A function $x(t)$ is periodic if $\exists T$ such that $\forall t, x(t-T)=x(t)$.

The fundamental period is the smallest such $T$ which satisfies the periodicity property in Definition 19

If $x(t)$ and $y(t)$ are functions with period $T_1$ and $T_2$ respectively, then $x(t)+y(t)$ is periodic if $\exists m, n \in \mathbb{Z}$ such that $mT_1 = nT_2$.

Given a periodic function $x(t)$ with fundamental period $T$ and fundamental frequency $\omega_0=\frac{2\pi}{T}$, the Fourier Series of $x$ is a weighted sum of the harmonic functions.

$x(t) = \sum_{k=-\infty}^{\infty}{a_ke^{jk\omega_0t}}$

To find the coefficients $a_k$:

$a_n = \frac{1}{T}\int_{T}{x(t)e^{-jn\omega_0t}dt}.$

For $a_0$, the DC offset term, this formula makes a lot of sense because it is just the average value of the function over one period.

$a_0 = \frac{1}{T}\int_{T}{x(t)dt}$

Because the Fourier Series is an infinite sum, there is a worry that for some functions $x(t)$, it will not converge. The **Dirichlet Convergent Requirements** tell us when the Fourier Series converges. More specificially, they tell us when

$\forall \tau, \ \lim_{M \rightarrow \infty}{x_M(\tau) = x(\tau)} \qquad x_M(t) = \sum_{k=-M}^{M}{a_k e^{jk\omega_0t}}$

The Fourier Series of a continuous time periodic function $x(t)$ will converge when $x$ is piecewise continuous and $\frac{d}{dt}x$is piecewise continuous.

If $x$ is continuous at $\tau$, $\lim_{M \rightarrow \infty}x_M(\tau) = x(\tau)$

If $x$ is discontinuous at $\tau$, then $\lim_{M\rightarrow \infty}x_M(\tau) = \frac{1}{2}(x(\tau^-) + x(\tau^+))$

A function $x[n]$ is periodic with period $N \in \mathbb{Z}$ if $\forall n, x[n+N]=x[n]$

However, there are some differences. For example, $x[n] = cos(\omega_0 n)$ is only periodic in discrete time if $\exists N, M \in \mathbb{Z}, \omega_0 N = 2 \pi M$.

$\phi_k[n] = e^{jk\frac{2\pi}{N}n}$

$\phi_k[n]$ is perioidic in n (i.e $\phi_k[n+N] = \phi_k[n]$)

$\phi_k[n]$ is periodic in k (i.e $\phi_{k+N}[n] = \phi_k[n]$)

$\phi_k[n]\cdot \phi_m[n] = \phi_{k + m}[n]$

Notice that with this basis, there are only N unique functions that we can use. An additional property of the $\phi_k[n]$ is that

$\sum_{n=<N>}{\phi_k[n]} = \begin{cases} N & \text{if } k = 0, \pm N, \pm 2N, \cdots\\ 0 & \text{otherwise.} \end{cases}$

Given a periodic discrete-time function $x[n]$ with period $N$, the Fourier series of the function is a weighted sum of the roots of unity basis functions.

$x[n] = \sum_{k=0}^{N-1}{a_k\phi_k[n]}$

In order to find the values of $a_k$, we can perform a similar process as in continuous time.

$\begin{aligned} x[n] &= \sum_{k=0}^{N-1}{a_k\phi_k[n]}\\ x[n]\phi_{-M}[n] &= \sum_{k=0}^{N-1}{a_k\phi_k[n]\phi_{-M}[n]}\\ \sum_{n=<N>}{x[n]\phi_{-M}[n]} &= \sum_{n=<N>}{\sum_{k=<N>}{a_k\phi_{k-M}[n]}} = \sum_{k=<N>}{a_k\sum_{n=<N>}{\phi_{k-M}[n]}}\\ \sum_{n=<N>}{x[n]\phi_{-M}[n]} &= a_MN\\ a_M &= \frac{1}{N}\sum_{n=<N>}{x[n]\phi_{-M}[n]}\end{aligned}$

**Linearity:** If $a_k$ and $b_k$ are the coefficients of the Fourier Series of $x(t)$ and $y(t)$ respectively, then $Aa_k + Bb_k$ are the coefficients of the Fourier series of $Ax(t)+By(t)$

**Time Shift:** If $a_k$ are the coefficients of the Fourier Series of $x(t)$, then $b_k = e^{-jk\frac{2\pi}{T}t_0}a_k$ are the coefficients of the Fourier Series of $\hat{x}(t)=x(t-t_0)$

**Time Reversal:** If $a_k$ are the coefficients of the Fourier Series of $x(t)$, then $b_k=a_{-k}$ are the coefficients of the Fourier Series of $x(-t)$

**Conjugate Symmetry:** If $a_k$ are the coefficients of the Fourier Series of $x(t)$, then $a_k^*$ are the coefficients of the Fourier Series of $x^*(t)$. This means that $x(t)$ is a real valued signal, then $a_k=a_{-k}^*$

$\textbf{Continuous Time: } \frac{1}{T}\int{|x(t)|^2dt} = \sum_{k=-\infty}^{\infty}{|a_k|^2}$

$\textbf{Discrete Time: } \frac{1}{N}\sum_{n=<N>}{|x[n]|^2} = \sum_{k=<N>}{|a_k|^2}$

A good way to interpret the Fourier Series is as a change of basis. In both the continuous and discrete case, we are projecting our signal $x$ onto a set of basis functions, and the coefficients $a_k$ are the coordinates of our signal in the new space.

Since in discrete time, signal is peroidic in $N$, we can turn any it into a vector $\vec{x}\in \mathbb{C}^N$.

$\vec{x} = \left[ \begin{array}{c} x[0]\\ x[1]\\ \vdots\\ x[N-1] \end{array} \right] \in \mathbb{C}^N$

We can use this to show that $\phi_k$ form an orthogonal basis. If we take two of them $\phi_k[n]$ and $\phi_M[n]$ ($k\ne M$) and compute their dot product of their vector forms, then

$\phi_k[n] \cdot \phi_M[n] = \phi_M^*\phi_k = \sum_{<n>}{\phi_{k-M}[n]} = 0$

That means that $\phi_k$ and $\phi_M$ are orthogonal, and they are $N$ of them, therefore they are a basis. If we compute their magnitudes, we see

$\phi_k \cdot \phi_k = ||\phi_k||^2 = N, \therefore ||\phi_k|| = \sqrt{N}$

Finally, if we compute $\vec{x}\phi_M$ where $\vec{x}$ is the vector form of an N-periodic signal,

$\vec{x}\cdot \vec{\phi_M} = \left(\sum_{i=0}^{N-1}{a_i\phi_i}\right)\cdot \phi_M = Na_m$

$a_m = \frac{1}{N}\vec{x}\cdot \phi_M$

This is exactly the equation we use for finding the Fourier Series coefficients, and notice that it is a projection since $N = ||\phi_m||^2$. This gives a nice geometric intution for Parseval’s theorem

$\frac{1}{N}\sum{|x[n]|^2} = \frac{1}{N}||\vec{x}||^2 = \sum{|a_k|^2}$

In continuous time, our bases functions are $\phi_k(t) = e^{jk\frac{2\pi}{T}t}$ for $k \in (-\infty, \infty)$ Since we can’t convert continuous functions into vectors, these $\phi_k$ are really a basis for the vector space of square integrable functions on the interval $[0, T]$. The inner product for this vector space is

$<x, y> = \int_{0}^{T}{x(t)y^*(t)}.$