The Fourier Series

Continuous Time

Definition 19

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

Definition 20

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

Theorem 1

If x(t)x(t) and y(t)y(t) are functions with period T1T_1 and T2T_2 respectively, then x(t)+y(t)x(t)+y(t) is periodic if m,nZ\exists m, n \in \mathbb{Z} such that mT1=nT2mT_1 = nT_2.

Definition 21

Given a periodic function x(t)x(t) with fundamental period TT and fundamental frequency ω0=2πT\omega_0=\frac{2\pi}{T}, the Fourier Series of xx is a weighted sum of the harmonic functions.

x(t)=k=akejkω0tx(t) = \sum_{k=-\infty}^{\infty}{a_ke^{jk\omega_0t}}

To find the coefficients aka_k:

\begin{aligned} x(t) \cdot e^{-jn\omega_0t} &= \sum_{k=-\infty}^{\infty}{a_ke^{j\omega_0t(k-n)}}\\ \int_{T}{x(t) \cdot e^{-jn\omega_0t}dt} &= \sum_{k=-\infty}^{\infty}{a_k\int_{T}{e^{j\omega_0t(k-n)}}dt}\\ &= \begin{cases} Ta_k & \text{if } $k=n$,\\ 0 & \text{else } \end{cases}\end{aligned}

Rearranging this, we can see that

an=1TTx(t)ejnω0tdt.a_n = \frac{1}{T}\int_{T}{x(t)e^{-jn\omega_0t}dt}.

For a0a_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.

a0=1TTx(t)dta_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)x(t), it will not converge. The Dirichlet Convergent Requirements tell us when the Fourier Series converges. More specificially, they tell us when

τ, limMxM(τ)=x(τ)xM(t)=k=MMakejkω0t\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.

Theorem 2

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

  • If xx is continuous at τ\tau, limMxM(τ)=x(τ)\lim_{M \rightarrow \infty}x_M(\tau) = x(\tau)

  • If xx is discontinuous at τ\tau, then limMxM(τ)=12(x(τ)+x(τ+))\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.

Discrete Time

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

Definition 22

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

However, there are some differences. For example, x[n]=cos(ω0n)x[n] = cos(\omega_0 n) is only periodic in discrete time if N,MZ,ω0N=2πM\exists N, M \in \mathbb{Z}, \omega_0 N = 2 \pi M.

Theorem 3

The sum of two discrete periodic signals is periodic

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.

ϕk[n]=ejk2πNn\phi_k[n] = e^{jk\frac{2\pi}{N}n}

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

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

  • ϕk[n]ϕm[n]=ϕk+m[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 ϕk[n]\phi_k[n] is that

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

Definition 23

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

x[n]=k=0N1akϕk[n]x[n] = \sum_{k=0}^{N-1}{a_k\phi_k[n]}

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

x[n]=k=0N1akϕk[n]x[n]ϕM[n]=k=0N1akϕk[n]ϕM[n]n=<N>x[n]ϕM[n]=n=<N>k=<N>akϕkM[n]=k=<N>akn=<N>ϕkM[n]n=<N>x[n]ϕM[n]=aMNaM=1Nn=<N>x[n]ϕM[n]\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}

Properties of the Fourier Series

Linearity: If aka_k and bkb_k are the coefficients of the Fourier Series of x(t)x(t) and y(t)y(t) respectively, then Aak+BbkAa_k + Bb_k are the coefficients of the Fourier series of Ax(t)+By(t)Ax(t)+By(t)

Time Shift: If aka_k are the coefficients of the Fourier Series of x(t)x(t), then bk=ejk2πTt0akb_k = e^{-jk\frac{2\pi}{T}t_0}a_k are the coefficients of the Fourier Series of x^(t)=x(tt0)\hat{x}(t)=x(t-t_0)

Time Reversal: If aka_k are the coefficients of the Fourier Series of x(t)x(t), then bk=akb_k=a_{-k} are the coefficients of the Fourier Series of x(t)x(-t)

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

Theorem 4 (Parseval's Theorem)

Continuous Time: 1Tx(t)2dt=k=ak2\textbf{Continuous Time: } \frac{1}{T}\int{|x(t)|^2dt} = \sum_{k=-\infty}^{\infty}{|a_k|^2}

Discrete Time: 1Nn=<N>x[n]2=k=<N>ak2\textbf{Discrete Time: } \frac{1}{N}\sum_{n=<N>}{|x[n]|^2} = \sum_{k=<N>}{|a_k|^2}

Interpreting the Fourier Series

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 xx onto a set of basis functions, and the coefficients aka_k are the coordinates of our signal in the new space.

Discrete Time

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

x=[x[0]x[1]x[N1]]CN\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 ϕk\phi_k form an orthogonal basis. If we take two of them ϕk[n]\phi_k[n] and ϕM[n]\phi_M[n] (kMk\ne M) and compute their dot product of their vector forms, then

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

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

ϕkϕk=ϕk2=N,ϕk=N\phi_k \cdot \phi_k = ||\phi_k||^2 = N, \therefore ||\phi_k|| = \sqrt{N}

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

xϕM=(i=0N1aiϕi)ϕM=Nam\vec{x}\cdot \vec{\phi_M} = \left(\sum_{i=0}^{N-1}{a_i\phi_i}\right)\cdot \phi_M = Na_m

am=1NxϕMa_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=ϕm2N = ||\phi_m||^2. This gives a nice geometric intution for Parseval’s theorem

1Nx[n]2=1Nx2=ak2\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.

Continuous Time

In continuous time, our bases functions are ϕk(t)=ejk2πTt\phi_k(t) = e^{jk\frac{2\pi}{T}t} for k(,)k \in (-\infty, \infty) Since we can’t convert continuous functions into vectors, these ϕk\phi_k are really a basis for the vector space of square integrable functions on the interval [0,T][0, T]. The inner product for this vector space is

<x,y>=0Tx(t)y(t).<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.

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