Question:

Consider a closed-loop control system with unity negative feedback and $K G(s)$ in the forward path, where the gain $K = 2$. The complete Nyquist plot of the transfer function $G(s)$ is shown in the figure. Note that the Nyquist contour has been chosen to have the clockwise sense. Assume $G(s)$ has no poles on the closed right-half of the complex plane. The number of poles of the closed-loop transfer function in the closed right-half of the complex plane is ____________.

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When $G(s)$ has no right-half-plane poles, the stability of the closed-loop system depends only on whether the Nyquist plot encircles $-1$. No encirclement means a stable system.
Updated On: Dec 15, 2025
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The Correct Option is A

Solution and Explanation

The Nyquist stability criterion states: \[ N = Z - P, \] where $N =$ number of clockwise encirclements of the point $-1 + 0j$, $P =$ number of poles of $G(s)$ in the right-half plane, $Z =$ number of closed-loop poles in the right-half plane.
Step 1: Determine $P$.
The question states that $G(s)$ has no poles in the closed right-half plane. Thus, \[ P = 0. \] Step 2: Determine $N$ from the Nyquist plot.
The Nyquist plot (clockwise sense) is shown. Gain $K = 2$ shifts the critical point to $-1$. Inspecting the plot, the Nyquist curve does not encircle the point $-1$. Thus, \[ N = 0. \] Step 3: Solve for $Z$.
Using the Nyquist equation: \[ N = Z - P \quad \Rightarrow \quad 0 = Z - 0 \Rightarrow Z = 0. \] Thus, there are no closed-loop poles in the right-half plane, meaning the system is stable.
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