SME
Control Systems · Problem type
Gain margin
Use this for “Find the gain margin of G(s) = 2 ÷ (s(s + 1)(s + 2)) in unity feedback.”
Diagram
Corners at a = 1 rad/s, b = 2 rad/s.
Phase
crossover
(−180°)
at
1.41
rad/s:
GM
=
1
0.333
=
3.00
(9.54
dB).
This is an example. Type your own values to start a new problem.
G(s)
=
K
s(s
+
)(s
+
)
ω₁₈₀
Phase crossover
calculated · rad/s
Kₘₐₓ
Largest stable gain
calculated
GM
Gain margin (ratio)
calculated
GMdB
Gain margin
calculated · dB
Formulas
ω₁₈₀
=
√
a
×
b
1.4142
=
√
1 × 2
Kₘₐₓ
=
a
×
b
×
(
a
+
b
)
6 = 1 × 2 × (1 + 2)
GM
=
Kₘₐₓ
K
3
=
6
2
GMdB = 20 × log₁₀(GM)
9.5424 = 20 × log₁₀(3)
Assumptions
•
Unity feedback around G(s) = K ÷ (s(s + a)(s + b)), with K, a and b positive.
•
The phase −90° − tan⁻¹(ω ÷ a) − tan⁻¹(ω ÷ b) reaches −180° where ω² = ab.
•
There |G| = K ÷ Kₘₐₓ, so GM = Kₘₐₓ ÷ K; Routh’s test gives the same Kₘₐₓ.
•
A gain margin above 1 (0 dB) means the closed loop is stable.
Step-by-step
WE KNOW
K = 2, a = 1 rad/s, b = 2 rad/s
FIND
phase crossover (ω₁₈₀), largest stable gain (Kₘₐₓ), gain margin (ratio) (GM), gain margin (GMdB)
Step 1 · Find phase crossover (ω₁₈₀)
Use ω₁₈₀ = √(a × b)
The
two
pole
angles
add
to
90°
when
ω
2
=
ab,
since
tan⁻¹
x
+
tan⁻¹(
1
x
)
=
90°.
ω₁₈₀
=
√
1 × 2
ω₁₈₀
=
1.4142
rad/s
Step 2 · Find largest stable gain (Kₘₐₓ)
Use Kₘₐₓ = a × b × (a + b)
At
ω₁₈₀
,
|jω(jω
+
a
)(jω
+
b
)|
=
ab(
a
+
b
):
the
gain
that
puts
|G|
at
exactly
1.
Kₘₐₓ = 1 × 2 × (1 + 2)
Kₘₐₓ = 2 × 3
Kₘₐₓ = 6
Step 3 · Find gain margin (ratio): GM
Use GM = Kₘₐₓ ÷ K
The
margin
is
1
÷
|G(jω₁₈₀)|:
how
many
times
K
can
grow
before
the
loop
is
unstable.
GM
=
6
2
GM = 3
Step 4 · Find gain margin (GMdB)
Use GMdB = 20 × log₁₀(GM)
A gain ratio in decibels is 20 log₁₀ of it.
GMdB = 20 × log₁₀(3)
GMdB = 20 × 0.4771
GMdB = 9.5424 dB
Check
Put the numbers back in. Both sides should match.
1.4142
=
√
1 × 2
✓
6 = 1 × 2 × (1 + 2) ✓
3
=
6
2
✓
9.5424 = 20 × log₁₀(3) ✓
Topic check
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