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1. Ch0-1矩阵的导数运算

1.1标量向量方程对向量求导，分母布局，分子布局

1.1.1 标量方程对向量的导数

• $y$ 为 一元向量 或 二元向量
  
• $y$为多元向量
  $$\vec{y}=\left[y_1,y_2,\cdots ,y_{\mathrm{n}}\right]\Rightarrow \frac{\partial f\left(\vec{y}\right)}{\partial \vec{y}}$$
  其中：$f\left(\vec{y}\right)$ 为标量 $1\times 1$, $\vec{y}$为向量 $1\times n$

1. 分母布局 Denominator Layout——行数与分母相同
   $\frac{\partial f\left(\vec{y}\right)}{\partial \vec{y}}={\left[\begin{array}{c}\frac{\partial f\left(\vec{y}\right)}{\partial y_1}\\ ⋮\\ \frac{\partial f\left(\vec{y}\right)}{\partial y_{\mathrm{n}}}\end{array}\right]}_{n\times 1}$
2. 分子布局 Nunerator Layout——行数与分子相同
   $\frac{\partial f\left(\vec{y}\right)}{\partial \vec{y}}={\left[\begin{array}{ccc}\frac{\partial f\left(\vec{y}\right)}{\partial y_1}& \cdots & \frac{\partial f\left(\vec{y}\right)}{\partial y_{\mathrm{n}}}\end{array}\right]}_{1\times n}$

1.1.2 向量方程对向量的导数

$\vec{f}\left(\vec{y}\right)={\left[\begin{array}{c}{\vec{f}}_1\left(\vec{y}\right)\\ ⋮\\ {\vec{f}}_{\mathrm{n}}\left(\vec{y}\right)\end{array}\right]}_{n\times 1},\vec{y}={\left[\begin{array}{c}{y}_1\\ ⋮\\ y_{\mathrm{m}}\end{array}\right]}_{\mathrm{m}\times 1}$
$\frac{\partial \vec{f}{\left(\vec{y}\right)}_{n\times 1}}{\partial {\vec{y}}_{\mathrm{m}\times 1}}={\left[\begin{array}{c}\frac{\partial \vec{f}\left(\vec{y}\right)}{\partial y_1}\\ ⋮\\ \frac{\partial \vec{f}\left(\vec{y}\right)}{\partial y_{\mathrm{m}}}\end{array}\right]}_{\mathrm{m}\times 1}={\left[\begin{array}{ccc}\frac{\partial f_1\left(\vec{y}\right)}{\partial y_1}& \cdots & \frac{\partial f_{\mathrm{n}}\left(\vec{y}\right)}{\partial y_1}\\ ⋮& \ddots & ⋮\\ \frac{\partial f_1\left(\vec{y}\right)}{\partial y_{\mathrm{m}}}& \cdots & \frac{\partial f_{\mathrm{n}}\left(\vec{y}\right)}{\partial y_{\mathrm{m}}}\end{array}\right]}_{\mathrm{m}\times \mathrm{n}}$, 为分母布局

若： $\vec{y}={\left[\begin{array}{c}{y}_1\\ ⋮\\ y_{\mathrm{m}}\end{array}\right]}_{\mathrm{m}\times 1},A=\left[\begin{array}{ccc}{a}_{11}& \cdots & a_{1\mathrm{n}}\\ ⋮& \ddots & ⋮\\ a_{\mathrm{m}1}& \cdots & a_{\mathrm{mn}}\end{array}\right]$, 则有：

• $\frac{\partial A\vec{y}}{\partial \vec{y}}=A^{\mathrm{T}}$(分母布局)
• $\frac{\partial {\vec{y}}^{\mathrm{T}}A\vec{y}}{\partial \vec{y}}=A\vec{y}+A^{\mathrm{T}}\vec{y}$, 当$A=A^{\mathrm{T}}$时, $\frac{\partial {\vec{y}}^{\mathrm{T}}A\vec{y}}{\partial \vec{y}}=2A\vec{y}$

