# Trouble with inverse Z-transform and calculating of samples

I have a little problem. I have to solve this task but I can't.

Z-transform of sequence $$\{x(k)\}$$ describe by the formula:

$$X(z) = \frac{2.5 -3.15z^{-1} + 1.2 z^{-2}}{1-2.3z^{-1} + 1.2z^{-2}}$$

Calculate of samples $$x(-2), x(-1), x(0), x(1)$$ and $$x(2)$$ double-sides sequence, which correspond this Z-transform.

Thank you very much.

The poles in the equation are Z=1.5,0.8;hence to be two sided the roc is $$0.8<=z<=1.5$$ so the 0.8 term is right sided and 1.5 term is left sided let us do partial fractions now $$X(Z)=1+\frac{1.5-0.85z^{-1}} {(1-0.8z^{-1})(1-1.5z^{-1})}$$ after doing some maths i got $$x(n)=\delta(n)-0.5(0.8)^nu(n)-2(1.5)^nu(-n-1)$$ so x(-2)=-0.88888,x(-1)=-1.3333,x(0)=0.5,x(1)=-0.4,x(2)=-0.32.

• Yes, I've got the same results besides x(0), I've got x(0) = 2.5 (why did you get x(0) = 0.5), I decribed my solve here, but answer isn't correct and I don't know why)) – Alexander Lyapunov Dec 28 '18 at 11:58
• I think partial fraction method is applicable if numerator degree is less than denominator degree so you have done it opposite hence our answers differ at x(0). I hope my point got trough. – Ch.Siva Ram Kishore Dec 28 '18 at 12:52
• Shouldn't the last term in your expression for $x[n]$ be multiplied with $u[-n-1]$ (instead of $u[-n+1]$)? – Matt L. Dec 28 '18 at 16:51
• @MattL. Yes I got confused. – Ch.Siva Ram Kishore Dec 28 '18 at 16:59
• Ch.Siva Ram Kishore, you're right! I wrote description there (link here). Thank you everyone! – Alexander Lyapunov Feb 9 '19 at 14:34

"Double-side" sequence I mean sequence like this:

And I'm trying to solve this task with MatLab, I've used function "iztrans()" (inverse z-transformation) and I've gotten this results (below MatLab code and results):

But sequence which I've gotten isn't correct. Could you say what is my mistake?