IWSR

The sixth lecture of Games101 covers the topic of Fourier transforms. After spending some time after class to understand it (even though I studied it in college, I had mostly forgotten), I wrote this article to record my learning insights.

I strongly recommend the series "Purely Practical Mathematical Derivation" by Teacher DR_CAN in the reference materials! Most of this article consists of notes from his course.

This article will derive from the following aspects in order:

• Orthogonality of trigonometric functions
• Fourier series expansion for functions with a period of 2 $\pi$
• Fourier series expansion for functions with a period of T
• Complex form of Fourier series
• Fourier series expansion for non-periodic functions

It should be noted that due to my limited knowledge, I cannot provide a proof for the Fourier series expansion formula, just as I cannot prove why G=mg. For truths, it may be simpler to use them directly. However, for readers who feel uncomfortable not understanding, I recommend the article written by Teacher Wang Jiliang, which provides a detailed introduction to how Fourier extracted the Fourier series from the heat conduction equation.

Fourier series expansion : $f(x) = \frac{a_0}{2} + \varSigma_{n=1}^{+\infty}(a_{n}\cos nx + b_{n}\sin nx)$

Orthogonality of Trigonometric Functions#

If there exist functions f(x) and g(x) such that $\int_{a}^{b} f(x)g(x) dx = 0$, then we say that the functions f(x) and g(x) are orthogonal on the interval [a, b].

For the trigonometric functions $0(\sin 0), 1(\cos 0), \sin x, \cos x, \sin 2x, \cos 2x..... \sin nx, \cos nx$, they are orthogonal on the interval [- $\pi$, $\pi$ ].

Copy$$\begin{cases} \int_{-\pi}^{\pi} \cos (nx) \sin (mx) dx = 0, \\ \int_{-\pi}^{\pi} \cos (nx) \cos (mx) dx = 0, & n \neq m \\ \int_{-\pi}^{\pi} \sin (nx) \sin (mx) dx = 0 & n \neq m \end{cases}$$

The above integrals can be proven using the product-to-sum formulas of trigonometric functions. Here, only one calculation process is provided for reference.

Copy$$\int_{-\pi}^{\pi} \cos (nx) \sin (mx) dx = \frac{1}{2} \int_{-\pi}^{\pi} \sin(n+m)x dx - \frac{1}{2} \int_{-\pi}^{\pi} \sin(n-m)x dx \\ = \frac{1}{2} [-\frac{1}{n+m} \cos (n+m)x |_{-\pi}^{\pi} + \frac{1}{n-m} \cos (n-m)x |_{-\pi}^{\pi}] \\ = 0$$

So what happens if n = m? For combinations like sin and cos, it doesn't matter, but for cos and cos or sin and sin, it is equivalent to integrating the product of two identical trigonometric functions. Below is the solution for the combination of $\cos mx$ and $\cos mx$.

Copy$$\int_{-\pi}^{\pi} \cos (mx) \cos (mx) dx = \int_{-\pi}^{\pi} \frac{1 + \cos (2m)x}{2} dx \\ = \frac{1}{2} (x + \frac{1}{2m} \sin{2mx})|_{-\pi}^{\pi} = \pi$$

Remember the result derived here, as it will be used in the next section.

Fourier Series Expansion for Functions with a Period of 2 $\pi$#

At the beginning, the relevant formulas for Fourier series were directly given. Next, we will introduce how to combine the orthogonality of trigonometric functions to find the coefficients in the formula.

Finding $a_{0}$#

Integrate both sides of the Fourier series expansion formula from $-\pi$ to $\pi$.

Copy$$\int_{-\pi}^{\pi} f(x) dx \\ = \int_{-\pi}^{\pi} (\frac{a_0}{2} + \varSigma_{n=1}^{+\infty}(a_{n}\cos nx + b_{n}\sin nx)) dx \\ = \int_{-\pi}^{\pi} \frac{a_0}{2} dx + a_{n} \int_{-\pi}^{\pi} \varSigma_{n=1}^{+\infty} 1 \cos nx dx + b_{n} \int_{-\pi}^{\pi} \varSigma_{n=1}^{+\infty} 1 \sin nx dx$$

After integrating both sides, it is obvious that the combinations of 1, $\cos nx$ and 1, $\sin nx$ can be eliminated based on the orthogonality of trigonometric functions, leading to the following equality.

