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1. 三角函数基本性质

本文主要用于复习一下傅里叶级数、傅里叶变换的基础
基本定理

三角函数的正交性: 频率不同的三角函数乘积在一个周期内的积分是0，即：

\int_{- π}^{π} s i n (m x \pm \frac{π}{2}) c o s (n x \pm \frac{π}{2}) d x = 0 m \neq n (0) \int_{0}^{T} s i n (x \pm \frac{π}{2}) d t = 0

$\int_{-\pi}^{\pi}sin(mx\pm\frac{\pi}{2})cos(nx\pm\frac{\pi}{2})dx = 0 \quad m \neq n \quad\quad (0) \\ \int_{0}^T{sin(x\pm\frac{\pi}{2})}dt = 0$

对于任意 $m n \quad m \geq 1 \quad n \geq 1$ ,

\int_{- π}^{π} s i n (m x) c o s (n x) d x = 0 (1)

$\int_{-\pi}^{\pi}sin(mx)cos(nx)dx = 0 \quad\quad(1)$

和差化积、积化和差公式
根据欧拉公式:

e^{i θ} = c o s (θ) + i s i n (θ) e^{i (θ + α)} = c o s (θ + α) + i s i n (θ + α) = e^{i α} e^{i α} = [c o s (θ) + i s i n (θ)] [c o s (α) + i s i n (α)] = c o s (θ) c o s (α) - s i n (θ) s i n (α) + i [c o s (θ) s i n (α) + s i n (θ) c o s (α)] \to c o s (θ + α) = c o s (θ) c o s (α) - s i n (θ) s i n (α) s i n (θ + α) = c o s (θ) s i n (α) + s i n (θ) c o s (α) c o s (2 x) = c o s (x)^{2} - s i n (x)^{2}

$e^{i\theta} = cos(\theta) + isin(\theta) \\ e^{i(\theta+\alpha)} = cos(\theta + \alpha) + isin(\theta + \alpha) = e^{i\alpha}e^{i\alpha} = \\ [cos(\theta) + isin(\theta)][cos(\alpha) + isin(\alpha)] = \\ cos(\theta)cos(\alpha) - sin(\theta)sin(\alpha) +i[cos(\theta)sin(\alpha) + sin(\theta)cos(\alpha)] \rightarrow\\ cos(\theta + \alpha) = cos(\theta)cos(\alpha) - sin(\theta)sin(\alpha) \\ sin(\theta + \alpha) = cos(\theta)sin(\alpha) + sin(\theta)cos(\alpha) \\ cos(2x) = cos(x)^2 - sin(x)^2$

由和差化积可以得到积化和差公式:

s i n (θ) s i n (α) = \frac{s i n (θ + α) - c o s (θ + α)}{2} . . . .

$sin(\theta)sin(\alpha) = \frac{sin(\theta + \alpha) - cos(\theta + \alpha)}{2}\\ ....$

2. 实数域傅里叶级数

满足狄利克雷条件的周期函数，可以展开成傅里叶级数，傅里叶级数表示为:

f (t) = c_{0} + \sum_{n = 1}^{\infty} c_{n} c o s (n ω t + ϕ) = c_{0} + \sum_{n = 1}^{\infty} c_{n} c o s (ϕ) c o s (n ω t) - c_{n} s i n (ϕ) s i n (n ω t)

$f(t) = c_0 + \sum_{n=1}^{\infty}{c_ncos(n\omega t + \phi)} = \\ c_0 + \sum_{n=1}^{\infty}{c_ncos(\phi)cos(n\omega t) - c_nsin(\phi)sin(n\omega t)}$

$c_0$ 是其中的直流分量, 令 $a_n=c_ncos(\phi), b_n=-c_nsin(\phi)$ , 上式写作:

f (t) = c_{0} + \sum_{n = 1}^{\infty} [a_{n} c o s (n ω t) + b_{n} s i n (n ω t)]

