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The Inverse Z-Transform

Module by: Benjamin Fite. E-mail the author

Summary: This module describes the inverse Z-transform.

Introduction

When using the z-transform

Xz= n =xnzn X z n x n z n
(1)
it is often useful to be able to find xn x n given Xz X z . There are at least 4 different methods to do this:

Inspection Method

This "method" is to basically become familiar with the z-transform pair tables and then "reverse engineer".

Example 1

When given Xz=zzα X z z z α with an ROC of |z|>α z α we could determine "by inspection" that xn=αnun x n α n u n

Partial-Fraction Expansion Method

When dealing with linear time-invariant systems the z-transform is often of the form

Xz=BzAz= k =0M b k zk k =0N a k zk X z B z A z k 0 M b k z k k 0 N a k z k
(2)
This can also expressed as
Xz= a 0 b 0 k =1M1 c k z-1 k =1N1 d k z-1 X z a 0 b 0 k 1 M 1 c k z k 1 N 1 d k z
(3)
where c k c k represents the nonzero zeros of Xz X z and d k d k represents the nonzero poles.

If M<N M N then Xz X z can be represented as

Xz= k =1N A k 1 d k z-1 X z k 1 N A k 1 d k z
(4)
This form allows for easy inversions of each term of the sum using the inspection method and the transform table. If the numerator is a polynomial, however, then it becomes necessary to use partial-fraction expansion to put Xz X z in the above form. If MN M N then Xz X z can be expressed as
Xz= r =0MN B r zr+ k =0N1 b k ' zk k =0N a k zk X z r 0 M N B r z r k 0 N 1 b k ' z k k 0 N a k z k
(5)

Example 2

Find the inverse z-transform of Xz=1+2z-1+z-213z-1+2z-2 X z 1 2 z z -2 1 -3 z 2 z -2 where the ROC is |z|>2 z 2 . In this case M=N=2 M N 2 , so we have to use long division to get Xz=12+12+72z-113z-1+2z-2 X z 1 2 1 2 7 2 z 1 -3 z 2 z -2 Next factor the denominator. Xz=2+-1+5z-1(12z-1)(1z-1) X z 2 -1 5 z 1 2 z 1 z Now do partial-fraction expansion. Xz=12+ A 1 12z-1+ A 2 1z-1=12+9212z-1+-41z-1 X z 1 2 A 1 1 2 z A 2 1 z 1 2 9 2 1 2 z -4 1 z Now each term can be inverted using the inspection method and the z-transform table. Thus, since the ROC is |z|>2 z 2 , xn=12δn+922nun4un x n 1 2 δ n 9 2 2 n u n -4 u n

Demonstration of Partial Fraction Expansion

Figure 1: Interactive experiment illustrating how the Partial Fraction Expansion method is used to solve a variety of numerator and denominator problems. (To view and interact with the simulation, download the free Mathematica player at http://www.wolfram.com/products/player/download.cgi)
A demonstration involving Partial Fraction Expansion

Figure 2: video from Khan Academy
Khan Lecture on Partial Fraction Expansion

Power Series Expansion Method

When the z-transform is defined as a power series in the form

Xz= n =xnzn Xz n x n z n
(6)
then each term of the sequence xn x n can be determined by looking at the coefficients of the respective power of zn z n .

Example 3

Now look at the z-transform of a finite-length sequence.

Xz=z2(1+2z-1)(112z-1)(1+z-1)=z2+52z+12+z-1 X z z 2 1 2 z 1 1 2 z 1 z z 2 5 2 z 1 2 z
(7)
In this case, since there were no poles, we multiplied the factors of Xz X z . Now, by inspection, it is clear that xn=δn+2+52δn+1+12δn+δn1 x n δ n 2 5 2 δ n 1 1 2 δ n δ n 1 .

One of the advantages of the power series expansion method is that many functions encountered in engineering problems have their power series' tabulated. Thus functions such as log, sin, exponent, sinh, etc, can be easily inverted.

Example 4

Suppose Xz=log n (1+αz-1) X z n 1 α z Noting that log n (1+x)= n =1-1n+1xnn n 1 x n 1 -1 n 1 x n n Then Xz= n =1-1n+1αnznn X z n 1 -1 n 1 α n z n n Therefore Xz={-1n+1αnn  if  n10  if  n0 X z -1 n 1 α n n n 1 0 n 0

Contour Integration Method

Without going in to much detail

xn=12πirXzzn1d z x n 1 2 z r X z z n 1
(8)
where r r is a counter-clockwise contour in the ROC of Xz X z encircling the origin of the z-plane. To further expand on this method of finding the inverse requires the knowledge of complex variable theory and thus will not be addressed in this module.

Demonstration of Contour Integration

Figure 3: Interactive experiment illustrating how the contour integral is applied on a simple example. For a more in-depth discussion of this method, some background in complex analysis is required. (To view and interact with the simulation, download the free Mathematica player at http://www.wolfram.com/products/player/download.cgi)
A demonstration involving Contour Integration

Conclusion

The Inverse Z-transform is very useful to know for the purposes of designing a filter, and there are many ways in which to calculate it, drawing from many disparate areas of mathematics. All nevertheless assist the user in reaching the desired time-domain signal that can then be synthesized in hardware(or software) for implementation in a real-world filter.

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