若为分子布局，则有：$\frac{\partial A\vec{y}}{\partial \vec{y}}=A$

1.2 案例分析，线性回归

• $\frac{\partial A\vec{y}}{\partial \vec{y}}=A^{\mathrm{T}}$(分母布局)
• $\frac{\partial {\vec{y}}^{\mathrm{T}}A\vec{y}}{\partial \vec{y}}=A\vec{y}+A^{\mathrm{T}}\vec{y}$, 当$A=A^{\mathrm{T}}$时, $\frac{\partial {\vec{y}}^{\mathrm{T}}A\vec{y}}{\partial \vec{y}}=2A\vec{y}$

Linear Regression 线性回归
$$\hat{z}=y_1+y_{2}x\Rightarrow J=\sum _{i=1}^n{\left[z_i-\left(y_1+y_2{x}_i\right)\right]}^2$$
找到 $y_1,y_2$ 使得 $J$最小

$\vec{z}=\left[\begin{array}{c}{z}_1\\ ⋮\\ z_{\mathrm{n}}\end{array}\right],\left[\vec{x}\right]=\left[\begin{array}{lc}1& x_1\\ ⋮& ⋮\\ 1& x_{\mathrm{n}}\end{array}\right],\vec{y}=\left[\begin{array}{c}{y}_1\\ y_2\end{array}\right]\Rightarrow \hat{\vec{z}}=\left[\vec{x}\right]\vec{y}=\left[\begin{array}{c}{y}_1+y_2{x}_1\\ ⋮\\ y_1+y_2{x}_{\mathrm{n}}\end{array}\right]$
$J={\left[\vec{z}-\hat{\vec{z}}\right]}^{\mathrm{T}}\left[\vec{z}-\hat{\vec{z}}\right]={\left[\vec{z}-\left[\vec{x}\right]\vec{y}\right]}^{\mathrm{T}}\left[\vec{z}-\left[\vec{x}\right]\vec{y}\right]=\vec{z}{\vec{z}}^{\mathrm{T}}-{\vec{z}}^{\mathrm{T}}\left[\vec{x}\right]\vec{y}-{\vec{y}}^{\mathrm{T}}{\left[\vec{x}\right]}^{\mathrm{T}}\vec{z}+{\vec{y}}^{\mathrm{T}}{\left[\vec{x}\right]}^{\mathrm{T}}\left[\vec{x}\right]\vec{y}$
其中： ${\left({\vec{z}}^{\mathrm{T}}\left[\vec{x}\right]\vec{y}\right)}^{\mathrm{T}}={\vec{y}}^{\mathrm{T}}{\left[\vec{x}\right]}^{\mathrm{T}}\vec{z}$， 则有：
$$J=\vec{z}{\vec{z}}^{\mathrm{T}}-2{\vec{z}}^{\mathrm{T}}\left[\vec{x}\right]\vec{y}+{\vec{y}}^{\mathrm{T}}{\left[\vec{x}\right]}^{\mathrm{T}}\left[\vec{x}\right]\vec{y}$$
进而：
$\frac{\partial J}{\partial \vec{y}}=0-2{\left({\vec{z}}^{\mathrm{T}}\left[\vec{x}\right]\right)}^{\mathrm{T}}+2{\left[\vec{x}\right]}^{\mathrm{T}}\left[\vec{x}\right]\vec{y}=\nabla \vec{y}⟹\frac{\partial J}{\partial {\vec{y}}^{\ast }}=0,{\vec{y}}^{\ast }={\left({\left[\vec{x}\right]}^{\mathrm{T}}\left[\vec{x}\right]\right)}^{-1}{\left[\vec{x}\right]}^{\mathrm{T}}\vec{z}$
其中：${\left({\left[\vec{x}\right]}^{\mathrm{T}}\left[\vec{x}\right]\right)}^{-1}$不一定有解，则${\vec{y}}^{\ast }$无法得到解析解——定义初始${\vec{y}}^{\ast }$， ${\vec{y}}^{\ast }={\vec{y}}^{\ast }-\alpha \nabla ,\alpha =\left[\begin{array}{cc}{\alpha }_1& 0\\ 0& {\alpha }_2\end{array}\right]$
其中： $\alpha$称为学习率，对$x$而言则需进行归一化

