自动控制原理Lecture7

Rule 10: Consider a point s on the root locus close to open loop pole pj .
Root Locus Diagrams - Rules 10 and 11
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Control III Lecture 7
Now apply the angle criterion Arg (s − pj ) = π +
Control III Lecture 7
3016 - Control III Lecture 7
Learning Objectives Root Locus Diagrams Imaginary-axis Crossings RLDS - Rules 10 and 11 Example Pole Placement Controller Summary
Learning Objectives Root Locus Diagrams Imaginary-axis Crossings RLDS - Rules 10 and 11 Example Pole Placement Controller Summary
Example 1 from Previous Lecture
Imaginary-axis Crossings from the Routh Array
Slide 4 of 24
Control III Lecture 7
If n − m > 2, at least one root locus crosses the imaginary axis. This is because there will be at least 3 open loop zeros at ∞ and at least one root locus asymptote must make an angle 0 ≤ θ < π/2 with the imaginary axis. We can find the imag axis crossings by determining the CLTF denominator polynomial D(s) + KN (s) We set s = iω and look for real solutions for ω and K . We can also use Routh’s test but this won’t be considered in this course.
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Example Pole Placement Controller Summary
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Root Locus Diagrams - Rules 10 and 11
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Control III Lecture 7
Imaginary-axis Crossings RLDS - Rules 10 and 11 Example Pole Placement Controller Summary
Root Locus Diagrams
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Control III Lecture 7
We’ve seen the following properties of RLDs:
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Control III Lecture 7
Now lets find the intersections with the imaginary axis. The closed-loop transfer function has denominator Q(s) = s3 + 8s2 + 32s + K (make sure you know how to derive this!). To look for imag axis crossings, let s = iω , then we must have 0 = −8ω 2 + K (real part) 0 = −ω 3 + 32ω (imag part) We thus have the 2 solutions √ω = 0, K = 0, the open loop pole at the origin, and ω = 32, K = 256. Thus the CL system is stable for 0 < K < 256.
Learning Objectives Root Locus Diagrams Imaginary-axis Crossings RLDS - Rules 10 and 11 Example Pole Placement Controller Summary
Example 1 from Previous Lecture
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Control III Lecture 7
The negative real axis is on the locus since the single pole at s = 0 is to the right of any point on the negative real axis (Rule 6). The asymptotes intersect the real axis at the point (0+(-4+4i)+(-4-4i)-0)/3= -8/3 (Rule 7). Any breakaway points must satisfy (since N (s) = 1) d 0 = D(s) = 3s2 + 16s + 32. ds This has no real solutions, which means there are no real axis breakaway points (Rule 9).
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Control III Lecture 7
The RLD is :
Root Locus 15
Learning Objectives Root Locus Diagrams
10 s = 5.66i 5 Imaginary Axis
Imaginary-axis Crossings RLDS - Rules 10 and 11
The angle of departure of a locus from a pole and the angle of arrival at a zero can be determined as follows. The angle made by a locus at an open loop pole pj is given by ∑ ∑ ϕj = π + Arg(pj − zk ) − Arg(pj − pk ).
Learning Objectives Root Locus Diagrams Imaginary-axis Crossings RLDS - Rules 10 and 11 Example Pole Placement Controller Summary
One more property will be examined - the angle of departure of a locus from an open loop pole, or arrival at an open loop zero (Rules 10 and 11).
k̸=j k̸=j
Pole Placement Controller Summary
Root Locus Diagrams - Rules 10 and 11
Slide 10 of 24
Control III Lecture 7
Learning Objectives
pj
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φ
Root Locus Diagrams Imaginary-axis Crossings RLDS - Rules 10 and 11 Example Pole Placement Controller Summary
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Control III Lecture 7
An example : Consider a unity feedback system with 1 G(s) = . s[(s + 4)2 + 16] There are three open loop poles, at s = 0, −4 ± 4i. Therefore there are 3 loci (Rule 1). The three loci start at these poles (Rule 3). There are no finite open loop zeros, so no loci end at finite zeros (Rule 4). There are n − m = 3 open loop zeros at ∞, and these zeros at infinity mean there are 3 asymptotes, at angles π, ±π/3 (Rule 5).
Where loci start and finish (Rules 3 and 4); Asymptotes for loci - angle and intersection point on the real axis (Rules 5 and 7); Parts of the real axis that are on loci (Rule 6); Breakaway points of loci from the real axis (Rule 9); Intersection of loci with the imaginary axis (to finish today).
Learning Objectives Root Locus Diagrams Imaginary-axis Crossings RLDS - Rules 10 and 11 Example Pole Placement Controller Summary
Example 1 from Previous Lecture
Wednesday, 16 March, 2001
Learning Objectives
Slide 2 of 24
Control III Lecture 7
Learning Objectives Root Locus Diagrams
Imaginary-axis crossings in Root Locus Diagrams (RLDs). Rules 10 and 11 for construction of RLDs. Using RLDs for control system design: Pole placement controllers.
k k̸=j
Learning Objectives Root Locus Diagrams Imaginary-axis Crossings RLDS - Rules 10 and 11 Example
The angle made by a locus at an open loop zero zj is given by ∑ ∑ ψj = π + Arg(zj − zk ) − Arg(zj − pk ).
Learning Objectives Root Locus Diagrams Imaginary-axis Crossings RLDS - Rules 10 and 11 Example Pole Placement Controller Summary
Example 1 from Previous Lecture
m ∑ k=1
Learning Objectives
Arg (s − zk ) −
Fra Baidu bibliotek
∑
k̸=j
Arg (s − pk ) .
Root Locus Diagrams Imaginary-axis Crossings RLDS - Rules 10 and 11 Example