Reply by Country_Chiel●November 25, 20042004-11-25
"��" <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!
>
>"��" <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.
Reply by Clay Turner●November 21, 20042004-11-21
"��" <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
Reply by ●November 20, 20042004-11-20
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.