GATE EC 2014 Set 1 — Question 31

The Nyquist stability criterion states that for a unity feedback system with open-loop transfer function $kG(s)$, the number of unstable closed-loop poles $Z$ is given by $Z = P - N$, where $P$ is the number of unstable open-loop poles (poles of $G(s)$ in the Right Half Plane, RHP) and $N$ is the number of clockwise encirclements of the critical point $(-1/k, 0)$ by the Nyquist plot of $G(s)$. For stability, $Z$ must be zero.Let's analyze the given Nyquist plot:

• If $k$ is small, $-1/k$ is a large negative number (e.g., if $A=0.5$, and $k=0.1$, then $-1/k = -10$). The plot crosses at $-0.5$. The point $(-10,0)$ is to the left of $-0.5$. The plot does not encircle $(-10,0)$, so $N=0$, and the system is stable.
• If $k$ is large, $-1/k$ is a small negative number, close to $0$ (e.g., if $A=0.5$, and $k=10$, then $-1/k = -0.1$). The plot crosses at $-0.5$. The point $(-0.1,0)$ is to the right of $-0.5$. The Nyquist plot does encircle $(-0.1,0)$ in the counter-clockwise direction (assuming the plot is for positive $\omega$ and the full plot includes the negative $\omega$ mirror image). If it encircles, $N \neq 0$. If $P=0$, then $Z = 0 - N$. If $N$ is a non-zero integer, $Z$ will be non-zero, indicating instability.

Let's evaluate the options:
(A) $G(s)$ is an all-pass filter: An all-pass filter has a constant magnitude response ($|G(j\omega)| = 1$). The Nyquist plot clearly shows that the magnitude varies (it starts at some value and goes to 0), so this is incorrect.
(B) $G(s)$ is a strictly proper transfer function: As observed, the plot approaches the origin as $\omega \to \infty$, which is characteristic of a strictly proper transfer function. This statement is likely true.
(C) $G(s)$ is a stable and minimum-phase transfer function: The Nyquist plot does not directly provide enough information to definitively conclude minimum-phase (no RHP zeros). While it appears open-loop stable (no poles at origin or RHP poles indicated by starting from infinity), we cannot be certain about minimum-phase from this plot alone.
(D) The closed-loop system is unstable for sufficiently large and positive $k$: Since the Nyquist plot crosses the negative real axis at a finite point (say, $-A$, where $A>0$), the gain margin $GM = 1/A$ is finite. According to the Nyquist criterion, if $k > GM$, the critical point $(-1/k, 0)$ will be encircled, leading to instability (assuming $P=0$). Therefore, for sufficiently large and positive $k$, the closed-loop system will become unstable. This is a definitive conclusion from the Nyquist plot.Comparing (B) and (D), (D) is a more specific and critical conclusion regarding the stability of the closed-loop system, which is a primary purpose of the Nyquist plot.The final answer is $\boxed{D}$