Chapter11.2: Mathematical Model and Stability Analysis of Discrete System

This series of blogs mainly describes the application of Matlab software in automatic control. If you have no theoretical basis for automatic control, please learn the series of blog posts on automatic control first. This series of blogs will not explain the theoretical knowledge of automatic control in detail.
Links related to the theoretical basis of automatic control: https://blog.csdn.net/qq_39032096/category_10287468.html?spm=1001.2014.3001.5482
Blog reference books: "MATLAB/Simulink and Control System Simulation".

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2. Mathematical model and stability analysis of discrete systems

2.1 Mathematical model of discrete system

2.1.1 Linear constant coefficient difference equation

For a general linear steady discrete system, kkOutput c ( k ) c(k)at time$k$$c\left(k\right)$ is not only related tokkInput at time$k$$r\left(k\right)$ is related, and is related tokkInput r ( k − 1 ) before time$k$$r(k-1)，r(k-2)，\dots ,$ related to, and also related to$k$ previous output$c(k-1),c(k-2),\dots ,$ related; this relationship can be used$The\; n$ -order backward difference equation description:
$$\begin{array}{rl}& c(k)+a_{1}c(k-1)+a_{2}c(k-2)+\cdots +a_{n-1}c(k-n+1)+a_{n}c(k-n)\\ & \\ =& b_{0}r(k)+b_{1}r(k-1)+\cdots +b_{m-1}r(k-m+1)+b_{m}r(k-m)\end{array}$$
亦可表示为：
$$c(k)=-\sum _{i=1}^n{a}_{i}c(k-i)+\sum _{j=0}^m{b}_{j}r(k-j)$$
其中：$a_i(i=1,2,\dots ,n)、b_j(j=0,1,2,\dots ,m)$ is a constant coefficient,$m\le n$；

The above formula is called $nth$ order linear constant coefficient difference equation;

The linear steady discrete system can also adopt the following $Nth$ order forward difference equation description:
$$\begin{array}{rl}& c(k+n)+a_{1}c(k+n-1)+a_{2}c(k+n-2)+\cdots +a_{n-1}c(k+1)+a_{n}c(k)\\ & \\ =& b_{0}r(k+m)+b_{1}r(k+m-1)+b_{2}r(k+m-2)+\cdots +b_{m-1}r(k+1)+b_{m}r(k)\end{array}$$
It can also be expressed as:
$$c(k+n)=-\sum _{i=1}^n{a}_{i}c(k+n-i)+\sum _{j=0}^m{b}_{j}r(k+m-j)$$

2.1.2 Pulse transfer function

1. Pulse transfer function definition

   The open-loop discrete system is assumed as shown in the figure below:

   

   If the initial condition of the system is zero, the input signal is $r\left(t\right)$ , after sampling$r^{\ast }(t)$的$The\; z$ transformation function is$R(z)$，系统连续部分的输出为$c(t)$，采样后$c^{\ast }(t)$的$z$变换函数为$C(z)$，则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，记为：
   $$G(z)=\frac{C(z)}{R(z)}=\frac{\sum _{n=0}^{\infty }c(nT)z^{-n}}{\sum _{n=0}^{\infty }r(nT)z^{-n}}$$

   如果已知$R(z)、G(z)$，则在零初始条件下，线性定常离散系统的输出采样信号为：
   $$c^{\ast }(t)=Z^{-1}[C(z)]=Z^{-1}[G(z)R(z)]$$
   

2. Calculation of Impulse Transfer Function

   Example1: Assuming the open-loop system is shown in the figure below, find the corresponding pulse transfer function $G(z)$；

   

   untie:

   Put $G\left(s\right)$ is expanded into a partial fraction:
   $$G(s)=\frac{1}{s}-\frac{1}{s+a}$$
   可得：
   $$G(z)=\frac{z}{z-1}-\frac{z}{z-{\mathrm{e}}^{-aT}}=\frac{z(1-e^{-aT})}{(z-1)(z-{\mathrm{e}}^{-aT})}$$

2.1.3 Pulse transfer function of open-loop system

1. Two important properties of sampled Laplace transform

   1. The Laplace transform of the sampling function is periodic, namely:
      $$G^{\ast }(s)=G^{\ast }(s+\mathrm{j}k{\omega }_s)$$
      where:${oh}_s$is the sampling angular frequency;

   2. If the Laplace transform of the sampling function $E^{\ast }\left(s\right)$and Laplace transform of continuous function$G\left(s\right)$ are multiplied and then discretized, then$E^{\ast }\left(s\right)$can be derived from discrete notation, namely:
      $$\left[G\right(s)E^{\ast }(s){]}^{\ast }=G^{\ast }(s)E^{\ast }(s)$$

2. Impulse transfer function of an open-loop system with a series link

   1. There is a sampling switch between the series links

      Assume that the open-loop discrete system is shown in the figure below, in two series connection links $G_1(s)、G_2\left(s\right)$ are separated by ideal sampling switches;

      

      The pulse transfer function of the open-loop system with sampling switches between the series links is:
      $$G(z)=\frac{C(z)}{R(z)}=G_1(z)G_2\left(z\right)$$
      The pulse transfer function of two linear continuous links separated by an ideal sampling switch in series is equal to the product of the respective pulse transfer functions of these two links, which can be extended to nnA series of$n\; links;$

