# Z – transform  (Page 2/2)

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$=\left(-1\text{.}2{\right)}^{n}$ $n>0$

## 1.3 z–transform pairs

Table 4.1 gives many useful z-transform pairs . All signals are causal (right-sided), except two which are anticausal (left-sided). Notice that a transform can be expresed equivalently as a function of ${z}^{-1}$ or z , for example

Table 4.1 : Common z-transform pairs

$u\left(n\right)\to X\left(z\right)=$ $\frac{1}{1-{z}^{-1}}$ or $\frac{z}{z-1}$

${a}^{n}u\left(n\right)\to X\left(z\right)=$ $\frac{1}{1-{\text{az}}^{-1}}$ or $\frac{z}{z-a}$

$\left(\text{cos}n{\Omega }_{0}\right)u\left(n\right)\to X\left(z\right)=$ $\frac{1-{z}^{-1}\text{cos}{\Omega }_{0}}{1-{2z}^{-1}\text{cos}{\Omega }_{0}+{z}^{-2}}$ or $\frac{z\left(z-\text{cos}{\Omega }_{0}\right)}{{z}^{2}-2z\text{cos}{\Omega }_{0}+1}$

Which form is more appropriate depending on what we would like to do with the transform. (see sections 4.1.6 , 4.3 and 4.6).

## 1.4 z–transform for systems

The z-transform applies to systems as well as signals because systems are represented by their impulse responses which are functions of index n (time or space …) just like signals. Remember that many other transforms (Laplace, Fourier …) have the same property, and due to this property that the transforms are useful in the analysis and design of systems because signals and systems interact.

Specifically , the z–transform of impulse response h(n) is

$H\left(z\right)=$ $\sum _{n=0}^{\infty }h\left(n\right){z}^{-n}$ (one–sided transform ) (4.8)

or

$H\left(z\right)=$ $\sum _{n=-\infty }^{\infty }h\left(n\right){z}^{-n}$ (two–sided transform )(4.9)

depending on whether the system is causal or noncausal .

$H\left(z\right)$ , the z–transform of $h\left(n\right)$ , is called transfer function or system function of the system .

## Example 4.1.3

A system has impulse respone

h(n) = [1 , 2 , 3 , 4 , 5 , 6]

Find the transfer function .

## Solution

The system is of noncausal FIR type . Its transfer function is

$H\left(z\right)=$ $\sum _{n=-\infty }^{\infty }h\left(n\right){z}^{-n}$ $=\sum _{n=-2}^{3}h\left(n\right){z}^{-n}$

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

$={z}^{-2}+{2z}^{-1}+3+{4z}^{-1}+{5z}^{-2}+{6z}^{-3}$

On the contrary, if $H\left(z\right)$ is known as above we can easily obtain $h\left(n\right)$ .

## 1.5 eigen-function and eigen-value

We know that if the frequency response of a system is $H\left(\omega \right)$ then for input $x\left(n\right)=$ ${e}^{\text{jn}\omega }$ , the output is $y\left(n\right)=$ ${e}^{\text{jn}\omega }$ $H\left(\omega \right)$ . Because of this , ${e}^{\text{jn}\omega }$ is the eigen-function , and $H\left(\omega \right)$ the eigen-value of the system.

Now , for input

$x\left(n\right)={z}^{n}$ (4.10)

the system output is

$y\left(n\right)=h\left(n\right)\ast x\left(n\right)=$ $\sum _{k=0}^{\infty }h\left(k\right){z}^{n-k}$ $={z}^{n}\left[\sum _{k=0}^{\infty }h\left(k\right){z}^{-k}\right]$

In the brackets is just H(z) , thus

 $y\left(n\right)={z}^{n}H\left(z\right)$ (4.11)

Hence in the z–transform domain , ${z}^{n}$ is eigen-function and $H\left(z\right)$ is eigen-value of the system.

## 1.6 transfer function in terms of filter coefficients

For , Let’s begin with the general filter difference equation which is repeated here

$y\left(n\right)=$ $\sum _{k=1}^{M}{a}_{k}y\left(n-k\right)$ $+\sum _{k=-N}^{N}{b}_{k}x\left(n-k\right)$ (4.12)

where ${a}_{k}$ and ${b}_{k}$ are the filter coefficients (constants) , and the limits M, N can be extended to infinity .

Now we make the replacement $x\left(n\right)={z}^{n}$ and $y\left(n\right)={z}^{n}$ $H\left(z\right)$ to get

${z}^{n}H\left(z\right)=$ $\sum _{k=1}^{M}{a}_{k}{z}^{n-\text{k}}H\left(z\right)$ $+\sum _{k=-N}^{N}{b}_{k}{z}^{n-k}$

From this we derive the expression of $H\left(z\right)$

$H\left(z\right)=$ $\frac{\sum _{k=-N}^{N}{b}_{k}{z}^{-k}}{1-\sum _{k=1}^{M}{a}_{k}{z}^{-k}}$ (4.13)

It is worthwhile to notice that the above transfer function is resulted from the filter equation (4.12) . Other authors write the filter equation differently (for example, all the y terms are on the left side of the equation), leading to a slightly different expression of $H\left(z\right)$ . For nonrecursive fillters the denominator is just 1 and $H\left(z\right)$ becomes that of nonrecursive filters as we know .

The idea here is that when the filter equation is given, we collect its coefficients to place into the expression above of $H\left(z\right)$ without the need to take the z-transform. Vice versa, if we know $H\left(z\right)$ that means we know the filler coefficients , hence the filter equation.

## Example 4.1.4

Given

(a) $H\left(z\right)=$ $\frac{{2z}^{2}-3z}{{z}^{2}+0\text{.}5z-0\text{.}8}$

(b) $H\left(z\right)=$ $\frac{-\text{20}{z}^{2}+5z}{\text{10}{z}^{3}+{5z}^{2}-8z-1}$

Find the filter difference equation.

## Solution

(a) Write $H\left(z\right)$ as a function of ${z}^{-1}$ by multiplying the numerator and denominator with ${z}^{-2}$ :

$H\left(z\right)=$ $\frac{2-{3z}^{-1}}{1+0\text{.}{5z}^{-1}-0\text{.}{8z}^{-2}}=\frac{2-{3z}^{-1}}{1-\left(-0\text{.}{5z}^{-1}+0\text{.}{8z}^{-2}\right)}$

The coefficients are

${a}_{0}=2$ ${a}_{1}=-3$ ${b}_{1}=-0\text{.}5$ ${b}_{2}=0\text{.}8$

Thus the filler equation is

$y\left(n\right)=-0\text{.}5y\left(n-1\right)-0\text{.}8y\left(n-2\right)+2x\left(n\right)-3x\left(n-1\right)$

(b) Multiply the numerator and denominator with $0\text{.}{1z}^{-3}$ to make $\text{10}{z}^{-3}$ in the denominator equal to 1 :

$H\left(z\right)=$ $\frac{-{2z}^{-1}+{5z}^{-2}}{1+0\text{.}{5z}^{-1}-0\text{.}{8z}^{-2}-0\text{.}{1z}^{-3}}$

Collect the coefficients:

${a}_{1}=-2$ ${a}_{2}=5$ ${b}_{1}=-0\text{.}5$ ${b}_{2}=0\text{.}8$ ${b}_{3}=-0\text{.}1$

Thus

$y\left(n\right)=-0\text{.}5y\left(n-1\right)+0\text{.}8y\left(n-2\right)+0\text{.}1y\left(n-3\right)-2x\left(n-1\right)+5x\left(n-2\right)$ y(n)

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