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.

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

Started by ●November 20, 2004

Reply by ●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 23, 20042004-11-23

�� Sun, 21 Nov 2004 09:50:03 -0500 ʱ, "Clay Turner" <physics@bellsouth.net> д��: --> >"��" <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. > >ClayThanks a lot. Now it seems like a piece of cake.

Reply by ●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!