Let's break down the solution step by step for designing a **third-order unity-gain Bessel high-pass filter** with a **corner frequency \( f_c = 2 \) kHz** using the coefficients from the table provided. --- ### **Step 1: General Transfer Function for High-Pass Filter** A normalized third-order low-pass Bessel filter has the transfer function: \[ H_{LP}(s) = \frac{b_0}{a_3 s^3 + a_2 s^2 + a_1 s + a_0} \] For a **high-pass filter**, the transfer function is: \[ H_{HP}(s) = \frac{a_0 s^3}{a_3 s^3 + a_2 s^2 + a_1 s + a_0} \] --- ### **Step 2: Extract Coefficients from Table** From your table (for \( n=3 \)): | \( n \) | \( a_1 \) | \( a_2 \) | \( a_3 \) | \( b_1 \) | \( b_2 \) | \( b_3 \) | |---------|-----------|-----------|-----------|-----------|-----------|-----------| | 3 | 1.0000 | 1.3617 | 0.6180 | | | | So, for third order: - \( a_3 = 1 \) - \( a_2 = 1.3617 \) - \( a_1 = 0.6180 \) - \( a_0 = 0 \) (for low-pass; for high-pass, numerator uses \( s^3 \)) --- ### **Step 3: Substitute and Normalize** For a normalized filter (\( f_c = 1\, \text{rad/s} \)), the high-pass transfer function is: \[ H_{HP, norm}(s) = \frac{s^3}{s^3 + 1.3617 s^2 + 0.6180 s + 1} \] --- ### **Step 4: Denormalize for Target Corner Frequency** Given \( f_c = 2\, \text{kHz} \): - \(\omega_c = 2\pi f_c = 2\pi \times 2000 = 4000\pi \approx 12566.37\, \text{rad/s}\) Let \( s' = \frac{s}{\omega_c} \): \[ H_{HP}(s) = \frac{(s/\omega_c)^3}{(s/\omega_c)^3 + 1.3617(s/\omega_c)^2 + 0.6180(s/\omega_c) + 1} \] \[ = \frac{s^3}{s^3 + 1.3617 \omega_c s^2 + 0.6180 \omega_c^2 s + \omega_c^3} \] --- ### **Step 5: Plug in \(\omega_c\) Value** \[ \omega_c = 12566.37 \] So, \[ H_{HP}(s) = \frac{s^3}{s^3 + 1.3617 \times 12566.37\, s^2 + 0.6180 \times (12566.37)^2\, s + (12566.37)^3} \] Calculate the coefficients: - \( a_2 = 1.3617 \times 12566.37 \approx 17118.5 \) - \( a_1 = 0.6180 \times (12566.37)^2 \approx 0.6180 \times 157913670 \approx 97588.7 \) - \( a_0 = (12566.37)^3 \approx 1.99 \times 10^{12} \) So, \[ H_{HP}(s) = \frac{s^3}{s^3 + 17118.5 s^2 + 97588.7 s + 1.99 \times 10^{12}} \] --- ### **Step 6: Final Answer** **The transfer function for a third-order unity-gain Bessel high-pass filter with \( f_c = 2\, \text{kHz} \) is:** \[ \boxed{ H_{HP}(s) = \frac{s^3}{s^3 + 17118.5\, s^2 + 97588.7\, s + 1.99 \times 10^{12}} } \] --- **Summary of Steps:** 1. Write general high-pass transfer function using Bessel coefficients. 2. Extract coefficients from the table. 3. Normalize and then denormalize by substituting \( s = \omega_c s' \). 4. Plug in \( f_c \) to get numerical coefficients. 5. Write the final transfer function. Let me know if you need the circuit realization as well!