Finding the Discrete Pole Location

This solution outlines the steps to convert a continuous-time transfer function H(s) to a discrete-time transfer function H(z) using the bilinear transform and identify the location of its pole.

Bilinear Transform Steps

1. Identify Given Parameters:

   • Continuous Transfer Function: $H(s) = \frac{1+s/10^6}{s}$
   • Sampling Frequency: $f_s = 100 \text{ MHz} = 100 \times 10^6 \text{ Hz}$
   • Transform Method: Bilinear Transform

2. Calculate Sampling Period (T):

   The sampling period $T$ is the reciprocal of the sampling frequency $f_s$.

   $T = \frac{1}{f_s} = \frac{1}{100 \times 10^6} = 10^{-8} \text{ s}$

3. Apply Bilinear Transform Formula:

   The bilinear transform substitutes $s$ in the continuous-time function with an expression involving $z$:

   $s = \frac{2}{T} \frac{z-1}{z+1}$

   First, rewrite $H(s)$ as:

   $H(s) = \frac{s + 10^6}{10^6 s}$

   Substitute the bilinear transform expression for $s$:

   $H(z) = \frac{\left( \frac{2}{T} \frac{z-1}{z+1} \right) + 10^6}{10^6 \left( \frac{2}{T} \frac{z-1}{z+1} \right)}$

4. Simplify the Discrete Transfer Function H(z):

   Multiply the numerator and denominator by $T(z+1)$ to eliminate fractions within fractions:

   $H(z) = \frac{T \left( \frac{2}{T} \frac{z-1}{z+1} \right) + T(10^6)}{10^6 T \left( \frac{2}{T} \frac{z-1}{z+1} \right)} \times \frac{T(z+1)}{T(z+1)}$ $H(z) = \frac{2(z-1) + 10^6 T (z+1)}{10^6 \cdot 2 (z-1)}$

   Expand and group terms:

   $H(z) = \frac{(2 + 10^6 T)z + (10^6 T - 2)}{2 \times 10^6 (z-1)}$

5. Locate the Pole of H(z):

   The poles of $H(z)$ are the values of $z$ that make the denominator equal to zero. Set the denominator to zero:

   $2 \times 10^6 (z-1) = 0$

   Solving for $z$:

   $z-1 = 0 \implies z = 1$

   Therefore, the pole of the discrete-time transfer function $H(z)$ is located at $z = 1$. This value falls within the range specified.