   2. No sampling switch between series links

      Assume that the open-loop discrete system is shown in the figure below, in two series connection links $G_1(s)、G_2\left(s\right)$ , no ideal sampling switch is separated;

      

      The pulse transfer function of the open-loop system without sampling switches between the series links is:
      $$G(z)=\frac{C\left(z\right)}{R\left(z\right)}=G_1{G}_2\left(z\right)$$
      The pulse transfer function of two linear continuous links connected in series without an ideal sampling switch is equal to the corresponding$z$ -transform, can be generalized tonnA series of$n\; links;$

      Example2: Suppose the open-loop discrete system is shown in the figure below, the input signal $r(t)=1\left(t\right)$ , find the system$\left(a\right)$ sum$\left(b\right)$ pulse transfer function$G\left(z\right)$ and the output$z$变换$C(z)$；

      

      untie:

      Input $r\left(t\right)=1(t)$的$z$变换为：
      $$R(z)=\frac{z}{z-1}$$
      For system $(\mathrm{a})$：
      $$G_1(z)=Z\left[\frac{1}{s}\right]=\frac{z}{z-1}，G_2\left(z\right)=Z\left[\frac{a}{s+a}\right]=\frac{the}{z-{\mathrm{e}}^{-aT}}$$
      有：
      $$G(z)=G_1(z)G_2\left(z\right)=\frac{the{z}^2}{(z-1)(z-{\mathrm{e}}^{-aT})}$$

      $$C(z)=G(z)R(z)=\frac{the{z}^3}{(z-1{)}^2(z-e^{-aT})}$$

      For system $(\mathrm{b})$：
      $$G_1(s)G_2(s)=\frac{a}{s(s+a)}$$
      有：
      $$G(z)=G_1{G}_2(z)=Z\left[\frac{a}{s(s+a)}\right]=\frac{z(1-e^{-aT})}{(z-1)(z-{\mathrm{e}}^{-aT})}$$

      $$C(z)=G(z)R(z)=\frac{z^2(1-e^{-aT})}{(z-1)(z-{\mathrm{e}}^{-aT})}$$

   3. The pulse transfer function of the open-loop system with a zero-order keeper

      An open-loop discrete system with a zero-order keeper is shown in the figure below:

      

      In the figure: $G_h\left(s\right)$ is the zero-order keeper transfer function,$G_0\left(s\right)$ is a continuous partial transfer function, and there is no synchronous sampling switch isolation between the two series links;

      When there is a zero-order keeper, the pulse transfer function of the open-loop system:
      $$G(z)=\frac{C(z)}{R(z)}=(1-z^{-1})Z\left[\frac{G_0(s)}{s}\right]$$

2.1.4 Pulse transfer function of closed-loop system

A common error sampling closed-loop discrete system structure diagram analysis:

It can be seen from the above figure that the continuous output signal and error signal Laplace transform is:
$$C(s)=G\left(s\right)E^{\ast }(s)，E(s)=R(s)-H\left(s\right)C\left(s\right)$$
thus:
$$E(s)=R(s)-H(s)G(s)E^{\ast }\left(s\right)$$
error sampling signal$e^{The\; Laplace\; transform\; of\; \ast }\left(t\right)$is:
$$E^{\ast }(s)=R^{\ast }\left(s\right)-H{G}^{\ast }(s)E^{\ast }\left(s\right)$$
can be arranged:
$$E^{\ast }(s)=\frac{R^{\ast }(s)}{1+H{G}^{\ast }\left(s\right)}$$
由于：
$$C^{\ast }\left(s\right)={\left[G^{\ast }(s)E^{\ast }(s)\right]}^{\ast }=G^{\ast }\left(s\right)E^{\ast }\left(s\right)=\frac{G^{\ast }\left(s\right)}{1+H{G}^{\ast }\left(s\right)}{R}^{\ast }(s)$$
因此有：
$$E(z)=\frac{1}{1+HG(z)}R(z)，C(z)=\frac{G(z)}{1+HG\left(z\right)}R\left(z\right)$$
defines the error pulse transfer function of the closed-loop discrete system for the input quantity:
$${Phi}_e(z)=\frac{E(z)}{R(z)}=\frac{1}{1+HG\left(z\right)}$$
Define the impulse transfer function of the closed-loop discrete system with respect to the input quantity:
$$\Phi \left(z\right)=\frac{C(z)}{R\left(z\right)}=\frac{G(z)}{1+HG(z)}$$
Closed-loop discrete system characteristic equation:
$$D(z)=1+HG\left(z\right)=0$$
Note: As long as the error signal e ( t ) e(t)There is no sampling switch at$e$$($$t$$)$$r^{\ast }\left(t\right)$does not exist. At this time, the pulse transfer function of the closed-loop discrete system for the input quantity cannot be obtained, only$z$ transformation function$C(z)$；

2.1.5 Typical closed-loop discrete system and output $z-$ transform function

1. The first four typical discrete systems are as follows:

   