Copy$$\int_{-\pi}^{\pi} f(x) dx = \int_{-\pi}^{\pi} \frac{a_0}{2} dx \\ = \frac{a_0}{2} \int_{-\pi}^{\pi} 1 dx \\ = \frac{a_0}{2} x|_{-\pi}^{\pi} \\ = a_{0} \pi$$

Thus, we obtain $a_{0}$ as

Copy$$a_{0} = \frac{1}{2}\int_{-\pi}^{\pi} f(x) dx$$

Finding $a_{n}$#

Multiply both sides of the Fourier series expansion by $\cos mx$ and integrate.

Copy$$\int_{-\pi}^{\pi} f(x) \cos mx dx \\ = \int_{-\pi}^{\pi} (\frac{a_0}{2} \cos mx + \varSigma_{n=1}^{+\infty}(a_{n}\cos nx \cos mx + b_{n}\sin nx \cos mx)) dx \\ = \frac{a_0}{2} \int_{-\pi}^{\pi} 1 \cos mx dx + a_{n} \int_{-\pi}^{\pi} \varSigma_{n=1}^{+\infty} \cos nx \cos mx dx + b_{n} \int_{-\pi}^{\pi} \varSigma_{n=1}^{+\infty} \sin nx \cos mx dx$$

At this point, we can directly eliminate $\frac{a_0}{2} \int_{-\pi}^{\pi} 1 \cos mx dx$ and $b_{n} \int_{-\pi}^{\pi} \varSigma_{n=1}^{+\infty} \sin nx \cos mx dx$ based on the orthogonality of trigonometric functions, as they will both equal 0. However, we cannot eliminate the middle term $a_{n} \int_{-\pi}^{\pi} \varSigma_{n=1}^{+\infty} \cos nx \cos mx dx$ entirely, because when $m = n$, its integral result is $\pi$.

Thus, we obtain

Copy$$\int_{-\pi}^{\pi} f(x) \cos mx dx = a_{n} \pi$$

Rearranging gives

Copy$$a_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos mx dx$$

Finding $b_{n}$#

Multiply both sides of the Fourier series expansion by $\sin mx$ and integrate.

Copy$$\int_{-\pi}^{\pi} f(x) \sin mx dx = \int_{-\pi}^{\pi} (\frac{a_0}{2} \sin mx + \varSigma_{n=1}^{+\infty}(a_{n}\cos nx \sin mx + b_{n}\sin nx \sin mx)) dx$$

The proof process is omitted, as it follows the same reasoning as above, yielding

Copy$$b_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin mx dx$$

Fourier Series Expansion for Functions with a Period of T#

The method for expanding functions with a period of T is somewhat clever; it essentially transforms the expression through a parameter change to force it into the form of $\pi$.

Let T = 2L, then we have f(t) = f(t + 2L).

Let x = $\frac{\pi}{L}$ t, then t = $\frac{Lx}{\pi}$, thus f(t) = f( $\frac{Lx}{\pi}$ ).

t	x
0	0
2L	2 $\pi$
4L	4 $\pi$

If we consider g(x) as f( $\frac{L}{\pi}$ (x)), the corresponding relationship from the above table shows that the graph of g(x) can correspond to f(t), for example, f(2L) = g(2 $\pi$). The period of g(x) is 2 $\pi$, and we have already obtained the coefficients for the Fourier expansion of functions with a period of 2 $\pi$ in the previous section.

Copy$$\begin{cases} a_{0} = \frac{1}{2}\int_{-\pi}^{\pi} g(x) dx \\ a_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} g(x) \cos nx dx \\ b_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} g(x) \sin nx dx \end{cases}$$

Next, we just need to substitute x = $\frac{\pi}{L}$ t into the equations:

Copy$$\begin{cases} \cos nx = \cos \frac{n \pi}{L}t \\ \sin nx = \sin \frac{n \pi}{L}t \\ \int_{-\pi}^{\pi} 1 dx = \int_{-L}^{L} 1 d\frac{\pi t}{L} \\ g(x) = f(t) \end{cases}$$

Thus, we obtain the Fourier series expansion for functions with a period of T:

Copy$$f(t) = \frac{a_0}{2} + \varSigma_{n=1}^{+\infty}(a_{n}\cos \frac{n \pi}{L}t + b_{n}\sin \frac{n \pi}{L}t)$$

Where (the calculation process is omitted, and the method is consistent with the previous section):

