Design a Butterworth highpass filter to meet the specifications :The Gp > -1 dB, < -20 dB cop=20 rad/sec and Ws = 10 rad/se

To design a Butterworth highpass filter that meets the given specifications, follow these steps:

1. Determine the filter order (n): The filter order can be determined using the given specifications for passband and stopband. The formula for the Butterworth filter order is:

   $n = \left\lceil \frac{\log_{10} \left( \frac{10^{\frac{-20}{10}} - 1}{10^{\frac{-1}{10}} - 1} \right)}{2 \log_{10} \left( \frac{20}{10} \right)} \right\rceil$

   Simplifying the terms inside the logarithms:

   $10^{\frac{-20}{10}} = 10^{-2} = 0.01$ $10^{\frac{-1}{10}} \approx 0.794$

   Plugging these values into the formula:

   $n = \left\lceil \frac{\log_{10} \left( \frac{0.01 - 1}{0.794 - 1} \right)}{2 \log_{10} \left( 2 \right)} \right\rceil$

   $n = \left\lceil \frac{\log_{10} \left( \frac{-0.99}{-0.206} \right)}{2 \log_{10} \left( 2 \right)} \right\rceil$

   $n = \left\lceil \frac{\log_{10} (4.806)}{2 \log_{10} (2)} \right\rceil$

   $n = \left\lceil \frac{0.6826}{0.3010} \right\rceil$

   $n = \left\lceil 2.27 \right\rceil$

   Therefore, the filter order $n = 3$.

2. Determine the cutoff frequency (ωc): The cutoff frequency for a highpass Butterworth filter is determined by the passband edge frequency (ωp) and the filter order (n). The cutoff frequency is given by:

   $\omega_c = \omega_p \left(10^{\frac{-Gp}{10}} - 1\right)^{-\frac{1}{2n}}$

   Given $\omega_p = 20$ rad/sec and $G_p = -1$ dB:

   $\omega_c = 20 \left(10^{\frac{1}{10}} - 1\right)^{-\frac{1}{6}}$

   $\omega_c = 20 \left(1.2589 - 1\right)^{-\frac{1}{6}}$

   $\omega_c = 20 \left(0.2589\right)^{-\frac{1}{6}}$

   $\omega_c \approx 20 \times 1.348$

   $\omega_c \approx 26.96 \text{ rad/sec}$

3. Design the transfer function: The transfer function for a Butterworth highpass filter of order $n$ is given by:

   $H(s) = \frac{s^n}{s^n + \sum_{k=0}^{n-1} a_k s^k}$

   For $n = 3$, the poles of the Butterworth filter are located on the left half of the s-plane and are given by:

   $s_k = \omega_c e^{j\left(\frac{\pi}{2} + \frac{(2k+1)\pi}{2n}\right)}, \quad k = 0, 1, 2$

   The poles for $n = 3$ are:

   $s_0 = 26.96 e^{j\left(\frac{\pi}{2} + \frac{\pi}{6}\right)}$ $s_1 = 26.96 e^{j\left(\frac{\pi}{2} + \frac{3\pi}{6}\right)}$ $s_2 = 26.96 e^{j\left(\frac{\pi}{2} + \frac{5\pi}{6}\right)}$

   Simplifying these:

   $s_0 = 26.96 e^{j\frac{2\pi}{3}}$ $s_1 = 26.96 e^{j\pi}$ $s_2 = 26.96 e^{j\frac{4\pi}{3}}$

   The transfer function $H(s)$ can be written as:

   $H(s) = \frac{s^3}{(s - s_0)(s - s_1)(s - s_2)}$

   Substituting the values of $s_0, s_1,$ and $s_2$:

   $H(s) = \frac{s^3}{(s - 26.96 e^{j\frac{2\pi}{3}})(s - 26.96 e^{j\pi})(s - 26.96 e^{j\frac{4\pi}{3}})}$

   This is the transfer function of the Butterworth highpass filter that meets the given specifications.