GATE 2018 IN Q.36: solution and answer

Official answer: 0Source: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Loop transfer function: $$G(s)H(s)=\frac{s^2+0.2s+100}{2\left(s^2-0.2s+100\right)}$$

2. Open-loop poles: $s^2-0.2s+100=0\Rightarrow s=0.1\pm j\sqrt{99.99}\approx 0.1\pm j9.9995$. Both poles are in the right half-plane, so $P=2$. No poles lie on the $j\omega$ axis, so the Nyquist contour needs no indentations.

3. Put $s=j\omega$: $$G(j\omega )H(j\omega )=\frac{1}{2}\cdot \frac{(100-{\omega }^2)+j0.2\omega }{(100-{\omega }^2)-j0.2\omega }$$ The numerator and denominator are complex conjugates of each other, so they have equal magnitude.

4. Therefore $|G(j\omega )H(j\omega )|=\frac{1}{2}$ for every $\omega$. If $\theta (\omega )$ is the angle of the numerator $(100-{\omega }^2)+j0.2\omega$, then $\angle G(j\omega )H(j\omega )=2\theta (\omega )$. As $\omega$ goes from $0$ to $\infty$, $\theta$ rises from $0$ to $\pi$ (it passes $\pi /2$ at $\omega =10$, where $GH=-\frac{1}{2}$), so the plot goes once round the circle. On the infinite arc, $G(s)H(s)\to \frac{1}{2}$. The whole Nyquist plot therefore lies on the circle of radius $0.5$ centred at the origin.

5. The point $-1+j0$ is at distance $1>0.5$ from the origin. It is outside that circle, so the plot never encircles it: $N=0$.

6. Cross-check with the Nyquist criterion $N=Z-P$ (clockwise encirclements). The characteristic equation $1+G(s)H(s)=0$ gives $2(s^2-0.2s+100)+(s^2+0.2s+100)=3s^2-0.2s+300=0$. The roots are $s=\frac{0.2\pm \sqrt{0.04-3600}}{6}$. Their real part is $+\frac{0.2}{6}>0$, so both are in the RHP and $Z=2$. Then $N=2-2=0$, which agrees. (The closed loop is unstable, but that does not change the encirclement count.)