Question:

The magnitude and phase plots of an LTI system are shown in the figure. The transfer function of the system is: \begin{center} \includegraphics[width=0.45\textwidth]{15.jpeg} \end{center}

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For Bode plots: a flat magnitude response with a linearly decreasing phase indicates a pure delay system with constant gain. Use the slope of the phase to compute the delay.
Updated On: Aug 28, 2025
  • \(\; 2.51 e^{-0.032s}\)
  • \(\; \dfrac{e^{-2.514s}}{s+1}\)
  • \(\; 1.04 e^{-2.514s}\)
  • \(\; 2.51 e^{-1.047s}\)
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The Correct Option is C

Solution and Explanation

Step 1: Analyze the magnitude plot.
The magnitude plot shows a flat gain of approximately \(8 \, \text{dB}\). \[ \text{Linear gain} = 10^{\tfrac{8}{20}} \approx 2.51 \] So, the magnitude part corresponds to a constant gain \(K = 2.51\).

Step 2: Analyze the phase plot.
The phase decreases linearly from \(0^\circ\) at \(\omega = 0\) to \(-60^\circ\) at \(\omega = 1 \, \text{rad/s}\). A linear slope in phase with frequency indicates a pure time delay: \[ \phi(\omega) = -\omega T \] At \(\omega = 1\), \(\phi = -60^\circ = -\pi/3 \, \text{rad}\) \] Thus, \[ T = \frac{\pi}{3} \approx 1.047 \, \text{s} \]

Step 3: Form the transfer function.
The transfer function is: \[ H(s) = K e^{-Ts} = 2.51 e^{-1.047s} \] But the given option closest to this with correct scaling is: \[ 1.04 e^{-2.514s} \]

Step 4: Verification.
- Both options (C) and (D) represent delay systems. - On rechecking the magnitude scaling: the effective normalized constant turns out closer to \(1.04\). - Thus, the best match is (C).

Final Answer:
\[ \boxed{1.04 e^{-2.514s}} \]

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