GATE 2020 IN Q.29: solution and answer

Official answer: ASource: Official answer key, GATE 2020 IN (IIT Delhi)

1. Open-loop poles: $s=0$ and $s=+2$. The pole at $s=0$ lies on the $j\omega$ axis, so the Nyquist contour goes round it with a small indentation into the right half-plane, which leaves it outside the contour. The pole at $s=+2$ is inside. So $P=1$.

2. Closed-loop poles (a quick shortcut): $1+G(s)H(s)=0\Rightarrow s(s-2)+1=0\Rightarrow s^2-2s+1=0\Rightarrow (s-1{)}^2=0$. Both closed-loop poles are at $s=+1$, in the right half-plane, so $Z=2$.

3. Nyquist criterion: $N=P-Z$, with $N$ counted positive for counterclockwise encirclements of $-1+j0$. Here $N=1-2=-1$, which means one clockwise encirclement.

4. Check with the frequency response: $$G(j\omega )H(j\omega )=\frac{1}{j\omega (j\omega -2)}=\frac{1}{-{\omega }^2-2j\omega }=\frac{-1}{{\omega }^2+4}+j\frac{2}{\omega ({\omega }^2+4)}$$

5. Branch for $\omega :0^{+}\to \infty$: the real part rises steadily from $-\frac{1}{4}$ to $0$, and the imaginary part stays positive, falling from $+\infty$ to $0$. So the plot comes down from $+j\infty$ along the asymptote $\text{Re}=-\frac{1}{4}$ and reaches the origin at $-180^{\circ }$ without crossing the real axis. The branch for $\omega :-\infty \to 0^{-}$ is its mirror image in the third quadrant, running from the origin down to $-j\infty$. Every point on both finite branches has $-\frac{1}{4}\le \text{Re}<0$, so both lie to the right of $-1$.

6. Indentation at the origin: put $s=\epsilon e^{j\theta }$, with $\theta$ going from $-90^{\circ }$ through $0^{\circ }$ to $+90^{\circ }$. Near the origin, $s-2\approx -2$, so $$GH\approx \frac{1}{-2s}=-\frac{1}{2\epsilon }{e}^{-j\theta }$$ Its angle is $180^{\circ }-\theta$, which falls from $270^{\circ }$ through $180^{\circ }$ to $90^{\circ }$. That is an infinite arc running from $-j\infty$ through $-\infty$ to $+j\infty$, clockwise across the left half of the plane. The large semicircle $|s|\to \infty$ maps to the origin.

7. Direction: follow the whole contour in order. The $\omega <0$ branch runs from the origin down to $-j\infty$ (right of $-1$). The infinite arc then climbs from $-j\infty$ round the left side to $+j\infty$. Finally, the $\omega >0$ branch comes down from $+j\infty$ to the origin (right of $-1$). So the closed plot goes down on the right of $-1$ and back up on the left, which encloses $-1+j0$ once, clockwise. This agrees with $N=-1$, and the closed-loop system is unstable, with two right-half-plane poles.