[Notes] 2.z transformation related concepts and properties

Sequence summation of geometric numbers
$$S(n)=\frac{a_1(1-q^n)}{1-q}$$
$$S(\infty )=\frac{a_1}{1-q}$$

Laplace transform
$X(s)=L[x(n)]={\int }_{-\infty }^{\infty }x(n)e^{-st}dt$

z-transform

$$z=e^{sT}$$
$$X(z)=\sum _{n=-\infty }^{\infty }x(n)z^{-n}$$
zero point: the z value that makes the value 0
pole: makes the value tend to$\infty$ z value

The region of convergence of the z-transform

For a given sequence $x\left(n\right)$ , making$X\left(z\right)$ , immediately |$X(z)$|< ∞ \infty Allzz$of\; \infty$$The\; set\; of\; z$ values ​​is called$X\left(z\right)$ coverage
$$|X(z)|=\sum _{n=-\infty }^{\infty }|x(n)z^{-n}|=M<\infty Note$$
: What you want should be∣ z ∣ |z|The range of$|$$z$$|$

The region of convergence of a typical z-transform

0. Finite sequence
   $$0<|z|<\infty$$

1. Right sequence
   $$R_{x_{-}}<|z|<\infty$$

2. Left sequence
   $$-\infty <|z|<R_{x_{+}}$$

3. Bilateral sequence
   $$R_{x_{-}}<|z|<R_{x_{+}}$$

$Properties\; of\; the\; z-$ transform

1. linear
2. shift
   • Bilateral sequence
     $$\mathcal{L}[x(n-m)]=z^{-m}X(z)$$
     $$L\left[x\right(n+m)]=z^{m}X(z)$$
   • unilateral
3. Scale transformability of z domain
   $$\mathcal{L}[x(n)]=X(z),R_{x_{-}}<|z|<R_{x_{+}}$$
   $$\mathcal{L}[a^{n}x(n)]=X(\frac{z}{a}),|a|R_{x_{-}}<|z|<|a|R_{x+}$$
4. Linear weighting of the sequence
   $$L\left[nx\right(n)]=-z\cdot \frac{d}{dz}X\left(z\right)$$
   5. Sequence conjugate
   $$\mathcal{L}[x^{\ast }(n)]=X^{\ast }(z^{\ast })$$
5. Sequence defolding
   $$\mathcal{L}[x(-n)]=X(\frac{1}{z})$$
   $$\mathcal{L}[x(-n-m)]=(\frac{1}{z}{)}^{-m}X(\frac{1}{z})=z^{m}X(\frac{1}{z})$$
6. initial value theorem

For causal sequences:
$$x(0)=li{m}_{x\to \infty }X\left(z\right)$$
8. Final value theorem
Condition:$x\left(n\right)$ is a causal sequence, and the poles of X(z) are within the unit circle |z|=1 (up to z = 1 z=1on the unit circle$z=1$ available one floor)
$$x(\infty )=li{m}_{x\to \infty }=li{m}_{z\to 1}[(z-1)X(z)]=Res[X(z){]}_{z=1}$$
Otherwise the final value theorem cannot be used