數位控制

Z-Transform

設

$x(k)$ 是

$x(t)$ 的離散時間，則

$x(k)$ 的單邊 Z-Transform 為

$$\begin{array}{}\text{(1)}& x(z)=Z\{x(k)\}=\sum _{k=0}^{\mathrm{\infty }}x(k)z^{-k},\end{array}$$

1. Z-Transform 是離散時間的函數做 Laplace transform 後，再做變數變換
   $Z=e^{sT}$ 得到的
2. $Z^{-k}$ 屬於時間延遲 (
   $Z=e^{sT}$ 中的 T 是取樣週期)

---

特性

• 離散時間函數乘上

  $a^k$ ( k 想像成是 t )
  假如
  $Z\{x(k)\}=x(z)$ ，則
  
  $$\begin{array}{}\text{(2)}& Z\{a^{k}x(k)\}=x(a^{-1}z),\end{array}$$

• 平移定理 (Shifting theorem)
  

  $$\begin{array}{}\text{(3)}& Z\{x(k-n)\}=z^{-n}x(z),\end{array}$$
  
  $$\begin{array}{}\text{(4)}& Z\{x(k+n)\}=z^n[x(z)-\sum _{k=0}^{n-1}x(k)],\end{array}$$

• 初值定理
  

  $$\begin{array}{}\text{(5)}& x(0)=\underset{z\to \mathrm{\infty }}{lim}x(z),\end{array}$$

• 終值定理
  

  $$\begin{array}{rl}\text{(6)}& \underset{k\to \mathrm{\infty }}{lim}x(k)& =\underset{z\to 1}{lim}(1-z^{-1})x(z),& =\underset{z\to 1}{lim}(z-1)x(z),\end{array}$$

---

Inverse Z-Transform

有三種方法

冪級數法 (Power series)

以降冪形式做長除法，得到的級數可以對應到

$\sum _{k=0}^{\mathrm{\infty }}x(k)z^{-k}$ ，級數的係數即為

$x(0),x(1),x(2)...$

部分分式展開法

先將分式做部分分式展開，再對每個簡單的分式 inverse z-transform 。展開前記得要先將分式除以

$z$ ，之後再乘回去
eg.

$E(z)=\frac{0.1z(z+1)}{(z-1{)}^2(z-0.6)}$ ，求

$E(z)$ 之 inverse z-transform

$$\begin{array}{rl}\frac{E(z)}{z}& =\frac{1}{z-0.6}-\frac{1}{z-1}+\frac{0.5}{(z-1{)}^2}\\ E(z)& =\frac{-1}{z-1}+\frac{1}{z-2}\\ \text{so}e(k)& =Z^{-1}\{E(z)\}=-1^{k-1}+2^{k-1}\end{array}$$

注意
若欲 inverse z-transform 的分式具有共軛複數之極點在

$z=P,P^{\ast }$ ，則將該分式無條件展開成

$\frac{K}{z-P}+\frac{K^{\ast }}{z-P^{\ast }}$ or

$\frac{Kz}{z-P}+\frac{K^{\ast }z}{z-P^{\ast }}$ 的型式 (分子有沒有

$z$ 取決於對象分式的分子有無

$z$ )
Ex:
Given

$E(z)=\frac{-3.894z}{z^2+0.6065}$ ， find

$e(k)$ .

$$\begin{array}{rl}E(z)& =\frac{-3.894z}{(z-j0.7788)(z+j0.7788)}\\ & =\frac{Kz}{(z-j0.7788)}+\frac{K^{\ast }z}{(z+j0.7788)}\\ \text{then}K& =\frac{z-j0.7788}{z}E(z){|}_{z=j0.7788}=2.5j=2.5{\epsilon }^{j\frac{\pi }{2}}\\ P& =j0.7788=0.7788{\epsilon }^{j\frac{\pi }{2}}\\ \text{so}E(z)& =2.5(\frac{z{\epsilon }^{j\frac{\pi }{2}}}{z-0.78{\epsilon }^{j\frac{\pi }{2}}}+\frac{z{\epsilon }^{-j\frac{\pi }{2}}}{z-0.78{\epsilon }^{-j\frac{\pi }{2}}})\\ e(k)& =Z^{-1}\{E(z)\}=2.5[{\epsilon }^{j\frac{\pi }{2}}(0.78{\epsilon }^{j\frac{\pi }{2}}{)}^k+{\epsilon }^{-j\frac{\pi }{2}}(0.78{\epsilon }^{j\frac{\pi }{2}}{)}^k]\\ & =5(0.78{)}^k(\frac{{\epsilon }^{j(\frac{\pi }{2}+\frac{k\pi }{2})}+{\epsilon }^{-j(\frac{\pi }{2}+\frac{k\pi }{2})}}{2})\\ & =5(0.78{)}^k\cos (\frac{\pi }{2}+\frac{k\pi }{2})\\ & =5(0.78{)}^k\sin (\frac{k\pi }{2})\end{array}$$

