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.

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

enter image description here enter image description here enter image description here

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):

enter image description here enter image description here

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


1 Answer 1


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.

  • $\begingroup$ 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)) $\endgroup$ Commented Dec 28, 2018 at 11:58
  • $\begingroup$ 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. $\endgroup$ Commented Dec 28, 2018 at 12:52
  • 1
    $\begingroup$ Shouldn't the last term in your expression for $x[n]$ be multiplied with $u[-n-1]$ (instead of $u[-n+1]$)? $\endgroup$
    – Matt L.
    Commented Dec 28, 2018 at 16:51
  • $\begingroup$ @MattL. Yes I got confused. $\endgroup$ Commented Dec 28, 2018 at 16:59
  • $\begingroup$ Ch.Siva Ram Kishore, you're right! I wrote description there (link here). Thank you everyone! $\endgroup$ Commented Feb 9, 2019 at 14:34

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