I'm given a specific function to find its inverse Z-transform. Specifically: $$X(z) = z^3 + \frac{1}{z - 2i} + \frac{1}{z+2i}, |z| > 2 $$ Notice the $|z| >2$. Now what concerns me is the term $z^3$. It's a standard $Z$-transform that $$ \delta[n-m] \stackrel{\mathcal{Z}}{\longleftrightarrow} z^{-m}, $$ with its ROC being every $z$ except $\infty$ if $m <0$, as is the case here. So, doesn't this transform contradict the given fact that |z|>2? Because If I were to use that transform that'd mean that I would change the given ROC of $X(z)$ not to include $\infty$.

Since these problems are designed from other people couldn't it be that they just forgot that tiny detail and should've changed the starting ROC to $2 < |z| < \infty$? This is basically what I'm asking. If that's not the case, does there exist an inverse $Z$-transform of $z^{3}$ so that it will also include $\infty$ and not contradict the given known ROC?


1 Answer 1


I agree with you, it does not converge when $z=\infty$ or when $z\le2$ if we assume the inverse z-transform solution is anti-causal (as it would need to be from the $z^3$ term).

There is no inverse z-transform that also includes $z=\infty$ for $X(z)=z^3$, and given the inverse z-transform of this component alone is $u[n+3]$, which is non-causal, it can only exist if the ROC is less than $\infty$. The next two components with poles at $\pm2j$ would have a causal inverse z-transform if the $ROC>2$.

Note there is also no Fourier Transform since the unit circle for the equation as given is not in the ROC.

  • $\begingroup$ Thanks a lot Dan, I remember you answering my dsp questions like 2.5 years ago! $\endgroup$
    – Nyquist-er
    Jan 11 at 19:16
  • $\begingroup$ ha! yup, this is where you can find me! $\endgroup$ Jan 11 at 19:51

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