基本概念

复数

复数$C$的定义为

$$C=R+jI\text{\hspace{1em}(4.3)}$$

$R,I$为实数，$R$是实部，$I$是虚部，$j=\sqrt{-1}$。复数的共轭表示为$C^{\ast }$

$$C^{\ast }=R-jI\text{\hspace{1em}(4.4)}$$

从几何角度来看，复数可视为平面（称为复平面）上的一个点，其横坐标是实轴，纵坐标是虚轴，也就是说$R+jI$是得平面直角坐标系中的点$(R,I)$

极坐标表示复数

$$C=|C|(\text{cos}\theta +\text{jsin}\theta )\text{\hspace{1em}(4.5)}$$
$|C|=\sqrt{R^2+I^2}$是从复平面的原点延伸到点$(R,I)$的向量长度，$\theta$是该向量与实轴的夹角。

使用欧拉公式
$$e^{j\theta }=\text{cos}\theta +\text{jsin}\theta \text{\hspace{1em}(4.6)}$$

可以给坐极坐标的复数表示为
$$C=|C|e^{j\theta }\text{\hspace{1em}(4.7)}$$

复函数：
$F(u)=R(u)+jI(u)$
$F^{\ast }(u)=R(u)-jI(u)$
幅值是
$|F(u)|=\sqrt{R(u{)}^2+I(u{)}^2}$

a = np.complex(1 + 2j)

radian = np.arctan2(a.imag, a.real)
radian

1.1071487177940904

degree = np.rad2deg(radian)
degree

63.43494882292201

傅里叶级数

周期为$T$的连续变量$t$的周期函数$f(t)$，可表示为乘以适当系数的正弦函数和余弦函数之和
$$f(t)=\sum _{n=-\infty }^{\infty }{c}_n{e}^{j\frac{2\pi n}{T}t}\text{\hspace{1em}(4.8)}$$

$$c_n=\frac{1}{T}{\int }_{-T/2}^{T/2}f(t)e^{-j\frac{2\pi n}{T}t}\text{\hspace{1em}(4.9)}$$

冲激函数及其取样（筛选）性质

连续变量$t$在$t=0$处的单位冲激表示为$\delta (t)$，定义是
$$\delta (t)=\{\begin{array}{ll}\infty ,& t=0\\ 0,& t\ne 0\end{array}\text{\hspace{1em}(4.10)}$$
它被限制为满足恒等式
$${\int }_{-\infty }^{\infty }\delta (t)dt=1\text{\hspace{1em}(4.11)}$$

自然地，将$t$解释为时间时，冲激就可视为幅度无限、持续时间为0、具有单位面积的尖峰信息。冲激具有关于积分的所谓取样性质

$${\int }_{-\infty }^{\infty }f(t)\delta (t)dt=f(0)\text{\hspace{1em}(4.12)}$$

冲激
冲激并不是通常意义上的函数，更准确的名称是分布或广义函数。

任意一点$t_0$的取样性质为：
$${\int }_{-\infty }^{\infty }f(t)\delta (t-t_0)dt=f(t_0)\text{\hspace{1em}(4.13)}$$

例如： 若$f(t)=\text{cos}(t)$，则使用冲激$\delta (t-\pi )$得到结果$f(\pi )=cos(\pi )=-1$

冲激串KaTeX parse error: \tag works only in display equations

离散冲激定义为
$$\delta (x)=\{\begin{array}{ll}1,& x=0\\ 0,& x\ne 0\end{array}\text{\hspace{1em}(4.15)}$$
$$\sum _{-\infty }^{\infty }\delta (x)=1\text{\hspace{1em}(4.16)}$$
$$\sum _{-\infty }^{\infty }f(x)\delta (x)=f(0)\text{\hspace{1em}(4.17)}$$
$$\sum _{-\infty }^{\infty }f(x)\delta (x-x_0)dt=f(x_0)\text{\hspace{1em}(4.18)}$$

def impulse(x, x0):return np.piecewise(x, [x==x0, x!=x0], [1, 0])

