Question:medium

The Nyquist plot and step response/transfer function is given. Match List-I (Nyquist plot) with List-II (corresponding step response/transfer function).

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Use the Nyquist stability criterion to quickly categorize plots: - \textbf{No encirclement of -1:} Stable (response decays, e.g., I). - \textbf{Encirclement of -1:} Unstable (response grows, e.g., II). - \textbf{Passes through -1:} Marginally stable (response has sustained oscillations, e.g., III).
Updated On: Feb 18, 2026
  • A - II, B - I, C - III, D - IV
  • A - I, B - III, C - II, D - IV
  • A - I, B - II, C - IV, D - III
  • A - III, B - IV, C - I, D - II
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Plot A Analysis.The Nyquist plot (A) originates on the positive real axis, encircles the origin, and terminates at the origin, without encircling -1. This indicates stability. The circular path implies decaying oscillations, aligning with the damped oscillatory step response in Plot (I).
Step 2: Plot C Analysis.The Nyquist plot (C) encircles -1, signifying instability. An unstable system exhibits an unbounded step response, potentially with growing oscillations, consistent with the growing oscillatory response in Plot (II).
Step 3: Plot B Analysis.The Nyquist plot (B) passes through -1, indicating marginal stability. The system possesses poles on the jw-axis. A marginally stable system's step response oscillates with a constant, non-decaying amplitude, matching the sustained oscillation in Plot (III).
Step 4: Plot D Analysis.The Nyquist plot (D) corresponds to the transfer function (IV), \( \frac{1}{(s+2)^2} \). This represents a critically damped or overdamped stable second-order system. Its Nyquist plot begins at \(1/4\) on the real axis for \(\omega=0\) and approaches the origin as \(\omega \to \infty\) without any encirclements, aligning with the characteristics of plot (D).
Step 5: Matching Summary.A \(\to\) I, B \(\to\) III, C \(\to\) II, D \(\to\) IV, which corresponds to option (B).
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