Correct Answer - Option 4 :
\(\sqrt {10}\) rad / s
Concept:
Lead and Lag compensators:
Gc(s) = (1 + aTs) / (1 + Ts) where a and T > 0, a > 1 (lead) & a < 1 (lag)
∠Gc(s) = ϕ = tan-1 ωaT – tan-1 ωT
ωm is the geometric mean of the two corner frequencies 1/T and 1/aT
ωm = \(\frac{1}{{T\sqrt a }}\)
Calculations:
Given Gc(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
ωm = 1 / (\(\sqrt {0.1} \))
ωm = \(\sqrt {10} \) rad/sec