# 按定义写的，不知道是否正确，如不正确，请指出，感谢，只做展示
x = np.arange(10)
plt.stem(x, impulse(x, 5), )
plt.show()

def impulse_serial(x):s = np.ones_like(x)return s

# 只做展示
x = np.arange(10)
plt.stem(impulse_serial(x))
plt.show()

连续单变量函数的傅里叶变换

傅里叶变换对

连续变量$t$的连续函数$f(t)$的傅里叶变换由J{f(t)}\mathfrak{J}\{f(t)\}J{f(t)}表示，定义为
J{f(t)}=∫−∞∞f(t)e−j2πμtdt(4.19)\mathfrak{J}\{f(t)\} = \int_{-\infty}^{\infty} f(t) e^{-j2\pi\mu t}dt \tag{4.19}J{f(t)}=∫−∞∞​f(t)e−j2πμtdt(4.19)
J{f(t)}=F(μ)\mathfrak{J}\{f(t)\} = F(\mu)J{f(t)}=F(μ)
所以有
$$F(\mu )={\int }_{-\infty }^{\infty }f(t)e^{-j2\pi \mu t}dt\text{\hspace{1em}(4.20)}$$
反变换：
$$f(t)={\int }_{-\infty }^{\infty }F(\mu )e^{-j2\pi \mu t}d\mu \text{\hspace{1em}(4.21)}$$

通常表示为$f(t)\iff F(\mu )$

使用欧拉公式，可以写为
$$F(\mu )={\int }_{-\infty }^{\infty }f(t)[\text{cos}(2\pi \mu t)-\text{jsin}(2\pi \mu t)]dt\text{\hspace{1em}(4.22)}$$

傅里叶变换是$f(t)$乘以正弦函数的展开式， 其中正弦函数的频率是由$\mu$值决定。因此积分后留下的唯一变量是频率，因此，我们说傅里叶变换域是频率域。

盒式函数的傅里叶变换

$\begin{array}{rl}F(\mu )& ={\int }_{-\infty }^{\infty }f(t)e^{-j2\pi \mu t}dt={\int }_{-W/2}^{W/2}f(t)A{e}^{-j2\pi \mu t}dt\\ & =\frac{-A}{j2\pi \mu }[e^{-j2\pi \mu t}{]}_{-W/2}^{W/2}=\frac{-A}{j2\pi \mu }[e^{-j\pi \mu W}-e^{j\pi \mu W}]\\ & =\frac{A}{j2\pi \mu }[e^{j\pi \mu W}-e^{-j\pi \mu W}]\\ & =AW\frac{sin(\pi \mu W)}{(\pi \mu W)}\end{array}$

最后的结果是$\text{sinc}$函数
$$\text{sinc}(m)=\frac{\text{sin}(\pi m)}{(\pi m)}\text{\hspace{1em}(4.23)}$$

通常傅里叶变换中包含复数项，这是为了显示变换的幅值（一个实量）的约定。这个幅值称为傅里叶频谱或频谱：
$$|F(\mu )|=AW\left|\frac{sin(\pi \mu W)}{(\pi \mu W)}\right|$$

def box_function(x, w):w_2 = w / 2y = np.where(x, x > -w_2, 0)y = np.where(x < w_2, y, 0)return y

import mpl_toolkits.axisartist as axisartist
def setup_axes(fig, rect):ax = axisartist.SubplotZero(fig, rect)fig.add_axes(ax)for direction in ["xzero", "yzero"]:# adds arrows at the ends of each axisax.axis[direction].set_axisline_style("-|>")# adds X and Y-axis from the originax.axis[direction].set_visible(True)for direction in ["left", "right", "bottom", "top"]:# hides bordersax.axis[direction].set_visible(False)return ax

# 盒式函数的傅里叶变换
x = np.arange(-5, 5, 0.1)
y = box_function(x, 6)fig = plt.figure(figsize=(15, 5))
ax_1 = setup_axes(fig, 131)
ax_1.plot(x, y), ax_1.set_title('f(t)', loc='center', y=1.05), ax_1.set_ylim([0, 2]), ax_1.set_yticks([])f_u = np.sinc(x)
ax_2 = setup_axes(fig, 132)
ax_2.plot(x, f_u), ax_2.set_title('F(u)', loc='center', y=1.05), ax_2.set_yticks([]), #ax_2.set_ylim([-1, 2]),f_u_absolute = abs(f_u)
ax_3 = setup_axes(fig, 133)
ax_3.plot(x, f_u_absolute), ax_3.set_title('|F(u)|', loc='center', y=1.05), ax_3.set_yticks([]), #ax_3.set_ylim([-1, 2]),plt.tight_layout()
plt.show()