   Above $C(z)$求解依次为：
   $$\begin{array}{rl}& C(z)=\frac{G(z)R(z)}{1+GH(z)}\\ & \\ & C\left(z\right)=\frac{R{G}_1(z)G_2(z)}{1+G_{2}H{G}_1(z)}\\ & \\ & C(z)=\frac{G(z)R(z)}{1+G(z)H(z)}\\ & \\ & C\left(z\right)=\frac{G_1\left(z\right)G_2(z)R(z)}{1+G_1\left(z\right)G_{2}H(z)}\end{array}$$

2. The latter four typical discrete systems are as follows:

   

   Above $C(z)$求解依次为：
   $$\begin{array}{rl}& C(z)=\frac{R{G}_1\left(z\right)G_2\left(z\right)G_3(z)}{1+G_2(z)G_1{G}_{3}H(z)}\\ & \\ & C\left(z\right)=\frac{RG(z)}{1+HG(z)}\\ & \\ & C(z)=\frac{G_1\left(z\right)G_2(z)R(z)}{1+G_1(z)G_2(z)H(z)}\\ & \\ & C\left(z\right)=\frac{R(z)G(z)}{1+G(z)H(z)}\end{array}$$

2.2 Stability analysis of discrete systems

2.2.1 $s$ domain tozzMapping of$the\; z\; domain$

in zzIn the$z$$z={\mathrm{e}}^{sT}$ gives$s$ domain tozzrelationship in$the\; z$ domain, ssAny point in the$s$$s=p+j\omega$, mapped to$z$域为：
$$z=e^{(\sigma +j\omega )T}=e^{\sigma T}{e}^{\mathrm{j}\omega T}$$
$s$ domain tozzThe basic mapping relation of$z$
$$|z|=e^{\sigma T}，\angle z=\omega T$$
令$p=0$ , equivalent to taking$The\; imaginary\; axis\; of\; the\; s$ -plane, when$oh\to -\infty$ becomes$\infty$ , mapped to$The\; trajectory\; of\; the\; z$ plane is a unit circle centered at the origin;

1. 等$\sigma$ line map

   s s Equal σ \sigmaon$the\; s\; plane$$\sigma$ vertical line, mapped tozzThe trajectory on the$z$$|z|={\mathrm{e}}^{\sigma T}$ is a circle of radius, where:$T$ is the sampling period;

   

   s sThe imaginary axis on$the\; s$ -plane maps to zzthe unit circle in the$z-$ plane, therefore, the left half ssEqual σ \sigmaon$the\; s\; plane$$The\; \sigma$ line maps to$Concentric\; circles\; in\; the\; z-$ plane, inside the unit circle; right halfssEqual σ \sigmaon$the\; s\; plane$$The\; \sigma$ line maps to$Concentric\; circles\; on\; the\; z$ plane, outside the unit circle;

2. etc.ω $\omega$ line map

   At a specific sampling period $In\; the\; case\; of\; T$ ,ssEquivalent ω \omegaon$the\; s\; plane$$\omega$ horizontal line maps to$The\; trajectory\; on\; the\; z$ plane is a cluster of rays starting from the origin, and its phase angle$\angle z=\omega T$ is measured from the positive real axis, as shown in the figure below:

   

   s s ω = ω s / 2 \omega=\omega_s/2on$the\; s\; plane$$oh={oh}_s/2$ horizontal line, at$The\; z$ plane is exactly mapped to the negative real axis;

3. εζ $Zeta$ line mapping

   s s Equivalent ζ \zetaon$the\; s\; -plane$$The\; \zeta$ line is described by the following formula:
   $$s=-Oh\tan b+j\omega$$
   where:$\beta$为ζ \zetaThe angle between the$\zeta$
   $$z={\mathrm{e}}^{sT}=e^{(oh\tan b+j\omega )T}$$
   等$zeta$ line from$s$ domain to$z$域的实方关于式式的：
   $$|z|={\mathrm{e}}^{-(2p/h_s)oh\tan b},\angle z=\frac{2}{{oh}_s}$$
   Except for $b=0°$ and$b=90°$ outside, whenβ \betaWhen$\beta$ is a constant, the left half ssEquivalent ζ \zetaon$the\; s\; -plane$$\zeta$ line, mapped tozzA cluster of convergent logarithmic spirals inside the unit circle on the$z$ plane, whose starting point is zz1 1of the positive real axis on the$z\; plane$$1$ place, end point is$the\; origin\; of\; the\; z$ -plane;

   

   set ssThe main bands on$the\; s$$z={\mathrm{e}}^{sT}$ transform, mapped tozzThe unit circle on the$z\; plane\; and\; the\; negative\; real\; axis\; inside\; the\; unit\; circle,\; as\; shown\; in\; the\; following\; figure:$

   

   s s All minor bands in the$s$$The\; z-$ plane is mapped to the same unit circle and the negative real axis within the unit circle;

2.2.2 Necessary and sufficient conditions for the stability of discrete systems

In a linear steady continuous system, the necessary and sufficient conditions for system stability are: the solution of the homogeneous differential equation of the system is convergent, or the roots of the characteristic equation of the system all have negative real parts, or the poles of the system transfer function are all located in the left half $s$ plane;

