SME
Control Systems · Problem type
Design from overshoot and settling time
Use this for “A loop must overshoot 10% and settle in 2 s. Find ζ and ωₙ.”
Chart
Underdamped step response: ζ = 0.5912, ωₙ = 3.383 rad/s
Peak
1.1
at
Tₚ
=
π
ωd
=
1.15
s:
OS
=
exp(−ζπ/
√
1 −
ζ
2
)
=
10%
Tₛ
=
4
ζωₙ
=
2
s:
the
envelope
enters
the
±2%
band.
This is an example. Type your own values to start a new problem.
ζ
=
−ln(
÷
100)
π²
+
ln²(
÷
100)
ωₙ (rad/s)
=
4
×
Tₛ (s)
σ
Decay rate
worked out (not typed) · s⁻¹
ωd
Damped frequency
worked out (not typed) · rad/s
Formulas
ζ
=
−ln(%
OS
100
)
÷
√(
π
2
+
(ln(%
OS
100
))²)
0.5912
=
−ln(
10
100
)
÷
√(
π
2
+
(ln(
10
100
))²)
ωₙ
=
4
ζ
× Tₛ
3.3832
=
4
0.5912 × 2
σ
=
4
Tₛ
2
=
4
2
ωd
=
ωₙ
×
√
1 −
ζ
2
2.7288
=
3.3832
×
√
1 −
0.5912
2
Assumptions
•
The closed loop acts as the standard ωₙ² ÷ (s² + 2ζωₙ s + ωₙ²), with no zeros.
•
Underdamped (0 < ζ < 1) with a 2% settling band, so Tₛ = 4 ÷ (ζωₙ).
•
The overshoot fixes ζ alone; the settling time then fixes ωₙ.
Step-by-step
WE KNOW
%OS = 10%, Tₛ = 2 s
FIND
damping ratio (ζ), natural frequency (ωₙ), decay rate (σ), damped frequency (ωd)
Step 1 · Find damping ratio (ζ)
Use ζ = −ln(%OS ÷ 100) ÷ √(π² + (ln(%OS ÷ 100))²)
Take
ln
of
%OS
=
100
e
−ζπ ÷
√
1 −
ζ
2
and
solve
for
ζ
:
less
overshoot
needs
more
damping.
ζ
=
−ln(
10
100
)
÷
√(
π
2
+
(ln(
10
100
))²)
ln(
10
100
)
=
−2.3026
ζ
=
2.3026
÷
√
9.8696 + 5.3019
ζ
=
2.3026
÷
√
15.1715
ζ
=
2.3026
3.8951
ζ
=
0.591155
Step 2 · Find natural frequency (ωₙ)
Use ωₙ = 4 ÷ (ζ × Tₛ)
Tₛ
=
4
ζωₙ
:
divide
4
by
ζ
and
the
settling
time.
ωₙ
=
4
0.591155 × 2
ωₙ
=
4
1.1823
ωₙ = 3.383207 rad/s
Step 3 · Find decay rate (σ)
Use σ = 4 ÷ Tₛ
The
poles’
real
part
is
−
σ
,
with
σ
=
ζωₙ
=
4
Tₛ
.
σ
=
4
2
σ
=
2
s⁻¹
Step 4 · Find damped frequency (ωd)
Use ωd = ωₙ × √(1 − ζ²)
The
poles’
imaginary
part,
so
the
poles
sit
at
−
σ
±
jωd.
ωd
=
3.383207
×
√
1 −
0.591155
2
ωd
=
3.3832
×
√
1 − 0.3495
ωd
=
3.3832
×
√
0.6505
ωd = 3.3832 × 0.8065
ωd = 2.7288 rad/s
Check
Put the numbers back in. Both sides should match.
0.591155
=
−ln(
10
100
)
÷
√(
π
2
+
(ln(
10
100
))²)
✓
3.383207
=
4
0.591155 × 2
✓
2
=
4
2
✓
2.7288
=
3.383207
×
√
1 −
0.591155
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.