# 0.1 Discrete-time signals  (Page 7/10)

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## Examples of the z-transform

A few examples together with the above properties will enable one to solve and understand a wide variety of problems. These use the unit stepfunction to remove the negative time part of the signal. This function is defined as

$u\left(n\right)\phantom{\rule{0.277778em}{0ex}}=\phantom{\rule{0.277778em}{0ex}}\left\{\begin{array}{cc}1\hfill & \phantom{\rule{4.pt}{0ex}}\text{if}\phantom{\rule{4.pt}{0ex}}n\ge 0\hfill \\ 0\hfill & \phantom{\rule{4.pt}{0ex}}\text{if}\phantom{\rule{4.pt}{0ex}}n<0\hfill \end{array}\right)$

and several bilateral z-transforms are given by

• $\mathcal{Z}\left\{\delta \left(n\right)\right\}\phantom{\rule{0.277778em}{0ex}}=\phantom{\rule{0.277778em}{0ex}}1$ for all $z$ .
• $\mathcal{Z}\left\{u\left(n\right)\right\}\phantom{\rule{0.277778em}{0ex}}=\phantom{\rule{0.277778em}{0ex}}\frac{z}{z-1}$ for $|z|>1$ .
• $\mathcal{Z}\left\{u\left(n\right){a}^{n}\right\}\phantom{\rule{0.277778em}{0ex}}=\phantom{\rule{0.277778em}{0ex}}\frac{z}{z-a}$ for $|z|>|a|$ .

Notice that these are similar to but not the same as a term of a partial fraction expansion.

## Inversion of the z-transform

The z-transform can be inverted in three ways. The first two have similar procedures with Laplace transformations and the third has no counter part.

• The z-transform can be inverted by the defined contour integral in the ROC of the complex $z$ plane. This integral can be evaluated using the residue theorem [link] , [link] .
• The z-transform can be inverted by expanding $\frac{1}{z}F\left(z\right)$ in a partial fraction expansion followed by use of tables for the first orsecond order terms.
• The third method is not analytical but numerical. If $F\left(z\right)\phantom{\rule{0.277778em}{0ex}}=\phantom{\rule{0.277778em}{0ex}}\frac{P\left(z\right)}{Q\left(z\right)}$ , $f\left(n\right)$ can be obtained as the coefficients of long division.

For example

$\frac{z}{z-a}\phantom{\rule{0.277778em}{0ex}}=\phantom{\rule{0.277778em}{0ex}}1+a\phantom{\rule{0.166667em}{0ex}}{z}^{-1}+{a}^{2}{z}^{-2}+\cdots$

which is $u\left(n\right)\phantom{\rule{0.166667em}{0ex}}{a}^{n}$ as used in the examples above.

We must understand the role of the ROC in the convergence and inversion of the z-transform. We must also see the difference between the one-sided andtwo-sided transform.

## Solution of difference equations using the z-transform

The z-transform can be used to convert a difference equation into an algebraic equation in the same manner that the Laplace converts a differential equation in to an algebraic equation. The one-sided transform isparticularly well suited for solving initial condition problems. The two unilateral shift properties explicitly use the initial values of theunknown variable.

A difference equation DE contains the unknown function $x\left(n\right)$ and shifted versions of it such as $x\left(n-1\right)$ or $x\left(n+3\right)$ . The solution of the equation is the determination of $x\left(t\right)$ . A linear DE has only simple linear combinations of $x\left(n\right)$ and its shifts. An example of a linear second order DE is

$a\phantom{\rule{0.166667em}{0ex}}x\left(n\right)+b\phantom{\rule{0.166667em}{0ex}}x\left(n-1\right)+c\phantom{\rule{0.166667em}{0ex}}x\left(n-2\right)\phantom{\rule{0.277778em}{0ex}}=\phantom{\rule{0.277778em}{0ex}}f\left(n\right)$

A time invariant or index invariant DE requires the coefficients not be a function of $n$ and the linearity requires that they not be a function of $x\left(n\right)$ . Therefore, the coefficients are constants.

This equation can be analyzed using classical methods completely analogous to those used with differential equations. A solution of the form $x\left(n\right)=K{\lambda }^{n}$ is substituted into the homogeneous difference equation resulting in a second order characteristic equation whose two roots givea solution of the form ${x}_{h}\left(n\right)\phantom{\rule{0.277778em}{0ex}}=\phantom{\rule{0.277778em}{0ex}}{K}_{1}{\lambda }_{1}^{n}+{K}_{2}{\lambda }_{2}^{n}$ . A particular solution of a form determined by $f\left(n\right)$ is found by the method of undetermined coefficients, convolution or some other means. Thetotal solution is the particular solution plus the solution of the homogeneous equation and the three unknown constants ${K}_{i}$ are determined from three initial conditions on $x\left(n\right)$ .

It is possible to solve this difference equation using z-transforms in a similar way to the solving of a differential equation by use of theLaplace transform. The z-transform converts the difference equation into an algebraic equation. Taking the ZT of both sides of the DE gives

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