1. Necessary and sufficient conditions for the stability of discrete systems in the time domain

   Let the linear constant difference equation be:
   $$c(k)=-\sum _{i=1}^n{a}_{i}c(k-i)+\sum _{j=0}^m{b}_{j}r(k-j)$$
   其齐次差分方程为：
   $$c(k)+\sum _{i=1}^n{a}_{i}c(k-i)=0$$
   差分方程的特征方程为：
   $${\alpha }^n+a_1{\alpha }^{n-1}+a_2{\alpha }^{n-2}+\cdots +a_n=0$$
   当特征方程的根$|{\alpha }_i|<1(i=1,2,\dots ,n)$时，必有：$\underset{k\to \infty }{\lim }c(k)=0$，因此，系统稳定的充分必要条件是：当且仅当差分方程所有特征根的模$|{\alpha }_i|<1(i=1,2,\dots ,n)$，相应的线性定常离散系统是稳定的；

2. $z$域中离散系统稳定的充分必要条件

   设离散系统特征方程为：
   $$D(z)=1+GH(z)=0$$
   $In\; the\; z$ domain, the necessary and sufficient condition for the stability of a linear steady discrete system: if and only if all the characteristic roots of the characteristic equation of the discrete system are distributed inzzIn the unit circle on the$z$$|z_i|<1(i=1,2,\dots ,n)$ , the corresponding linear steady discrete system is stable;

2.2.3 Stability criteria for discrete systems

1. $w$ Transformation and Routh Stability Criterion

   令
   $$z=\frac{w+1}{w-1}\Rightarrow w=\frac{z+1}{z-1}$$
   complex variable $z$ give$w$ is a linear transformation of each other,$The\; w$ transformation is called a bilinear transformation;

   $z$ plane and$The\; corresponding\; relationship\; of\; the\; w$ plane is shown in the figure below:

   

   w wThe imaginary axis of$the\; w$ plane corresponds to zzThe unit on the$z$$The\; w$ plane corresponds tothe zzThe area inside the unit circle on the$z$$The\; w$ plane corresponds tozzThe area outside the unit circle on the$z\; plane;$

   Necessary and sufficient conditions for the stability of discrete systems: from the characteristic equation $1+GH(z)=$All roots of$0$ are at zzIn the unit circle on the$z$$1+GH(w)=$All roots of$0$$w$ plane;

2. Juli Stability Criterion

   Jules criterion is based on the closed-loop characteristic equation of the discrete system $D(z)=The\; coefficient\; of\; 0$ , to determine whether its root is located inzzIn the unit circle on the$z\; plane,\; so\; as\; to\; judge\; whether\; the\; discrete\; system\; is\; stable.$

   Let the discrete system $The\; n$ -order closed-loop characteristic equation is:
   $$D(z)=a_0+a_{1}z+a_2{z}^2+\cdots +a_n{z}^n=0,a_n>0$$
   is constructed as follows$(2n-3)$ row,$(n+1)$ Column Julia array.

   Number of lines	$z^0$	$z^1$	$z^2$	$z^3$	$\dots$	$z^{n-k}$	$\dots$	$z^{n-1}$	$z^n$
   $1$	$a_0$	$a_1$	$a_2$	$a_3$	$\dots$	$a_{n-k}$	$\dots$	$a_{n-1}$	$a_n$
   $2$	$a_n$	$a_{n-1}$	$a_{n-2}$	$a_{n-3}$	$\dots$	$a_k$	$\dots$	$a_1$	$a_0$
   $3$	$b_0$	$b_1$	$b_2$	$b_3$	$\dots$	$b_{n-k}$	$\dots$	$b_{n-1}$
   $4$	$b_{n-1}$	$b_{n-2}$	$b_{n-3}$	$b_{n-4}$	$\dots$	$b_{k-1}$	$\dots$	$b_0$
   $5$	$c_0$	$c_1$	$c_2$	$c_3$	$\dots$	$c_{n-2}$
   $6$	$c_{n-2}$	$c_{n-3}$	$c_{n-4}$	$c_{n-5}$	$\dots$	$c_0$
   $⋮$	$⋮$	$⋮$	$⋮$	$⋮$
   $2n-5$	$p_0$	$p_1$	$p_2$	$p_3$
   $2n-4$	$p_3$	$p_2$	$p_1$	$p_0$
   $2n-3$	$q_0$	$q_1$	$q_2$

   其中：
   $$\begin{array}{rlr}& b_k=\left|\begin{array}{cc}{a}_0& a_{n-k}\\ a_n& a_k\end{array}\right|，& k=0,1,2,\dots ,n-1\\ & & \\ & c_k=\left|\begin{array}{cc}{b}_0& b_{n-k-1}\\ b_{n-1}& b_k\end{array}\right|，& k=0,1,2,\dots ,n-2\\ & & \\ & d_k=\left|\begin{array}{cc}{c}_0& c_{n-k-2}\\ c_{n-2}& c_k\end{array}\right|，& k=0,1,2,\dots ,n-3\\ & & \\ & ⋮& \\ & & \\ & q_0=\left|\begin{array}{cc}{p}_0& p_3\\ p_3& p_0\end{array}\right|，q_1=\left|\begin{array}{cc}{p}_0& p_2\\ p_3& p_1\end{array}\right|，q_2=\left|\begin{array}{cc}{p}_0& p_1\\ p_3& p_2\end{array}\right|，& \end{array}$$
   朱利稳定判据：

