北京理工大学810自动控制原理考研课件7

G(s)
Gc(s)
Gc(s)
+
G(s) H(s)
(d) Input compensation
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2. Phase-lead design using the root locus Adding a single zero moves root locus to the left and achieves the higher stability.
s 4.5 Gc ( s ) s 11.6 K v 22.7
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3. Phase-lag design using the root locus The compensator is s 1 sz Gc ( s) , 1 or z p 0 s 1 s p
2
A compensator is an additional component or circuit that is inserted into a control system to compensate for a deficient performance.
R(s) + Gc(s) H(s) (a) Cascade compensation R(s) + H(s) (c) Output compensation G(s) Y(s) R(s) + Gc(s) H ( s) (b) Feedback compensation Y(s) R(s) Y(s) G(s) Y(s)
If the compensator pole and zero appear relatively close together and near the origin of the s-plane compared to n, it can increase the error constant by the factor while altering the root location very slightly.
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j
180 90 (180 arctan 2) 2 p p 38
2 tan p 1 p p 3.6
p
-p -1 0

Therefore, the compensator is
s 1 Gc ( s ) s 3.6
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5.6 Compensation by Using Root Locus Method 1. Introduction The design of a control system is concerned with the arrangement, or the plan, of the system structure and the selection of suitable components and parameters. The alteration or adjustment of a control system in order to provide a suitable performance is called compensation.
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Root Locus 4 3 2 1
unstable
K G( s) 2 s ( s 2)
Imag Axis
0 -1 -2 -3 -4 -6 -5 -4 -3 -2 Real Axis -1 0 1 2
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j
0
20

1. Obtain the root locus of the uncompensated system. 2. Determine the transient performance for the system and locate suitable dominant root locations on the uncompensated root locus that will satisfy the specifications. 3. Calculate the loop gain at the desired root location and the system error constant. 4. Compare the uncompensated error constant with the desired error constant, and calculate the necessary increase that must result from the pole-zero ratio of the compensator. 5. Locate the compensator pole and zero near the origin of the s-plane in comparison to n.
Example 5.15 Lead compensator using root locus
The uncompensated loop transfer function is K1 L( s ) 2 s The specifications for the system are
Ts ( 2%) 4s, P.O. 35%
Chapter 5 Root Locus Method
5.1 5.2 5.3 5.4 5.5 The Root Locus Concept The Root Locus Procedure Examples for Drawing Root Locus Parameter Design by the Root Locus Method Relationship between Performance and the distributing of close-loop zeros and poles 5.6 Compensation by Using Root Locus Method 5.7 Summary 5.8 Three-term (PID) Controllers
Root Locus 4
stable
3
K ( s 0.5) G( s) 2 s ( s 2)
2
1
Imag Axis
0
-1
-2
-3
-4 -2 -1.8 -1.6 -1.4 -1.2 -1 Real Axis -0.8 -0.6 -0.4 -0.2 0
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Shortcomings
It is difficult to realize
The compensated loop transfer function is
K1 ( s 1) L( s) Gc ( s)G( s) 2 s ( s 3.6)
and at the root location
(2.23) 2 (3.25) K1 8.1 2
the acceleration constant is
n Kv
2 n T ( s) 2 2 s 2n s n
2 n n G( s) Kv s(s 2n ) 2
Select the real part of the desired roots as n=4, where n=8.89 for =0.45.
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Place the zero of the compensator at s=-4. We get
p 50 p 10.6
The compensated loop transfer function is
K ( s 4) L( s) Gc (s )G (s ) s( s 2)(s 10.6)
j
0

0Βιβλιοθήκη Baidu

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1. List the system specification and translate them into a desired root location for the dominant roots. 2. Sketch the uncompensated root locus, and determine whether the desired root location can be realized. 3. If a compensator is necessary, place the zero of the phase-lead network directly below the desired root location (or to the left of the first two real poles). 4. Determine the pole location so that the total angle at the desired root location is 180° and therefore is on the compensated root locus. 5. Evaluate the total system gain at the desired root location and then calculate the error constant. 6. If the error constant is not satisfactory, a Phase-lag compensator is needed. 10
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If the natural frequency of the roots is large, the velocity constant should be reasonably large.
The desired closed-loop system is It’s open-loop transfer function is If be constant,
Therefore P.O. e
100% 0.32 4 Ts ( 2%) n 1
1 2
n
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Choose a desired dominant root location as
0.45 r1, r1 1 j 2
Place the zero of the compensator directly below the desired root location at s=-1.
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Example 5.16 Lead compensator for a type-one system The uncompensated loop transfer function is K L( s ) s( s 2) The specifications for the system are 0.45 and Kv 20 The gain of the uncompensated system must be K=40, and the roots are 2 s 2s 40 ( s 1 j 6.25)( s 1 j6.25) 0.16
Gc (s) 1 Td s
The noise is amplified, especially, the higher frequency noise.
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Phase-lead network
s 1 s z Gc ( s) s 1 s p
j
1, or p z 0
K1 8.1 Ka 2.25 3.6 3.6
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A computer simulation of the compensated system is
Ts ( 2%) 3.8s, P.O. 46%
These values compare moderately well with the specified values of 35% and 4s.
and at the root location the velocity constant is
9(8.25)(10.4) K 96.5 8
K (4) Kv 18.2 2(10.6)
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The velocity constant is less than 20, so we should repeat the design procedure with a new n=10. We have