Question:

The Nyquist plot of a strictly stable \( G(s) \), having the numerator polynomial as \( (s - 3) \), encircles the critical point \(-1\) once in the anti-clockwise direction. Which one of the following statements on the closed-loop system shown in the figure is correct?

Show Hint

When the open-loop system is stable and the Nyquist plot encircles \(-1\) in the anti-clockwise direction \( N>0 \), the closed-loop system will have right-half plane poles and is thus unstable.
Updated On: Apr 16, 2025
  • The system stability cannot be ascertained.
  • The system is marginally stable.
  • The system is stable.
  • The system is unstable.
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

According to the Nyquist Stability Criterion, the number of encirclements \( N \) of the point \( -1 + j0 \) by the Nyquist plot of the open-loop transfer function \( G(s)H(s) \) is related to the number of open-loop poles \( P \) in the right-half of the complex plane and the number of closed-loop poles \( Z \) in the right-half plane by the formula: \[ N = Z - P \] Given: \( G(s) \) is strictly stable \( \Rightarrow P = 0 \) (no open-loop poles in RHP) Nyquist plot encircles \( -1 \) once in anti-clockwise direction \( \Rightarrow N = +1 \) Then: \[ 1 = Z - 0 \Rightarrow Z = 1 \] So, the closed-loop system has one pole in the right-half plane \( \Rightarrow \) System is unstable.
Was this answer helpful?
0
0

Top Questions on Digital Signal Processing

View More Questions

Questions Asked in GATE EE exam

View More Questions