SME
Process Dynamics & Control
Control-loop design
Use this for “A process of three equal lags, each τ = 2 min, has Kₚ = 0.5. Find the ultimate gain and period under P control.”
Diagram
3
equal
lags
under
P
control
go
unstable
when
KcKₚ
=
8,
so
Kc,u
=
8
0.5
=
16.
There
ωᵤ
=
tan(180° ÷ 3)
τ
=
0.866
rad/min
and
Pᵤ
=
2π
ωᵤ
=
7.26
min.
Crossover frequency ωᵤ = 0.866 rad/min
This is an example. Type your own values to start a new problem.
ωᵤ (rad/min)
=
3
÷
τ (min)
Pᵤ (min)
=
2π
÷
ωᵤ (rad/min)
Kc,u
=
8
÷
Kₚ
Put s = iω into 1 + KcKₚ ÷ (τs + 1)³ = 0. The imaginary part gives ωuτ = √3, and the real part gives KcKₚ = 3(ωuτ)² − 1 = 8.
Formulas
Kc,u
=
(3
×
(ωᵤ ×
τ
)
2
−
1)
÷
Kₚ
16
=
(3
×
(0.866 × 2)
2
−
1)
÷
0.5
ωᵤ
=
√
3
τ
0.866
=
√
3
2
Pᵤ
=
2
×
π
ωᵤ
7.255
=
2
×
π
0.866
Assumptions
•
Three equal first-order lags, Kₚ ÷ (τs + 1)³, under proportional (P) control.
•
Ideal valve and sensor (gain 1); the loop goes unstable when KcKₚ reaches 8 (Routh or s = iω).
•
τ and Pᵤ in minutes, so ωᵤ is in radians per minute.
Step-by-step
WE KNOW
τ = 2 min, Kₚ = 0.5
FIND
crossover frequency (ωᵤ), ultimate period (Pᵤ), ultimate gain (Kc,u)
Step 1 · Find crossover frequency (ωᵤ)
Use ωᵤ = √3 ÷ τ
Put
s
=
iω
in
τ
3
s
3
+
3
τ
²
s
2
+
3τs
+
1
+
KcKₚ
=
0:
the
imaginary
part
is
0
when
ωτ
=
√
3
.
ωᵤ
=
√
3
2
ωᵤ
=
1.732
2
ωᵤ = 0.866 rad/min
Step 2 · Find ultimate period (Pᵤ)
Use Pᵤ = 2 × π ÷ ωᵤ
At the ultimate gain the loop cycles steadily; one cycle is 2π radians.
Pᵤ
=
2
×
π
0.866
Pᵤ = 7.255 min
Step 3 · Find ultimate gain (Kc,u)
Use Kc,u = (3 × (ωᵤ × τ)² − 1) ÷ Kₚ
Real
part
of
1
+
KcKₚ
÷
(iωτ + 1)
3
=
0:
with
ωuτ
=
√
3
the
loop
gain
KcKₚ
is
3
×
3
−
1
=
8.
Kc,u
=
(3
×
(0.866 × 2)
2
−
1)
÷
0.5
Kc,u
=
(3
×
(
√
3
)
2
−
1)
÷
0.5
Kc,u
=
3 × 3 − 1
0.5
Kc,u
=
8
0.5
Kc,u = 16
Check
Put the numbers back in. Both sides should match.
16
=
(3
×
(0.866 × 2)
2
−
1)
÷
0.5
✓
0.866
=
√
3
2
✓
7.255
=
2
×
π
0.866
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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