Question

A continuous time transfer function \( H(s) = \dfrac{1 + s/10^6}{s} \) is converted to a discrete time transfer function \( H(z) \) using a bilinear transform at a sampling rate of 100 MHz. The pole of \( H(z) \) is located at \( z = \) \(\underline{\hspace{2cm}}\).

Hint:

Use bilinear transform for converting continuous-time transfer functions to discrete-time by using \( s = \frac{2}{T} \left( \frac{1 - z^{-1}}{1 + z^{-1}} \right) \).

Answer 1

Correct Answer: 1

Bilinear transform formula: \[ z = \frac{1 + sT/2}{1 - sT/2} \] Where: \[ T = \frac{1}{100 \times 10^6} = 10^{-8}\ \text{seconds} \] Thus for \( s = 10^6 \) (from the given transfer function): \[ z = \frac{1 + 10^6 \times 10^{-8}/2}{1 - 10^6 \times 10^{-8}/2} = \frac{1 + 0.05}{1 - 0.05} = \frac{1.05}{0.95} \] Thus: \[ z = 1.105 \] \[ \boxed{1} \]