php制作招聘网站,php wordpress单本小说网站源码+采集,10个不愁销路的小型加工厂,网站开发项目团队本文仅供学习使用 本文参考#xff1a; B站#xff1a;DR_CAN Dr. CAN学习笔记-Ch00 - 数学知识基础 1. Ch0-1矩阵的导数运算1.1标量向量方程对向量求导#xff0c;分母布局#xff0c;分子布局1.1.1 标量方程对向量的导数1.1.2 向量方程对向量的导数 1.2 案例分析#xf… 本文仅供学习使用 本文参考 B站DR_CAN Dr. CAN学习笔记-Ch00 - 数学知识基础 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1.1.2 向量方程对向量的导数 1.2 案例分析线性回归1.3 矩阵求导的链式法则 2. Ch0-2 特征值与特征向量2.1 定义2.1.1 线性变换2.1.2 求解特征值特征向量2.1.3 应用对角化矩阵——解耦Decouple 2.2 Summary 3. Ch0-3线性化Linearization3.1 线性系统 Linear System 与 叠加原理 Superposition3.2 线性化Taylor Series3.3 Summary 4. Ch0-4线性时不变系统中的冲激响应与卷积4.1 LIT SystemLinear Time Invariant4.2 卷积 Convolution4.3 单位冲激 Unit Impulse——Dirac Delta 5. Ch0-5Laplace Transform of Convolution卷积的拉普拉斯变换6. Ch0-6复数Complex Number7. Ch0-7欧拉公式的证明8. Ch0-8Matlab/Simulink传递函数Transfer Function9. Ch0-9阈值选取-机器视觉中应用正态分布和6-sigma 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 y y 为 一元向量 或 二元向量 y y y为多元向量 y ⃗ [ y 1 , y 2 , ⋯ , y n ] ⇒ ∂ f ( y ⃗ ) ∂ y ⃗ \vec{y}\left[ y_1,y_2,\cdots ,y_{\mathrm{n}} \right] \Rightarrow \frac{\partial f\left( \vec{y} \right)}{\partial \vec{y}} y [y1,y2,⋯,yn]⇒∂y ∂f(y ) 其中 f ( y ⃗ ) f\left( \vec{y} \right) f(y ) 为标量 1 × 1 1\times 1 1×1, y ⃗ \vec{y} y 为向量 1 × n 1\times n 1×n
分母布局 Denominator Layout——行数与分母相同 ∂ f ( y ⃗ ) ∂ y ⃗ [ ∂ f ( y ⃗ ) ∂ y 1 ⋮ ∂ f ( y ⃗ ) ∂ y n ] n × 1 \frac{\partial f\left( \vec{y} \right)}{\partial \vec{y}}\left[ \begin{array}{c} \frac{\partial f\left( \vec{y} \right)}{\partial y_1}\\ \vdots\\ \frac{\partial f\left( \vec{y} \right)}{\partial y_{\mathrm{n}}}\\ \end{array} \right] _{n\times 1} ∂y ∂f(y ) ∂y1∂f(y )⋮∂yn∂f(y ) n×1分子布局 Nunerator Layout——行数与分子相同 ∂ f ( y ⃗ ) ∂ y ⃗ [ ∂ f ( y ⃗ ) ∂ y 1 ⋯ ∂ f ( y ⃗ ) ∂ y n ] 1 × n \frac{\partial f\left( \vec{y} \right)}{\partial \vec{y}}\left[ \begin{matrix} \frac{\partial f\left( \vec{y} \right)}{\partial y_1} \cdots \frac{\partial f\left( \vec{y} \right)}{\partial y_{\mathrm{n}}}\\ \end{matrix} \right] _{1\times n} ∂y ∂f(y )[∂y1∂f(y )⋯∂yn∂f(y )]1×n
