Z-Transfer Function z 传递函数
G(z)&=\dfrac{Y(z)}{U(z)}= \dfrac{\mathscr{Z}[y(kT)]}{\mathscr{Z}[u(kT)]} =\dfrac{\sum\limits_{k=0}^{\infty}y(nT)z^{-k}}{\sum\limits_{k=0}^{\infty}u(nT)z^{-k}} \end{align}$$ ### 一、开环控制 #### 串联环节  1. 串联环节有采样开关:总的传递函数为每个环节传递函数之积 $$\begin{align} Y(s)=G^{*}_{1}(s)G^{*}_{2}(s)U^{*}(s)\; {\color{red}\Rightarrow} \; \dfrac{Y(z)}{U(z)}=G_{1}(z)G_{2}(z) \end{align}$$ 2. 串联环节无采样开关:$G_{1}G_{2}(z)$ 表示先将传递函数相乘,再进行 $z$ 变换 $$\begin{align} Y(s)=[G_{1}(s)G_{2}(s)]^{*}U^{*}(s)\; {\color{red}\Rightarrow} \; \dfrac{Y(z)}{U(z)}=G_{1}G_{2}(z) \end{align}$$ 3. **ZOH 串联的广义对象**: [[信号保持#二、零阶保持器 ZOH\|信号保持#二、零阶保持器 ZOH]] $$\begin{align} G_{h}(s)=\dfrac{1-e^{ -Ts }}{s}\quad G(z)= \dfrac{Y(z)}{U(z)}=\mathscr{Z}\left[(1-e^{ -Ts })\cdot \dfrac{G_{o}(s)}{s}\right]=(1-z^{-1})\mathscr{Z}[\dfrac{G_{o}(s)}{s}] \end{align}$$ #### 并联环节 ### 二、闭环控制  $$\begin{align} & Y(s)=\dfrac{G_{1}(s)G_{2}(s)}{1+G_{1}(s)G_{2}(s)F(s)}R(s) \\ \\ \; {\color{red}\Rightarrow} \; & Y^{*}(s)=\dfrac{[G_{1}(s)G_{2}(s)]^{*}}{1+[G_{1}(s)G_{2}(s)F(s)]^{*}}R(s) \\ \\ \; {\color{red}\Rightarrow} \; & \dfrac{Y(z)}{R(z)}=\dfrac{G_{1}G_{2}(z)}{1+G_{1}G_{2}F(z)} \end{align}$$  $$\begin{align} & Y(s)= \dfrac{G_{1}(s)G_{2}(s)G_{3}(s)}{1+G_{1}(s)G_{2}(s)G_{3}(s)H(s)} R(s) \\ \\ \; {\color{red}\Rightarrow} \; & Y^{*}(s)= \dfrac{\left[G_{1}(s)R(s)\right]^{*}\; G_{2}^{*}(s)\; G_{3}^{*}(s)}{1+G^{*}_{2}(s)\;[G_{1}(s)G_{3}(s)H(s)]^{*}} \\ \\ \; {\color{red}\Rightarrow} \; & \dfrac{Y(z)}{R(z)}= \dfrac{G_{2}(z)G_{3}(z) {\color{red}G_{1}R(z)}}{1+G_{2}(z)G_{3}HG_{1}(z)} \dfrac{1}{\color{red}R(z)} \end{align}$$ > [!important] > > 一定要注意采样开关的位置!!!线性定常离散系统的脉冲传递函数定义为:系统输出采样信号的 z 变换与输入采样信号的 z 变换 之比。用于描述系统的差分方程: