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Process Dynamics & Control · Problem type
IMC PI tuning
Use this for “A process has K = 2, τ = 10 min and θ = 2 min. Find IMC PI settings for τc = 4 min.”
Chart
y(t)
=
2(1
−
exp(−
t − θ
τ
)),
τ
=
10
min
Dead time θ = 2 min: nothing moves until then.
After θ + τ the response has made 63.2% of its change; after 5τ, 99.3%.
Closed-loop time constant τc = 4 min
Controller gain Kc = 0.8333
Integral time τI = 10 min
This is an example. Type your own values to start a new problem.
Kc
=
τ (min)
÷
(
K
×
(
τc (min)
+
θ (min)
))
τI (min)
=
τ (min)
Formulas
Kc
=
τ
K
× (τc +
θ
)
0.8333
=
10
2 × (4 + 2)
τI
=
τ
10 = 10
Assumptions
•
The process is first order plus dead time (FOPDT): gain K, time constant τ, dead time θ.
•
IMC (internal model control) tuning picks the closed-loop time constant τc: smaller is faster but less robust.
•
A common start is τc near θ or larger; the controller is PI, Kc(1 + 1 ÷ (τI s)).
Step-by-step
WE KNOW
K = 2, τ = 10 min, θ = 2 min, τc = 4 min
FIND
controller gain (Kc), integral time (τI)
Step 1 · Find controller gain (Kc)
Use Kc = τ ÷ (K × (τc + θ))
The
IMC
rule:
a
slow
process
(big
τ
)
takes
more
gain;
a
big
gain,
delay
or
τc
takes
less.
Kc
=
10
2 × (4 + 2)
Kc
=
10
2 × 6
Kc
=
10
12
Kc = 0.8333
Step 2 · Find integral time (τI)
Use τI = τ
The integral time equals the process time constant, so the controller’s zero cancels the process lag.
τI = 10 min
Check
Put the numbers back in. Both sides should match.
0.8333
=
10
2 × (4 + 2)
✓
10 = 10 ✓
Topic check
See what you know: 5 questions on this topic
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