GATE 2024 EE Q.57: solution and answer

Official answer: 4 to 8Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Open-loop poles: $s^2+2s+5=0\Rightarrow p_{1,2}=-1\pm j2$. Open-loop zeros: $s^2-2s+2=0\Rightarrow z_{1,2}=1\pm j$. Both zeros are in the right half-plane. The loop has unity negative feedback and $K\ge 0$, so the $180^{\circ }$ angle condition applies: $\sum \angle (s-z_i)-\sum \angle (s-p_j)=180^{\circ }$ (mod $360^{\circ }$).

2. Apply the angle condition at a point just off $p_1=-1+j2$ to get the angle of departure: $${\varphi }_d=180^{\circ }+\sum \angle (p_1-z_i)-\sum _{j\ne 1}\angle (p_1-p_j)$$

3. Angle from $z_1=1+j$: $p_1-z_1=-2+j1$ lies in the second quadrant, so its angle is $180^{\circ }-{\tan }^{-1}\frac{1}{2}=180^{\circ }-26.57^{\circ }=153.43^{\circ }$.

4. Angle from $z_2=1-j$: $p_1-z_2=-2+j3$ lies in the second quadrant, so its angle is $180^{\circ }-{\tan }^{-1}\frac{3}{2}=180^{\circ }-56.31^{\circ }=123.69^{\circ }$.

5. Angle from the other pole $p_2=-1-j2$: $p_1-p_2=j4$, which is $90^{\circ }$.

6. $${\varphi }_d=180^{\circ }+(153.43^{\circ }+123.69^{\circ })-90^{\circ }=367.13^{\circ }$$

7. Bring it into $[0^{\circ },360^{\circ })$: $367.13^{\circ }-360^{\circ }=7.13^{\circ }\approx 7^{\circ }$.

8. Independent check: the characteristic equation is $s^2+2s+5+K(s^2-2s+2)=0$. Near $K=0$, the branch leaves $p_1$ in the direction of $\frac{ds}{dK}=-\frac{s^2-2s+2}{2s+2}$ evaluated at $s=-1+j2$. There $s^2=-3-j4$, so $s^2-2s+2=1-j8$ and $2s+2=j4$. This gives $\frac{ds}{dK}=-\frac{1-j8}{j4}=2+j0.25$, whose angle is ${\tan }^{-1}(0.125)=7.13^{\circ }$. That agrees with the result above.