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1. Ch0-1矩陣的導(dǎo)數(shù)運(yùn)算

1.1標(biāo)量向量方程對向量求導(dǎo)，分母布局，分子布局

1.1.1 標(biāo)量方程對向量的導(dǎo)數(shù)

• $y$ 為 一元向量 或 二元向量
  
• $y$為多元向量
  $$\vec{y}=\left[y_1,y_2,?,y_{\mathrm{n}}\right]?\frac{?f\left(\vec{y}\right)}{?\vec{y}}$$
  其中：$f\left(\vec{y}\right)$ 為標(biāo)量 $1\times 1$, $\vec{y}$為向量 $1\times n$

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

1.1.2 向量方程對向量的導(dǎo)數(shù)

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

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

• $\frac{?A\vec{y}}{?\vec{y}}=A^{\mathrm{T}}$(分母布局)
• $\frac{?{\vec{y}}^{\mathrm{T}}A\vec{y}}{?\vec{y}}=A\vec{y}+A^{\mathrm{T}}\vec{y}$, 當(dāng)$A=A^{\mathrm{T}}$時, $\frac{?{\vec{y}}^{\mathrm{T}}A\vec{y}}{?\vec{y}}=2A\vec{y}$

若為分子布局，則有：$\frac{?A\vec{y}}{?\vec{y}}=A$

1.2 案例分析，線性回歸

• $\frac{?A\vec{y}}{?\vec{y}}=A^{\mathrm{T}}$(分母布局)
• $\frac{?{\vec{y}}^{\mathrm{T}}A\vec{y}}{?\vec{y}}=A\vec{y}+A^{\mathrm{T}}\vec{y}$, 當(dāng)$A=A^{\mathrm{T}}$時, $\frac{?{\vec{y}}^{\mathrm{T}}A\vec{y}}{?\vec{y}}=2A\vec{y}$

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

$\vec{z}=\left[\begin{array}{c}{z}_1\\ ?\\ z_{\mathrm{n}}\end{array}\right],\left[\vec{x}\right]=\left[\begin{array}{lc}1& x_1\\ ?& ?\\ 1& x_{\mathrm{n}}\end{array}\right],\vec{y}=\left[\begin{array}{c}{y}_1\\ y_2\end{array}\right]?\hat{\vec{z}}=\left[\vec{x}\right]\vec{y}=\left[\begin{array}{c}{y}_1+y_2{x}_1\\ ?\\ y_1+y_2{x}_{\mathrm{n}}\end{array}\right]$
$J={\left[\vec{z}?\hat{\vec{z}}\right]}^{\mathrm{T}}\left[\vec{z}?\hat{\vec{z}}\right]={\left[\vec{z}?\left[\vec{x}\right]\vec{y}\right]}^{\mathrm{T}}\left[\vec{z}?\left[\vec{x}\right]\vec{y}\right]=\vec{z}{\vec{z}}^{\mathrm{T}}?{\vec{z}}^{\mathrm{T}}\left[\vec{x}\right]\vec{y}?{\vec{y}}^{\mathrm{T}}{\left[\vec{x}\right]}^{\mathrm{T}}\vec{z}+{\vec{y}}^{\mathrm{T}}{\left[\vec{x}\right]}^{\mathrm{T}}\left[\vec{x}\right]\vec{y}$
其中： ${\left({\vec{z}}^{\mathrm{T}}\left[\vec{x}\right]\vec{y}\right)}^{\mathrm{T}}={\vec{y}}^{\mathrm{T}}{\left[\vec{x}\right]}^{\mathrm{T}}\vec{z}$， 則有：
$$J=\vec{z}{\vec{z}}^{\mathrm{T}}?2{\vec{z}}^{\mathrm{T}}\left[\vec{x}\right]\vec{y}+{\vec{y}}^{\mathrm{T}}{\left[\vec{x}\right]}^{\mathrm{T}}\left[\vec{x}\right]\vec{y}$$
進(jìn)而：
$\frac{?J}{?\vec{y}}=0?2{\left({\vec{z}}^{\mathrm{T}}\left[\vec{x}\right]\right)}^{\mathrm{T}}+2{\left[\vec{x}\right]}^{\mathrm{T}}\left[\vec{x}\right]\vec{y}=?\vec{y}?\frac{?J}{?{\vec{y}}^{?}}=0,{\vec{y}}^{?}={\left({\left[\vec{x}\right]}^{\mathrm{T}}\left[\vec{x}\right]\right)}^{?1}{\left[\vec{x}\right]}^{\mathrm{T}}\vec{z}$
其中：${\left({\left[\vec{x}\right]}^{\mathrm{T}}\left[\vec{x}\right]\right)}^{?1}$不一定有解，則${\vec{y}}^{?}$無法得到解析解——定義初始${\vec{y}}^{?}$， ${\vec{y}}^{?}={\vec{y}}^{?}?\alpha ?,\alpha =\left[\begin{array}{cc}{\alpha }_1& 0\\ 0& {\alpha }_2\end{array}\right]$
其中： $\alpha$稱為學(xué)習(xí)率，對$x$而言則需進(jìn)行歸一化

1.3 矩陣求導(dǎo)的鏈?zhǔn)椒▌t

標(biāo)量函數(shù)： $J=f\left(y\left(u\right)\right),\frac{?J}{?u}=\frac{?J}{?y}\frac{?y}{?u}$

