arrow
arrow
arrow
​A control engineer designs a lag-lead compensator where the poles are at s=−βT1,−1βT2&nbs
Question

A control engineer designs a lag-lead compensator where the poles are at s=βT1,1βT2 and zeros are at s=1T1,1T2. What condition is true for the parameter β in a lag-lead compensator?\textbf{} \text{A control engineer designs a lag-lead compensator where the poles are at } s = -\frac{\beta}{T_1}, -\frac{1}{\beta T_2} \text{ and zeros are at } s = -\frac{1}{T_1}, -\frac{1}{T_2}. \text{ What condition is true for the parameter } \beta \text{ in a lag-lead compensator?}​​

A.

β<0

B.

β>0

C.

β<1

D.

β>1

Correct option is C

Conceptlag–lead compensator combines the characteristics of both lag and lead networks.Its transfer function is generally expressed as:G(s)=K(1+sT1)(1+sT2)(1+sβT1)(1+sβT2)The parameters T1,T2 define the zero locations, while the factor β shifts the poles relative to the zeros.Condition for parametersFor the lead section:The pole must lie to the right of its zero on the real axis (more positive), meaning the zero is placed further left.This occurs when:β<1For the lag section:The pole must lie closer to the origin than the zero, providing phase lag.This requirement is satisfied when:β>1\textbf{Concept} \\[4pt]\text{A \textit{lag–lead compensator} combines the characteristics of both lag and lead networks.} \\[4pt]\text{Its transfer function is generally expressed as:} \\[6pt]G(s) = K \frac{(1 + sT_1)(1 + sT_2)}{(1 + s\beta T_1)(1 + s\beta T_2)} \\[6pt]\text{The parameters } T_1, T_2 \text{ define the zero locations, while the factor } \beta \text{ shifts the poles relative to the zeros.} \\[10pt]\textbf{Condition for parameters} \\[6pt]\textbf{For the lead section:} \\[4pt]\text{The pole must lie to the right of its zero on the real axis (more positive), meaning the zero is placed further left.} \\[4pt]\text{This occurs when:} \\[4pt]\beta < 1 \\[10pt]\textbf{For the lag section:} \\[4pt]\text{The pole must lie closer to the origin than the zero, providing phase lag.} \\[4pt]\text{This requirement is satisfied when:} \\[4pt]\beta > 1

Interpretation in a lag–lead networkSince a lag–lead compensator contains both lead and lag parts: The lag network operates with β>1 The lead network operates with β<1Thus, the β associated with the lead portion must meet the condition:β<1The role of β is to determine how far apart the pole and zero are placed on the real axis.For a lead compensator, moving the zero to the left of the pole increases phase lead, which happens when:β<1\textbf{Interpretation in a lag–lead network} \\[6pt]\text{Since a lag–lead compensator contains both lead and lag parts:} \\[4pt]\quad \bullet \ \text{The \textit{lag network} operates with } \beta > 1 \\[2pt]\quad \bullet \ \text{The \textit{lead network} operates with } \beta < 1 \\[6pt]\text{Thus, the } \beta \text{ associated with the \textit{lead portion} must meet the condition:} \\[6pt]\boxed{\beta < 1} \\[6pt]\text{The role of } \beta \text{ is to determine how far apart the pole and zero are placed on the real axis.} \\[4pt]\text{For a lead compensator, moving the zero to the left of the pole increases phase lead, which happens when:} \\[6pt]\beta < 1​​

test-prime-package

Access ‘BEL’ Mock Tests with

  • 60000+ Mocks and Previous Year Papers
  • Unlimited Re-Attempts
  • Personalised Report Card
  • 500% Refund on Final Selection
  • Largest Community
students-icon
354k+ students have already unlocked exclusive benefits with Test Prime!
test-prime-package

Access ‘BEL’ Mock Tests with

  • 60000+ Mocks and Previous Year Papers
  • Unlimited Re-Attempts
  • Personalised Report Card
  • 500% Refund on Final Selection
  • Largest Community
students-icon
354k+ students have already unlocked exclusive benefits with Test Prime!
Our Plans
Monthsup-arrow