Question:
In the Nyquist plot of the open-loop transfer function
\[
G(s)H(s) = \frac{3s+5}{s-1}
\]
corresponding to the feedback loop shown in the figure, the infinite semi-circular arc of the Nyquist contour in the \(s\)-plane is mapped into a point at:
\begin{center}
\includegraphics[width=0.5\textwidth]{05.jpeg}
\end{center}
In the Nyquist plot of the open-loop transfer function
\[
G(s)H(s) = \frac{3s+5}{s-1}
\]
corresponding to the feedback loop shown in the figure, the infinite semi-circular arc of the Nyquist contour in the \(s\)-plane is mapped into a point at:
\begin{center}
\includegraphics[width=0.5\textwidth]{05.jpeg}
\end{center}
Show Hint
For Nyquist plots, the high-frequency asymptote is obtained by dividing the leading coefficients of numerator and denominator.
Updated On: Aug 28, 2025
- \(G(s)H(s) = \infty\)
- \(G(s)H(s) = 0\)
- \(G(s)H(s) = 3\)
- \(G(s)H(s) = -5\)
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The Correct Option is C
Solution and Explanation
Step 1: Behavior along infinite semicircle.
On Nyquist contour, for large \(|s| \to \infty\):
\[
G(s)H(s) = \frac{3s+5}{s-1} \approx \frac{3s}{s} = 3.
\]
Step 2: Mapping to a point.
The infinite arc in \(s\)-plane is mapped to constant point \(3\) in the GH-plane.
Final Answer: \[ \boxed{3} \]
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