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The transfer function of the lag compensator is:

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  1. \(\mathrm{G}_{\mathrm{c}}(\mathrm{s})=\frac{\mathrm{s}+\frac{1}{\mathrm{T}}}{\mathrm{s}+\frac{1}{\alpha \mathrm{T}}} \) where α < 1
  2. \(\mathrm{G}_{\mathrm{c}}(\mathrm{s})=\frac{\mathrm{s}-\frac{1}{\mathrm{~T}}}{\mathrm{s}-\frac{1}{\alpha \mathrm{T}}}\) where α > 1
  3. \(\mathrm{G}_{\mathrm{c}}(\mathrm{s})=\frac{\mathrm{s}+\frac{1}{\mathrm{~T}}}{\mathrm{s}+\frac{1}{\alpha \mathrm{T}}}\) where α > 1
  4. \(\mathrm{G}_{\mathrm{c}}(\mathrm{s})=\frac{\mathrm{s}-\frac{1}{\mathrm{~T}}}{\mathrm{s}-\frac{1}{\alpha \mathrm{T}}} \) where α < 1

Answer (Detailed Solution Below)

Option 3 : \(\mathrm{G}_{\mathrm{c}}(\mathrm{s})=\frac{\mathrm{s}+\frac{1}{\mathrm{~T}}}{\mathrm{s}+\frac{1}{\alpha \mathrm{T}}}\) where α > 1
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Detailed Solution

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Concept:

A lag compensator is used in control systems to improve steady-state accuracy while having minimal impact on transient response.

It is characterized by a transfer function of the form:

This ensures that the pole is closer to the origin than the zero, which causes the phase to lag and improves the low-frequency gain.

Final Answer: Option C

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