Question

A phase lead compensator has its transfer function, \(T/F = \frac{{1 + 0.5\;s}}{{1 + 0.05\;s}}\). The maximum phase lead and the corresponding frequency, respectively are nearly 
1. sin^-1(0.9)and 6 r/s
2. sin^-1(0.82) and 4 r/s
3. sin^-1(0.9) and 4 r/s
4. sin^-1(0.82) and 6 r/s

Answer 1

Correct Answer - Option 4 : sin^-1(0.82) and 6 r/s

Concept:

The standard T/F of the compensator is 

\(\frac{{1 + aTs}}{{1 + Ts}}\) 

Maximum phase lead

\( {\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\\ \)

Maximum phase lead frequency, 

\({\omega _m} = \frac{1}{{T\sqrt a }}\)

Calculation:

The given transfer function is,

\(T/F = \frac{{1 + 0.5\;s}}{{1 + 0.05\;s}}\)

By comparing both transfer functions,

aT = 0.5

T = 0.05

a = 10

Maximum phase lead

\(\begin{array}{l} {\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\\ = {\sin ^{ - 1}}\left( {\frac{{10 - 1}}{{10 + 1}}} \right)\\ = {\sin ^{ - 1}}\left( {\frac{9}{{11}}} \right) \end{array}\)

= sin^-1 (0.82)

Maximum phase lead frequency is:

\({\omega _m} = \frac{1}{{T\sqrt a }}\)

\(= \frac{1}{{0.05\sqrt {10} }}\)

ωm = 6.66 rad/sec