GATE 2023 IN Q.25: solution and answer

Official answer: 2Source: Official answer key, GATE 2023 IN (IIT Kanpur)

1. Argument principle about the origin: as $s$ goes clockwise around the Nyquist contour (up the $j\omega$ axis from $-j\infty$ to $+j\infty$, then around the infinite semicircle enclosing the right half-plane), the plot of $G(s)H(s)$ circles the origin $N=Z-P$ times clockwise. Here $Z$ counts the right-half-plane zeros of $G(s)H(s)$ itself (not of $1+G(s)H(s)$), and $P$ counts its right-half-plane poles. No pole lies on the $j\omega$ axis, so the contour needs no indentation.

2. Zeros are at $s=1,2$, both in the RHP, so $Z=2$. Poles are at $s=-1,-2$, both in the LHP, so $P=0$. That gives $N=2-0=2$ clockwise encirclements.

3. Check with the magnitude. Because $|j\omega -1|=|j\omega +1|$ and $|j\omega -2|=|j\omega +2|$, $$|G(j\omega )H(j\omega )|=\frac{1}{2}\text{for all }\omega ,$$ so the plot lies on a circle of radius $\frac{1}{2}$ centred at the origin. It is an all-pass function.

4. Check with the phase. For $\omega >0$, the vector $j\omega -1$ lies in the second quadrant, so its angle is $180^{\circ }-{\tan }^{-1}\omega$; similarly for $j\omega -2$. Hence $$\angle G(j\omega )H(j\omega )=\left(180^{\circ }-{\tan }^{-1}\omega \right)+\left(180^{\circ }-{\tan }^{-1}\frac{\omega }{2}\right)-{\tan }^{-1}\omega -{\tan }^{-1}\frac{\omega }{2}=360^{\circ }-2{\tan }^{-1}\omega -2{\tan }^{-1}\frac{\omega }{2}$$

5. As $\omega$ goes from $0$ to $\infty$, the phase falls steadily from $360^{\circ }$ to $360^{\circ }-180^{\circ }-180^{\circ }=0^{\circ }$, starting and ending at the point $G(0)H(0)=\frac{1}{2}$. That is one full clockwise turn around the origin.

6. The branch for $\omega$ from $-\infty$ to $0$ is the complex conjugate (mirror image in the real axis) of the $\omega >0$ branch, traversed in the opposite order of $\omega$. Reflection and reversal together keep the sense of rotation, so it adds a second clockwise turn. The infinite semicircle maps to the single point $G(\infty )H(\infty )=\frac{1}{2}$ and adds nothing. Total: 2 encirclements (clockwise), which agrees with $N=Z-P$.