1.1.2 向量方程对向量的导数 f ⃗ ( y ⃗ ) [ f ⃗ 1 ( y ⃗ ) ⋮ f ⃗ n ( y ⃗ ) ] n × 1 , y ⃗ [ y 1 ⋮ y m ] m × 1 \vec{f}\left( \vec{y} \right) \left[ \begin{array}{c} \vec{f}_1\left( \vec{y} \right)\\ \vdots\\ \vec{f}_{\mathrm{n}}\left( \vec{y} \right)\\ \end{array} \right] _{n\times 1},\vec{y}\left[ \begin{array}{c} y_1\\ \vdots\\ y_{\mathrm{m}}\\ \end{array} \right] _{\mathrm{m}\times 1} f (y ) f 1(y )⋮f n(y ) n×1,y y1⋮ym m×1 ∂ f ⃗ ( y ⃗ ) n × 1 ∂ y ⃗ m × 1 [ ∂ f ⃗ ( y ⃗ ) ∂ y 1 ⋮ ∂ f ⃗ ( y ⃗ ) ∂ y m ] m × 1 [ ∂ f 1 ( y ⃗ ) ∂ y 1 ⋯ ∂ f n ( y ⃗ ) ∂ y 1 ⋮ ⋱ ⋮ ∂ f 1 ( y ⃗ ) ∂ y m ⋯ ∂ f n ( y ⃗ ) ∂ y m ] m × n \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}\\ \vdots\\ \frac{\partial \vec{f}\left( \vec{y} \right)}{\partial y_{\mathrm{m}}}\\ \end{array} \right] _{\mathrm{m}\times 1}\left[ \begin{matrix} \frac{\partial f_1\left( \vec{y} \right)}{\partial y_1} \cdots \frac{\partial f_{\mathrm{n}}\left( \vec{y} \right)}{\partial y_1}\\ \vdots \ddots \vdots\\ \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{matrix} \right] _{\mathrm{m}\times \mathrm{n}} ∂y m×1∂f (y )n×1 ∂y1∂f (y )⋮∂ym∂f (y ) m×1 ∂y1∂f1(y )⋮∂ym∂f1(y )⋯⋱⋯∂y1∂fn(y )⋮∂ym∂fn(y ) m×n, 为分母布局
若 y ⃗ [ y 1 ⋮ y m ] m × 1 , A [ a 11 ⋯ a 1 n ⋮ ⋱ ⋮ a m 1 ⋯ a m n ] \vec{y}\left[ \begin{array}{c} y_1\\ \vdots\\ y_{\mathrm{m}}\\ \end{array} \right] _{\mathrm{m}\times 1}, A\left[ \begin{matrix} a_{11} \cdots a_{1\mathrm{n}}\\ \vdots \ddots \vdots\\ a_{\mathrm{m}1} \cdots a_{\mathrm{mn}}\\ \end{matrix} \right] y y1⋮ym m×1,A a11⋮am1⋯⋱⋯a1n⋮amn , 则有 ∂ A y ⃗ ∂ y ⃗ A T \frac{\partial A\vec{y}}{\partial \vec{y}}A^{\mathrm{T}} ∂y ∂Ay AT(分母布局) ∂ y ⃗ T A y ⃗ ∂ y ⃗ A y ⃗ A T y ⃗ \frac{\partial \vec{y}^{\mathrm{T}}A\vec{y}}{\partial \vec{y}}A\vec{y}A^{\mathrm{T}}\vec{y} ∂y ∂y TAy Ay ATy , 当 A A T AA^{\mathrm{T}} AAT时, ∂ y ⃗ T A y ⃗ ∂ y ⃗ 2 A y ⃗ \frac{\partial \vec{y}^{\mathrm{T}}A\vec{y}}{\partial \vec{y}}2A\vec{y} ∂y ∂y TAy 2Ay 若为分子布局则有 ∂ A y ⃗ ∂ y ⃗ A \frac{\partial A\vec{y}}{\partial \vec{y}}A ∂y ∂Ay A 1.2 案例分析线性回归 ∂ A y ⃗ ∂ y ⃗ A T \frac{\partial A\vec{y}}{\partial \vec{y}}A^{\mathrm{T}} ∂y ∂Ay AT(分母布局) ∂ y ⃗ T A y ⃗ ∂ y ⃗ A y ⃗ A T y ⃗ \frac{\partial \vec{y}^{\mathrm{T}}A\vec{y}}{\partial \vec{y}}A\vec{y}A^{\mathrm{T}}\vec{y} ∂y ∂y TAy Ay ATy , 当 A A T AA^{\mathrm{T}} AAT时, ∂ y ⃗ T A y ⃗ ∂ y ⃗ 2 A y ⃗ \frac{\partial \vec{y}^{\mathrm{T}}A\vec{y}}{\partial \vec{y}}2A\vec{y} ∂y ∂y TAy 2Ay Linear Regression 线性回归 z ^ y 1 y 2 x ⇒ J ∑ i 1 n [ z i − ( y 1 y 2 x i ) ] 2 \hat{z}y_1y_2x\Rightarrow J\sum_{i1}^n{\left[ z_i-\left( y_1y_2x_i \right) \right] ^2} z^y1y2x⇒Ji1∑n[zi−(y1y2xi)]2 找到 y 1 , y 2 y_1,y_2 y1,y2 使得 J J