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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.

Consider the LTI system with the input x1(n), the unit sample response x2(n), and the output y(n), where x1(n) = (1/6)^n * u(n-1) and x2(n) = [1 + (1/4)^n] * u(n). Find the Region of Convergence (ROC).

This problem involves calculating the Region of Convergence (ROC) for an LTI system with the given input sequences x1(n) = (1/6)^n * u(n-1) and x2(n) = [1 + (1/4)^n] * u(n). Learn how to find the ROC using Z-transform and analyze the stability of the system.

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