冲激和冲激串的傅里叶变换

J{δ(t)}=F(μ)=∫−∞∞δ(t)e−j2πμtdt=∫−∞∞e−j2πμtδ(t)dt=e−j2πμ0=e0=1\begin{aligned} \mathfrak{J}\{\delta(t)\} & = F(\mu) = \int_{-\infty}^{\infty} \delta(t) e^{-j2\pi\mu t}dt \\ & = \int_{-\infty}^{\infty} e^{-j2\pi\mu t} \delta(t)dt \\ & = e^{-j2\pi\mu_0} \\ & = e^0 = 1 \end{aligned}J{δ(t)}​=F(μ)=∫−∞∞​δ(t)e−j2πμtdt=∫−∞∞​e−j2πμtδ(t)dt=e−j2πμ0​=e0=1​

J{δ(t−t0)}=F(μ)=∫−∞∞δ(t−t0)e−j2πμtdt=∫−∞∞e−j2πμtδ(t−t0)dt=e−j2πμt0\begin{aligned} \mathfrak{J}\{\delta(t - t_0)\} & = F(\mu) = \int_{-\infty}^{\infty} \delta(t - t_0) e^{-j2\pi\mu t}dt \\ & = \int_{-\infty}^{\infty} e^{-j2\pi\mu t} \delta(t - t_0)dt \\ & = e^{-j2\pi\mu t_0} \\ \end{aligned}J{δ(t−t0​)}​=F(μ)=∫−∞∞​δ(t−t0​)e−j2πμtdt=∫−∞∞​e−j2πμtδ(t−t0​)dt=e−j2πμt0​​

冲激串的傅里叶变换

S(μ)=J{SΔT(t)}=J{1Δ∑n=−∞∞ej2πnΔTt}=1ΔTJ{∑n=−∞∞ej2πnΔTt}=1ΔT∑n=−∞∞δ(μ−nΔT)\begin{aligned} S(\mu) & = \mathfrak{J}\{S_{\Delta T}(t)\} = \mathfrak{J}\{\frac{1}{\Delta } \sum_{n=-\infty}^{\infty} e^{j\frac{2\pi n}{\Delta T}t}\} \\ & = \frac{1}{\Delta T} \mathfrak{J}\{\sum_{n=-\infty}^{\infty} e^{j\frac{2\pi n}{\Delta T}t}\} \\ & = \frac{1}{\Delta T} \sum_{n=-\infty}^{\infty} \delta(\mu - \frac{n}{\Delta T}) \\ \end{aligned}S(μ)​=J{SΔT​(t)}=J{Δ1​n=−∞∑∞​ejΔT2πn​t}=ΔT1​J{n=−∞∑∞​ejΔT2πn​t}=ΔT1​n=−∞∑∞​δ(μ−ΔTn​)​

卷积

$$(f\star h)(t)={\int }_{-\infty }^{\infty }f(\tau )h(t-\tau )d\tau \text{\hspace{1em}(4.24)}$$

J{(f⋆h)(t)}=∫−∞∞[∫−∞∞f(τ)h(t−τ)dτ]e−j2πμtdt=∫−∞∞f(τ)[∫−∞∞h(t−τ)e−j2πμtdt]dτ=H(μ)∫−∞∞f(τ)e−j2πμτdτ=H(μ)F(μ)=(H∙F)(μ)\begin{aligned} \mathfrak{J}\{(f \star h)(t) \} & = \int_{-\infty}^{\infty} \bigg[\int_{-\infty}^{\infty} f(\tau) h(t - \tau) d\tau \bigg] e^{-j2\pi \mu t} dt\\ & = \int_{-\infty}^{\infty} f(\tau) \bigg[\int_{-\infty}^{\infty} h(t - \tau) e^{-j2\pi \mu t} dt\bigg] d\tau \\ & = H(\mu) \int_{-\infty}^{\infty} f(\tau) e^{-j2\pi \mu \tau} d\tau \\ & = H(\mu)F(\mu) = (H \bullet F)(\mu) \end{aligned}J{(f⋆h)(t)}​=∫−∞∞​[∫−∞∞​f(τ)h(t−τ)dτ]e−j2πμtdt=∫−∞∞​f(τ)[∫−∞∞​h(t−τ)e−j2πμtdt]dτ=H(μ)∫−∞∞​f(τ)e−j2πμτdτ=H(μ)F(μ)=(H∙F)(μ)​

• 傅里叶正变换
  $$(f\star h)(t)\iff (H\bullet F)(\mu )\text{\hspace{1em}(4.25)}$$

• 傅里叶反变换
  $$(f\bullet h)(t)\iff (H\star F)(\mu )\text{\hspace{1em}(4.26)}$$