ISRO EC 2023 Q.60: solution and answer

Official answer: CSource: ISRO's revised and final answer key of 16 February 2024

1. The Nyquist criterion rests on the argument principle, which follows from Cauchy's residue theorem. For a function $G(s)$, the integral that is evaluated around a closed contour is that of its logarithmic derivative, $\frac{G^{\prime }(s)}{G(s)}$. The options write the integrand loosely as $G(s)$, so each option is read as the value of this integral, $\oint \frac{G^{\prime }(s)}{G(s)}ds$, around the closed contour.

2. At a zero of order 1 at $s=a$, write $G(s)=(s-a)h(s)$ with $h(a)\ne 0$. Then $\frac{G^{\prime }(s)}{G(s)}=\frac{1}{s-a}+\frac{h^{\prime }(s)}{h(s)}$, which has a simple pole at $s=a$ with residue $+1$.

3. At a simple pole of $G$ at $s=b$, write $G(s)=\frac{h(s)}{s-b}$ with $h(b)\ne 0$. Then $\frac{G^{\prime }(s)}{G(s)}=-\frac{1}{s-b}+\frac{h^{\prime }(s)}{h(s)}$, so the residue is $-1$. A pole or zero of order $k$ gives $\mp k$ and $+k$ in the same way.

4. Cauchy's residue theorem says the contour integral equals $2\pi j$ times the sum of the residues inside the contour. The $n$ zeros contribute $+n$ and the $m$ poles contribute $-m$, so $$\oint \frac{G^{\prime }(s)}{G(s)}ds=2\pi j(n-m).$$

5. Since $\frac{G^{\prime }(s)}{G(s)}ds=d(\ln G)$, the integral equals $2\pi j$ times the number of counter-clockwise encirclements of the origin by the image of the contour in the $G(s)$ plane. This number is therefore $n-m$, that is, zeros minus poles.

6. Compare with the options. Option A has no count of poles or zeros. Option B, $2\pi j(m-n)$, has the sign reversed. Option D, $2\pi jm$, counts only the poles and gives them a positive sign. Only option C, $2\pi j(n-m)$, matches.