SME
Control Systems · Problem type
Root locus: asymptotes and breakaway
Use this for “Sketch the root locus of K/(s(s + 2)(s + 4)): find the centroid and the breakaway point.”
Diagram
Poles (×) at s = 0, −2, −4.
σₐ
=
−
6
3
=
−2;
asymptotes
at
60°,
180°,
300°.
Breakaway at s = −0.8453, where dK/ds = 0.
At K = 15, the closed-loop poles are s = −0.5 ± j1.658, −5.
This is an example. Type your own values to start a new problem.
G(s)H(s)
=
K
(s
−
)(s
−
(
)
)(s
−
(
)
)
σₐ
Centroid
worked out (not typed)
sb
Breakaway point
worked out (not typed)
Kₘₐₓ
Largest stable gain
worked out (not typed)
GM
Gain margin
calculated
Formulas
σₐ
=
p₁
+
p₂
+
p₃
3
−2
=
0 + (−2) + (−4)
3
3
×
sb
2
−
2
×
(
p₁
+
p₂
+
p₃
)
×
sb
+
(
p₁
×
p₂
+
p₁
×
p₃
+
p₂
×
p₃
)
=
0
3
×
(−0.8453)
2
−
2
×
(0
+
(−2)
+
(−4))
×
(−0.8453)
+
(0
×
(−2)
+
0
×
(−4)
+
(−2)
×
(−4))
=
0
Kₘₐₓ
=
−(
p₁
+
p₂
+
p₃
)
×
(
p₁
×
p₂
+
p₁
×
p₃
+
p₂
×
p₃
)
+
p₁
×
p₂
×
p₃
48 = −(0 + (−2) + (−4)) × (0 × (−2) + 0 × (−4) + (−2) × (−4)) + 0 × (−2) × (−4)
GM
=
Kₘₐₓ
K
3.2
=
48
15
Assumptions
•
G(s)H(s) = K ÷ ((s − p₁)(s − p₂)(s − p₃)) with no zeros: three branches, all off to infinity.
•
The three asymptotes leave the centroid σₐ = (p₁ + p₂ + p₃) ÷ 3 at 60°, 180° and 300°.
•
The breakaway is where dK/ds = 0, between the two right-hand poles.
•
All three poles are real numbers in s⁻¹, p₁ > p₂ > p₃, with p₁ at or left of 0. The closed loop is s³ + a₂s² + a₁s + a₀ + K = 0, with a₂ = −(p₁ + p₂ + p₃), a₁ = p₁p₂ + p₁p₃ + p₂p₃ and a₀ = −p₁p₂p₃.
Step-by-step
WE KNOW
p₁ = 0, p₂ = −2, p₃ = −4, K = 15
FIND
centroid (σₐ), breakaway point (sb), largest stable gain (Kₘₐₓ), gain margin (GM)
Step 1 · Find centroid (σₐ)
Use σₐ = (p₁ + p₂ + p₃) ÷ 3
The centre of the poles: their sum over the 3 asymptotes (3 poles, no zeros).
σₐ
=
0 + (−2) + (−4)
3
σₐ
=
−
6
3
σₐ = −2
Step 2 · Find breakaway point (sb)
Use 3 × sb² − 2 × (p₁ + p₂ + p₃) × sb + (p₁ × p₂ + p₁ × p₃ + p₂ × p₃) = 0
Set dK/ds = 0 and take the root between the two right-hand poles.
sb
=
((
p₁
+
p₂
+
p₃
)
+
√(
(
p₁
+
p₂
+
p₃
)
2
−
3
×
(
p₁
×
p₂
+
p₁
×
p₃
+
p₂
×
p₃
)))
÷
3
sb = ((0 + (−2) + (−4)) + √((0 + (−2) + (−4))² − 3 × (0 × (−2) + 0 × (−4) + (−2) × (−4)))) ÷ 3
sb
=
−6 +
√
12
3
sb
=
−6 + 3.4641
3
sb = −0.8453
Step 3 · Find largest stable gain (Kₘₐₓ)
Use Kₘₐₓ = −(p₁ + p₂ + p₃) × (p₁ × p₂ + p₁ × p₃ + p₂ × p₃) + p₁ × p₂ × p₃
Routh
on
s
3
+
a₂
s
2
+
a₁s
+
a₀
+
K
:
stable
while
a₂a₁
>
a₀
+
K
,
so
the
branches
cross
the
jω
axis
at
Kₘₐₓ.
Kₘₐₓ = −(0 + (−2) + (−4)) × (0 × (−2) + 0 × (−4) + (−2) × (−4)) + 0 × (−2) × (−4)
Kₘₐₓ = −(−2 + (−4)) × 8 + 0 × (−2) × (−4)
Kₘₐₓ = −(−2 + (−4)) × 8 + 0 × (−4)
Kₘₐₓ = −(−2 + (−4)) × 8
Kₘₐₓ = −(−6) × 8
Kₘₐₓ = 6 × 8
Kₘₐₓ = 48
Step 4 · Find gain margin (GM)
Use GM = Kₘₐₓ ÷ K
How
many
times
K
can
grow
before
the
poles
cross
into
the
right
half-plane.
GM
=
48
15
GM = 3.2
Check
Put the numbers back in. Both sides should match.
−2
=
0 + (−2) + (−4)
3
✓
3
×
(−0.8453)
2
−
2
×
(0
+
(−2)
+
(−4))
×
(−0.8453)
+
(0
×
(−2)
+
0
×
(−4)
+
(−2)
×
(−4))
=
0
✓
48 = −(0 + (−2) + (−4)) × (0 × (−2) + 0 × (−4) + (−2) × (−4)) + 0 × (−2) × (−4) ✓
3.2
=
48
15
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.