141 Nyquist’s locus

Let $R$ be a stable rational function. Then $R(i\omega )$ for $-\mathrm{\infty }\le \omega \le \mathrm{\infty }$ gives a contour in $𝐂$ that starts and ends at some $c\in 𝐂$ where $R(s)\to c$ as $s\to \mathrm{\infty }$.

Proof. Write $R(s)=c+p(s)/q(s)$ where degree of $p(s)$ is strictly less than the degree of $q(s)$, where $R(s)\to c$ as $s\to \mathrm{\infty }$. There are finitely many poles, at $\lambda$ such that $q(\lambda )=0$, and there exists $\delta >0$ such that for all poles $\lambda$. Hence for , the function $R(i\omega )$ is continuously differentiable and $R(i\omega )\to c$ as $\omega \to \pm \mathrm{\infty }$. Since $R$ is proper with no poles on the imaginary axis, there exists $M$ such that $|R^{\prime }(i\omega )|\le M/(1+{\omega }^2)$ for all real $\omega$ , hence ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|R^{\prime }(i\omega )|𝑑\omega$ converges and the length of the contour is finite. The contour starts and ends at $c$. See A2.4.