The given system involves an LTI (Linear Time-Invariant) system, where you are asked to determine the Region of Convergence (ROC) for two sequences: $x_1(n)$ and $x_2(n)$.

The sequences are: $x_1(n) = \left( \frac{1}{6} \right)^n u(n-1)$ $x_2(n) = \left[ 1 + \left( \frac{1}{4} \right)^n \right] u(n)$ where $u(n)$ is the unit step function.

Analyzing $x_1(n)$:

1. $\left( \frac{1}{6} \right)^n$ is an exponentially decaying sequence.
2. The term $u(n-1)$ shifts the unit step function, meaning $x_1(n)$ is nonzero for $n \geq 1$.

For the Z-transform of $x_1(n)$: $X_1(z) = \sum_{n=1}^{\infty} \left( \frac{1}{6} \right)^n z^{-n}$ The series converges when $|z| > \frac{1}{6}$, as the ROC of an exponentially decaying term $a^n$ is $|z| > a$.

Thus, the ROC for $x_1(n)$ is $( \frac{1}{6}, \infty )$.

Analyzing $x_2(n)$:

1. The term $\left( \frac{1}{4} \right)^n$ is also exponentially decaying.
2. $u(n)$ means the sequence is valid for $n \geq 0$.

For the Z-transform of $x_2(n)$: $X_2(z) = \sum_{n=0}^{\infty} \left[ 1 + \left( \frac{1}{4} \right)^n \right] z^{-n}$ This can be split into two parts:

• The constant term $1$ contributes to a sequence with no decay, so its ROC is the entire z-plane except $z = 0$.
• The term $\left( \frac{1}{4} \right)^n$ has an ROC $|z| > \frac{1}{4}$.

Therefore, the ROC for $x_2(n)$ is $( \frac{1}{4}, \infty )$.

Summary of ROCs:

• ROC of $x_1(n)$: $( \frac{1}{6}, \infty )$
• ROC of $x_2(n)$: $( \frac{1}{4}, \infty )$

Would you like more details on the Z-transform process or have any questions?

Here are 5 related questions for further exploration:

1. How do you calculate the Z-transform of a shifted sequence like $u(n-1)$?
2. What happens to the ROC when sequences are multiplied or convolved?
3. How does the ROC relate to the stability of a system?
4. Can the ROC include points on the unit circle?
5. What is the impact of poles on the ROC?

Tip: Always check the ROC carefully when analyzing system stability or reconstructing a signal from its Z-transform.