$f(t) = c_0 + \sum_{n=1}^{\infty}{[a_ncos(n\omega t) + b_nsin(n\omega t)]}$

对上述级数的系数求解方式如下
令 $K(t) = f(t)sin(k\omega t)$ ，则：

\int_{0}^{T} K (t) d t = \int_{0}^{T} c_{0} s i n (k ω t) d t + \int_{0}^{T} s i n (k ω t) \sum_{n = 1}^{\infty} [a_{n} c o s (ω t) + b_{n} s i n (ω t)] d t

$\int_{0}^{T}K(t)dt = \\ \int_{0}^{T}c_0sin(k\omega t)dt + \int_{0}^{T}{sin(k\omega t) \sum_{n=1}^{\infty}[a_ncos(\omega t) + b_nsin(\omega t)]}dt$

根据第一节的定理，可以得知:
$\int_{0}^{T}c_0sin(k\omega t)dt = 0$

\int_{0}^{T} s i n (k ω t) \sum_{n = 1}^{\infty} [a_{n} c o s (n ω t) + b_{n} s i n (n ω t)] d t = \int_{0}^{T} \sum_{0}^{\infty} [s i n (k ω t) a_{n} c o s (n ω t)] d t + \int_{0}^{T} \sum_{0}^{\infty} [s i n (k ω t) b_{n} s i n (n ω t)] d t = \sum_{n = 1}^{\infty} \int_{0}^{T} [s i n (k ω t) a_{n} c o s (n ω t)] d t + \sum_{n = 1}^{\infty} \int_{0}^{T} [s i n (k ω t) b_{n} s i n (n ω t)] d t = a (t) + b (t)

$\int_{0}^{T}{sin(k\omega t) \sum_{n=1}^{\infty}[a_ncos(n\omega t) + b_nsin(n\omega t)]}dt = \\ \int_{0}^{T}{\sum_0^{\infty}[sin(k\omega t)a_ncos(n\omega t)]}dt + \int_{0}^{T}{\sum_0^{\infty}[sin(k\omega t)b_nsin(n\omega t)]}dt = \\ \sum_{n=1}^{\infty}{\int_0^{T}[sin(k\omega t)a_ncos(n\omega t)]dt} + \sum_{n=1}^{\infty}{\int_0^{T}[sin(k\omega t)b_nsin(n\omega t)]dt} = \\ a(t) + b(t)$

根据(1), $a(t) = 0$ , 根据(0), 当 $k=n，b(t) \neq 0$ ，此时继续推导:

\sum_{n = 1}^{\infty} \int_{0}^{T} [s i n (k ω t) b_{n} s i n (n ω t)] d t = b_{n} \int {s i n (ω t)}^{2} d t

$\sum_{n=1}^{\infty}{\int_0^{T}[sin(k\omega t)b_nsin(n\omega t)]dt} = b_n\int {sin(\omega t)}^2 dt$

根据:

s i n (x)^{2} + c o s (x)^{2} = 1 c o s (2 x) = c o s (x)^{2} - s i n (x)^{2} = (1 - s i n (x)^{2}) - s i n (x)^{2} s i n^{2} (x) = \frac{1 - c o s (2 x)}{2} b_{n} \int {s i n (ω t)}^{2} d t = b_{n} \int \frac{1 - c o s (2 t)}{2} d t = \frac{b_{n} T}{2} b_{n} = \frac{2}{T} \int f (t) s i n (n ω t) d t

$sin(x)^2 + cos(x)^2 = 1\\ cos(2x) = cos(x)^2 - sin(x)^2 = (1-sin(x)^2) - sin(x)^2 \\ sin^2(x) = \frac{1-cos(2x)}{2} \\ b_n\int {sin(\omega t)}^2 dt = b_n \int{ \frac{1-cos(2t)}{2}}dt=\frac{b_nT}{2} \\ b_n = \frac{2}{T}\int{f(t)sin(n\omega t)}dt$

类似的，求出 $a_n$