Let $$x\left[ n \right] = {\left( { – {1 \over 9}} \right)^n}u\left( n \right) – {\left( { – {1 \over 3}} \right)^n}u\left( { – n – 1} \right).$$ The Region of Convergence (ROC) of the z-transform of x[n]

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The correct answer is $\boxed{{1 \over 3} > \left| z \right| > {1 \over 9}}$.

The region of convergence (ROC) of the z-transform of a sequence $x[n]$ is the set of values of $z$ for which the z-transform converges. The z-transform of $x[n]$ is defined as

$$X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}$$

The ROC of $X(z)$ is the set of values of $z$ for which the infinite series in the definition of $X(z)$ converges.

In this case, $x[n] = {\left( { – {1 \over 9}} \right)^n}u\left( n \right) – {\left( { – {1 \over 3}} \right)^n}u\left( { – n – 1} \right)$.

The z-transform of $x[n]$ is

$$X(z) = {1 \over 1 – {1 \over 9} z^{-1}} – {1 \over 1 + {1 \over 3} z^{-1}}$$

The poles of $X(z)$ are at $z = 9$ and $z = -3$.

The ROC of $X(z)$ is the set of values of $z$ for which the distance between $z$ and any pole is greater than 1.

In this case, the distance between $z$ and $z = 9$ is $\sqrt{100} = 10$. The distance between $z$ and $z = -3$ is $\sqrt{10} = 3.16$.

Therefore, the ROC of $X(z)$ is the set of values of $z$ such that $10 > \left| z \right| > 3.16$, or $${1 \over 3} > \left| z \right| > {1 \over 9}$$

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