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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 ( 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

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Source:  OpenStax, Brief notes on signals and systems. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col10565/1.7
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