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Inverse z-Transform

Module by: Mark A. Davenport. E-mail the author

Inverse zz-transform

Up to this point, we have ignored how to actually invert a z-transform to find x[n]x[n] from X(z)X(z). Doing so is very different from inverting a DTFT. We will consider three main techniques:

  1. Inspection (look it up in a table)
  2. Partial fraction expansion
  3. Power series expansion

One can also use contour integration combined with the Cauchy Residue Theorem. See Oppenheim and Schafer for details.

Inspection

Basically, become familiar with the zz-transform pairs listed in tables, and “reverse engineer”

Example 1

Suppose that

X ( z ) = z z - a , | z | > | a | . X ( z ) = z z - a , | z | > | a | .
(1)

By now you should be able to recognize that x[n]=anu[n]x[n]=anu[n].

Partial fraction expansion

If X(z)X(z) is rational, break it up into a sum of elementary forms, each of which can be inverted by inspection.

Example 2

Suppose that

X ( z ) = 1 + 2 z - 1 + z - 2 1 - 3 2 z - 1 + 1 2 z - 2 , | z | > 1 . X ( z ) = 1 + 2 z - 1 + z - 2 1 - 3 2 z - 1 + 1 2 z - 2 , | z | > 1 .
(2)

By computing a partial fraction expansion we can decompose X(z)X(z) into

X ( z ) = 8 1 - z - 1 - 9 1 - 1 2 z - 1 + 2 , X ( z ) = 8 1 - z - 1 - 9 1 - 1 2 z - 1 + 2 ,
(3)

where each term in the sum can be inverted by inspection.

Power Series Expansion

Recall that

X ( z ) = ∑ n = - ∞ ∞ x [ n ] z - n = ... x [ - 2 ] z 2 + x [ - 1 ] z + x [ 0 ] + x [ 1 ] z - 1 + x [ 2 ] z - 2 + ... . X ( z ) = ∑ n = - ∞ ∞ x [ n ] z - n = ... x [ - 2 ] z 2 + x [ - 1 ] z + x [ 0 ] + x [ 1 ] z - 1 + x [ 2 ] z - 2 + ... .
(4)

If we know the coefficients for the Laurent series expansion of X(z)X(z), then these coefficients give us the inverse zz-transform.

Example 3

Suppose

X(z)=z21-12z-11+z-11-z-1=z2-12z-1+12z-1X(z)=z21-12z-11+z-11-z-1=z2-12z-1+12z-1
(5)

Then

x[n]=δ[n+2]-12δ[n+1]-δ[n]+12δ[n-1].x[n]=δ[n+2]-12δ[n+1]-δ[n]+12δ[n-1].
(6)

Example 4

Suppose

X(z)=log(1+az-1),|z|>|a|X(z)=log(1+az-1),|z|>|a|
(7)

where loglog denotes the complex logarithm. Recalling the Laurent series expansion

log(1+x)=∑n=1∞(-1)n+1xnnlog(1+x)=∑n=1∞(-1)n+1xnn
(8)

we can write

X(z)=∑n=1∞(-1)n+1annz-n.X(z)=∑n=1∞(-1)n+1annz-n.
(9)

Thus we can infer that

x[n]=(-1)n+1annn≥10n≤0.x[n]=(-1)n+1annn≥10n≤0.
(10)

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