GATE 2020 EE Q.24: solution and answer

Official answer: 2Source: Official answer key, GATE 2020 EE (IIT Delhi)

1. Open-loop poles are $s=-a$, $s=+b$ and $s=-c$. Since $a,b,c>0$, exactly one of them ($s=b$) lies in the open right half-plane, and none lies on the imaginary axis. So $P=1$.

2. Write $$1+G(s)=\frac{(s+a)(s-b)(s+c)+K}{(s+a)(s-b)(s+c)}.$$ At any root of the denominator the numerator equals $K\ne 0$, so nothing cancels: $1+G(s)$ has exactly the same poles as $G(s)$, and therefore also $P=1$ in the RHP.

3. The zeros of $1+G(s)$ are the roots of $(s+a)(s-b)(s+c)+K=0$, which is the closed-loop characteristic equation. Let $Z$ be the number of these zeros inside the Nyquist contour. Because the plot never passes through the origin, $1+G(j\omega )\ne 0$ for every $\omega$, so no zero sits on the imaginary axis and $Z$ counts zeros in the open RHP only.

4. Nyquist criterion (argument principle): with the contour traversed clockwise and clockwise encirclements of the origin of the $(1+G)$-plane counted as positive, $$N=Z-P.$$

5. One clockwise encirclement is given, so $N=+1$: $$Z=N+P=1+1=2$$

6. The closed-loop transfer function is $$\frac{G(s)}{1+G(s)}=\frac{K}{(s+a)(s-b)(s+c)+K}.$$ Its numerator is the non-zero constant $K$, so no pole-zero cancellation is possible. Its poles are exactly the zeros of $1+G(s)$, and 2 of them lie in the open right half of the $s$-plane.