Z – transform

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1.1 definition: the z-transform x(z) of a causal discrete – time signal x(n) is defined as

$X\left(z\right)=$ $\sum _{n=0}^{\infty }x\left(n\right){z}^{-n}$ $\left(4\text{.}1\right)$

z is a complex variable of the transform domain and can be considered as the complex frequency. Remember index n can be time or space or some other thing, but is usually taken as time. As defined above , $X\left(z\right)$ is an integer power series of ${z}^{-1}$ with corresponding $x\left(n\right)$ as coefficients. Let’s expand $X\left(z\right)$ :

$X\left(z\right)=$ $\sum _{n=-\infty }^{\infty }x\left(n\right){z}^{-n}$ $=$ $x\left(0\right)+x\left(1\right){z}^{-1}+x\left(2\right){z}^{-2}+\text{.}\text{.}\text{.}$ (4.2)

In general one writes

$X\left(z\right)=$ $Z\left[x\left(n\right)\right]$ (4.3)

In Eq.(4.1) the summation is taken from $n=0$ to $\infty$ , ie , $X\left(z\right)$ is not at all related to the past history of $x\left(n\right)$ . This is one–sided or unilateral z-transform . Sometime the one–sided z-transform has to take into account the initial conditions of $x\left(n\right)$ (see section 4.7).

In general , signals exist at all time , and the two-sided or bilateral z–transform is defined as

$H\left(z\right)=$ $\sum _{n=-\infty }^{\infty }h\left(n\right){z}^{-n}$

$=x\left(-2\right){z}^{2}+x\left(-1\right)z+x\left(0\right)+x\left(1\right){z}^{-1}+x\left(2\right){z}^{-2}+\text{.}\text{.}\text{.}$ (4.4)

Because $X\left(z\right)$ is an infinite power series of ${z}^{-1}$ , the transform only exists at values where the series converges (i.e. goes to zero as $n\to \infty$ or - $\infty$ ). Thus the z-transform is accompanied with its region of convergence (ROC) where it is finite (see section 4.4).

A number of authors denote ${X}^{+}\left(z\right)$ for one-side z-transform.

Example 4.1.1

Find the z–transform of the two signals of Fig.4.1

Solution

(a) Notice the signal is causal and monotically decreasing and its value is just $0\text{.}{8}^{n}$ for $n\ge 0$ . So we write

$x\left(n\right)=0\text{.}{8}^{n}u\left(n\right)$

and use the transform $\left(4\text{.}1\right)$

$X\left(z\right)=$ $\sum _{n=0}^{\infty }x\left(n\right){z}^{-n}$

$=1+0\text{.}{8z}^{-1}+0\text{.}\text{64}{z}^{-2}+0\text{.}\text{512}{z}^{-3}+\text{.}\text{.}\text{.}$

$=1+\left(0\text{.}{8z}^{-1}\right)+\left(0\text{.}{8z}^{-1}{\right)}^{2}+\left(0\text{.}{8z}^{-1}{\right)}^{3}+\text{.}\text{.}\text{.}$

Applying the formula of infinite geometric series which is repeated here

$1+a+{a}^{2}+{a}^{3}+\text{.}\text{.}\text{.}=$ $\sum _{n=0}^{\infty }{a}^{n}$ $=\frac{1}{1-a}$ $\mid a\mid <1$ (4.5)

to obtain

$X\left(z\right)=$ $\frac{1}{1-0\text{.}{8z}^{-1}}$ $=\frac{z}{z-0\text{.}8}$

The result can be left in either of the two forms .