   特征方程$D(z)=0$的根，全部位于$z$平面上单位圆内的充分必要条件是：
   $$D(1)>0，D(-1)\{\begin{array}{l}>0，当n为偶数时\\ <0，当n为奇数时\end{array}$$
   及以下$n-1$个约束条件成立：
   $$|a_0|<a_n,|b_0|>|b_{n-1}|,|c_0|>|c_{n-2}|,|d_0|>|d_{n-3}|,\dots ,|q_0|>|q_2|$$

2.2.4 离散系统的稳态误差

设单位反馈误差采样系统如下图所示：

其中：$G(s)$为连续部分的传递函数，$e(t)$为系统连续误差信号，$e^{\ast }(t)$为系统采样误差信号，其$z$变换为：
$$E(z)=R(z)-C(z)=[1-\Phi (z)]R(z)={\Phi }_e(z)R(z)$$
其中：
$${\Phi }_e(z)=\frac{E(z)}{R(z)}=\frac{1}{1+G(z)}$$
如果${\Phi }_e(z)$的极点全部位于$z$平面上的单位圆内，即若离散系统是稳定的，可用$z$变换的终值定理求出采样瞬时的稳态误差：
$$e_{ss}(\infty )=\underset{t\to \infty }{\lim }{e}^{\ast }(t)=\underset{z\to 1}{\lim }(1-z^{-1})E(z)=\underset{z\to 1}{\lim }\frac{(z-1)R(z)}{z[1+G(z)]}$$

2.2.5 离散系统的型别与静态误差系数

在离散系统中，把开环脉冲传递函数$G(z)$具有$z=1$的极点数$\nu$作为划分离散系统型别的标准，把$G(z)$中$\nu =0,1,2,\dots$的系统，称为$0$型、Ⅰ型和Ⅱ型离散系统等；

1. 单位阶跃输入时的稳态误差

   当系统输入为单位阶跃函数$r(t)=1(t)$时，$z$变换函数为：
   $$R(z)=\frac{z}{z-1}$$
   稳态误差为：
   $$e_{ss}(\infty )=\underset{z\to 1}{\lim }\frac{1}{1+G(z)}=\frac{1}{\underset{z\to 1}{\lim }[1+G(z)]}=\frac{1}{K_p}$$
   其中静态位置误差系数为：
   $$K_p=\underset{z\to 1}{\lim }[1+G(z)]$$
   在单位阶跃函数作用下，$0$型离散系统在采样瞬时存在位置误差，Ⅰ型及Ⅰ型以上的离散系统，在采样瞬时没有位置误差；

2. 单位斜坡输入时的稳态误差

   当系统输入为单位斜坡函数$r(t)=t$时，$The\; z$ transformation function is:
   $$R(z)=\frac{Tz}{(z-1{)}^2}$$
   稳态误差为：
   $$e_{ss}(\infty )=\underset{z\to 1}{\lim }\frac{T}{(z-1)[1+G(z)]}=\frac{T}{\underset{z\to 1}{\lim }(z-1)G(z)}=\frac{T}{K_v}$$
   其中静态速度误差系数为：
   $$K_v=\underset{z\to 1}{\lim }(z-1)G(z)$$
   $0$型离散系统不能承受单位斜坡函数作用，Ⅰ型离散系统在单位斜坡函数作用下存在速度误差，Ⅱ型及Ⅱ型以上离散系统在单位斜坡函数作用下不存在稳态误差；

3. 单位加速度输入时的稳态误差

   当系统输入为单位加速度函数$r(t)=t^2/2$时，$z$变换函数为：
   $$R(z)=\frac{T^{2}z(z+1)}{2(z-1{)}^3}$$
   稳态误差为：
   $$e_{ss}(\infty )=\underset{z\to 1}{\lim }\frac{T^2(z+1)}{2(z-1{)}^2[1+G(z)]}=\frac{T^2}{\underset{z\to 1}{\lim }(z-1{)}^{2}G(z)}=\frac{T^2}{K_a}$$
   其中静态加速度误差系数为：
   $$K_a=\underset{z\to 1}{\lim }(z-1{)}^{2}G(z)$$
   $0$型和Ⅰ型离散系统不能承受单位加速度函数作用，Ⅱ型离散系统在单位加速度函数作用下存在加速度误差，Ⅲ型及Ⅲ型以上的离散系统在单位加速度函数作用下，不存在采样瞬时的稳态误差；

   单位反馈离散系统的稳态误差小结：

   系统类型	位置误差$r(t)=1(t)$	速度误差$r(t)=t$	加速度误差$r(t)=\frac{1}{2}{t}^2$
   $0$型	$\frac{1}{K_p}$	$\infty$	$\infty$
   Ⅰ型	$0$	$\frac{T}{K_v}$	$\infty$
   Ⅱ型	$0$	$0$	$\frac{T^2}{K_a}$
   Ⅲ型	$0$	$0$	$0$

