SME
Control Systems
Frequency response
Use this for “Find the phase margin of G(s) = 20 ÷ (s(s + 3)) in unity feedback.”
Diagram
Corner at a = 3 rad/s.
Gain crossover at 4.00 rad/s: PM = 180° − 143° = 36.9°.
This is an example. Type your own values to start a new problem.
G(s)
=
K
s(s
+
)
ωc
Gain crossover
calculated · rad/s
φc
Phase at ωc
calculated · °
PM
Phase margin
calculated · °
Formulas
ωc
2
=
−
a
2
+
√
a
4
+ 4 ×
K
2
2
4
2
=
−
3
2
+
√
3
4
+ 4 ×
20
2
2
φc
=
−90
−
tan⁻¹(
ωc
a
)
−143.1301
=
−90
−
tan⁻¹(
4
3
)
PM = 180 + φc
36.8699 = 180 + (−143.1301)
Assumptions
•
Unity feedback around G(s) = K ÷ (s(s + a)), with K and a positive and ω in rad/s.
•
The gain crossover ωc is where |G(jω)| = 1, which is 0 dB on the Bode plot.
•
PM = 180° + ∠G(jωc); a phase margin above about 45° means little overshoot.
Step-by-step
WE KNOW
K = 20, a = 3 rad/s
FIND
gain crossover (ωc), phase at ωc (φc), phase margin (PM)
Step 1 · Find gain crossover (ωc)
Use ωc² = (−a² + √(a⁴ + 4 × K²)) ÷ 2
Setting
K
ω
√
ω
2
+
a
2
=
1
gives
ω
4
+
a
2
ω
2
−
K
2
=
0;
its
positive
root
for
ω
2
,
written
2
K
²
a
2
+
√
a
4
+ 4
K
²
so
nothing
cancels.
ωc
=
√(2
×
K
2
a
2
+
√
a
4
+ 4 ×
K
2
)
ωc
=
√(2
×
20
2
3
2
+
√
3
4
+ 4 ×
20
2
)
ωc
=
√(2
×
20
2
3
2
+
√
81 + 4 × 400
)
ωc
=
√(2
×
20
2
3
2
+
√
81 + 1,600
)
ωc
=
√(2
×
20
2
3
2
+
√
1
,681
)
ωc
=
√(2
×
20
2
9 + 41
)
ωc
=
√(2
×
20
2
50
)
ωc
=
√(2
×
400
50
)
ωc
=
√(
800
50
)
ωc
=
√
16
ωc = 4 rad/s
Step 2 · Find phase at ωc (φc)
Use φc = −90 − tan⁻¹(ωc ÷ a)
The
integrator
1
s
adds
−90°
and
the
pole
at
−a
adds
−tan⁻¹(
ω
a
).
φc
=
−90
−
tan⁻¹(
4
3
)
φc = −143.1301°
Step 3 · Find phase margin (PM)
Use PM = 180 + φc
The margin is how far the phase at crossover stays above −180°.
PM = 180 + (−143.1301)
PM = 36.8699°
Check
Put the numbers back in. Both sides should match.
4
2
=
−
3
2
+
√
3
4
+ 4 ×
20
2
2
✓
−143.1301
=
−90
−
tan⁻¹(
4
3
)
✓
36.8699 = 180 + (−143.1301) ✓
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Bode asymptotes and the exact gain
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