1.3 矩阵求导的链式法则

标量函数： $J=f\left(y\left(u\right)\right),\frac{\partial J}{\partial u}=\frac{\partial J}{\partial y}\frac{\partial y}{\partial u}$

标量对向量求导：$J=f\left(\vec{y}\left(\vec{u}\right)\right),\vec{y}={\left[\begin{array}{c}{y}_1\left(\vec{u}\right)\\ ⋮\\ y_{\mathrm{m}}\left(\vec{u}\right)\end{array}\right]}_{m\times 1},\vec{u}={\left[\begin{array}{c}{\vec{u}}_1\\ ⋮\\ {\vec{u}}_{\mathrm{n}}\end{array}\right]}_{\mathrm{n}\times 1}$

分析：${\frac{\partial J_{1\times 1}}{\partial u_{\mathrm{n}\times 1}}}_{\mathrm{n}\times 1}={\frac{\partial J}{\partial y_{m\times 1}}}_{m\times 1}{\frac{\partial y_{m\times 1}}{\partial u_{\mathrm{n}\times 1}}}_{\mathrm{n}\times \mathrm{m}}$ 无法相乘

$\vec{y}={\left[\begin{array}{c}{y}_1\left(\vec{u}\right)\\ y_2\left(\vec{u}\right)\end{array}\right]}_{2\times 1},\vec{u}={\left[\begin{array}{c}{\vec{u}}_1\\ {\vec{u}}_2\\ {\vec{u}}_3\end{array}\right]}_{3\times 1}$
$$\begin{gathered} J=f\left(\vec{y}\left(\vec{u}\right)\right),\frac{\partial J}{\partial \vec{u}}={\left[\begin{array}{c}\frac{\partial J}{\partial {\vec{u}}_1}\\ \frac{\partial J}{\partial {\vec{u}}_2}\\ \frac{\partial J}{\partial {\vec{u}}_3}\end{array}\right]}_{3\times 1}⟹\begin{array}{c}\frac{\partial J}{\partial {\vec{u}}_1}=\frac{\partial J}{\partial y_1}\frac{\partial y_1\left(\vec{u}\right)}{\partial {\vec{u}}_1}+\frac{\partial J}{\partial y_2}\frac{\partial y_2\left(\vec{u}\right)}{\partial {\vec{u}}_1}\\ \frac{\partial J}{\partial {\vec{u}}_2}=\frac{\partial J}{\partial y_1}\frac{\partial y_1\left(\vec{u}\right)}{\partial {\vec{u}}_2}+\frac{\partial J}{\partial y_2}\frac{\partial y_2\left(\vec{u}\right)}{\partial {\vec{u}}_2}\\ \frac{\partial J}{\partial {\vec{u}}_3}=\frac{\partial J}{\partial y_1}\frac{\partial y_1\left(\vec{u}\right)}{\partial {\vec{u}}_3}+\frac{\partial J}{\partial y_2}\frac{\partial y_2\left(\vec{u}\right)}{\partial {\vec{u}}_3}\end{array} \\ ⟹\frac{\partial J}{\partial \vec{u}}={\left[\begin{array}{lc}\frac{\partial y_1\left(\vec{u}\right)}{\partial {\vec{u}}_1}& \frac{\partial y_2\left(\vec{u}\right)}{\partial {\vec{u}}_1}\\ \frac{\partial y_1\left(\vec{u}\right)}{\partial {\vec{u}}_2}& \frac{\partial y_2\left(\vec{u}\right)}{\partial {\vec{u}}_2}\\ \frac{\partial y_1\left(\vec{u}\right)}{\partial {\vec{u}}_3}& \frac{\partial y_2\left(\vec{u}\right)}{\partial {\vec{u}}_3}\end{array}\right]}_{3\times 2}{\left[\begin{array}{c}\frac{\partial J}{\partial y_1}\\ \frac{\partial J}{\partial y_2}\end{array}\right]}_{2\times 2}=\frac{\partial \vec{y}\left(\vec{u}\right)}{\partial \vec{u}}\frac{\partial J}{\partial \vec{y}} \end{gathered}$$

