For the following network to work as lag compensator, the value of R₂ would be

Lag compensator:

Transfer function:

If it is in the form of \(\frac{{1 + aTs}}{{1 + Ts}}\), then a < 1

If it is in the form of \(\frac{{s + a}}{{s + b}}\), then a > b

Maximum phase lag frequency: \({\omega _m} = \frac{1}{{T\sqrt a }}\)

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

ϕm is negative

Pole zero plot:

The pole is nearer to the origin.

Filter: It is a low pass filter (LPF)

Effect on the system:

• Rise time and settling time increases and Bandwidth decreases
• The transient response becomes slower
• The steady-state response is improved
• Stability decreases

Application:

The equivalent Laplace transform network for the given network is,

By applying voltage division,

\({V_2}\left( s \right) = {V_1}\left( s \right)\left( {\frac{{{R_2} + \frac{1}{{Cs}}}}{{{R_1} + {R_2} + \frac{1}{{Cs}}}}} \right)\)

\( ⇒ \frac{{{V_2}\left( s \right)}}{{{V_1}\left( s \right)}} = \frac{{{R_2}Cs + 1}}{{\left( {{R_1} + {R_2}} \right)Cs + 1}}\)

\( = \frac{{{R_2}Cs + 1}}{{\left( {\frac{{{R_1} + {R_2}}}{{{R_2}}}} \right){R_2}Cs + 1}}\)

\( ⇒ \frac{{{V_2}\left( s \right)}}{{{V_1}\left( s \right)}} = \frac{{\tau s + 1}}{{\beta \tau s + 1}},\;\tau = {R_2}C,\;\beta = \frac{{{R_1} + {R_2}}}{{{R_2}}}\)

The above system to be lag compensator, β > 1

\(\beta = \frac{{{R_1} + {R_2}}}{{{R_2}}} > 1\)

⇒ R₁ > 0

Therefore, at any value of R₂ the given system acts as lag compensator.