Let G(s)H(s) be the open loop transfer function (OLTF)
for which a bode plot is drawn. Then
Gain
cross-over frequency ( Wgc )
Gain cross-over frequency is the frequency at which
OLTF magnitude is equal to 1 or 0dB.
i.e
|G(jw)H(jw)|=1 or 0dB
, where W=Wgc
Phase
cross-over frequency
( Wpc )
Phase cross-over frequency is the frequency at which phase
of OLTF is equal to -1800.
i.e ∠G(jw)H(jw) = -1800 , where W=Wpc
Gain
Margin (GM)
Gain Margin is the factor by which the gain factor K
can be multiplied before the closed loop system reaches to the verge of
instability.
It is calculated as
GM = -20 log X
(in dB)
where,
X=|G(jw)H(jw)| , at
W=Wpc
Phase
Margin (PM)
Phase Margin is defined as the allowable phase lag
before the closed loop system reaches to the verge of instability.
It is calculated as
PM = 180+ Ф
where, Ф = ∠G(jw)H(jw) , at W=Wgc
Stability Rules of Bode Plot
(1) If GM and
PM both are positive, or , Wgc <
Wpc , the closed loop sytem is stable.
(2) If GM and
PM both are negative, or , Wgc >
Wpc , the closed loop sytem is unstable.
(3) If GM and
PM both are zero, or , Wgc =
Wpc , the closed loop sytem is marginally stable.
Gain Margin Phase Margin Solved Example
Q. Construct bode plot for the open loop transfer
function
and find
(a) Gain cross-over frequency
(b) Phase
cross-over frequency
(c) Gain Margin
(d) Phase Margin
(e) Closed loop stability of the system
Sol :- The bode plot comes out to be as shown below
(a) |G(jw)H(jw)|=1 at W=2.5 rad/sec
therefore, gain cross-over frequency = 2.5 rad/sec
(b) ∠G(jw)H(jw)
= -1800 at
W=5 rad/sec
therefore, phase cross-over frequency = 5 rad/sec
(c) GM = -20 log |G(jw)H(jw)| at W=Wpc=5 rad/sec
We get
GM=11.21 dB
(d) PM = 180+ Ф
where, Ф = ∠G(jw)H(jw) at W=Wgc=2.5 rad/sec
we get PM= 1800-1530
= 270
(e) Since both GM and PM are positive , therefore,
closed loop system is stable.
Related :-
Bode Plot step-by-step
What is Root locus | solved examples
Routh Stability Criterion
Related :-
Bode Plot step-by-step
What is Root locus | solved examples
Routh Stability Criterion