$$\frac{\partial J}{\partial \vec{u}}=\frac{\partial \vec{y}\left(\vec{u}\right)}{\partial \vec{u}}\frac{\partial J}{\partial \vec{y}}$$

eg:
$$\vec{x}\left[k+1\right]=A\vec{x}\left[k\right]+B\vec{u}\left[k\right],J={\vec{x}}^{\mathrm{T}}\left[k+1\right]\vec{x}\left[k+1\right]$$
$$\frac{\partial J}{\partial \vec{u}}=\frac{\partial \vec{x}\left[k+1\right]}{\partial \vec{u}}\frac{\partial J}{\partial \vec{x}\left[k+1\right]}=B^{\mathrm{T}}\cdot 2\vec{x}\left[k+1\right]=2B^{\mathrm{T}}\vec{x}\left[k+1\right]$$

2. Ch0-2 特征值与特征向量

2.1 定义

$$A\vec{v}=\lambda \vec{v}$$
对于给定线性变换 $A$，特征向量eigenvector $\vec{v}$ 在此变换后仍与原来的方向共线，但长度可能会发生改变，其中 $\lambda$ 为标量，即缩放比例，称其为特征值eigenvalue

2.1.1 线性变换

2.1.2 求解特征值，特征向量

$$A\vec{v}=\lambda \vec{v}\Rightarrow \left(A-\lambda E\right)\vec{v}=0\Rightarrow \left|A-\lambda E\right|=0$$

2.1.3 应用：对角化矩阵——解耦Decouple

$P=\left[{\vec{v}}_1,{\vec{v}}_2\right]$—— coordinate transformation matrix

$$\begin{gathered} AP=A\left[\begin{array}{cc}{\vec{v}}_1& {\vec{v}}_2\end{array}\right]=\left[\begin{array}{cc}A\left[\begin{array}{c}{v}_{11}\\ v_{12}\end{array}\right]& A\left[\begin{array}{c}{v}_{21}\\ v_{22}\end{array}\right]\end{array}\right]=\left[\begin{array}{cc}{\lambda }_1{v}_{11}& {\lambda }_2{v}_{21}\\ {\lambda }_1{v}_{12}& {\lambda }_2{v}_{22}\end{array}\right]=\left[\begin{array}{cc}{v}_{11}& v_{21}\\ v_{12}& v_{22}\end{array}\right]\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right]=P\Lambda \\ \Rightarrow AP=P\Lambda \Rightarrow P^{-1}AP=\Lambda \end{gathered}$$

• 微分方程组 state-space rep
  

2.2 Summary

1. $A\vec{v}=\lambda \vec{v}$ 在一条直线上
2. 求解方法： $\left|A-\lambda E\right|=0$
3. $P^{-1}AP=\Lambda ,P=\left[\begin{array}{ccc}{\vec{v}}_1& {\vec{v}}_2& \cdots \end{array}\right],\Lambda =\left[\begin{array}{ccc}{\lambda }_1& & \\ & {\lambda }_2& \\ & & \ddots \end{array}\right]$
4. $\dot{x}=Ax,x=Py,\dot{y}=\Lambda y$

