# How to get the inverse Z-transform of the following expression?

Started by November 20, 2004
```                       1
H(Z) = ---------------------------------
( 1+0.81*Z^(-2) )( 1+0.81Z^(2) )

So far, only could I get is the inverse of
1
H1(Z) = --------------------
( 1+0.81*Z^(-2) )

or
1
H2(Z) = --------------------
( 1+0.81*Z^(2) )

Any ideas? thanks.

```
```"&#2013266100;&#2013281911;" <abc@none.com> wrote in message
news:apqvp092s5rlp7p51marnfajt3osehbmga@news.cn99.com...
>                        1
> H(Z) = ---------------------------------
>         ( 1+0.81*Z^(-2) )( 1+0.81Z^(2) )
>
> So far, only could I get is the inverse of
>                1
> H1(Z) = --------------------
>         ( 1+0.81*Z^(-2) )
>
> or
>                1
> H2(Z) = --------------------
>         ( 1+0.81*Z^(2) )
>
> Any ideas? thanks.

Yes,

Use partial fraction decomposition to split the fraction into two parts.
Each of them can then be easily invere Z- transformed.

Clay

```
```&#2013265940;&#2013265946; Sun, 21 Nov 2004 09:50:03 -0500 &#689;, "Clay Turner" <physics@bellsouth.net>
&#1076;&#2013266113;&#2013265931;:
--

>
>"&#2013266100;&#2013281911;" <abc@none.com> wrote in message
>news:apqvp092s5rlp7p51marnfajt3osehbmga@news.cn99.com...
>>                        1
>> H(Z) = ---------------------------------
>>         ( 1+0.81*Z^(-2) )( 1+0.81Z^(2) )
>>
>> So far, only could I get is the inverse of
>>                1
>> H1(Z) = --------------------
>>         ( 1+0.81*Z^(-2) )
>>
>> or
>>                1
>> H2(Z) = --------------------
>>         ( 1+0.81*Z^(2) )
>>
>> Any ideas? thanks.
>
>Yes,
>
>Use partial fraction decomposition to split the fraction into two parts.
>Each of them can then be easily invere Z- transformed.
>
>Clay

Thanks a lot. Now it seems like a piece of cake.
```
```"&#2013266100;&#2013281911;" <abc@none.com> wrote in message
news:apqvp092s5rlp7p51marnfajt3osehbmga@news.cn99.com...
>                        1
> H(Z) = ---------------------------------
>         ( 1+0.81*Z^(-2) )( 1+0.81Z^(2) )
>
> So far, only could I get is the inverse of
>                1
> H1(Z) = --------------------
>         ( 1+0.81*Z^(-2) )
>
> or
>                1
> H2(Z) = --------------------
>         ( 1+0.81*Z^(2) )
>
> Any ideas? thanks.
>
>

Careful, this symmetrical expression could be interpreted as a causal and
anticausal system in cascade - well more of a power spectrum expression so
it could have terms going forward and backwards in time.Or it could be
wholly causal..Personally, I feel if somebody uses z^-1 terms then they
should stick to it and not confuse matters by using positive powers of z
too!

```