Question

The maximum phase shift that can be obtained by using a lead compensator with transfer function \(G_c (s)=\frac{4(1+0.15s)}{(1+0.05 s) }\) equal to
1. 15° 
2. 30° 
3. 45° 
4. 60° 

Answer 1

Correct Answer - Option 2 : 30° 

Concept:

The standard T/F of the compensator is 

\(\frac{{1 + aTs}}{{1 + Ts}}\)

Maximum phase lead

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

Maximum phase lead frequency, 

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

Calculation:

The given transfer function is,

\(T/F = 4(\frac{{1 + 0.15s}}{{1 + 0.05s}})\)

By comparing both transfer functions,

aT = 0.15

T = 0.05

a = 3

Maximum phase lead

\(\begin{array}{l} {ϕ _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\\ = {\sin ^{ - 1}}\left( {\frac{{3 - 1}}{{3 + 1}}} \right)\\ = {\sin ^{ - 1}}\left( {\frac{1}{{2}}} \right) \end{array}\)

= sin-1 (0.5)

ϕ_m = 30°