A continuous-time first-order system with its pole at s = −100 s⁻¹ is discretized with a sampling period Ts = 1 ms. Which of the following is closest to the discrete-time pole given by the relation z = e^(sTs)?

Answer: B. About 0.905

z = e^(−100 × 0.001) = e^(−0.1) ≈ 0.905. The mapping z = e^(sTs) takes the left half of the s-plane to the inside of the unit circle of the z-plane, so a stable continuous-time pole becomes a discrete-time pole with magnitude less than 1. When the sampling period is much shorter than the time constant (10 ms here), the pole approaches 1. Approximating with the backward difference s ≈ (1 − z⁻¹)/Ts gives 1/(1 + 0.1) ≈ 0.909, almost the same value.

• A. This uses the value of sTs directly. The pole is e^(sTs).
• B. Correct. Its magnitude is less than 1, so the discrete-time system is also stable.
• C. This is e^(0.1), with the sign of the exponent reversed; it lies outside the unit circle and would be an unstable pole.