3. Ch0-3线性化Linearization

3.1 线性系统 Linear System 与 叠加原理 Superposition

$$\dot{x}=f\left(x\right)$$

1. $x_1,x_2$ 是解
2. $x_3=k_1{x}_1+k_2{x}_2,k_1,k_2\in \mathbb{R}$
3. $x_3$ 是解

eg:
$$\begin{gathered} \ddot{x}+2\dot{x}+\sqrt{2}x=0\surd \\ \ddot{x}+2\dot{x}+\sqrt{2}{x}^2=0\times \\ \ddot{x}+\sin \dot{x}+\sqrt{2}x=0\times \end{gathered}$$

3.2 线性化：Taylor Series

$$f\left(x\right)=f\left(x_0\right)+\frac{f^{\prime }\left(x_0\right)}{1!}\left(x-x_0\right)+\frac{{f^{\prime }}^{\prime }\left(x_0\right)}{2!}{\left(x-x_0\right)}^2+\cdots +\frac{f^n\left(x_0\right)}{n!}{\left(x-x_0\right)}^n$$

若$x-x_0\to 0,{\left(x-x_0\right)}^n\to 0$，则有：$\Rightarrow f\left(x\right)=f\left(x_0\right)+f^{\prime }\left(x_0\right)\left(x-x_0\right)\Rightarrow f\left(x\right)=k_1+k_{2}x-k_3{x}_0\Rightarrow f\left(x\right)=k_{2}x+b$

eg1:

eg2:

eg3:

3.3 Summary

1. $f\left(x\right)=f\left(x_0\right)+\frac{f^{\prime }\left(x_0\right)}{1!}\left(x-x_0\right),x-x_0\to 0$
2. $\left[\begin{array}{c}{\dot{x}}_{1\mathrm{d}}\\ {\dot{x}}_{2\mathrm{d}}\end{array}\right]={\left[\begin{array}{cc}\frac{\partial f_1}{\partial x_1}& \frac{\partial f_1}{\partial x_2}\\ \frac{\partial f_2}{\partial x_1}& \frac{\partial f_2}{\partial x_2}\end{array}\right]|}_{\mathrm{x}={\mathrm{x}}_0}\left[\begin{array}{c}{x}_{1\mathrm{d}}\\ x_{2\mathrm{d}}\end{array}\right]$

4. Ch0-4线性时不变系统中的冲激响应与卷积

4.1 LIT System：Linear Time Invariant

• 运算operator : O { ⋅ } O\left\{ \cdot \right\} O{⋅}
  I n p u t O { f ( t ) } = o u t p u t x ( t ) \begin{array}{c} Input\\ O\left\{ f\left( t \right) \right\}\\ \end{array}=\begin{array}{c} output\\ x\left( t \right)\\ \end{array} InputO{f(t)}​=outputx(t)​

• 线性——叠加原理superpositin principle：
  { O { f 1 ( t ) + f 2 ( t ) } = x 1 ( t ) + x 2 ( t ) O { a f 1 ( t ) } = a x 1 ( t ) O { a 1 f 1 ( t ) + a 2 f 2 ( t ) } = a 1 x 1 ( t ) + a 2 x 2 ( t ) \begin{cases} O\left\{ f_1\left( t \right) +f_2\left( t \right) \right\} =x_1\left( t \right) +x_2\left( t \right)\\ O\left\{ af_1\left( t \right) \right\} =ax_1\left( t \right)\\ O\left\{ a_1f_1\left( t \right) +a_2f_2\left( t \right) \right\} =a_1x_1\left( t \right) +a_2x_2\left( t \right)\\ \end{cases} ⎩ ⎨ ⎧​O{f1​(t)+f2​(t)}=x1​(t)+x2​(t)O{af1​(t)}=ax1​(t)O{a1​f1​(t)+a2​f2​(t)}=a1​x1​(t)+a2​x2​(t)​

• 时不变Time Invariant：
  O { f ( t ) } = x ( t ) ⇒ O { f ( t − τ ) } = x ( t − τ ) O\left\{ f\left( t \right) \right\} =x\left( t \right) \Rightarrow O\left\{ f\left( t-\tau \right) \right\} =x\left( t-\tau \right) O{f(t)}=x(t)⇒O{f(t−τ)}=x(t−τ)