Inversion-formula method (利用餘數定理)

不要用，在

$s=0$ 會錯

$eg.E(z)=\frac{z}{(z-1{)}^2},e(k)=?$

for multiple poles at

$z=a$

$(residue{)}_{z=a}=\frac{1}{(m-1)!}\frac{{\mathrm{d}}^{m-1}}{\mathrm{d}{z}^{m-1}}[(z-a{)}^{m}E(z)z^{k-1}]{|}_{z=a}$

$\begin{array}{rl}& E(z)z^{k-1}=\frac{z^k}{(z-1{)}^2}\\ & e(k)=(residue{)}_{z=1}\\ & =\frac{1}{(2-1)!}\frac{{\mathrm{d}}^{2-1}}{\mathrm{d}{z}^{2-1}}[(z-1{)}^2\frac{z^k}{(z-1{)}^2}]{|}_{z=1}\\ & =\frac{\mathrm{d}}{\mathrm{dz}}{z}^k{|}_{z=1}\\ & =k\end{array}$

狀態方程式 State equation 的 Z-Transform

求 Transfer function

給一 State equation 如下:

$\begin{array}{rl}x(k+1)=Ax(k)+Bu(k)& (1)\\ y(k)=Cx(k)+Du(k)& (2)\end{array}$

• 對 (1) 做 z-transform ，假設

  $x(0)=0$
  
  $\begin{array}{rl}zx(z)-zx(0)& =Ax(z)+Bu(z)\\ (zI-A)x(z)& =Bu(z)\\ \Rightarrow x(z)& =(zI-A{)}^{-1}Bu(z)\end{array}$

• 對 (2) 做 z-transform
  

  $\begin{array}{rl}y(z)& =Cx(z)+Du(z)\end{array}$

• $G(z)=\frac{Y(z)}{U(z)}=C[zI-A{]}^{-1}B+D$

State equation 的解

$\begin{array}{rl}x(k+1)=Ax(k)+Bu(k)& (1)\\ y(k)=Cx(k)+Du(k)& (2)\end{array}$

• 遞迴解法
  

  $$\begin{gathered} x(1)=Ax(0)+Bu(0) \\ x(2)=Ax(1)+Bu(1)=A^{2}x(0)+ABu(0)+Bu(1) \\ x(3)=A^{3}x(0)+A^{2}Bu(0)+ABu(1)+Bu(2) \end{gathered}$$

• z-transform 解法

define the fundamental matrix

$\begin{array}{rl}\mathrm{\Phi }=A^k& \Rightarrow x(k)=\mathrm{\Phi }(k)x(0)+\sum _{j=0}^{k-1}\mathrm{\Phi }(k-1-j)Bu(j)\\ & \Rightarrow y(k)=C\mathrm{\Phi }(k)x(0)+C\sum _{j=0}^{k-1}\mathrm{\Phi }(k-1-j)Bu(j)+Du(k)\end{array}$

$eg.A=\left[\begin{array}{cc}0& 1\\ -2& 3\end{array}\right],\mathrm{\Phi }(k)=?$

$$\begin{gathered} zI-A=\left[\begin{array}{cc}z& -1\\ 2& z+3\end{array}\right],det(zI-A)=(z+2)(z+1) \\ z(zI-A{)}^{-1}=\left[\begin{array}{cc}\frac{2z}{z+1}+\frac{-z}{z+2}& \frac{z}{z+1}+\frac{-3}{z+2}\\ \frac{2z}{z+1}+\frac{2z}{z+2}& \frac{-z}{z+1}+\frac{2z}{z+2}\end{array}\right]=z\mathrm{\Phi }(k) \\ \Rightarrow \mathrm{\Phi }(k)=z^{-1}z(zI-A{)}^{-1}=\left[\begin{array}{cc}2(-1{)}^k-(-2{)}^k& (-1{)}^k-(-2{)}^k\\ -2(-1{)}^k+(-2{)}^k& -((-1{)}^k+2(-2{)}^k)\end{array}\right] \end{gathered}$$