GATE EC 2025 Set 1 — Question 26

The given system is a unity-negative-feedback system. The open-loop transfer function is $KG(s)$.
From Figure (ii), the open-loop poles (marked with X) are at $p_1 = 0$, $p_2 = -1$, $p_3 = -2$, and $p_4 = -3$. There are no open-loop zeros explicitly shown.
Thus, we can write $G(s) = \frac{1}{s(s+1)(s+2)(s+3)}$.For a point $s_0$ to be on the root locus, it must satisfy the angle condition:
$\sum \angle (s_0 - z_i) - \sum \angle (s_0 - p_j) = (2k+1)180^\circ$, for $K > 0$.We need to check if the point $s_0 = -1 + j1$ lies on the root locus. Let's calculate the angles from each pole to $s_0$: Vector: $s_0 - p_1 = (-1 + j1) - 0 = -1 + j1$
Angle: $\theta_1 = \angle(-1 + j1) = 135^\circ$ Vector: $s_0 - p_2 = (-1 + j1) - (-1) = j1$
Angle: $\theta_2 = \angle(j1) = 90^\circ$ Vector: $s_0 - p_3 = (-1 + j1) - (-2) = 1 + j1$
Angle: $\theta_3 = \angle(1 + j1) = 45^\circ$ Vector: $s_0 - p_4 = (-1 + j1) - (-3) = 2 + j1$
Angle: $\theta_4 = \angle(2 + j1) = \arctan(\frac{1}{2}) \approx 26.565^\circ$Since there are no zeros, the sum of angles from poles is the negative of the angle of $G(s_0)$.
Sum of angles from poles = $\theta_1 + \theta_2 + \theta_3 + \theta_4 = 135^\circ + 90^\circ + 45^\circ + 26.565^\circ = 296.565^\circ$.For the point $s_0 = -1 + j1$ to be on the root locus, the angle condition requires the sum of angles from poles to be an odd multiple of $180^\circ$ (i.e., $180^\circ, 540^\circ, -180^\circ$, etc.).
However, $296.565^\circ$ is not an odd multiple of $180^\circ$.Therefore, the point $s = -1 + j1$ does not lie on the root locus for the given system.
This implies that there is no positive value of $K$ for which the system will have a pole at $-1 + j1$.The final answer is $\boxed{\text{C}}$