SME
Control Systems
Transfer functions
Use this for “Find the percent overshoot and settling time of 25/(s² + 4s + 25).”
Chart
Underdamped step response: ζ = 0.4, ωₙ = 5 rad/s
Peak
1.25
at
Tₚ
=
π
ωd
=
0.6856
s:
OS
=
exp(−ζπ/
√
1 −
ζ
2
)
=
25.38%
Tₛ
=
4
ζωₙ
=
2
s:
the
envelope
enters
the
±2%
band.
This is an example. Type your own values to start a new problem.
T(s)
=
ωₙ
2
s²
+
2(
)(
)s
+
ωₙ
2
σ
Decay rate
calculated · s⁻¹
ωd
Damped frequency
calculated · rad/s
%OS
Percent overshoot
calculated · %
Tₚ
Peak time
calculated
Tₛ
Settling time
calculated
Formulas
σ
=
ζ
×
ωₙ
2 = 0.4 × 5
ωd
=
ωₙ
×
√
1 −
ζ
2
4.5826
=
5
×
√
1 −
0.4
2
ωₙ
=
√
σ
2
+
ωd
2
5
=
√
2
2
+
4.5826
2
%OS
=
100
×
e
−
ζ
× π ÷
√
1 −
ζ
2
25.3827
=
100
×
e
−0.4 × π ÷
√
1 −
0.4
2
Tₚ
=
π
ωd
0.6856
=
π
4.5826
Tₛ
=
4
σ
2
=
4
2
Assumptions
•
The system is ωₙ² ÷ (s² + 2ζωₙ s + ωₙ²): no zeros, a DC gain of 1 and a unit step input.
•
Underdamped (0 < ζ < 1): the poles are the complex pair −σ ± jωd in the left half-plane.
•
Tₛ = 4 ÷ σ is when the envelope e−σt enters the 2% band around the final value.
Step-by-step
WE KNOW
ωₙ = 5 rad/s, ζ = 0.4
FIND
decay rate (σ), damped frequency (ωd), percent overshoot (%OS), peak time (Tₚ), settling time (Tₛ)
Step 1 · Find decay rate (σ)
Use σ = ζ × ωₙ
The
poles’
real
part
is
−ζωₙ:
σ
is
how
fast
the
ringing
dies
away.
σ
=
0.4
×
5
σ
=
2
s⁻¹
Step 2 · Find damped frequency (ωd)
Use ωd = ωₙ × √(1 − ζ²)
The poles’ imaginary part: the system rings at ωd, a little slower than ωₙ.
ωd
=
5
×
√
1 −
0.4
2
ωd
=
5
×
√
1 − 0.16
ωd
=
5
×
√
0.84
ωd = 5 × 0.9165
ωd = 4.58258 rad/s
Step 3 · Find percent overshoot (%OS)
Use %OS = 100 × e−ζ × π ÷ √(1 − ζ²)
The
overshoot
depends
on
ζ
alone:
the
less
damping,
the
higher
the
first
peak.
%OS
=
100
×
e
−0.4 × π ÷
√
1 −
0.4
2
%OS
=
100
×
e
−0.4π ÷
√
1 − 0.16
%OS
=
100
×
e
−0.4π ÷
√
0.84
%OS
=
100
×
e
−0.4π ÷ 0.91652
%OS
=
100
×
e
−1.3711
%OS = 100 × 0.25383
%OS = 25.3827%
Step 4 · Find peak time (Tₚ)
Use Tₚ = π ÷ ωd
The first peak comes half a damped period after the step.
Tₚ
=
π
4.58258
Tₚ = 0.6856 s
Step 5 · Find settling time (Tₛ)
Use Tₛ = 4 ÷ σ
e
−σt
falls
to
about
2%
(
e
-4
=
0.018)
when
σt
=
4.
Tₛ
=
4
2
Tₛ = 2 s
Check
Put the numbers back in. Both sides should match.
2 = 0.4 × 5 ✓
4.58258
=
5
×
√
1 −
0.4
2
✓
5
=
√
2
2
+
4.58258
2
✓
25.3827
=
100
×
e
−0.4 × π ÷
√
1 −
0.4
2
✓
0.6856
=
π
4.58258
✓
2
=
4
2
✓
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Design from overshoot and settling time
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