<< Chapter < Page Chapter >> Page >

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 ( n ) = 1 if n 0 0 if n < 0

and several bilateral z-transforms are given by

  • Z { δ ( n ) } = 1 for all z .
  • Z { u ( n ) } = z z - 1 for | z | > 1 .
  • Z { u ( n ) a n } = 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 1 z F ( z ) 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 ( z ) = P ( z ) Q ( z ) , f ( n ) can be obtained as the coefficients of long division.

For example

z z - a = 1 + a z - 1 + a 2 z - 2 +

which is u ( n ) 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 ( n ) and shifted versions of it such as x ( n - 1 ) or x ( n + 3 ) . The solution of the equation is the determination of x ( t ) . A linear DE has only simple linear combinations of x ( n ) and its shifts. An example of a linear second order DE is

a x ( n ) + b x ( n - 1 ) + c x ( n - 2 ) = f ( n )

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 ( n ) . 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 ( n ) = K λ 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 ( n ) = K 1 λ 1 n + K 2 λ 2 n . A particular solution of a form determined by f ( n ) 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 ( n ) .

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

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Brief notes on signals and systems. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col10565/1.7
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Brief notes on signals and systems' conversation and receive update notifications?

Ask