Question

The transfer function of first order high pass digital Butterworth filter that has 3dB cut off frequency \( \omega = 0.15\pi \) using bilinear transformation with T=1s

• A. \( H(z) = \frac{1}{1 - 0.24 \left( \frac{z + 1}{z - 1} \right)} \)
• B. \( H(z) = \frac{1}{1 + 0.15 \left( \frac{z + 1}{z - 1} \right)} \)
• C. \( H(z) = \frac{1}{1 + 0.24 \left( \frac{z - 1}{z + 1} \right)} \)
• D. \( H(z) = \frac{1}{1 + 0.48 \left( \frac{z + 1}{z - 1} \right)} \)

Hint:

When using bilinear transformation, always substitute \( s = \frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}} \) and remember the general expression for the filter type that you are using, in this case, it is first order high pass filter.

Answer 1

The Correct Option is D

Step 1: Given cutoff frequency is \( \omega_c = 0.15\pi \), and \( T=1\). Using bilinear transformation \( s = \frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}} \) \[ s = 2\frac{1-z^{-1}}{1+z^{-1}} \] Step 2: A first-order high pass filter has a transfer function as: \[ H(s) = \frac{s}{s+\omega_c} \] Step 3: Substitute \(s\) and \( \omega_c \) \[ H(z) = \frac{2\frac{1-z^{-1}}{1+z^{-1}}}{2\frac{1-z^{-1}}{1+z^{-1}} + 0.15\pi} \] After solving we get: \[ H(z) = \frac{1-z^{-1}}{1 + \frac{0.15\pi}{2} + (1-\frac{0.15\pi}{2})z^{-1}} = \frac{1-z^{-1}}{1+0.235 + (1-0.235)z^{-1}} \] \[ H(z) \approx \frac{1-z^{-1}}{1+0.48z^{-1}} \]