SME
Control Systems
Laplace-domain modeling
Use this for “A system with G(s) = 2/(0.5s + 1) gets a step of 3. Find y after 1 s.”
Chart
y(t) = 6(1 − exp(−t/τ)), τ = 0.5 s
After τ the response has made 63.2% of its change; after 5τ, 99.3%.
At t = 1 s: y = 5.188.
This is an example. Type your own values to start a new problem.
G(s)
=
K
s
+
1
y∞
=
K
×
u
y
=
y∞
(1
−
e
−
/
)
Read K and τ off G(s). A step of size u settles at K × u, and the output climbs toward it with time constant τ.
Formulas
y∞
=
K
×
u
6 = 2 × 3
y
=
y∞
×
(1
−
e
−
t
÷
τ
)
5.188
=
6
×
(1
−
e
−1 ÷ 0.5
)
Assumptions
•
G(s) = K ÷ (τs + 1) stands for τ dy/dt + y = Ku, with y(0) = 0.
•
The step of size u starts at t = 0; from the transform table, y = y∞(1 − e−t/τ).
•
The output and the input carry the system’s own units; the DC gain K = G(0) converts one to the other.
Step-by-step
WE KNOW
K = 2, τ = 0.5 s, u = 3, t = 1 s
FIND
final value (y∞), output at t (y)
Step 1 · Find final value (y∞)
Use y∞ = K × u
Final-value
theorem:
y∞
=
G(0)
×
u
,
and
G(0)
=
K
0 + 1
=
K
.
y∞ = 2 × 3
y∞ = 6
Step 2 · Find output at t (y)
Use y = y∞ × (1 − e−t ÷ τ)
The
share
of
the
climb
made
so
far:
63.2%
after
one
τ
,
98.2%
after
4
τ
.
y
=
6
×
(1
−
e
−1 ÷ 0.5
)
y
=
6
×
(1
−
e
-2
)
y
=
6
×
(1
−
0.1353)
y
=
6
×
0.8647
y
=
5.188
Check
Put the numbers back in. Both sides should match.
6 = 2 × 3 ✓
5.188
=
6
×
(1
−
e
−1 ÷ 0.5
)
✓
Topic check
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