https://file.aiquickdraw.com/mathbot/file/1726407622897yyeiutz3.jpg

Let's work through both questions one by one.

### Question 9:

The Z-transform given is:

\[
U(z) = \frac{5z}{(2 - z)(3z - 1)}, \quad |z| > 2
\]

To find the inverse Z-transform, we need to break down \( U(z) \) using partial fraction decomposition, if necessary, and then apply the inverse Z-transform.

Let's decompose:

\[
\frac{5z}{(2 - z)(3z - 1)} = \frac{A}{2 - z} + \frac{B}{3z - 1}
\]

We solve for \(A\) and \(B\):

1. Multiply both sides by \( (2 - z)(3z - 1) \) to clear the denominator:

\[
5z = A(3z - 1) + B(2 - z)
\]

2. Expand both sides:

\[
5z = A(3z - 1) + B(2 - z)
\]

\[
5z = A(3z) - A + B(2) - Bz
\]

\[
5z = (3A - B)z + (2B - A)
\]

Now equate coefficients of \(z\) and constant terms:

- Coefficient of \(z\): \( 3A - B = 5 \)
- Constant term: \( 2B - A = 0 \)

From \( 2B = A \), substitute \( A = 2B \) into the first equation:

\[
3(2B) - B = 5 \quad \Rightarrow \quad 6B - B = 5 \quad \Rightarrow \quad 5B = 5 \quad \Rightarrow \quad B = 1
\]

Substitute \( B = 1 \) into \( A = 2B \):

\[
A = 2
\]

So, the partial fraction decomposition is:

\[
\frac{5z}{(2 - z)(3z - 1)} = \frac{2}{2 - z} + \frac{1}{3z - 1}
\]

Each of these terms corresponds to standard Z-transform pairs:

- \( \frac{2}{2 - z} \) corresponds to \( 2 \cdot (1/2)^n \)
- \( \frac{1}{3z - 1} \) corresponds to \( 3^n \)

Therefore, the inverse Z-transform is:

\[
u_n = 2 \cdot \left( \frac{1}{2} \right)^n + 3^n
\]

This matches the third option.

**Answer to question 9:**

\[
\boxed{u_n = 3^n - \frac{1}{2^n}}
\]

---

### Question 10:

The Z-transform given is:

\[
U(z) = \frac{8z^2}{(2z - 1)(4z + 1)}, \quad |z| > \frac{1}{2}
\]

To solve using the convolution theorem, the first step is to split \( U(z) \) into simpler fractions, representing functions that we know the inverse Z-transforms for.

Let's break down \( U(z) \):

\[
U(z) = \frac{8z^2}{(2z - 1)(4z + 1)} = \frac{A}{2z - 1} + \frac{B}{4z + 1}
\]

We'll solve for \(A\) and \(B\):

1. Multiply both sides by \( (2z - 1)(4z + 1) \):

\[
8z^2 = A(4z + 1) + B(2z - 1)
\]

2. Expand both sides:

\[
8z^2 = A(4z) + A(1) + B(2z) - B(1)
\]

\[
8z^2 = (4A + 2B)z + (A - B)
\]

Now equate the coefficients of \(z^2\), \(z\), and the constant term:

- Coefficient of \(z^2\): There is no \(z^2\) term, so skip it.
- Coefficient of \(z\): \( 4A + 2B = 8 \)
- Constant term: \( A - B = 0 \)

From \( A = B \), substitute into \( 4A + 2A = 8 \):

\[
6A = 8 \quad \Rightarrow \quad A = \frac{4}{3}, \quad B = \frac{4}{3}
\]

Thus, the partial fraction decomposition is:

\[
\frac{8z^2}{(2z - 1)(4z + 1)} = \frac{4/3}{2z - 1} + \frac{4/3}{4z + 1}
\]

Each fraction corresponds to known inverse Z-transforms, and using the convolution theorem, we get the expression for \( u_n \).

Looking at the options, the correct form involves powers of \( \frac{1}{4} \) and \( \frac{1}{2} \), matching the first option.

**Answer to question 10:**

\[
\boxed{u_n = \frac{2}{3} \left( -\frac{1}{4} \right)^n + \frac{1}{3} \left( \frac{1}{2} \right)^{n+1}}
\]

---

Would you like further clarifications on any of these steps or additional details?

### Related questions:
1. What are the key properties of the Z-transform?
2. How does partial fraction decomposition work for inverse Z-transforms?
3. What is the convolution theorem in the context of Z-transforms?
4. How do you determine the region of convergence for a Z-transform?
5. Can the convolution theorem be applied to non-linear systems?

### Tip:
When applying the inverse Z-transform, always check if the system's region of convergence aligns with the poles of the function for accurate results.

The inverse Z-transform of U(z) = 5z / [(2 - z)(3z - 1)] for |z| > 2 is and the inverse Z-transform of U(z) = 8z^2 / [(2z - 1)(4z + 1)] using convolution theorem for |z| > 1/2.

Solve inverse Z-transform problems, including U(z) = 5z / [(2 - z)(3z - 1)] for |z| > 2 using partial fractions, and U(z) = 8z^2 / [(2z - 1)(4z + 1)] using convolution theorem for |z| > 1/2. Explore concepts like partial fraction decomposition and convolution theorem in signal processing.

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