Copy$$\begin{cases} a_{0} = \frac{1}{L}\int_{-L}^{L} f(t) dt \\ a_{n} = \frac{1}{L} \int_{-L}^{L} f(t) \cos \frac{n \pi}{L} t dt \\ b_{n} = \frac{1}{L} \int_{-L}^{L} f(t) \sin \frac{n \pi}{L} t dt \end{cases}$$

If we replace L with T, it will become:

Copy$$f(t) = \frac{a_0}{2} + \varSigma_{n=1}^{+\infty}(a_{n}\cos \frac{2n \pi}{T}t + b_{n}\sin \frac{2n \pi}{T}t)$$

Copy$$\begin{cases} a_{0} = \frac{2}{T}\int_{0}^{T} f(t) dt \\ a_{n} = \frac{2}{T} \int_{0}^{T} f(t) \cos \frac{2 n \pi}{T} t dt \\ b_{n} = \frac{2}{T} \int_{0}^{T} f(t) \sin \frac{2 n \pi}{T} t dt \end{cases}$$

Complex Form of Fourier Series#

This section will use Euler's formula, and I will write another article later to prove Euler's formula (the proof requires Taylor expansion, although Euler himself did not derive it this way).

Copy$$e^{i\theta} = \cos \theta + i \sin \theta$$

Copy$$\begin{cases} e^{i\theta} = \cos \theta + i \sin \theta \\ e^{i-\theta} = \cos -\theta + i \sin -\theta = \cos \theta - i \sin \theta \end{cases}$$

From the above, we can derive:

Copy$$\begin{cases} \sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i} = -i \frac{e^{i \theta} - e^{-i \theta}}{2} \\ \cos \theta = \frac{e^{i \theta} + e^{- i \theta}}{2} \end{cases}$$

Let $x = \frac{2 \pi t}{T}$ and substitute this into the Fourier series to obtain:

Copy$$f(x) = \frac{a_0}{2} + \varSigma_{n=1}^{+\infty}(a_{n} \frac{e^{i nx} + e^{- i nx}}{2} -i b_{n} \frac{e^{i nx} - e^{-i nx}}{2}) \\ = \frac{a_0}{2} + \varSigma_{n=1}^{+\infty} \frac{a_{n} - i b_{n}}{2} e^{i nx} + \varSigma_{n=1}^{+\infty} \frac{a_{n}+i b_{n}}{2} e^{-i nx}$$

The subsequent transformations may be a bit confusing, but once understood, it will make sense.

Treating $\frac{a_{0}}{2}$ as $\varSigma_{n=0}^{0} \frac{a_{0}}{2} e^{i nx}$, since when n = 0, $e^{i nx}$ also equals 1, thus it holds.

The transformation of $\varSigma_{n=1}^{+\infty} e^{-i nx}$ is also interesting; treating n as -1 to - $\infty$, the expression becomes $\varSigma_{n=-1}^{-\infty} e^{i nx}$.

Thus, we obtain:

Copy$$f(x) = \varSigma_{n = -\infty}^{+\infty} C_{n} e^{i nx}$$

Replacing x gives:

Copy$$f(t) = \varSigma_{n = -\infty}^{+\infty} C_{n} e^{i \frac{2 n \pi}{T} t}$$

Where:

Copy$$Cn = \begin{cases} \frac{a_{0}}{2}, & n = 0 \\ \frac{a_{n} - i b_{n}}{2} & n = 1,2,3... \\ \frac{a_{-n}+i b_{-n}}{2} & n = -1, -2, -3... \end{cases}$$

Since we have already derived the Fourier series expansion for functions with a period of T in the previous section:

Copy$$\begin{cases} a_{0} = \frac{2}{T}\int_{0}^{T} f(t) dt \\ a_{n} = \frac{2}{T} \int_{0}^{T} f(t) \cos \frac{2 n \pi}{T} t dt \\ b_{n} = \frac{2}{T} \int_{0}^{T} f(t) \sin \frac{2 n \pi}{T} t dt \end{cases}$$

Substituting gives:

Copy$$Cn = \begin{cases} \frac{1}{T}\int_{0}^{T} f(t) dt, & n = 0 \\ \frac{1}{T} \int_{0}^{T} f(t) (\cos \frac{2n\pi}{T} t - i \sin \frac{2n\pi}{T} t) dt & n > 0 \\ \frac{1}{T} \int_{0}^{T} f(t) (\cos \frac{2n\pi}{T} t - i \sin \frac{2n\pi}{T} t) dt & n < 0 \end{cases}$$

