• Fourier Series

• Theorem
(references: Second Order Linear Partial Differential Equations, 复数形式傅立叶变换的物理意义，相位究竟指的是什么？)

$$\text{Suppose } f(x) \text{ is a periodic function with period } T \text{ and is an integrable function on } [0, T]. \\ \text{Then, the Fourier Series of } f(x) \text{ can be written as }$$

\begin{align} f(x) & = \frac{c_0}{2} + \sum_{n=1}^{\infty} c_ncos(n \cdot \frac{2\pi}{T} \cdot x + \varphi_n) \\ &= \frac{c_0}{2} + \sum_{n=1}^{\infty} c_ncos(\varphi _n)cos(n \cdot \frac{2\pi}{T} \cdot x)+ (-c_n)sin(\varphi _n)sin(n \cdot \frac{2\pi}{T} \cdot x) \\ &\text{( let } a_0 = c_0, \;a_n = c_ncos(\varphi _n) \text{ and } b_n = (-c_n)sin(\varphi _n) \;) \\ &= \frac{a_0}{2} + \sum_{n=1}^{\infty} a_ncos(n \cdot \frac{2\pi}{T} \cdot x)+ b_nsin(n \cdot \frac{2\pi}{T} \cdot x) \\ \\ \text{where } c_n &= \sqrt{a_n^2 + b_n^2} = \sqrt{c_n^2(cos^2(\varphi _n) + sin^2(\varphi _n))} = \sqrt{c_n^2} \;\; (Amplitude)\\ \varphi_n &= tan^{-1}(-\frac{b_n}{a_n}) \;\; (Phase)\\ a_0 &= \frac{1}{T}\int_{0}^{T}f(x)dx \\ a_n &= \frac{1}{T}\int_{0}^{T}f(x) \cdot cos(n \cdot \frac{2\pi}{T} \cdot x)dx \\ b_n &= \frac{1}{T}\int_{0}^{T}f(x) \cdot sin(n \cdot \frac{2\pi}{T} \cdot x)dx \end{align}

• Derivation for the Fourier Coefficients, $$a_0$$, $$a_n$$, $$b_n$$:
(references: Khan - Fourier Series introduction, 8 Fourier Series)
Using the facts that
\begin{align} \int_{0}^{2 \pi} sin(mx)dx &= 0\; , \forall \text{ integer } m \\ \int_{0}^{2 \pi} cos(mx)dx &= 0\; , \forall \text{ non-zero integer } m \\ \int_{0}^{2 \pi} sin(mx)cos(nx)dx &= 0\; , \forall \text{ integer } m \text{, } n\\ \int_{0}^{2 \pi} cos(mx)cos(nx)dx &= \left\{ \begin{matrix} 0\;, &\;\; m \neq n \\ 2\pi\;, &\;\; m = n = 0\\ \pi\;, &\;\; m = n \neq 0 \end{matrix}\right. \\ \int_{0}^{2 \pi} sin(mx)sin(nx)dx &= \left\{ \begin{matrix} 0 \;, &\;\; m \neq n\\ \pi \;, &\;\; m = n \neq0 \end{matrix}\right. \end{align}

to calculate $$\int_{0}^{2 \pi}f(x)dx$$, $$\int_{0}^{2 \pi}f(x)cos(mx)dx$$ and $$\int_{0}^{2 \pi}f(x)sin(mx)dx$$.
Then obtain the Fourier Coefficients,

\begin{align} a_0 &= \frac{1}{\pi}\int_{0}^{2\pi}f(x)dx \\ a_n &= \frac{1}{\pi}\int_{0}^{2\pi}f(x) \cdot cos(n \cdot x)dx \\ b_n &= \frac{1}{\pi}\int_{0}^{2\pi}f(x) \cdot sin(n \cdot x)dx \end{align}

Let $$x = \frac{2\pi}{T} \cdot y$$ and $$dx = \frac{2\pi}{T} \cdot dy$$,
then

\begin{align} a_0 &= \frac{1}{T}\int_{0}^{T}f(x)dx \\ a_n &= \frac{1}{T}\int_{0}^{T}f(x) \cdot cos(n \cdot \frac{2\pi}{T} \cdot x)dx \\ b_n &= \frac{1}{T}\int_{0}^{T}f(x) \cdot sin(n \cdot \frac{2\pi}{T} \cdot x)dx \end{align}

