# Fourier Series

## Fourier series

The Fourier series is used to analyze arbitrary periodic functions by showing them as a composite of sines and cosines.

A function is considered periodic if $x(t) = x(t+T)\,$ for $T \neq 0$.

The exponential form of the Fourier series is defined as $x(t) = \sum_{n=-\infty}^\infty \alpha_n e^{{j2\pi nt}/T} \,$

## Determining the coefficient $\alpha_n \,$

$x(t) = \sum_{n=-\infty}^\infty \alpha_n e^{{j2\pi nt}/T} \,$

• The definition of the Fourier series

$\int_{-T/2}^{T/2} x(t)\, dt = \sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2} e^{{j2\pi nt}/T} dt$

• Integrating both sides for one period. The range of integration is arbitrary, but using $\int_{-T/2}^{T/2}$ instead of $\int_{0}^{T}$scales nicely when extending the Fourier series to a non-periodic function

$\int_{-T/2}^{T/2} x(t) e^{{-j2\pi mt}/T} dt = \sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2} e^{{j2\pi nt}/T}e^{{-j2\pi mt}/T} dt = \sum_{n=-\infty}^\infty \alpha_n \int_{-T/2}^{T/2} e^{{j2\pi (n-m)t}/T} dt$

• Multiply by the complex conjugate

$\int_{-T/2}^{T/2} x(t) e^{{-j2\pi mt}/T} dt = \sum_{n=-\infty}^\infty \alpha_n \frac{Te^{{j2\pi (n-m)t}/T}}{{j2\pi (n-m)}} \bigg|_{-T/2}^{T/2} = \sum_{n=-\infty}^\infty \alpha_n T\delta_{n,m} = T\alpha_m$

• $\frac{Te^{{j2\pi (n-m)t}/T}}{{j2\pi (n-m)}} \bigg|_{-T/2}^{T/2} = T\frac{e^{j\pi(n-m)}-e^{-j\pi(n-m)}}{j2\pi(n-m)} = T \frac{\sin\pi(n-m)}{\pi(n-m)} = \begin{Bmatrix} T, n=m \\ 0, n\ne m \end{Bmatrix} = T\delta_{n,m}$
• Using L'Hopitals to evaluate the $\frac{T\cdot 0}{0}$ case. Note that n & m are integers

$\alpha_m = \frac{1}{T}\int_{-T/2}^{T/2} x(t) e^{{-j2\pi mt}/T} dt$

## Linear Time Invariant Systems

Must meet the following criteria

• Time independance
• Linearity
• Scaling (homogeneity)

## The Dot Product, Complex Conjugates, and Orthogonality

Geometrically, the dot product is a scalar projection of a onto b

• $\vec a \cdot \vec b = \left | a \right \vert \left | b \right \vert \cos \theta$

Arthimetically, multiply like terms and add

• $(3,2,1)\cdot(5,6,7)=3\cdot5^*+2\cdot6^*+1\cdot7^*$

Lets imagine that we are only have one dimension

• $(a+jb)\hat i \cdot (a+jb)\hat i \ne a^2+b^2$

In order to get the real parts and imaginary parts to multiply as like terms, we need to take the complex conjugate of one of the terms

• $(a+jb)\hat i \cdot (a-jb)\hat i = a^2+b^2$

To test for orthogonality, take the complex conjugate of one of the vectors and multiply.

• $\int_{-\infty}^{\infty} \phi_n (t) \phi_m^* (t) dt = 0$

## Changing Basis Functions

We'd like to change from $\sum_{n=-\infty}^{\infty} \alpha_n e^{j2\pi nt/T}$ to $\sum_{m=0}^{\infty} c_m \cos \left (\frac{2\pi mt}{T}+\Theta_m \right)$

$x(t) = \sum_{n=-\infty}^{\infty} \alpha_n e^{j2\pi nt/T} = \underbrace{ \sum_{n=-\infty}^{-1} \alpha_n e^{j2\pi nt/T} }_{n'=-n} + \; \alpha_0 \; + \sum_{n=1}^{\infty} \alpha_n e^{j2\pi nt/T} = \underbrace{\sum_{n'=1}^{\infty} \alpha_n e^{j2\pi nt/T}}_{m=n'} + \; \alpha_0 \; + \underbrace{\sum_{n=1}^{\infty} \alpha_n e^{j2\pi nt/T}}_{m=n}$

$= \alpha_0 + \sum_{m=1}^{\infty} \left (\alpha_m e^{j2\pi mt/T} + \alpha_{-m} e^{-j2\pi mt/T}\right)$

If we assume $x(t) \in \Re \ \forall \ m$, then to make the imaginary parts cancel out

• $\alpha_{-m} = \alpha_m^*$
• $u + u^* = 2 \Re [u]$
• $\alpha_{m} = | \alpha_m |e^{j\phi m} \,$

$= \alpha_0 + \sum_{m=1}^{\infty} 2 \Re \left[\alpha_m e^{j2\pi mt/T}\right] = \alpha_0 + \sum_{m=1}^{\infty} 2 \Re \left[|\alpha_m| e^{j\phi m}e^{j2\pi mt/T}\right] = \alpha_0 + \sum_{m=1}^{\infty} |\alpha_m|2 \Re \left[e^{j(2\pi mt/T+\phi m)}\right]= \alpha_0 + \sum_{m=1}^{\infty} |\alpha_m|2\cos \left (\frac{2\pi mt}{T} + \phi_m \right )$

Changing variables

• $c_0 = \alpha_0 \,$
• $c_m = 2 | \alpha_m| \,$
• $\Theta_m = \phi_m \,$

$= \sum_{m=0}^{\infty} c_m \cos \left (\frac{2\pi mt}{T}+\Theta_m \right)$

## Identities

$e^{j \theta} = \cos \theta + j \sin \theta \,$ Euler's identity linking rectangular and polar coordinates

$\sin x = \frac{e^{jx}-e^{-jx}}{2j} \,$

$\cos x = \frac{e^{jx}+e^{-jx}}{2} \,$

$ = \int_{-\infty}^\infty u^*(x) v(x) dx$

$\alpha_{-m} = \alpha^* \,$

The dirac delta has an infinite height and an area of 1