Z – transform  (Page 22/2)

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

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

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 , $X(z)$ is an integer power series of $z^{-1}$ with corresponding $x(n)$ as coefficients. Let’s expand $X(z)$ :

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

In general one writes

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

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

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

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

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

Because $X(z)$ 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 $X^{+}(z)$ 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

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

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

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

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

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

Applying the formula of infinite geometric series which is repeated here

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

to obtain

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

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

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

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

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

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

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

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

1.2 the inverse z-transform

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

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

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

$X(n)\leftrightarrow Z(z)$ (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. $X(z)=$ $\frac{z}{z-0\text{.}8}$
2. $\frac{1}{z+1\text{.}2}$

Solution

(a) Let’s write

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

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

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

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

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

or

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

(b) Let’s write

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

Next , let’s expand $X(z)$ :

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

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

Thus

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

or

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

That is

$x(n)=0$