Inverse Z transform by inversion integral method

Sep 2009
135
1
Hi guys i am stuck with this question.
I have to calculate inverse Z transform by inversion integral method of

X(Z) = 1 / (1+ z^ -2) (z raise to power -2)

I have done it in matlab and by computation menthod and answer comes to be same as

[1 0 -1 0 1 0 -1 - - - - - ]

but when i do it with inversion integral my answer comes to be i^k ( i= iota)

if i put values of k as 0 1 2 3 my answers does not match
plz help me
 

mr fantastic

MHF Hall of Fame
Dec 2007
16,948
6,768
Zeitgeist
Hi guys i am stuck with this question.
I have to calculate inverse Z transform by inversion integral method of

X(Z) = 1 / (1+ z^ -2) (z raise to power -2)

I have done it in matlab and by computation menthod and answer comes to be same as

[1 0 -1 0 1 0 -1 - - - - - ]

but when i do it with inversion integral my answer comes to be i^k ( i= iota)

if i put values of k as 0 1 2 3 my answers does not match
plz help me
I get the MATLAB answer (which is consistent with the series expansion of X(Z)). Please post all your working.
 
Sep 2009
135
1
I cannot post my working here as i have done it with hand and it will take a lot time to type in Microsoft word for me

Kindly if u can do this question by inversion integral and upload it
the poles come to be + iota and - iota
 

mr fantastic

MHF Hall of Fame
Dec 2007
16,948
6,768
Zeitgeist
I cannot post my working here as i have done it with hand and it will take a lot time to type in Microsoft word for me

Kindly if u can do this question by inversion integral and upload it
the poles come to be + iota and - iota
So what you're saying is that it's too much work for you to type out here but not too much work for me.

Wrong reply.