標(biāo)量對向量求導(dǎo)：$J=f\left(\vec{y}\left(\vec{u}\right)\right),\vec{y}={\left[\begin{array}{c}{y}_1\left(\vec{u}\right)\\ ?\\ y_{\mathrm{m}}\left(\vec{u}\right)\end{array}\right]}_{m\times 1},\vec{u}={\left[\begin{array}{c}{\vec{u}}_1\\ ?\\ {\vec{u}}_{\mathrm{n}}\end{array}\right]}_{\mathrm{n}\times 1}$

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

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

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

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

2. Ch0-2 特征值與特征向量

2.1 定義

$$A\vec{v}=\lambda \vec{v}$$
對于給定線性變換 $A$，特征向量eigenvector $\vec{v}$ 在此變換后仍與原來的方向共線，但長度可能會發(fā)生改變，其中 $\lambda$ 為標(biāo)量，即縮放比例，稱其為特征值eigenvalue

2.1.1 線性變換

2.1.2 求解特征值，特征向量

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

2.1.3 應(yīng)用：對角化矩陣——解耦Decouple

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

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

• 微分方程組 state-space rep
  

2.2 Summary

1. $A\vec{v}=\lambda \vec{v}$ 在一條直線上
2. 求解方法： $\left|A?\lambda E\right|=0$
3. $P^{?1}AP=\Lambda ,P=\left[\begin{array}{ccc}{\vec{v}}_1& {\vec{v}}_2& ?\end{array}\right],\Lambda =\left[\begin{array}{ccc}{\lambda }_1& & \\ & {\lambda }_2& \\ & & ?\end{array}\right]$
4. $\dot{x}=Ax,x=Py,\dot{y}=\Lambda y$

3. Ch0-3線性化Linearization

3.1 線性系統(tǒng) Linear System 與 疊加原理 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 線性化：Taylor Series

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

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

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. $\left[\begin{array}{c}{\dot{x}}_{1\mathrm{d}}\\ {\dot{x}}_{2\mathrm{d}}\end{array}\right]={\left[\begin{array}{cc}\frac{?f_1}{?x_1}& \frac{?f_1}{?x_2}\\ \frac{?f_2}{?x_1}& \frac{?f_2}{?x_2}\end{array}\right]|}_{\mathrm{x}={\mathrm{x}}_0}\left[\begin{array}{c}{x}_{1\mathrm{d}}\\ x_{2\mathrm{d}}\end{array}\right]$

4. Ch0-4線性時不變系統(tǒng)中的沖激響應(yīng)與卷積

4.1 LIT System：Linear Time Invariant

• 運(yùn)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)?

• 線性——疊加原理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)?

• 時不變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 卷積 Convolution

4.3 單位沖激 Unit Impulse——Dirac Delta

LIT系統(tǒng)，h(t)可以完全定義系統(tǒng)

5. Ch0-5Laplace Transform of Convolution卷積的拉普拉斯變換

線性時不變系統(tǒng) ： LIT System
沖激響應(yīng)：Impluse Response
卷積：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 : $x\left(t\right)?g\left(t\right)={\int }_0^{t}x\left(\tau \right)g\left(t?\tau \right)\mathrm{d}\tau$

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

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

6. Ch0-6復(fù)數(shù)Complex Number

$x^2?2x+2=0?x=1\pm i$

• 代數(shù)表達(dá)：$z=a+bi,\mathrm{Re}\left(z\right)=a,\mathrm{Im}\left(z\right)=b$, 分別稱為實(shí)部與虛部
• 幾何表達(dá)：$z=\left|z\right|\cos \theta +\left|z\right|\sin \theta i=\left|z\right|\left(\cos \theta +\sin \theta i\right)$
  
• 指數(shù)表達(dá)：$z=\left|z\right|e^{i\theta }$

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

共軛：$z_1=a_1+b_{1}i,z_2=a_2?b_{2}i?z_1={\bar{z}}_2$

7. Ch0-7歐拉公式的證明

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

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

• 當(dāng) $b=0$ 時，$\sin a=c$ 不成立（所設(shè)$a,b\in \mathbb{R}$）
• 當(dāng) $\cos a=0$ 時，$\frac{1}{2}\left(e^{?b}+e^b\right)=\pm c?1+e^{2b}\pm 2c{e}^b=0$
  設(shè)$u=e^b>0$ ，則有：$u=\pm c\pm \sqrt{c^2?1}$
  $\therefore b=\ln \left(c\pm \sqrt{c^2?1}\right)$
  $?z=\frac{\pi }{2}+2k\pi +\ln \left(c\pm \sqrt{c^2?1}\right)i=\frac{\pi }{2}+2k\pi +\ln \left(2\pm \sqrt{3}\right)i$

8. Ch0-8Matlab/Simulink傳遞函數(shù)Transfer Function

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

9. Ch0-9閾值選取-機(jī)器視覺中應(yīng)用正態(tài)分布和6-sigma

5M1E——造成產(chǎn)品質(zhì)量波動的六因素
人 Man Manpower
機(jī)器 Machine
材料 Material
方法 Method
測量 Measurment
環(huán)境 Envrionment

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

隨機(jī)變量與正態(tài)分布 Normal Distribution
$$X=\left(\mu ,{\sigma }^2\right)$$
$\mu$ : 期望（平均值）， ${\sigma }^2$：方差

6σ與實(shí)際應(yīng)用