# Inverse Z transform by inversion integral method

• Jul 2nd 2010, 08:17 PM
moonnightingale
Inverse Z transform by inversion integral method
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
• Jul 2nd 2010, 09:09 PM
mr fantastic
Quote:

Originally Posted by moonnightingale
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.
• Jul 3rd 2010, 02:47 AM
moonnightingale
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
• Jul 3rd 2010, 04:47 AM
mr fantastic
Quote:

Originally Posted by moonnightingale
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.