J最小 z ⃗ [ z 1 ⋮ z n ] , [ x ⃗ ] [ 1 x 1 ⋮ ⋮ 1 x n ] , y ⃗ [ y 1 y 2 ] ⇒ z ⃗ ^ [ x ⃗ ] y ⃗ [ y 1 y 2 x 1 ⋮ y 1 y 2 x n ] \vec{z}\left[ \begin{array}{c} z_1\\ \vdots\\ z_{\mathrm{n}}\\ \end{array} \right] ,\left[ \vec{x} \right] \left[ \begin{array}{l} 1 x_1\\ \vdots \vdots\\ 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_1y_2x_1\\ \vdots\\ y_1y_2x_{\mathrm{n}}\\ \end{array} \right] z z1⋮zn ,[x ] 1⋮1x1⋮xn ,y [y1y2]⇒z ^[x ]y y1y2x1⋮y1y2xn J [ z ⃗ − z ⃗ ^ ] T [ z ⃗ − z ⃗ ^ ] [ z ⃗ − [ x ⃗ ] y ⃗ ] T [ z ⃗ − [ x ⃗ ] y ⃗ ] z ⃗ z ⃗ T − z ⃗ T [ x ⃗ ] y ⃗ − y ⃗ T [ x ⃗ ] T z ⃗ y ⃗ T [ x ⃗ ] T [ x ⃗ ] y ⃗ 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} J[z −z ^]T[z −z ^][z −[x ]y ]T[z −[x ]y ]z z T−z T[x ]y −y T[x ]Tz y T[x ]T[x ]y 其中 ( z ⃗ T [ x ⃗ ] y ⃗ ) T y ⃗ T [ x ⃗ ] T z ⃗ \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} (z T[x ]y )Ty T[x ]Tz 则有 J z ⃗ z ⃗ T − 2 z ⃗ T [ x ⃗ ] y ⃗ y ⃗ T [ x ⃗ ] T [ x ⃗ ] y ⃗ 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} Jz z T−2z T[x ]y y T[x ]T[x ]y 进而 ∂ J ∂ y ⃗ 0 − 2 ( z ⃗ T [ x ⃗ ] ) T 2 [ x ⃗ ] T [ x ⃗ ] y ⃗ ∇ y ⃗ ⟹ ∂ J ∂ y ⃗ ∗ 0 , y ⃗ ∗ ( [ x ⃗ ] T [ x ⃗ ] ) − 1 [ x ⃗ ] T z ⃗ \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}\Longrightarrow \frac{\partial J}{\partial \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} ∂y ∂J0−2(z T[x ])T2[x ]T[x ]y ∇y ⟹∂y ∗∂J0,y ∗([x ]T[x ])−1[x ]Tz 其中 ( [ x ⃗ ] T [ x ⃗ ] ) − 1 \left( \left[ \vec{x} \right] ^{\mathrm{T}}\left[ \vec{x} \right] \right) ^{-1} ([x ]T[x ])−1不一定有解则 y ⃗ ∗ \vec{y}^* y ∗无法得到解析解——定义初始 y ⃗ ∗ \vec{y}^* y ∗ y ⃗ ∗ y ⃗ ∗ − α ∇ , α [ α 1 0 0 α 2 ] \vec{y}^*\vec{y}^*-\alpha \nabla ,\alpha \left[ \begin{matrix} \alpha _1 0\\ 0 \alpha _2\\ \end{matrix} \right] y ∗y ∗−α∇,α[α100α2] 其中 α \alpha α称为学习率对 x x 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 ( y ( u ) ) , ∂ J ∂ u ∂ J ∂ y ∂ y ∂ u Jf\left( y\left( u \right) \right) ,\frac{\partial J}{\partial u}\frac{\partial J}{\partial y}\frac{\partial y}{\partial u} Jf(y(u)),∂u∂J∂y∂J∂u∂y
标量对向量求导 J f ( y ⃗ ( u ⃗ ) ) , y ⃗ [ y 1 ( u ⃗ ) ⋮ y m ( u ⃗ ) ] m × 1 , u ⃗ [ u ⃗ 1 ⋮ u ⃗ n ] n × 1 Jf\left( \vec{y}\left( \vec{u} \right) \right) ,\vec{y}\left[ \begin{array}{c} y_1\left( \vec{u} \right)\\ \vdots\\ y_{\mathrm{m}}\left( \vec{u} \right)\\ \end{array} \right] _{m\times 1},\vec{u}\left[ \begin{array}{c} \vec{u}_1\\ \vdots\\ \vec{u}_{\mathrm{n}}\\ \end{array} \right] _{\mathrm{n}\times 1} Jf(y (u )),y y1(u )⋮ym(u ) m×1,u u 1⋮u n n×1