4.2 卷积 Convolution

4.3 单位冲激 Unit Impulse——Dirac Delta

LIT系统，h(t)可以完全定义系统

5. Ch0-5Laplace Transform of Convolution卷积的拉普拉斯变换

线性时不变系统 ： LIT System
冲激响应：Impluse Response
卷积：Convolution

Laplace Transform : $X\left(s\right)=\mathcal{L}\left[x\left(t\right)\right]={\int }_0^{\infty }x\left(t\right)e^{-st}\mathrm{d}t$

Convolution : $x\left(t\right)\ast g\left(t\right)={\int }_0^{t}x\left(\tau \right)g\left(t-\tau \right)\mathrm{d}\tau$

证明：$\mathcal{L}\left[x\left(t\right)\ast g\left(t\right)\right]=X\left(s\right)G\left(s\right)$
$\mathcal{L}\left[x\left(t\right)\ast g\left(t\right)\right]={\int }_0^{\infty }{\int }_0^{t}x\left(\tau \right)g\left(t-\tau \right)\mathrm{d}\tau e^{-st}\mathrm{d}t={\int }_0^{\infty }{\int }_{\tau }^{\infty }x\left(\tau \right)g\left(t-\tau \right)e^{-st}\mathrm{d}t\mathrm{d}\tau$
>令：$u=t-\tau ,t=u+\tau ,\mathrm{d}t=\mathrm{d}u+\mathrm{d}\tau ,t\in \left[\tau ,+\infty \right)\Rightarrow u\in \left[0,+\infty \right)$
$\mathcal{L}\left[x\left(t\right)\ast g\left(t\right)\right]={\int }_0^{\infty }{\int }_0^{\infty }x\left(\tau \right)g\left(u\right)e^{-s\left(u+\tau \right)}\mathrm{d}u\mathrm{d}\tau ={\int }_0^{\infty }x\left(\tau \right)e^{-s\tau }\mathrm{d}\tau {\int }_0^{\infty }g\left(u\right)e^{-su}\mathrm{d}u=X\left(s\right)G\left(s\right)$

$$\mathcal{L}\left[x\left(t\right)\ast g\left(t\right)\right]=\mathcal{L}\left[x\left(t\right)\right]\mathcal{L}\left[g\left(t\right)\right]=X\left(s\right)G\left(s\right)$$

6. Ch0-6复数Complex Number

$x^2-2x+2=0\Rightarrow x=1\pm i$

• 代数表达：$z=a+bi,\mathrm{Re}\left(z\right)=a,\mathrm{Im}\left(z\right)=b$, 分别称为实部与虚部
• 几何表达：$z=\left|z\right|\cos \theta +\left|z\right|\sin \theta i=\left|z\right|\left(\cos \theta +\sin \theta i\right)$
  
• 指数表达：$z=\left|z\right|e^{i\theta }$

$z_1=\left|z_1\right|e^{i{\theta }_1},z_2=\left|z_2\right|e^{i{\theta }_2}\Rightarrow z_1\cdot z_2=\left|z_1\right|\left|z_2\right|e^{i\left({\theta }_1+{\theta }_2\right)}$

共轭：$z_1=a_1+b_{1}i,z_2=a_2-b_{2}i\Rightarrow z_1={\bar{z}}_2$

7. Ch0-7欧拉公式的证明

更有意思的版本
$$e^{i\theta }=\cos \theta +\sin \theta i,i=\sqrt{-1}$$
证明：
$$\begin{gathered} f\left(\theta \right)=\frac{e^{i\theta }}{\cos \theta +\sin \theta i} \\ {f}^{\prime }\left(\theta \right)=\frac{i{e}^{i\theta }\left(\cos \theta +\sin \theta i\right)-e^{i\theta }\left(-\sin \theta +\cos \theta i\right)}{{\left(\cos \theta +\sin \theta i\right)}^2}=0 \\ \Rightarrow f\left(\theta \right)=\mathrm{cons}\tan \mathrm{t} \\ f\left(\theta \right)=f\left(0\right)=\frac{e^{i0}}{\cos 0+\sin 0i}=1\Rightarrow \frac{e^{i\theta }}{\cos \theta +\sin \theta i}=1 \\ \Rightarrow e^{i\theta }=\cos \theta +\sin \theta i \end{gathered}$$

