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$X(z)=$ $\sum _{n=0}^{\infty}x(n){z}^{-n}$ $(4\text{.}1)$
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.
Find the z–transform of the two signals of Fig.4.1
(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
$1+a+{a}^{2}+{a}^{3}+\text{.}\text{.}\text{.}=$ $\sum _{n=0}^{\infty}{a}^{n}$ $=\frac{1}{1-a}$ $\mid a\mid <1$ (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 $(4\text{.}1)$
$X(z)=$ $\sum _{n=0}^{\infty}x(n){z}^{-n}$
$=0+1\text{.}0({z}^{-1})-1\text{.}2({z}^{-1}{)}^{2}+1\text{.}\text{44}({z}^{-1}{)}^{3}-1\text{.}\text{718}({z}^{-1}{)}^{4}+\text{.}\text{.}\text{.}$
$={z}^{-1}[1+(-1\text{.}{\mathrm{2z}}^{-1})+(-1\text{.}{\mathrm{2z}}^{-1}{)}^{2}+(-1\text{.}{\mathrm{2z}}^{-1}{)}^{3}+\text{.}\text{.}\text{.}]$
$={z}^{-1}\frac{1}{1+1\text{.}{\mathrm{2z}}^{-1}}=\frac{{z}^{-1}}{1+1\text{.}{\mathrm{2z}}^{-1}}=\frac{1}{z+1\text{.}2}$
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
Find the inverse z-transform of the following
(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
$x(n)=0$ $n\le 0$
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