However, the expression is somewhat awkward, so we simplify using Euler's formula:

Copy$$\cos \frac{2n\pi}{T} - i \sin \frac{2n\pi}{T} = \cos -\frac{2n\pi}{T} + i \sin -\frac{2n\pi}{T} = e^{-i \frac{2n\pi}{T}}$$

Copy$$Cn = \begin{cases} \frac{1}{T}\int_{0}^{T} f(t) dt, & n = 0 \\ \frac{1}{T} \int_{0}^{T} f(t) e^{-i \frac{2n\pi}{T} t} dt & n > 0 \\ \frac{1}{T} \int_{0}^{T} f(t) e^{-i \frac{2n\pi}{T} t} dt & n < 0 \end{cases}$$

Notice that the results for n > 0 and n < 0 are the same. Also, for n = 0, we can see that its expression is also the same as for n > 0 and n < 0 (after substituting n=0, $e^{-i \frac{ 2n \pi }{T}}$ results in 1).

Thus, the expression for $C_{n}$ can be unified as:

Copy$$C_{n} = \frac{1}{T} \int_{0}^{T} f(t) e^{-i \frac{ 2n \pi }{T} t} dt$$

At this point, we let $\omega_{0}$ be $\frac{ 2 \pi }{T}$ (the angular frequency in engineering), yielding:

Copy$$C_{n} = \frac{1}{T} \int_{0}^{T} f(t) e^{-i n \omega_{0} t} dt$$

Thus, we have obtained the complex form of the Fourier series:

Copy$$\begin{cases} f(t) = \varSigma_{n = -\infty}^{+\infty} C_{n} e^{i n \omega_{0} t} \\ C_{n} = \frac{1}{T} \int_{0}^{T} f(t) e^{-i n \omega_{0} t} dt \end{cases}$$

Next, let's make a small transformation.

Replace the limits of integration with [-T/2, T/2], and then substitute back into the original expression (the definite integral of periodic functions will yield the same result as long as the difference in limits is one period):

Copy$$\begin{cases} f(t) = \varSigma_{n = -\infty}^{+\infty} C_{n} e^{i n \omega_{0} t} \\ C_{n} = \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(t) e^{-i n \omega_{0} t} dt \end{cases}$$

Fourier Series Expansion for Non-Periodic Functions#

The treatment for expanding non-periodic functions into Fourier series is also quite clever. The general idea is to treat the non-periodic function as a periodic function with an infinite period, and then use the expansion formula for periodic functions for transformation.

Let $\frac{1}{T}$ be $f$, and since T = $\infty$, $f$ approaches zero, leading to:

Copy$$f(t) = \varSigma_{n = -\infty}^{+\infty} (\int_{-\infty}^{\infty} f(t) e^{-i n \omega_{0} t} dt) e^{i n \omega_{0} t} f$$

Doesn't that look familiar? (the definition of definite integrals). Since $f$ approaches 0 infinitely, and $\omega_{0} = \frac{ 2 \pi }{T} = 2 \pi f$, we can apply the sum-to-integral transformation to obtain:

Copy$$f(t) = \int_{-\infty}^{\infty} (\int_{-\infty}^{\infty} f(t) e^{-i 2 \pi f t} dt) e^{i 2 \pi f t} df$$

However, I think everyone might be more familiar with the following form:

Copy$$f(t) = \frac{1}{ 2 \pi } \int_{-\infty}^{\infty} (\int_{-\infty}^{\infty} f(t) e^{-i \omega t} dt) e^{i \omega t} d \omega$$

Both expressions are essentially the same, just one is from the perspective of frequency and the other from angular velocity.

Where:

Copy$$F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t} dt$$

is the Fourier transform.

And:

Copy$$f(t) = \frac{1}{ 2 \pi } \int_{-\infty}^{\infty} F(\omega) e^{i \omega t} d \omega$$

is the inverse Fourier transform.

Reference Materials#

Purely Practical Mathematical Derivation_Fourier Series and Fourier Transform_Part5_Deriving Fourier Transform from Fourier Series

Frontier Practices in the Internet of Things - Fourier Series

First release on Bilibili! A 【Fourier Transform】 explanation that even a paramecium can understand, taught by Teacher Li Yongle from Tsinghua University, helping you understand Fourier Transform and distinguish the principles of beautification and voice transformation...

Visual demonstration of Fourier Transform