The gain cross-over frequency and phase margin of the transfer function  are 

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

Concept:

Phase margin is the difference between the phase and 180°, for an output signal (relative to its input) at zero dB gain.

i.e. PM = 180° + ϕg, where

ϕg = Phase angle at the gain crossover frequency (ωg)

-ωg = frequency at which the magnitude of G(s) H(s) becomes 1).

Calculation:

\(G\left( s \right)H\left( s \right) = \frac{1}{{s\left( {s + 1} \right)}}\)

Substituting s with jω

\(\Rightarrow G\left( {j{\rm{ω }}} \right)H\left( {j{\rm{ω }}} \right) = \frac{1}{{j{\rm{ω }}\left( {j{\rm{ω }} + 1} \right)}}\)

Magnitude (M) \(= \left| {G\left( {j{\rm{ω }}} \right)H\left( {j{\rm{ω }}} \right)} \right| = \frac{1}{{{\rm{ω }}\sqrt {1 + {{\rm{ω }}^2}} }}\) 

At the gain cross over frequency ωg, M = 1.

\(\Rightarrow 1 = \frac{1}{{{ω _g}\sqrt {ω _g^2 + 1} }} \Rightarrow 1 = \frac{1}{{{\rm{ω }}_{\rm{g}}^2\left( {{\rm{ω }}_g^2 + 1} \right)}}\)

⇒ ωg2 (ωg2 + 1) = 1

⇒ ωg4 + ωg2 – 1 = 0     …1)

Let ωg2 = x

Equation 1), becomes x2 + x – 1 = 0. Solving for ‘x’, we get:

\(x = \frac{{ - 1 \pm \sqrt {1 + 4} }}{2} = \frac{{ - 1 \pm \sqrt 5 }}{2}\)

\(\Rightarrow {\rm{ω }}_{\rm{g}}^2 = \frac{{ - 1 + \sqrt 5 }}{2} \)

\({ω _g} = \sqrt {\frac{{ - 1 + \sqrt 5 }}{2}} \)

ω_g = 0.786 = 1 rad/sec

Now, the phase at gain crossover frequency ⇒ ϕg

\({\phi _g} = - \frac{\pi }{2} - {\tan ^{ - 1}}\left( {{ω _g}} \right)\)

\(= - \frac{\pi }{2} - {\tan ^{ - 1}}\left( {1} \right)\)

= -135°