分析 ∂ J 1 × 1 ∂ u n × 1 n × 1 ∂ J ∂ y m × 1 m × 1 ∂ y m × 1 ∂ u n × 1 n × m \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}} ∂un×1∂J1×1n×1∂ym×1∂Jm×1∂un×1∂ym×1n×m 无法相乘 y ⃗ [ y 1 ( u ⃗ ) y 2 ( u ⃗ ) ] 2 × 1 , u ⃗ [ u ⃗ 1 u ⃗ 2 u ⃗ 3 ] 3 × 1 \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} y [y1(u )y2(u )]2×1,u u 1u 2u 3 3×1 J f ( y ⃗ ( u ⃗ ) ) , ∂ J ∂ u ⃗ [ ∂ J ∂ u ⃗ 1 ∂ J ∂ u ⃗ 2 ∂ J ∂ u ⃗ 3 ] 3 × 1 ⟹ ∂ J ∂ u ⃗ 1 ∂ J ∂ y 1 ∂ y 1 ( u ⃗ ) ∂ u ⃗ 1 ∂ J ∂ y 2 ∂ y 2 ( u ⃗ ) ∂ u ⃗ 1 ∂ J ∂ u ⃗ 2 ∂ J ∂ y 1 ∂ y 1 ( u ⃗ ) ∂ u ⃗ 2 ∂ J ∂ y 2 ∂ y 2 ( u ⃗ ) ∂ u ⃗ 2 ∂ J ∂ u ⃗ 3 ∂ J ∂ y 1 ∂ y 1 ( u ⃗ ) ∂ u ⃗ 3 ∂ J ∂ y 2 ∂ y 2 ( u ⃗ ) ∂ u ⃗ 3 ⟹ ∂ J ∂ u ⃗ [ ∂ y 1 ( u ⃗ ) ∂ u ⃗ 1 ∂ y 2 ( u ⃗ ) ∂ u ⃗ 1 ∂ y 1 ( u ⃗ ) ∂ u ⃗ 2 ∂ y 2 ( u ⃗ ) ∂ u ⃗ 2 ∂ y 1 ( u ⃗ ) ∂ u ⃗ 3 ∂ y 2 ( u ⃗ ) ∂ u ⃗ 3 ] 3 × 2 [ ∂ J ∂ y 1 ∂ J ∂ y 2 ] 2 × 2 ∂ y ⃗ ( u ⃗ ) ∂ u ⃗ ∂ J ∂ y ⃗ Jf\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}\Longrightarrow \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} \\ \Longrightarrow \frac{\partial J}{\partial \vec{u}}\left[ \begin{array}{l} \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}} Jf(y (u )),∂u ∂J ∂u 1∂J∂u 2∂J∂u 3∂J 3×1⟹∂u 1∂J∂y1∂J∂u 1∂y1(u )∂y2∂J∂u 1∂y2(u )∂u 2∂J∂y1∂J∂u 2∂y1(u )∂y2∂J∂u 2∂y2(u )∂u 3∂J∂y1∂J∂u 3∂y1(u )∂y2∂J∂u 3∂y2(u )⟹∂u ∂J ∂u 1∂y1(u )∂u 2∂y1(u )∂u 3∂y1(u )∂u 1∂y2(u )∂u 2∂y2(u )∂u 3∂y2(u ) 3×2[∂y1∂J∂y2∂J]2×2∂u ∂y (u )∂y ∂J ∂ J ∂ u ⃗ ∂ y ⃗ ( u ⃗ ) ∂ u ⃗ ∂ J ∂ y ⃗ \frac{\partial J}{\partial \vec{u}}\frac{\partial \vec{y}\left( \vec{u} \right)}{\partial \vec{u}}\frac{\partial J}{\partial \vec{y}} ∂u ∂J∂u ∂y (u )∂y ∂J
eg: x ⃗ [ k 1 ] A x ⃗ [ k ] B u ⃗ [ k ] , J x ⃗ T [ k 1 ] x ⃗ [ k 1 ] \vec{x}\left[ k1 \right] A\vec{x}\left[ k \right] B\vec{u}\left[ k \right] ,J\vec{x}^{\mathrm{T}}\left[ k1 \right] \vec{x}\left[ k1 \right] x [k1]Ax [k]Bu [k],Jx T[k1]x [k1] ∂ J ∂ u ⃗ ∂ x ⃗ [ k 1 ] ∂ u ⃗ ∂ J ∂ x ⃗ [ k 1 ] B T ⋅ 2 x ⃗ [ k 1 ] 2 B T x ⃗ [ k 1 ] \frac{\partial J}{\partial \vec{u}}\frac{\partial \vec{x}\left[ k1 \right]}{\partial \vec{u}}\frac{\partial J}{\partial \vec{x}\left[ k1 \right]}B^{\mathrm{T}}\cdot 2\vec{x}\left[ k1 \right] 2B^{\mathrm{T}}\vec{x}\left[ k1 \right] ∂u ∂J∂u ∂x [k1]∂x [k1]∂JBT⋅2x [k1]2BTx [k1]
