# ROC of Z Transform of $x(n) = 2(3)^nu(-n)$

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?

• But sir, $3^nu(-n)$ is NOT the time reversed version of $3^nu(n)$. So we cant apply Time reversal property here. Right? Dec 12, 2019 at 15:03
• the ROC should be |z|<3 because pole is at |z|=3. Dec 12, 2019 at 15:18

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 $$|z|<3$$

• That was useful sir. Thank you Dec 12, 2019 at 17:49
• Great! Did it answer your question? You can "Accept answer" if it did Dec 12, 2019 at 19:12

or using the time-domain inversion property:

input: $$f[n]$$ , so its z-transform is $$F(z)$$;

inversion property: input $$f[-n]$$, its z-transform is $$F(1/z)$$