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Using definition, I got its Z transform as $X(z) = \dfrac{2}{1-\dfrac{z}{3}}$ and the summation converges only when $|z|<\frac{1}{3}$. So its ROC is $|z|<\frac{1}{3}$.

But my question is: for such a left sided signal $x(n)$, its ROC should be inner to the innermost pole. But the pole of $\dfrac{2}{1-\dfrac{z}{3}}$ is $z=3$ and not $z= \frac{1}{3}$. So the ROC I found is incorrect, right?

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    $\begingroup$ But sir, $3^nu(-n)$ is NOT the time reversed version of $3^nu(n)$. So we cant apply Time reversal property here. Right? $\endgroup$ – Shehin Dec 12 '19 at 15:03
  • $\begingroup$ the ROC should be |z|<3 because pole is at |z|=3. $\endgroup$ – Ch.Siva Ram Kishore Dec 12 '19 at 15:18
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Using the definition, and letting $k=-n$:

$X(z)=2 \sum_{n=-\infty}^03^nz^{-n}=2\sum_{k=0}^{\infty}\big(\frac{1}{3}z\big)^k$

Now we can use the geometric series (https://en.wikipedia.org/wiki/Geometric_series) formula only if we require $|\frac{1}{3}z|<1$, or equivalently $|z|<3$

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  • $\begingroup$ That was useful sir. Thank you $\endgroup$ – Shehin Dec 12 '19 at 17:49
  • $\begingroup$ Great! Did it answer your question? You can "Accept answer" if it did $\endgroup$ – Engineer Dec 12 '19 at 19:12

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