Root Locus로 damping ration가 정해졌을 경우, 목표점을 구하기
Control Systems Engineering 예제
Root Locus Design
Root Locus · Damping Ratio · Integral Compensator · Pole-Zero Compensation
1. Root Locus Design
G(s)
=
K
/
[(s+1)(s+2)(s+10)]
+ 1
R(s) ------->(Σ)---->[ K ]---->[ ----------- ]-----> C(s)
- ^ (s+1)(s+2)(s+10) |
| |
+--------------------------------------+
ζ = 0.174
2. 감쇠비와 목표 Pole
s
=
-σ
±
jωd
ζ
=
σ / ωn
ωn
=
√(σ2 + ωd2)
ζ = cos θ
Im
↑
|
| ● -σ + jωd
| /|
| / |
| / | ωd
| /θ |
------------+-----+--------------------→ Re
-σ
s
≈
-0.69
±
j3.92
3. Root Locus Angle Condition
θp1
+
θp2
+
θp3
=
180°
θp1
=
tan-1
[
ωd/(10-σ)
]
θp2
=
tan-1
[
ωd/(2-σ)
]
θp3
=
tan-1
[
ωd/(1-σ)
]
4. Magnitude Condition
|KG(s)| = 1
K
=
|
(s+10)(s+2)(s+1)
|
K ≈ 164
5. Integral Compensator
Gc(s)
=
K
(s+0.1)
/
s
GOL(s)
=
K
(s+0.1)
/
[
s(s+1)(s+2)(s+10)
]
+ K(s+0.1) 1
R(s) ------->(Σ)---->[ ------- ]---->[ ----------- ]-----> C(s)
- ^ s (s+1)(s+2)(s+10) |
| |
+---------------------------------------------+
6. Pole-Zero 배치
Re axis
×----------------×------×------○----×--------------→
-10 -2 -1 -0.1 0
× : Pole
○ : Zero
Pole:
s = 0
Zero:
s = -0.1
7. 보상 후 Gain
K
=
|
s(s+1)(s+2)(s+10)
/
(s+0.1)
|
K ≈ 163.6
보상 전 K ≈ 164
보상 후 K ≈ 163.6
보상 후 K ≈ 163.6
Root Locus, Damping Ratio, Integral Compensator,
Pole-Zero Compensation, Angle Condition, Magnitude Condition