2. Ch0-2 特征值与特征向量
2.1 定义 A v ⃗ λ v ⃗ A\vec{v}\lambda \vec{v} Av λv 对于给定线性变换 A A A特征向量eigenvector v ⃗ \vec{v} 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 λ \lambda λ 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 v ⃗ λ v ⃗ ⇒ ( A − λ E ) v ⃗ 0 ⇒ ∣ A − λ E ∣ 0 A\vec{v}\lambda \vec{v}\Rightarrow \left( A-\lambda E \right) \vec{v}0\Rightarrow \left| A-\lambda E \right|0 Av λv ⇒(A−λE)v 0⇒∣A−λE∣0
2.1.3 应用对角化矩阵——解耦Decouple P [ v ⃗ 1 , v ⃗ 2 ] P\left[ \vec{v}_1,\vec{v}_2 \right] P[v 1,v 2]—— coordinate transformation matrix A P A [ v ⃗ 1 v ⃗ 2 ] [ A [ v 11 v 12 ] A [ v 21 v 22 ] ] [ λ 1 v 11 λ 2 v 21 λ 1 v 12 λ 2 v 22 ] [ v 11 v 21 v 12 v 22 ] [ λ 1 0 0 λ 2 ] P Λ ⇒ A P P Λ ⇒ P − 1 A P Λ APA\left[ \begin{matrix} \vec{v}_1 \vec{v}_2\\ \end{matrix} \right] \left[ \begin{matrix} 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{matrix} \right] \left[ \begin{matrix} \lambda _1v_{11} \lambda _2v_{21}\\ \lambda _1v_{12} \lambda _2v_{22}\\ \end{matrix} \right] \left[ \begin{matrix} v_{11} v_{21}\\ v_{12} v_{22}\\ \end{matrix} \right] \left[ \begin{matrix} \lambda _1 0\\ 0 \lambda _2\\ \end{matrix} \right] P\varLambda \\ \Rightarrow APP\varLambda \Rightarrow P^{-1}AP\varLambda APA[v 1v 2][A[v11v12]A[v21v22]][λ1v11λ1v12λ2v21λ2v22][v11v12v21v22][λ100λ2]PΛ⇒APPΛ⇒P−1APΛ
微分方程组 state-space rep
2.2 Summary A v ⃗ λ v ⃗ A\vec{v}\lambda \vec{v} Av λv 在一条直线上求解方法 ∣ A − λ E ∣ 0 \left| A-\lambda E \right|0 ∣A−λE∣0 P − 1 A P Λ , P [ v ⃗ 1 v ⃗ 2 ⋯ ] , Λ [ λ 1 λ 2 ⋱ ] P^{-1}AP\varLambda , P\left[ \begin{matrix} \vec{v}_1 \vec{v}_2 \cdots\\ \end{matrix} \right] , \varLambda \left[ \begin{matrix} \lambda _1 \\ \lambda _2 \\ \ddots\\ \end{matrix} \right] P−1APΛ,P[v 1v 2⋯],Λ λ1λ2⋱ x ˙ A x , x P y , y ˙ Λ y \dot{x}Ax, xPy,\dot{y}\varLambda y x˙Ax,xPy,y˙Λy
3. Ch0-3线性化Linearization
3.1 线性系统 Linear System 与 叠加原理 Superposition x ˙ f ( x ) \dot{x}f\left( x \right) x˙f(x) x 1 , x 2 x_1,x_2 x1,x2 是解 x 3 k 1 x 1 k 2 x 2 , k 1 , k 2 ∈ R x_3k_1x_1k_2x_2,k_1,k_2\in \mathbb{R} x3k1x1k2x2,k1,k2∈R x 3 x_3 x3 是解 eg: x ¨ 2 x ˙ 2 x 0 √ x ¨ 2 x ˙ 2 x 2 0 × x ¨ sin x ˙ 2 x 0 × \ddot{x}2\dot{x}\sqrt{2}x0 √ \\ \ddot{x}2\dot{x}\sqrt{2}x^20 × \\ \ddot{x}\sin \dot{x}\sqrt{2}x0 × x¨2x˙2 x0√x¨2x˙2 x20×x¨sinx˙2 x0×
