Fourier Series as exactly a shifted sum of sinusoids
Key idea: every periodic function with period \(L\) can be represented as a sum of sinusoids
\begin{equation} f(t) = A_0 + \sum_{i=1}^{\infty} B_{j} \sin \qty(k \omega t + \phi_{j}) \end{equation}
where \(\omega = \frac{2\pi}{T}\). notice! without the \(A_0\) shift, our thing would integrate to \(0\) for every \(L\); hence, to bias the mean, we change \(A_0\).
Now, we ideally really want to get rid of that shift term \(\phi\), applying the sin sum formula:
\begin{align} f(t) &= A_0 + \sum_{i=1}^{\infty} B_{j} \sin \qty(k_{j} \omega t + \phi_{j}) \\ &= A_0 + \sum_{j=1}^{\infty } A_{j} \cos \qty(\phi_{j}) \sin \qty(k_{j}\omega t) + B_{j} \sin \qty (\phi_{j}) \cos \qty(k_{j} \omega t) \\ &= b_0 + \sum_{j=1}^{\infty} a_{j} \sin \qty(\omega k_{j} t) + \sum_{j=1}^{\infty} b_{j} \cos \qty(k \omega t) \end{align}
we can move back and fourth before the representation as follows:
\begin{equation} \begin{cases} a_{j} = A_{j} \cos \qty(\phi_{j}) \\ b_{j} = A_{j} \sin \qty(\phi_{j}) \\ b_{0} = A_{0} \\ A_{j}^{2} = a_{j}^{2} + b_{j}^{2} \\ \tan \qty(\phi_{j}) = \frac{b_{j}}{a_{j}} \end{cases} \end{equation}
in a sense, this is a polar representation of the sum of sinusoids of system. Recall to get the actual coefficients, see General Fourier Decomposition.
signal representation
KEY IDEA: we can approximate all values of a function with just specifying the parameters of the sine function.
In particular, any signal \(f\) is uniquely specified by specifying only its Fourier representation:
\begin{equation} \left\{A_{j}, \phi_{j}\right\}_{0}^{\infty} \cup \{A_0\} \end{equation}
The smallest \(f_{1} = \frac{1}{T}\), called the fundamental frequency of this system, and any higher are harmonics; in particular, \(f_{j} = \frac{j}{T}\) are called the jth harmonic.
To represent finite-duration signal, we just create the finite-periodic extension of this signal by coping it over and over.
Key thing to remember: remember that odd/even extensions have period *2T!!!*