Question

If r = 1 in the \(G(s) = \frac {1 + 0.1rs}{1 + rs}\), then the compensator can give the minimum phase at a frequency of
1. \(\sqrt {0.1}\) rad / s
2. 0.777 rad / s
3. 1 rad / s
4. \(\sqrt {10}\) rad / s

Answer 1

Correct Answer - Option 4 : \(\sqrt {10}\) rad / s

Concept:

Lead and Lag compensators:

G_c(s) = (1 + aTs) / (1 + Ts) where a and T > 0, a > 1 (lead) & a < 1 (lag)

∠G_c(s) = ϕ = tan^-1 ωaT – tan^-1 ωT

ω_m is the geometric mean of the two corner frequencies 1/T and 1/aT

ω_m = \(\frac{1}{{T\sqrt a }}\)

Calculations:

Given G_c(s) = \(\frac{{1 + 0.1rs}}{{1 + rs}}\)

r = 1

By substituting r value we get \({G_c}\left( s \right) = \frac{{1\; + \;0.1s}}{{1\ + \;s}}\)

Here T = 1 and a = 0.1

ω_m = 1 / (\(\sqrt {0.1} \))

ω_m = \(\sqrt {10} \) rad/sec