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The problem involves finding the frequency response \( H(\omega) \) for a second-order system characterized by specific poles and zeros. Here's the breakdown of the solution.

### Problem Breakdown

1. **Poles**: \( p_1 = p_2 = 0.6 \), so there’s a double pole at \( |p| = 0.6 \).
2. **Zeros**: 
   - \( z_1 = e^{j \frac{5 \pi}{8}} \)
   - \( z_2 = e^{-j \frac{5 \pi}{8}} \)
   
   These zeros are complex conjugates and lie on the unit circle.

3. **Magnitude Condition**: The magnitude \( |H(\omega)| \) is normalized to 1 at \( \omega = 0 \).

### Steps to Determine \( H(\omega) \)

Given the zeros, the transfer function \( H(z) \) in terms of \( z \)-transform can be expressed as:
\[
H(z) = \frac{(z - z_1)(z - z_2)}{(z - p_1)^2}
\]

For simplicity, evaluating the frequency response \( H(\omega) \) is typically done by substituting \( z = e^{j \omega} \). 

#### Frequency Response \( H(\omega) \)

1. **Zeros Contribution**:
   Since \( z_1 = e^{j \frac{5 \pi}{8}} \) and \( z_2 = e^{-j \frac{5 \pi}{8}} \), the product \( (e^{j \omega} - z_1)(e^{j \omega} - z_2) \) simplifies to:
   \[
   |H(\omega)| = 2 \cos\left(\omega - \frac{5 \pi}{8}\right)
   \]


The transfer function H(z) of a second-order system has a double pole at p1, p2 = 0.6, and two zeros at z1 = e^(j5π/8), z2 = e^(-j5π/8). The magnitude of the frequency response H(ω) is normalized as |H(ω)|=1 at ω=0. Determine the frequency response, H(ω).

Learn how to determine the frequency response H(ω) for a second-order system with a double pole at 0.6 and zeros at e^(j5π/8) and e^(-j5π/8). This solution covers key concepts in frequency response and pole-zero analysis, suitable for undergraduate electrical engineering students.

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