3.2 线性化Taylor Series f ( x ) f ( x 0 ) f ′ ( x 0 ) 1 ! ( x − x 0 ) f ′ ′ ( x 0 ) 2 ! ( x − x 0 ) 2 ⋯ f n ( x 0 ) n ! ( x − x 0 ) n 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 f(x)f(x0)1!f′(x0)(x−x0)2!f′′(x0)(x−x0)2⋯n!fn(x0)(x−x0)n
若 x − x 0 → 0 , ( x − x 0 ) n → 0 x-x_0\rightarrow 0,\left( x-x_0 \right) ^n\rightarrow 0 x−x0→0,(x−x0)n→0则有 ⇒ f ( x ) f ( x 0 ) f ′ ( x 0 ) ( x − x 0 ) ⇒ f ( x ) k 1 k 2 x − k 3 x 0 ⇒ f ( x ) k 2 x b \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_1k_2x-k_3x_0\Rightarrow f\left( x \right) k_2xb ⇒f(x)f(x0)f′(x0)(x−x0)⇒f(x)k1k2x−k3x0⇒f(x)k2xb eg1: eg2: eg3:
3.3 Summary f ( x ) f ( x 0 ) f ′ ( x 0 ) 1 ! ( x − x 0 ) , x − x 0 → 0 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\rightarrow 0 f(x)f(x0)1!f′(x0)(x−x0),x−x0→0 [ x ˙ 1 d x ˙ 2 d ] [ ∂ f 1 ∂ x 1 ∂ f 1 ∂ x 2 ∂ f 2 ∂ x 1 ∂ f 2 ∂ x 2 ] ∣ x x 0 [ x 1 d x 2 d ] \left[ \begin{array}{c} \dot{x}_{1\mathrm{d}}\\ \dot{x}_{2\mathrm{d}}\\ \end{array} \right] \left. \left[ \begin{matrix} \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{matrix} \right] \right|_{\mathrm{x}\mathrm{x}_0}\left[ \begin{array}{c} x_{1\mathrm{d}}\\ x_{2\mathrm{d}}\\ \end{array} \right] [x˙1dx˙2d][∂x1∂f1∂x1∂f2∂x2∂f1∂x2∂f2] xx0[x1dx2d]
4. Ch0-4线性时不变系统中的冲激响应与卷积
4.1 LIT SystemLinear 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{a1f1(t)a2f2(t)}a1x1(t)a2x2(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 ( s ) L [ x ( t ) ] ∫ 0 ∞ x ( t ) e − s t d t X\left( s \right) \mathcal{L} \left[ x\left( t \right) \right] \int_0^{\infty}{x\left( t \right) e^{-st}}\mathrm{d}t X(s)L[x(t)]∫0∞x(t)e−stdt
Convolution : x ( t ) ∗ g ( t ) ∫ 0 t x ( τ ) g ( t − τ ) d τ x\left( t \right) *g\left( t \right) \int_0^t{x\left( \tau \right) g\left( t-\tau \right)}\mathrm{d}\tau x(t)∗g(t)∫0tx(τ)g(t−τ)dτ 证明 L [ x ( t ) ∗ g ( t ) ] X ( s ) G ( s ) \mathcal{L} \left[ x\left( t \right) *g\left( t \right) \right] X\left( s \right) G\left( s \right) L[x(t)∗g(t)]X(s)G(s) L [ x ( t ) ∗ g ( t ) ] ∫ 0 ∞ ∫ 0 t x ( τ ) g ( t − τ ) d τ e − s t d t ∫ 0 ∞ ∫ τ ∞ x ( τ ) g ( t − τ ) e − s t d t d τ \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 L[x(t)∗g(t)]∫0∞∫0tx(τ)g(t−τ)dτe−stdt∫0∞∫τ∞x(τ)g(t−τ)e−stdtdτ 令 u t − τ , t u τ , d t d u d τ , t ∈ [ τ , ∞ ) ⇒ u ∈ [ 0 , ∞ ) ut-\tau ,tu\tau ,\mathrm{d}t\mathrm{d}u\mathrm{d}\tau ,t\in \left[ \tau ,\infty \right) \Rightarrow u\in \left[ 0,\infty \right) ut−τ,tuτ,dtdudτ,t∈[τ,∞)⇒u∈[0,∞) L [ x ( t ) ∗ g ( t ) ] ∫ 0 ∞ ∫ 0 ∞ x ( τ ) g ( u ) e − s ( u τ ) d u d τ ∫ 0 ∞ x ( τ ) e − s τ d τ ∫ 0 ∞ g ( u ) e − s u d u X ( s ) G ( s ) \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}uX\left( s \right) G\left( s \right) L[x(t)∗g(t)]∫0∞∫0∞x(τ)g(u)e−s(uτ)dudτ∫0∞x(τ)e−sτdτ∫0∞g(u)e−suduX(s)G(s) L [ x ( t ) ∗ g ( t ) ] L [ x ( t ) ] L [ g ( t ) ] X ( s ) G ( s ) \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) L[x(t)∗g(t)]L[x(t)]L[g(t)]X(s)G(s)
6. Ch0-6复数Complex Number x 2 − 2 x 2 0 ⇒ x 1 ± i x^2-2x20\Rightarrow x1\pm i x2−2x20⇒x1±i
代数表达 z a b i , R e ( z ) a , I m ( z ) b zabi,\mathrm{Re}\left( z \right) a,\mathrm{Im}\left( z \right) b zabi,Re(z)a,Im(z)b, 分别称为实部与虚部几何表达 z ∣ z ∣ cos θ ∣ z ∣ sin θ i ∣ z ∣ ( cos θ sin θ i ) z\left| z \right|\cos \theta \left| z \right|\sin \theta i\left| z \right|\left( \cos \theta \sin \theta i \right) z∣z∣cosθ∣z∣sinθi∣z∣(cosθsinθi) 指数表达 z ∣ z ∣ e i θ z\left| z \right|e^{i\theta} z∣z∣eiθ z 1 ∣ z 1 ∣ e i θ 1 , z 2 ∣ z 2 ∣ e i θ 2 ⇒ z 1 ⋅ z 2 ∣ z 1 ∣ ∣ z 2 ∣ e i ( θ 1 θ 2 ) 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)} z1∣z1∣eiθ1,z2∣z2∣eiθ2⇒z1⋅z2∣z1∣∣z2∣ei(θ1θ2) 共轭 z 1 a 1 b 1 i , z 2 a 2 − b 2 i ⇒ z 1 z ˉ 2 z_1a_1b_1i,z_2a_2-b_2i\Rightarrow z_1\bar{z}_2 z1a1b1i,z2a2−b2i⇒z1zˉ2 7. Ch0-7欧拉公式的证明
更有意思的版本 e i θ cos θ sin θ i , i − 1 e^{i\theta}\cos \theta \sin \theta i,i\sqrt{-1} eiθcosθsinθi,i−1 证明 f ( θ ) e i θ cos θ sin θ i f ′ ( θ ) i e i θ ( cos θ sin θ i ) − e i θ ( − sin θ cos θ i ) ( cos θ sin θ i ) 2 0 ⇒ f ( θ ) c o n s tan t f ( θ ) f ( 0 ) e i 0 cos 0 sin 0 i 1 ⇒ e i θ cos θ sin θ i 1 ⇒ e i θ cos θ sin θ i f\left( \theta \right) \frac{e^{i\theta}}{\cos \theta \sin \theta i} \\ f^{\prime}\left( \theta \right) \frac{ie^{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 f(θ)cosθsinθieiθf′(θ)(cosθsinθi)2ieiθ(cosθsinθi)−eiθ(−sinθcosθi)0⇒f(θ)constantf(θ)f(0)cos0sin0iei01⇒cosθsinθieiθ1⇒eiθcosθsinθi 求解 sin x 2 \sin x2 sinx2 令 sin z 2 c , z ∈ C \sin z2c,z\in \mathbb{C} sinz2c,z∈C { e i z cos z sin z i e i ( − z ) cos z − sin z i ⇒ e i z − e − i z 2 sin z i \begin{cases} e^{iz}\cos z\sin zi\\ e^{i\left( -z \right)}\cos z-\sin