2.3 离散系统的数字校正

2.3.1 数字控制器的脉冲传递函数

设离散系统如下图所示：

其中：$D(z)$为数字控制器的脉冲传递函数，$G(z)$为保持器与被控对象的传递函数，$H(s)$为反馈测量装置的传递函数；

设$H(s)=1$，$G(s)$的$z$变换为$G(z)$，可得系统的闭环脉冲传递函数：
$$\Phi (z)=\frac{D(z)G(z)}{1+D(z)G(z)}=\frac{C(z)}{R(z)}$$
误差脉冲传递函数：
$${\Phi }_e(z)=\frac{1}{1+D(z)G(z)}=\frac{E(z)}{R(z)}$$
可得数字控制器的脉冲传递函数为：
$$D(z)=\frac{\Phi (z)}{G(z)[1-\Phi (z)]}=\frac{1-{\Phi }_e(z)}{G(z){\Phi }_e(z)}，{\Phi }_e(z)=1-\Phi (z)$$

2.3.2 最少拍系统设计

最少拍系统：指在典型输入作用下，能以有限拍结束响应过程，且在采样时刻上无稳态误差的离散系统；

常见的典型输入：单位阶跃函数、单位速度函数、单位加速度函数；
$$\begin{array}{rl}& Z\left[1(t)\right]=\frac{z}{z-1}=\frac{1}{1-z^{-1}}\\ & \\ & Z\left[t\right]=\frac{Tz}{(z-1{)}^2}=\frac{T{z}^{-1}}{(1-z^{-1}{)}^2}\\ & \\ & Z\left[\frac{1}{2}{t}^2\right]=\frac{T^{2}z(z+1)}{2(z-1{)}^3}=\frac{\frac{1}{2}{T}^2{z}^{-1}(1+z^{-1})}{(1-z^{-1}{)}^3}\end{array}$$
典型输入可表示为一般形式：
$$R(z)=\frac{A(z)}{(1-z^{-1}{)}^m}$$
其中：$A(z)$是不含$(1-z^{-1})$因子的$z^{-1}$多项式；

最少拍系统的设计原则：若系统广义被控对象$G(z)$无延迟且在$z$平面单位圆上及单位圆外无零极点，要求选择闭环脉冲传递函数$\Phi (z)$，使系统在典型输入作用下，经最少采样周期后能使输出序列在各采样时刻的稳态误差为零，达到完全跟踪的目的，从而确定所需要的数字控制器的脉冲传递函数$D(z)$；

误差信号$e(t)$的$z$变换为：
$$E(z)={\Phi }_e(z)R(z)=\frac{{\Phi }_e(z)A(z)}{(1-z^{-1}{)}^m}$$
离散系统的稳态误差为：
$$e_{ss}(\infty )=\underset{z\to 1}{\lim }(1-z^{-1})E(z)=\underset{z\to 1}{\lim }(1-z^{-1})\frac{A(z)}{(1-z^{-1}{)}^m}{\Phi }_e(z)$$
使$e_{ss}(\infty )$为零的条件是：${\Phi }_e(z)$中包含$(1-z^{-1}{)}^m$的因子，即：
$${\Phi }_e(z)=(1-z^{-1}{)}^{m}F(z)$$
其中：$F(z)$不含$(1-z^{-1})$因子的多项式，为了$D(z)$简单，阶数最低，取$F(z)=1$。

1. 单位阶跃输入

   由$r(t)=1(t)$时有$m=1,A(z)=1$可得：
   $${\Phi }_e(z)=1-z^{-1},\Phi (z)=z^{-1}$$
   数字控制器脉冲传递函数为：
   $$D(z)=\frac{z^{-1}}{(1-z^{-1})G(z)}，E(z)=\frac{A(z)}{(1-z^{-1}{)}^m}{\Phi }_e(z)=1$$
   $e(0)=1,e(T)=e(2T)=\cdots =0$，可见，最少拍系统经过一拍即可完全跟踪输入$r(t)=1(t)$，这样的离散系统称为一拍系统，$t_s=T$；

2. 单位斜坡输入

   由$r(t)=t$时，有$m=2,A(z)=T{z}^{-1}$，可得：
   $${\Phi }_e(z)=(1-z^{-1}{)}^{m}F(z)=(1-z^{-1}{)}^2，\Phi (z)=1-{\Phi }_e(z)=2z^{-1}-z^{-2}$$
   数字控制器脉冲传递函数为：
   $$D(z)=\frac{\Phi (z)}{G(z){\Phi }_e(z)}=\frac{z^{-1}(2-z^{-1})}{(1-z^{-1}{)}^{2}G(z)}，E(z)=\frac{A(z)}{(1-z^{-1}{)}^m}{\Phi }_e(z)=T{z}^{-1}$$
   $e(0)=0,e(T)=T,e(2T)=e(3T)=\cdots =0$，最少拍系统经过二拍即可完全跟踪输入$r(t)=t$，这样的离散系统称为二拍系统，其调节时间$t_s=2T$；

