice that $\displaystyle\int_0^T e^{j(k-n)w_0t}\dt = \int_0^T \cos(k-n)\omega_0 t + j \int_0^T \sin(k-n)\omega_0 t$, LHS indicates that for $k=n$ the integral equals to $T$, and RHS indicates that for $k\neq n$ the integral equals to $0$, since the integral length $T$ must be the multiple of period of $\cos(k-n)\omega t$ and $\sin(k-n)\omega t$. i.e.

\[ \int_0^T e^{j(k-n)w_0t}\dt = \left\{\begin{array}{ll}\begin{align} T, ~~~~ & k = n \\ 0, ~~~~ & k \neq n \end{align}\end{array}\right. \]

Hence, only $Ta_n$ left on the RHS of the equation:

\[ \int_0^T x(t)e^{-jn\omega_0t} \dt = Ta_n \]

which concludes the theorem.

Specially, if the signal $x(t)$ is a real signal, we can represent it by trigonometric function, specifically, if $a_k = A_ke^{j\theta_k} = B_k + jC_k$, we can rewrite $x(t)$ as following:

\[ x(t) = a_0 + 2\sum_{k=1}^{\infty} A_k\cos{(k\omega_0t+\theta_k)} = a_0 + 2\sum_{k=1}^{\infty}\left(B_k\cos{k\omega_0 t} - C_k\sin{k\omega_0 t}\right) \]

Proof. since $x(t)$ is real signal, we have $x(t) = x^*(t)$, a.k.a. x(t) is always the same with its conjugate.

\[ \begin{align} x(t) &= x^*(t) \\ \sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t} &= \sum_{k=-\infty}^{\infty}a^*_k e^{-jk\omega_0 t} = \sum_{k=-\infty}^{\infty}a^*_{-k} e^{jk\omega_0 t} \end{align} \]

comparing both sides, we have $a_k = a^*_{-k}$, which could be used to simplify the expression of $x(t)$: \[ \begin{align} x(t) &= \sum_{k=-\infty}^{\infty}a^*_{k} e^{jk\omega_0 t} \\ &= a_0 + \sum_{k=1}^{\infty} (a_{k} e^{jk\omega t} + a_{-k} e^{-jk\omega_0 t}) \\ &= a_0 + \sum_{k=1}^{\infty} (a_{k} e^{jk\omega t} + a^*_{k} e^{-jk\omega_0 t}) \\ \end{align} \]

Notice that $a_k = \overline{a^*_{k}}$, and $e^{jk\omega t} = \overline{e^{-jk\omega t}}$, hence \[ \begin{align} x(t) = a_0 + \sum_{k=1}^{\infty} 2\mathcal{Re}\{{a_k e^{jk\omega t}}\} \end{align} \]

Denoted $a_k = A_k e^{j\theta_k}$, with the Euler Theorem applied,

\[ x(t) = a_0 + 2\sum_{k=1}^{\infty} A_k\cos{(k\omega_0t+\theta_k)} \]

Or denoted $a_k = B_k + jC_k$, where $B_k$ and $C_k$ are both real number, with the Euler Theorem applied,

\[ x(t) = a_0 + 2\sum_{k=1}^{\infty}\left(B_k\cos{k\omega t} - C_k\sin{k\omega t}\right) \]

Property of Fouries Transformation

Linearity. For any two signals $x(t), y(t)$ which corresponding to fouries series $a_k, b_k$, the fouries series of the linear combination of them will also be the same linear combination of the fouries series:

\[ z(t) = Ax(t) + By(t) \xrightarrow[]{\mathcal{FS}} c_k = Aa_k + Bb_k \]