SME
Process Dynamics & Control · Problem type
Overshoot, decay ratio and period
Use this for “A process has damping ratio 0.3 and natural frequency 0.5 per min. Find its overshoot, decay ratio and period.”
Chart
Underdamped step response: ζ = 0.3, ωₙ = 0.5 min⁻¹
Peak
1.37
at
Tₚ
=
π
ωd
=
6.59
min:
OS
=
exp(−ζπ/
√
1 −
ζ
2
)
=
37.23%
Tₛ
=
4
ζωₙ
=
26.7
min:
the
envelope
enters
the
±2%
band.
This is an example. Type your own values to start a new problem.
OS
=
100
×
e
−π
×
÷
1
−
2
DR
=
(
÷
100)
2
P (min)
=
2π
÷
(
ωₙ (min⁻¹)
×
1
−
ζ
2
)
Formulas
OS
=
100
×
e
−π ×
ζ
÷
√
1 −
ζ
2
37.23
=
100
×
e
−π × 0.3 ÷
√
1 −
0.3
2
DR
=
(
OS
100
)
2
0.1386
=
(
37.23
100
)
2
P
=
2π
ωₙ ×
√
1 −
ζ
2
13.17
=
2π
0.5 ×
√
1 −
0.3
2
Assumptions
•
Underdamped second order, 0 < ζ < 1, after a step in its input; ωₙ = 1 ÷ τ.
•
OS is the first peak’s rise past the final value, as a percent of the whole change.
•
Decay ratio: the second overshoot over the first; period P: the time from peak to peak.
Step-by-step
WE KNOW
ζ = 0.3, ωₙ = 0.5 min⁻¹
FIND
overshoot (OS), decay ratio (DR), period (P)
Step 1 · Find overshoot (OS)
Use OS = 100 × e−π × ζ ÷ √(1 − ζ²)
The
closed-form
step
response
peaks
first
at
this
share
past
the
final
value;
it
depends
on
ζ
alone.
OS
=
100
×
e
−π × 0.3 ÷
√
1 −
0.3
2
OS
=
100
×
e
−π × 0.3 ÷
√
1 − 0.09
OS
=
100
×
e
−π × 0.3 ÷
√
0.91
OS
=
100
×
e
−π × 0.3 ÷ 0.9539
OS
=
100
×
e
−0.9425 ÷ 0.9539
OS
=
100
×
e
−0.988
OS = 100 × 0.3723
OS = 37.23%
Step 2 · Find decay ratio (DR)
Use DR = (OS ÷ 100)²
Each overshoot is OS times the one before, so the second over the first is OS².
DR
=
(
37.23
100
)
2
DR
=
0.3723
2
DR = 0.1386
Step 3 · Find period (P)
Use P = 2π ÷ (ωₙ × √(1 − ζ²))
The
response
swings
at
the
damped
frequency
ωₙ
√
1 −
ζ
2
;
one
full
swing
takes
2π
over
it.
P
=
2π
0.5 ×
√
1 −
0.3
2
P
=
2π
0.5 ×
√
0.91
P
=
6.283
0.477
P
=
13.17
min
Check
Put the numbers back in. Both sides should match.
37.23
=
100
×
e
−π × 0.3 ÷
√
1 −
0.3
2
✓
0.1386
=
(
37.23
100
)
2
✓
13.17
=
2π
0.5 ×
√
1 −
0.3
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.