求解：$\sin x=2$
令：$\sin z=2=c,z\in \mathbb{C}$
$\{\begin{array}{l}{e}^{iz}=\cos z+\sin zi\\ e^{i\left(-z\right)}=\cos z-\sin zi\end{array}\Rightarrow e^{iz}-e^{-iz}=2\sin zi$
$\therefore \sin z=\frac{e^{iz}-e^{-iz}}{2i}=c\Rightarrow \frac{e^{ai-b}-e^{b-ai}}{2i}=\frac{e^{ai}{e}^{-b}-e^b{e}^{-ai}}{2i}=c$
且有：$\{\begin{array}{l}{e}^{ia}=\cos a+\sin ai\\ e^{i\left(-a\right)}=\cos a-\sin ai\end{array}$
$$\begin{gathered} \Rightarrow \frac{e^{-b}\left(\cos a+\sin ai\right)-e^b\left(\cos a-\sin ai\right)}{2i}=\frac{\left(e^{-b}-e^b\right)\cos a-\left(e^{-b}+e^b\right)\sin ai}{2i}=c \\ \Rightarrow \frac{1}{2}\left(e^b-e^{-b}\right)\cos ai+\frac{1}{2}\left(e^{-b}+e^b\right)\sin a=c=c+0i \end{gathered}$$
$\Rightarrow \{\begin{array}{l}\frac{1}{2}\left(e^{-b}+e^b\right)\sin a=c\\ \frac{1}{2}\left(e^b-e^{-b}\right)\cos a=0\end{array}$

• 当 $b=0$ 时，$\sin a=c$ 不成立（所设$a,b\in \mathbb{R}$）
• 当 $\cos a=0$ 时，$\frac{1}{2}\left(e^{-b}+e^b\right)=\pm c\Rightarrow 1+e^{2b}\pm 2c{e}^b=0$
  设$u=e^b>0$ ，则有：$u=\pm c\pm \sqrt{c^2-1}$
  $\therefore b=\ln \left(c\pm \sqrt{c^2-1}\right)$
  $\Rightarrow z=\frac{\pi }{2}+2k\pi +\ln \left(c\pm \sqrt{c^2-1}\right)i=\frac{\pi }{2}+2k\pi +\ln \left(2\pm \sqrt{3}\right)i$

8. Ch0-8Matlab/Simulink传递函数Transfer Function

$$\begin{gathered} {\mathcal{L}}^{-1}\left[a_{0}Y\left(s\right)+sY\left(s\right)\right]={\mathcal{L}}^{-1}\left[b_{0}U\left(s\right)+b_{1}sU\left(s\right)\right] \\ \Rightarrow a_{0}y\left(t\right)+\dot{y}\left(t\right)=b_{0}u\left(t\right)+b_1\dot{u}\left(t\right) \\ \Rightarrow \dot{y}-b_1\dot{u}=b_{0}u-y \end{gathered}$$

9. Ch0-9阈值选取-机器视觉中应用正态分布和6-sigma

5M1E——造成产品质量波动的六因素
人 Man Manpower
机器 Machine
材料 Material
方法 Method
测量 Measurment
环境 Envrionment

DMAIC —— 6σ管理中的流程改善
定义 Define
测量 Measure
分析 Analyse
改善 Improve
控制 Control

随机变量与正态分布 Normal Distribution
$$X=\left(\mu ,{\sigma }^2\right)$$
$\mu$ : 期望（平均值）， ${\sigma }^2$：方差

6σ与实际应用