Reply by Country_Chiel November 25, 20042004-11-25
"&#4294967295;&#4294967295;" <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!
Reply by November 23, 20042004-11-23
&#4294967295;&#4294967295; Sun, 21 Nov 2004 09:50:03 -0500 &#689;, "Clay Turner" <physics@bellsouth.net>
&#1076;&#4294967295;&#4294967295;:
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> >"&#4294967295;&#4294967295;" <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
"&#4294967295;&#4294967295;" <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.