SME
Control Systems
Stability and root locus
Use this for “For what K is s³ + 6s² + 8s + K = 0 stable?”
Diagram
First column 1, 6, 4.667, 20: each sign change is a pole in the right half-plane.
Tap a worked-out cell to see its cross product.
The
s
1
row
is
6 × 8 − K
6,
so
stable
for
0
<
K
<
48.
At
K
=
48
the
s
1
row
is
0;
6s²
+
48
=
0
gives
ω
=
√
8
=
2.828
rad/s.
This is an example. Type your own values to start a new problem.
s³
+
a₂
s²
+
a₁
s
+
K
=
0
b₁
First-column entry of the s¹ row
worked out (not typed)
Kₘₐₓ
Largest stable gain
calculated
ωc
Crossing frequency
calculated · rad/s
Formulas
b₁
=
a₂
×
a₁
−
K
a₂
4.6667
=
6 × 8 − 20
6
Kₘₐₓ
=
a₂
×
a₁
48 = 6 × 8
ωc
=
√
a₁
2.8284
=
√
8
Assumptions
•
The Routh first column is 1, a₂, (a₂a₁ − K) ÷ a₂, K; the system is stable when all four are positive.
•
At Kₘₐₓ the s¹ row is 0, and the auxiliary polynomial a₂s² + K = 0 gives the crossing frequency ωc.
•
K > 0 is the controller gain; a₂ and a₁ come from the plant and are positive.
Step-by-step
WE KNOW
a₂ = 6, a₁ = 8, K = 20
FIND
first-column entry of the s¹ row (b₁), largest stable gain (Kₘₐₓ), crossing frequency (ωc)
Step 1 · Find first-column entry of the s¹ row (b₁)
Use b₁ = (a₂ × a₁ − K) ÷ a₂
The
s
1
entry
is
the
2
×
2
cross
product
of
the
two
rows
above,
over
the
s
2
row’s
lead.
b₁
=
6 × 8 − 20
6
b₁
=
48 − 20
6
b₁
=
28
6
b₁
=
4.6667
Step 2 · Find largest stable gain (Kₘₐₓ)
Use Kₘₐₓ = a₂ × a₁
The
s
1
entry
a₂a₁
−
K
a₂
stays
positive
while
K
is
under
a₂
a₁
.
Kₘₐₓ = 6 × 8
Kₘₐₓ = 48
Step 3 · Find crossing frequency (ωc)
Use ωc = √a₁
At
Kₘₐₓ,
a₂
s
2
+
a₂
a₁
=
0
gives
s
=
±j√
a₁
:
the
poles
cross
the
jω
axis
there.
ωc
=
√
8
ωc = 2.8284 rad/s
Check
Put the numbers back in. Both sides should match.
4.6667
=
6 × 8 − 20
6
✓
48 = 6 × 8 ✓
2.8284
=
√
8
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.