# The Fourier Series

To find the coefficients

$a_k$

:

Rearranging this, we can see that

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

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

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

will converge.

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

These convergence requirements are for pointwise convergence only. They do not necessarily imply that the graphs of the Fourier Series and the original function will look the same.

The definition for periodicity in discrete time is the exact same as the definition in continuous time.

However, there are some differences. For example,

$x[n] = cos(\omega_0 n)$

is only periodic in discrete time if $\exists N, M \in \mathbb{Z}, \omega_0 N = 2 \pi M$

.Theorem 3 is not always true in continuous time but it is in discrete time.

The Fourier Series in discrete time is the same idea as the Fourier series in continuous time: to express every signal as a linear combination of complex exponentials. The discrete time basis that we use are the Nth roots of unity.

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

- $\phi_k[n]$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

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

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 $a_k=a_{-k}^*$

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

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

because we know the norms of two vectors in different bases must be equal.

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

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

We can use this inner product to conduct the same proof as we did in discrete time.

Last modified 11mo ago