SME
Control Systems
PID control
Use this for “A plant 2/(4s + 1) under proportional control Kₚ = 4.5: find the steady-state error and time constant.”
Chart
y(t) = 0.9(1 − exp(−t/τ)), τ = 0.4 s
After τ the response has made 63.2% of its change; after 5τ, 99.3%.
This is an example. Type your own values to start a new problem.
G(s)
=
K
s
+
1
K
p
=
Kₚ
L
Loop gain
calculated
Gcl
Closed-loop DC gain
calculated
τcl
Closed-loop time constant
calculated
eₛₛ
Steady-state error
calculated
Formulas
L
=
K
×
Kₚ
9 = 2 × 4.5
Gcl
=
L
1 +
L
0.9
=
9
1 + 9
τcl
=
τ
1 +
L
0.4
=
4
1 + 9
eₛₛ
=
1
1 +
L
0.1
=
1
1 + 9
Assumptions
•
Unity feedback and a unit step in the set point; the actuator doesn’t saturate.
•
The plant is first order, K ÷ (τs + 1), and the controller multiplies the error by Kₚ.
•
The closed loop is first order too: L ÷ (τs + 1 + L), so it settles at L ÷ (1 + L).
•
The dashed curve is the plant alone, with no controller.
Step-by-step
WE KNOW
K = 2, τ = 4 s, Kₚ = 4.5
FIND
loop gain (L), closed-loop DC gain (Gcl), closed-loop time constant (τcl), steady-state error (eₛₛ)
Step 1 · Find loop gain (L)
Use L = K × Kₚ
Around the loop the two gains multiply.
L
=
2
×
4.5
L
=
9
Step 2 · Find closed-loop DC gain (Gcl)
Use Gcl = L ÷ (1 + L)
The
output
settles
where
y
=
L
(1
−
y),
so
y
=
L
1 +
L
.
Gcl
=
9
1 + 9
Gcl
=
9
10
Gcl = 0.9
Step 3 · Find closed-loop time constant (τcl)
Use τcl = τ ÷ (1 + L)
Feedback
moves
the
pole
from
−
1
τ
to
−
1 +
L
τ
:
the
loop
is
1
+
L
times
faster.
τcl
=
4
1 + 9
τcl
=
4
10
τcl = 0.4 s
Step 4 · Find steady-state error (eₛₛ)
Use eₛₛ = 1 ÷ (1 + L)
What
the
output
falls
short
of
the
unit
step:
1
−
L
1 +
L
.
eₛₛ
=
1
1 + 9
eₛₛ
=
1
10
eₛₛ = 0.1
Check
Put the numbers back in. Both sides should match.
9 = 2 × 4.5 ✓
0.9
=
9
1 + 9
✓
0.4
=
4
1 + 9
✓
0.1
=
1
1 + 9
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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