The network having transfer function, \(\frac{1+\frac{s}{4}}{1+\frac{s}{25}}\) will provide maximum phase lead at a frequency of:

Concept:

The standard phase lead compensating network is given as:

\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{\alpha \left( {1 + sT} \right)\;}}{{\left( {1 + \alpha sT} \right)}}\) ------(1)

The maximum phase Occurs at

\(ω_m=\frac{1}{T√{\alpha }}\)

Maximum phase (ϕm)

\(sin{\phi _m} = \frac{{1 - \alpha }}{{1 + \alpha }}\)

Calculation:

Given:

\(G(s)=\frac{1+\frac{s}{4}}{1+\frac{s}{25}}\)

Compare it with the standard transfer function i.e. equation 1

\(T=\frac{1}{4}, \ \alpha T=\frac{1}{25}\)

The maximum phase Occurs at

\(ω_m=\frac{1}{T√{\alpha }}=√{\frac{1}{\alpha T}\frac{1}{T}}\)

ω_m = √(100) = 10 rad/sec