Every periodic function with period T can be written as a linear combination:
\begin{equation} f(t) = b_{0} + \sum_{j=1}^{\infty}a_{j} \sin \left( 2\pi \frac{j}{T} t\right) + b_{j} \cos \left(2\pi \frac{j}{T} t\right) \end{equation}
Finite-Bandwidth Signal If the summation here is finite, we call this representation as finite-bandwidth. You can draw out two separate stem plots, representing the \sin term frequencies and the \cos term frequencies.
Bandwidth For a particular signal, identify the largest and smallest frequency corresponding to non-zero coefficients, then our bandwidth is defined by:
\begin{equation} BW = f_{\max} - f_{\min} \end{equation}
if there is a nonzero shift for a cosine series, we consider f_{\min} = 0.
Discrete Fourier Transform To represent a sinusoidal sequence, we sample a point for every \frac{t}{n} times, obtaining a list of n samples of a sinusoid; with this, we obtain a sequence:
\begin{align} [y(t_0) \dots y(t_1) \dots y(t_{n-1})] \end{align}
We’ll then represent this sequence as a weighted sum of discrete time sinusoids sampled again n times:
\begin{equation} \left\{\sin \left(2 \pi \frac{1}{n} k\right): k \in 0,1, \dots, n-1 \right\} \end{equation}
\begin{equation} \left\{\cos \left(2 \pi \frac{1}{n} k\right): k \in 0,1, \dots, n-1 \right\} \end{equation}
Then, we slowly increase the sampling frequency:
\begin{equation} \left\{\sin \left(2 \pi \frac{2}{n} k\right): k \in 0,1, \dots, n-1 \right\} \end{equation}
\begin{equation} \left\{\cos \left(2 \pi \frac{2}{n} k\right): k \in 0,1, \dots, n-1 \right\} \end{equation}
notice how each of these frequencies yields a list of n values; by adding them all up together with appropriate coefficients, we can represent each element of our signal.
\begin{equation} y(t_{k}) = b_0 + \sum_{j=1}^{\infty} a_{j} \sin \left(2 \pi \frac{j}{n} k\right) + b_{j} \cos \left(2 \pi \frac{j}{n} k\right) \end{equation}
for a discrete input list with n elements. IMPORTANTNLY, however:
\begin{equation} \sin \left(2\pi \frac{(j+n+i)}{n} k\right) = \sin \left(2 \pi \frac{j}{n} k + 2\pi k + 2 \pi \frac{i}{n} k\right) = \sin \left(2 \pi \frac{(j+i)}{n} k + 2\pi k\right) \end{equation}
(and because \sin is 2\pi periodic, that term goes away).
Because all things beyond n is a repeat of stuff below n, as shown above:
\begin{equation} y(t_{k}) = b_0 + \sum_{j=1}^{n} a_{j} \sin \left(2 \pi \frac{j}{n} k\right) + b_{j} \cos \left(2 \pi \frac{j}{n} k\right) \end{equation}
further, at j=n, the sin term becomes 0 and the cos term becomes a constant, this allows us to write:
\begin{equation} y(t_{k}) = b_0 + \sum_{j=1}^{n-1} a_{j} \sin \left(2 \pi \frac{j}{n} k\right) + b_{j} \cos \left(2 \pi \frac{j}{n} k\right) \end{equation}
Now, let’s for a bit consider a sine-only series; let’s break our j into:
\begin{equation} j = 1, 2, 3 \dots, \frac{N-1}{2}, \frac{N-1}{2} + 1, \frac{N-1}{2} + 2 \dots N-1 \end{equation}
Consider the \frac{N-1}{2} + 1 term, we have, for some k:
\begin{equation} \sin \left(2 \pi \frac{\left(\frac{N-1}{2}+1\right)}{N} k\right) = -\sin \left(2 \pi \frac{\frac{N-1}{2}}{N} k\right) \end{equation}
Meaning, it collapses in the negative of the previous term. This patterns continue:
\begin{equation} \sin \left(2 \pi \frac{\left(\frac{N-1}{2}+2\right)}{N} k\right) = -\sin \left(2 \pi \frac{\left(\frac{N-1}{2}-1\right)}{N} k\right) \end{equation}
this means that all terms after \frac{N-1}{2} is irrelevant!
The above actually holds for cosine as well. This gives:
\begin{equation} y(t_{k}) = b_0 + \sum_{j=1}^{\frac{n-1}{2}} a_{j} \sin \left(2 \pi \frac{j}{n} k\right) + b_{j} \cos \left(2 \pi \frac{j}{n} k\right) \end{equation}
for t_{k} \in [0, …, n-1]
Notice! We have FIXED coefficients for all n times. This in total gives us 2\left(\frac{n-1}{2}\right) + 1 = n unknowns!
This is n equations and n unknowns, we can solve it as a linear system by collapsing it into a linear system:
for
\begin{equation} Y = \mqty(y(t_0) \\ \dots \\ y(t_{n-1})) \end{equation}
and formulate a map F of sinusoidal bases