DIFFERENCE between the phase and 180°, for an output signal (relative to its input) at zero dB gain.i.e. PM = 180° + ϕg, whereϕg = 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 = xEquation 1), becomes x2 + x – 1 = 0. Solving for ‘x’, we get:\(x = \frac{{ - 1 \pm \sqrt {1 + 4} }}{2} = \frac{{ - 1 \pm \sqrt 5 }}{2}\)\(\Rightarrow {\rm{ω }}_{\rm{g}}^2 = \frac{{ - 1 + \sqrt 5 }}{2} \)\({ω _g} = \sqrt {\frac{{ - 1 + \sqrt 5 }}{2}} \)ωg = 0.786 = 1 rad/secNow, 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°So, PM (Phase margin) = 180 + ϕg = 180 – 135° = 45°