0.1 Discrete-time signals  (Page 7/10)

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

and several bilateral z-transforms are given by

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

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 $F\left(z\right)=\frac{P\left(z\right)}{Q\left(z\right)}$ , $f\left(n\right)$ can be obtained as the coefficients of long division.

For example

which is $u\left(n\right)a^n$ 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 $x(n-1)$ or $x(n+3)$ . 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 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 $x_h\left(n\right)=K_1{\lambda }_1^n+K_2{\lambda }_2^n$ . 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