(b) The signal is alternatively positive and negative with increasing value .The signal is divergent . We can put the signal in the form

$x\left(n\right)=\left(-1\text{.}2{\right)}^{n-1}u\left(n-1\right)$

which is $\left(-1\text{.}2{\right)}^{n}u\left(n\right)$ delayed one index(sample) . Let’s use the transform $\left(4\text{.}1\right)$

$X\left(z\right)=$ $\sum _{n=0}^{\infty }x\left(n\right){z}^{-n}$

$=0+1\text{.}0\left({z}^{-1}\right)-1\text{.}2\left({z}^{-1}{\right)}^{2}+1\text{.}\text{44}\left({z}^{-1}{\right)}^{3}-1\text{.}\text{718}\left({z}^{-1}{\right)}^{4}+\text{.}\text{.}\text{.}$

$={z}^{-1}\left[1+\left(-1\text{.}{2z}^{-1}\right)+\left(-1\text{.}{2z}^{-1}{\right)}^{2}+\left(-1\text{.}{2z}^{-1}{\right)}^{3}+\text{.}\text{.}\text{.}\right]$

$={z}^{-1}\frac{1}{1+1\text{.}{2z}^{-1}}=\frac{{z}^{-1}}{1+1\text{.}{2z}^{-1}}=\frac{1}{z+1\text{.}2}$

1.2 the inverse z-transform

The inverse z-transform is denoted by ${Z}^{-1}$ :

$x\left(n\right)={Z}^{-1}\left[X\left(z\right)\right]$ (4.6)

The signal $x\left(n\right)$ and its transform constitutes a transform pair

$X\left(n\right)↔Z\left(z\right)$ (4.7)

One way to find the inverse transform , whenever possible , is to utilize just the z-transform definition. General methods of the inverse z-transform are discursed in section 4.5 and 4.6

Example 4.1.2

Find the inverse z-transform of the following

1. $X\left(z\right)=$ $\frac{z}{z-0\text{.}8}$
2. $\frac{1}{z+1\text{.}2}$

Solution

(a) Let’s write

$X\left(z\right)=$ $\frac{z}{\text{z-}0\text{.}8}$ $=$ $\frac{1}{1-0\text{.}{8z}^{-1}}$

$=1+\left(0\text{.}{8z}^{-1}\right)+\left(0\text{.}{8z}^{-1}{\right)}^{2}+\left(0\text{.}{8z}^{-1}{\right)}^{3}+\text{.}\text{.}\text{.}$

$=1+0\text{.}{8z}^{-1}+0\text{.}\text{64}{z}^{-2}+0\text{.}\text{512}{z}^{-3}+\text{.}\text{.}\text{.}$

By comparing term by term with Equation $\left(4\text{.}2\right)$ we get

$x\left(n\right)=\left[1,0\text{.}8,0\text{.}\text{64},0\text{.}\text{512};\text{.}\text{.}\text{.}\right]$

or

$x\left(n\right)=$ $0\text{.}{8}^{n}u\left(n\right)$

(b) Let’s write

$X\left(z\right)=$ $\frac{1}{z+1,2}$ $=\frac{{z}^{-1}}{1+1,2{z}^{-1}}$ $={z}^{-1}\frac{1}{1+1,2{z}^{-1}}$

Next , let’s expand $X\left(z\right)$ :

$X\left(z\right)$ $={z}^{-1}\left[1+\left(-1\text{.}{2z}^{-1}\right)+\left(-1\text{.}{2z}^{-1}{\right)}^{2}+\left(-1\text{.}{2z}^{-1}{\right)}^{3}+\text{.}\text{.}\text{.}\right]$

$=0+1\text{.}{0z}^{-1}-1\text{.}{2z}^{-2}+1\text{.}\text{44}{z}^{-3}-1\text{.}\text{728}{z}^{-4}+\text{.}\text{.}\text{.}$

Thus

$x\left(n\right)=\left[0,1\text{.}0,-1\text{.}2,1\text{.}\text{44},-1\text{.}\text{728},\text{.}\text{.}\text{.}\right]$

or

$x\left(n\right)=\left(-1\text{.}2{\right)}^{n-1}u\left(n-1\right)$

That is

$x\left(n\right)=0$ $n\le 0$

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