3. 单位加速度输入

   由于$r(t)=t^2/2$时，有$m=3,A(z)=\frac{1}{2}{T}^2{z}^{-1}(1+z^{-1})$，可得：
   $${\Phi }_e(z)=(1-z^{-1}{)}^3，\Phi (z)=3z^{-1}-3z^{-2}+z^{-3}$$
   数字控制脉冲传递函数为：
   $$D(z)=\frac{z^{-1}(3-3z^{-1}+z^{-2})}{(1-z^{-1}{)}^{3}G(z)}，E(z)=A(z)=\frac{1}{2}{T}^2{z}^{-1}+\frac{1}{2}{T}^2{z}^{-2}$$
   输出脉冲序列为：
   $$C(z)=\Phi (z)R(z)=\frac{3}{2}{T}^2{z}^{-2}+\frac{9}{2}{T}^2{z}^{-3}+\cdots +\frac{n^2}{2}{T}^2{z}^{-n}+\cdots +$$
   有：
   $$\begin{array}{rl}& e(0)=0,e(T)=\frac{1}{2}{T}^2,e(2T)=\frac{1}{2}{T}^2,e(3T)=e(4T)=\cdots =0\\ & \\ & c(0)=c(T)=0,c(2T)=1.5T^2,c(3T)=4.5T^2,\cdots \end{array}$$
   最少拍系统经过三拍即可完全跟踪输入$r(t)=t^2/2$，这样的离散系统称为三拍系统，调节时间为$t_s=3T$；

4. 最少拍系统的设计结果

   

   小结：

   • 从快速性而言，按单位斜坡输入设计的最少拍系统，在各种典型输入作用下，其动态过程均为二拍；
   • 从准确性而言，系统对单位阶跃输入和单位斜坡输入，在采样时刻均无稳态误差，对单位加速度输入，采样时刻上的稳态误差为常量$T$；
   • 从动态性能而言，系统对单位斜坡输入下的响应性能较好，因为系统本身就是针对此设计的，但系统对单位阶跃输入响应性能较差，有$100\%$的超调量，因此按某种典型输入设计的最少拍系统，适应性较差；
   • 从平稳性而言，在各种典型输入作用下系统进入稳态后，在非采样时刻一般均存在纹波，从而增加系统的机械磨损；

5. 实例分析

   Example8： 设单位反馈线性定常离散系统的连续部分和零阶保持器的传递函数分别为：
   $$G_0(s)=\frac{10}{s(s+1)}，G_h(s)=\frac{1-{\mathrm{e}}^{-sT}}{s}$$
   其中采样周期为$T=1s$；要求系统在单位斜坡输入时实现最少拍控制，求数字控制器脉冲传递函数$D(z)$。

   解：

   系统开环传递函数：
   $$G(s)=G_0(s)G_h(s)=\frac{10(1-e^{-sT})}{s^2(s+1)}$$
   因为：
   $$Z\left[\frac{1}{s^2(s+1)}\right]=\frac{Tz}{(z-1{)}^2}-\frac{(1-{\mathrm{e}}^{-T})z}{(z-1)(z-{\mathrm{e}}^{-T})}$$
   因此有：
   $$G(z)=10(1-z^{-1})\left[\frac{Tz}{(z-1{)}^2}-\frac{(1-{\mathrm{e}}^{-T})z}{(z-1)(z-{\mathrm{e}}^{-T})}\right]=\frac{3.68z^{-1}(1+0.717z^{-1})}{(1-z^{-1})(1-0.368z^{-1})}$$
   其中，$r(t)=t$，闭环脉冲传递函数和误差脉冲传递函数：
   $$\Phi (z)=2z^{-1}(1-0.5z^{-1})，{\Phi }_e(z)=(1-z^{-1}{)}^2$$
   可得：
   $$D(z)=\frac{\Phi (z)}{G(z){\Phi }_e(z)}=\frac{0.543(1-0.368z^{-1})(1-0.5z^{-1})}{(1-z^{-1})(1+0.717z^{-1})}$$

2.3.3 无纹波最少拍系统设计

无纹波最少拍系统的设计要求：在某一种典型输入作用下设计的系统，其输出响应经过尽可能少的采样周期后，不仅在采样时刻输出可以完全跟踪输入，且在非采样时刻不存在纹波；

1. 无纹波最少拍系统的必要条件

   为了在稳态过程中获得无纹波的平滑输出$c^{\ast }(t)$，被控对象$G_0(s)$必须有能力给出与输入$r(t)$相同的平滑输出$c(t)$；

   若针对单位斜坡输入$r(t)=t$设计最少拍系统，则$G_0(s)$的稳态输出也必须是斜坡函数，因此，$G_0(s)$必须至少有一个积分环节，使被控对象在零阶保持器常值输出信号作用下，稳态输出为等速变化量；若针对单位加速度输入$r(t)=t^2/2$设计最少拍系统，则$G_0(s)$至少应包括两个积分环节；

   若输入信号为：
   $$r(t)=R_0+R_{1}t+\frac{1}{2}{t}^2+\cdots +\frac{1}{(q-1)!}{R}_{q-1}{t}^{q-1}$$
   无纹波最少拍系统的必要条件：被控对象传递函数$G_0(s)$中，至少包含$(q-1)$个积分环节；

2. 无纹波最少拍系统的附加条件
   $$D(z)=\frac{\Phi (z)}{G(z){\Phi }_e(z)}\Rightarrow D(z){\Phi }_e(z)=\frac{\Phi (z)}{G(z)}$$
   设广义对象脉冲传递函数为：
   $$G(z)=\frac{P(z)}{Q(z)}$$
   其中：$P(z)$为$G(z)$的零点多项式，$Q(z)$为$G(z)$的极点多项式；

