The transfer function of the lead compensator is:

Question

TuteeHUB · Accepted Answer

${G_c}\left( s \right) = \frac{1}{\beta }\frac{{s - \frac{1}{T}}}{{s - \frac{1}{{\beta T}}}}$ Where β > 1

TuteeHUB · Answer

${G_c}\left( s \right) = \frac{1}{\beta }\frac{{s + \frac{1}{T}}}{{s + \frac{1}{{\beta T}}}}$ Where β < 1

TuteeHUB · Answer

${G_c}\left( s \right) = \frac{1}{\beta }\frac{{s + \frac{1}{T}}}{{s + \frac{1}{{\beta T}}}}$ Where β > 1

TuteeHUB · Answer

${G_c}\left( s \right) = \frac{1}{\beta }\frac{{s - \frac{1}{T}}}{{s - \frac{1}{{\beta T}}}}$ Where β < 1

1.	The transfer function of the lead compensator is:
A.	\({G_c}\left( s \right) = \frac{1}{\beta }\frac{{s + \frac{1}{T}}}{{s + \frac{1}{{\beta T}}}}\) Where β < 1
B.	\({G_c}\left( s \right) = \frac{1}{\beta }\frac{{s - \frac{1}{T}}}{{s - \frac{1}{{\beta T}}}}\) Where β > 1
C.	\({G_c}\left( s \right) = \frac{1}{\beta }\frac{{s + \frac{1}{T}}}{{s + \frac{1}{{\beta T}}}}\) Where β > 1
D.	\({G_c}\left( s \right) = \frac{1}{\beta }\frac{{s - \frac{1}{T}}}{{s - \frac{1}{{\beta T}}}}\) Where β < 1
Answer» B. \({G_c}\left( s \right) = \frac{1}{\beta }\frac{{s - \frac{1}{T}}}{{s - \frac{1}{{\beta T}}}}\) Where β > 1

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