zi\\ \end{cases}\Rightarrow e^{iz}-e^{-iz}2\sin zi {eizcoszsinziei(−z)cosz−sinzi⇒eiz−e−iz2sinzi ∴ sin z e i z − e − i z 2 i c ⇒ e a i − b − e b − a i 2 i e a i e − b − e b e − a i 2 i c \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^be^{-ai}}{2i}c ∴sinz2ieiz−e−izc⇒2ieai−b−eb−ai2ieaie−b−ebe−aic 且有 { e i a cos a sin a i e i ( − a ) cos a − sin a i \begin{cases} e^{ia}\cos a\sin ai\\ e^{i\left( -a \right)}\cos a-\sin ai\\ \end{cases} {eiacosasinaiei(−a)cosa−sinai ⇒ e − b ( cos a sin a i ) − e b ( cos a − sin a i ) 2 i ( e − b − e b ) cos a − ( e − b e b ) sin a i 2 i c ⇒ 1 2 ( e b − e − b ) cos a i 1 2 ( e − b e b ) sin a c c 0 i \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 acc0i ⇒2ie−b(cosasinai)−eb(cosa−sinai)2i(e−b−eb)cosa−(e−beb)sinaic⇒21(eb−e−b)cosai21(e−beb)sinacc0i ⇒ { 1 2 ( e − b e b ) sin a c 1 2 ( e b − e − b ) cos a 0 \Rightarrow \begin{cases} \frac{1}{2}\left( e^{-b}e^b \right) \sin ac\\ \frac{1}{2}\left( e^b-e^{-b} \right) \cos a0\\ \end{cases} ⇒{21(e−beb)sinac21(eb−e−b)cosa0 当 b 0 b0 b0 时 sin a c \sin ac sinac 不成立所设 a , b ∈ R a,b\in \mathbb{R} a,b∈R当 cos a 0 \cos a0 cosa0 时 1 2 ( e − b e b ) ± c ⇒ 1 e 2 b ± 2 c e b 0 \frac{1}{2}\left( e^{-b}e^b \right) \pm c\Rightarrow 1e^{2b}\pm 2ce^b0 21(e−beb)±c⇒1e2b±2ceb0 设 u e b 0 ue^b0 ueb0 则有 u ± c ± c 2 − 1 u\pm c\pm \sqrt{c^2-1} u±c±c2−1 ∴ b ln ( c ± c 2 − 1 ) \therefore b\ln \left( c\pm \sqrt{c^2-1} \right) ∴bln(c±c2−1 ) ⇒ z π 2 2 k π ln ( c ± c 2 − 1 ) i π 2 2 k π ln ( 2 ± 3 ) i \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 ⇒z2π2kπln(c±c2−1 )i2π2kπln(2±3 )i 8. Ch0-8Matlab/Simulink传递函数Transfer Function L − 1 [ a 0 Y ( s ) s Y ( s ) ] L − 1 [ b 0 U ( s ) b 1 s U ( s ) ] ⇒ a 0 y ( t ) y ˙ ( t ) b 0 u ( t ) b 1 u ˙ ( t ) ⇒ y ˙ − b 1 u ˙ b 0 u − y \mathcal{L} ^{-1}\left[ a_0Y\left( s \right) sY\left( s \right) \right] \mathcal{L} ^{-1}\left[ b_0U\left( s \right) b_1sU\left( s \right) \right] \\ \Rightarrow a_0y\left( t \right) \dot{y}\left( t \right) b_0u\left( t \right) b_1\dot{u}\left( t \right) \\ \Rightarrow \dot{y}-b_1\dot{u}b_0u-y L−1[a0Y(s)sY(s)]L−1[b0U(s)b1sU(s)]⇒a0y(t)y˙(t)b0u(t)b1u˙(t)⇒y˙−b1u˙b0u−y
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 ( μ , σ 2 ) X\left( \mu ,\sigma ^2 \right) X(μ,σ2) μ \mu μ : 期望平均值 σ 2 \sigma ^2 σ2方差
6σ与实际应用
