A continuous-time plant has a stable pole at s = −5. A designer samples this system with period T = 0.1 s and claims the corresponding z-domain pole must be at z = −0.5 on the negative real axis. Is this correct?
AYes, because negative real s-plane poles map directly to the negative real z-axis
BNo, the pole maps to z = e^{−5 × 0.1} = e^{−0.5} ≈ 0.607, which lies on the positive real axis inside the unit circle
CNo, the pole maps to the unit circle at z = e^{j5} because negative s-values correspond to oscillatory modes
DYes, because the bilinear transform maps negative real s-values to negative real z-values
The mapping between the s-plane and z-plane is z = e^{sT}. For s = −5 and T = 0.1, z = e^{−0.5} ≈ 0.607 — a positive real number well inside the unit circle, confirming stability. Negative real s-axis poles always map to positive real z values in (0, 1) because the exponent e^{negative real} is real and positive. Negative real z-axis poles (z ∈ (−1, 0)) correspond to s-plane poles near ±jπ/T — oscillatory modes near the Nyquist frequency, not stable real poles. The common misconception of 'negative maps to negative' ignores the exponential relationship.
Question 2 Multiple Choice
Why is the zero-order hold (ZOH) included when computing the discrete-time equivalent of a continuous-time plant?
ATo add differentiation that compensates for the derivative-like effect of the analog-to-digital converter
BTo model the digital-to-analog conversion that holds each computed control value constant until the next sample instant
CTo cancel Nyquist-frequency aliasing introduced by the sampler at the system input
DTo convert the z-domain transfer function back into an equivalent continuous-time transfer function for analysis
After the digital controller computes a control value at sample k, a digital-to-analog converter must produce a physical voltage or current that actuates the plant. The ZOH holds this value constant until the next sample at k+1. This holding behavior introduces dynamics: the ZOH adds a lag of T/2 to the effective plant and affects the frequency response. The ZOH-equivalent pulse transfer function G(z) = (1 − z^{−1})·Z{G(s)/s} captures both the continuous plant dynamics and the holding effect in a single discrete-time model.
Question 3 True / False
A discrete-time closed-loop system with all poles strictly inside the unit circle (|z| < 1) is guaranteed to be stable.
TTrue
FFalse
Answer: True
Stability in the z-domain is directly analogous to the left half-plane criterion in the s-domain. The mapping z = e^{sT} transforms the imaginary axis (σ = 0) to the unit circle and the stable left half-plane (σ < 0) to the interior of the unit circle. Any pole with |z| < 1 corresponds to a mode that decays geometrically — z^k → 0 as k → ∞. A pole at |z| > 1 corresponds to exponential growth and instability. The unit circle is the exact boundary between stable and unstable discrete-time behavior.
Question 4 True / False
Sampling a continuous-time control system at a higher rate usually improves closed-loop controller performance.
TTrue
FFalse
Answer: False
This is a common misconception. While too slow a sampling rate causes aliasing, intersample errors, and degraded phase margin (the ZOH lag T/2 worsens with large T), excessively fast sampling introduces its own problems: quantization noise is amplified because the control effort changes very little between samples yet the quantization error remains the same; computational latency becomes a larger fraction of the sampling period; and fixed-point arithmetic precision limits become relevant. Beyond roughly 10–20× the closed-loop bandwidth, further rate increases typically provide no meaningful performance improvement and may degrade it.
Question 5 Short Answer
Explain why the z-transform is not simply the Laplace transform with z substituted for s, and what relationship actually connects the two domains.
Think about your answer, then reveal below.
Model answer: The Laplace transform is defined on continuous-time signals x(t); the z-transform is defined on discrete sequences x[k]. They are mathematically separate tools. The connection is the mapping z = e^{sT}: when you sample a continuous signal every T seconds and take the z-transform of the resulting sequence, the result relates to the Laplace transform of the original signal through this exponential. The mapping wraps the s-plane's vertical structure periodically onto the z-plane — the imaginary axis maps to the unit circle, the left half-plane maps to the interior of the unit circle. Simply substituting z for s gives wrong pole locations and invalidates stability analysis.
Understanding z = e^{sT} as the fundamental relationship prevents numerous errors: why negative real s-poles map to positive real z values, why the stability boundary changes from an axis to a circle, why frequency aliasing occurs (the mapping from Ω to ω is not one-to-one), and why the bilinear transform w = (z−1)/(z+1) is a valid approximation to e^{sT}−1 for small sT. All discrete-time analysis follows from this single exponential relationship.