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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+1annn10n0.x[n]=(-1)n+1annn10n0.
(10)

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