SME
Control Systems · Problem type
Closing the loop on a first-order plant
Use this for “G(s) = 4/(s + 1) has unity feedback. Find the closed-loop pole, the DC gain and the time constant.”
Diagram
Poles (×) at s = −1.
At KH = 4, the closed-loop poles are s = −5.
This is an example. Type your own values to start a new problem.
G(s)
=
K
s
+
H
=
p
Closed-loop pole
calculated · s⁻¹
T(0)
Closed-loop DC gain
calculated
τ
Closed-loop time constant
calculated
Formulas
p
=
−(
a
+
K
×
H
)
−5 = −(1 + 4 × 1)
T(0)
=
K
a
+
K
×
H
0.8
=
4
1 + 4 × 1
τ
=
1
a
+
K
×
H
0.2
=
1
1 + 4 × 1
Assumptions
•
Negative feedback: T = G ÷ (1 + GH) = K ÷ (s + a + KH), so the one pole moves from −a to −(a + KH).
•
The plant is stable (a > 0); K and a are in s⁻¹, while H and T(0) are plain ratios (unity feedback means H = 1).
•
A pole at −σ decays as e−σt, so the time constant is 1 ÷ σ.
Step-by-step
WE KNOW
K = 4 s⁻¹, a = 1 s⁻¹, H = 1
FIND
closed-loop pole (p), closed-loop DC gain (T(0)), closed-loop time constant (τ)
Step 1 · Find closed-loop pole (p)
Use p = −(a + K × H)
The
closed
loop’s
denominator
is
s
+
a
+
KH;
it
is
zero
at
s
=
−(
a
+
KH).
p
=
−(1
+
4
×
1)
p
=
−5
s⁻¹
Step 2 · Find closed-loop DC gain (T(0))
Use T(0) = K ÷ (a + K × H)
Put s = 0 in T(s): the steady output per unit of step input.
T(0)
=
4
1 + 4 × 1
T(0)
=
4
5
T(0) = 0.8
Step 3 · Find closed-loop time constant (τ)
Use τ = 1 ÷ (a + K × H)
The
pole
sits
at
−(
a
+
KH),
and
its
response
decays
as
e
−(
a
+
KH
)t
.
τ
=
1
1 + 4 × 1
τ
=
1
5
τ
=
0.2
s
Check
Put the numbers back in. Both sides should match.
−5 = −(1 + 4 × 1) ✓
0.8
=
4
1 + 4 × 1
✓
0.2
=
1
1 + 4 × 1
✓
Topic check
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