• Complex Form:
(references: wiki-Euler’s formula, Complex Fourier Series)
$$f(x) = \sum_{n=-\infty}^{\infty} d_ne^{i(n\frac{2\pi}{T}x)}$$

By using Euler’s formula, $$e^{ix} = cos(x) + isin(x)$$, we know \begin{align} cos(x) &= \frac{e^{ix} + e^{-ix}}{2} \\ sin(x) &= \frac{e^{ix} - e^{-ix}}{2i} = \frac{-ie^{ix} + ie^{-ix}}{2} \end{align}

The $$f(x)$$ can be rewritten as,

\begin{align} f(x) &= \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n(\frac{1}{2}e^{i(n\frac{2\pi}{T}x)} + \frac{1}{2}e^{-i(n\frac{2\pi}{T}x)}) + b_n(\frac{1}{2}ie^{i(n\frac{2\pi}{T}x)} - \frac{1}{2}ie^{-i(n\frac{2\pi}{T}x)}) \\ &= \frac{a_0}{2} + \sum_{n=1}^{\infty} \frac{(a_n - ib_n)}{2}e^{i(n\frac{2\pi}{T}x)} + \frac{(a_n + ib_n)}{2}e^{-i(n\frac{2\pi}{T}x)} \\ &= \sum_{n=-\infty}^{\infty} d_ne^{i(n\frac{2\pi}{T}x)} \end{align}

where the $$d_n$$ is called the Complex Fourier Coefficients, \begin{align} d_n = \left\{\begin{matrix} \frac{a_n - ib_n}{2}&, \;\;\;n\geq 1\\ \frac{a_0}{2}&, \;\;\;n= 0\\ \frac{a_{\left | n \right |} + ib_{\left | n \right |}}{2}&, \;\;\;n\leq 1 \end{matrix}\right. \end{align}

• Fourier Transformation
It is a process to convert a time function, $$f(x)$$ from time domain ($$x$$) to frequency domain ($$n$$).

Since we know that

\begin{align} \frac{1}{T}\int_{x = 0}^{T}e^{i(n\frac{2\pi}{T}x)}dx &= \frac{1}{T}\int_{x = 0}^{T}cos(n\frac{2\pi}{T}x) + isin(n\frac{2\pi}{T}x)dx \\ &= \left\{\begin{matrix} 0, \;\;\;n \neq 0\\ 1, \;\;\;n = 0 \end{matrix}\right. \end{align}

Then, we can get

\begin{align} \frac{1}{T}\int_{x = 0}^{T}e^{-i(n\frac{2\pi}{T}x)}f(x)dx &= \frac{1}{T}\int_{x = 0}^{T}e^{-i(n\frac{2\pi}{T}x)}\sum_{r=-\infty}^{\infty} d_re^{i(r\frac{2\pi}{T}x)}dx \\ &= \sum_{r=-\infty}^{\infty}\int_{x = 0}^{T}d_re^{i(r-n)\frac{2\pi}{T}x}dx \\ &= d_n \\ &= \left\{\begin{matrix} \frac{a_n - ib_n}{2}&, \;\;\;n\geq 1\\ \frac{a_0}{2}&, \;\;\;n= 0\\ \frac{a_{\left | n \right |} + ib_{\left | n \right |}}{2}&, \;\;\;n\leq 1 \end{matrix}\right. \end{align}

As we can see here, we have converted $$f(x)$$ from time domain ($$x$$) to the frequency domain ($$n$$). This process is called Fourier Transformation. Given a certain n, we know its $$d_n$$ which is composed of $$a_n$$ and $$b_n$$. By doing some calculation, we then get the $$c_n$$ and $$\varphi_n$$ which give us enough information to create a function to approach the orginal function $$f(x)$$.

• Examples in R
http://www.di.fc.ul.pt/~jpn/r/fourier/fourier.html