   则有：
   $$D(z){\Phi }_e(z)=\frac{\Phi (z)Q(z)}{P(z)}$$
   $D(z){\Phi }_e(z)$成为$z^{-1}$有限多项式的条件是：$\Phi (z)$的零点应抵消$G(z)$的全部零点，即有：
   $$\Phi (z)=P(z)M(z)$$
   其中：$M(z)$为待定$z^{-1}$多项式；上式为无纹波最少拍系统的附加条件。

3. 无纹波最少拍系统设计

   Example9： 设单位反馈线性定常离散系统的连续部分和零阶保持器的传递函数分别为：
   $$G_0(s)=\frac{10}{s(s+1)}，G_h(s)=\frac{1-{\mathrm{e}}^{-sT}}{s}$$
   其中采样周期为$T=1s$；要求系统在单位斜坡输入时实现无纹波最少拍控制，求数字控制器脉冲传递函数$D(z)$。

   解：

   系统开环传递函数：
   $$G(s)=G_0(s)G_h(s)=\frac{10(1-{\mathrm{e}}^{-sT})}{s^2(s+1)}$$
   因为：
   $$Z\left[\frac{1}{s^2(s+1)}\right]=\frac{Tz}{(z-1{)}^2}-\frac{(1-{\mathrm{e}}^{-T})z}{(z-1)(z-{\mathrm{e}}^{-T})}$$
   因此有：
   $$G(z)=10(1-z^{-1})\left[\frac{Tz}{(z-1{)}^2}-\frac{(1-e^{-T})z}{(z-1)(z-e^{-T})}\right]=\frac{3.68z^{-1}(1+0.717z^{-1})}{(1-z^{-1})(1-0.368z^{-1})}$$
   可见，$G(z)$有一个零点$z=-0.717$，有一个延迟因子$z^{-1}$，且在单位圆上有一个极点$z=1$；

   零点补偿，根据无纹波附加条件，$G(z)$中$z=-0.717$零点应被$\Phi (z)$零点对消；

   因此，令$M(z)=a+b{z}^{-1}$，其中$a、b$待定，选择：
   $$\Phi (z)=z^{-1}(1+0.717z^{-1})(a+b{z}^{-1})$$
   由最少拍条件下，在单位斜坡输入下，
   $${\Phi }_e(z)=(1-z^{-1}{)}^2，\Phi (z)=2z^{-1}(1-0.5z^{-1})$$
   因无纹波时，要求$\Phi (z)$比有纹波时增加一阶，选择：
   $${\Phi }_e(z)=(1-z^{-1}{)}^2(1+c{z}^{-1})$$
   其中：$c$待定；

   可得：
   $$1-{\Phi }_e(z)=(2-c)z^{-1}+(2c-1)z^{-2}-c{z}^{-3}$$

   $$\Phi (z)=z^{-1}(1+0.717z^{-1})(a+b{z}^{-1})=a{z}^{-1}+(b+0.717a)z^{-2}+0.717b{z}^{-3}$$

   对应系数相等，可得：
   $$a=1.408,b=-0.826,c=0.592$$
   因此：
   $$\Phi (z)=1.408z^{-1}(1+0.717z^{-1})(1-0.587z^{-1})$$

   $${\Phi }_e(z)=(1-z^{-1}{)}^2(1+0.592z^{-1})$$

   可得：
   $$D(z)=\frac{\Phi (z)}{G(z){\Phi }_e(z)}=\frac{0.383(1-0.368z^{-1})(1-0.587z^{-1})}{(1-z^{-1})(1+0.592z^{-1})}$$

2.3.4 PID数字控制器的实现

$\mathrm{PID}$控制器的传递函数为：
$$D(s)=\frac{U(s)}{X(s)}=K_1+\frac{K_2}{s}+K_{3}s$$
将其中的微分项和积分项进行离散化处理，可以得到$\mathrm{PID}$控制器的数字实现；
u ( k T ) = d x d t ∣ t = k T = 1 T { x ( k T ) − x [ ( k − 1 ) T ] } u(kT)=\left.\frac{ {\rm d}x}{ {\rm d}t}\right|_{t=kT}=\frac{1}{T}\left\{x(kT)-x[(k-1)T]\right\} u(kT)=dtdx​ ​t=kT​=T1​{ x(kT)−x[(k−1)T]}
即：
$$U(z)=\frac{(1-z^{-1})}{T}X(z)=\frac{z-1}{Tz}X(z)$$

$$u(kT)=u[(k-1)T]+Tx(kT)$$

即：
$$U(z)=z^{-1}U(z)+TX(z)$$
整理可得：
$$U(z)=\frac{Tz}{z-1}X(z)$$
$\mathrm{PID}$控制器在$z$域的传递函数为：
$$D(z)=\frac{U(z)}{X(z)}=K_1+K_2\frac{Tz}{z-1}+K_3\frac{z-1}{Tz}$$
记$x(kT)=x(k)$，可得$\mathrm{PID}$控制器的差分方程：
$$\begin{array}{rl}u(k)& =K_{1}x(k)+K_2[u(k-1)+Tx(k)]+\frac{K_3}{T}[x(k)-x(k-1)]\\ & \\ & =\left[K_1+K_{2}T+\frac{K_3}{T}\right]x(k)-\frac{K_3}{T}x(k-1)+K_{2}u(k-1)\end{array}$$