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Chapter5 Root Locus Method


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Chapter 5 Root Locus Method
5.1 The Root Locus Concept 5.2 The Root Locus Procedure 5.3 Ex

amples for Drawing Root Locus 5.4 Relationship between Performance and the distributing of closed-loop zeros and poles 5.5 Compensation by using Root Locus Method 5.6 Summary

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?Relative stability and transient performance are directly related to the location of the roots. ? It is necessary to adjust one or more system parameters to obtain suitable root locations. The above two facts make it worthwhile to determine how the roots migrate about the s-plane as the parameters are varied.

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The root locus method was introduced by Evans in 1948 and has been developed and utilized extensively in control engineering. ? It provides the engineer with a measure of the

sensitivity of the roots to a variation in the parameter. ? It can be used to great advantage in conjunction with the Routh-Hurwitz criterion. ? As it provides graphical information, an approximate sketch can be used to obtain qualitative information concerning the stability and performance of the system. ? If the root locations are not satisfied, the necessary parameter adjustments often can be readily ascertained from the root locus. So, the root locus method can be used to analyze and design a system.

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5.1 The Root Locus Concept
Definition of Root Locus The root locus is the path of the roots of the characteristic equation traced out in the s-plane as a system parameter is changed from 0 to infinity.

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C (s) G( s) ? R( s) 1 ? G ( s ) H ( s )

1 ? G( s ) H ( s ) ? 0

G(s) H (s) ? ?1
G( s) H ( s) ? K ? (s ? z j )
i ?1 m j ?1 n

? ?1

? ( s ? pi )

? s ? f (K )

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Example: A second-order system K G ( s) ? ; H ( s ) ? 1; s ( s ? 2)

?n C (s) K ? 2 ? 2 R( s ) s ? 2s ? K s ? 2?? n s ? ? n 2
2

s1, 2 ? ?1 ? 1 ? K

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Location of roots for the characteristic equation: K 0 0.5 0.75 1.0 2.0 3.0 50.0 S1 0 -0.3 -0.5 -1.0 -1.0+j1.0 -1.0+j1.4 -1.0+j7.0 S2 -2 -1.7 -1.5 -1.0 -1.0-j1.0 -1.0-j1.4 -1.0-j7.0

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Analysis:

K ?? ? ?;

? n ?; ? d ?; ? ?
Ka The root locus is a vertical line for K >Ka(1.0) .

return

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5.2 The Root Locus Procedure
5.2.1 Magnitude and Angle Requirement 5.2.2 Rules for drawing root locus

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5.2.1 Magnitude and Angle Requirement

Root locus (negative feedback)

G(s) H (s) ? ?1 1 ? G( s ) H ( s ) ? 0 magnitude requirement

G(s) H (s) ? 1
angle requirement

argG( s) H ( s) ? ?G( s) H ( s) ? (2k ? 1)? , k ? 0,?1,?2,.....

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Note: When plotting the root locus, only the angle requirement is the sufficient and necessary condition (because s is vary with K ). ? The magnitude requirement is used to determine K for a given root s1. ? The angle requirement is used to verify a test point s1 as a root location.

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K (s ? z1 )?( s ? zm ) G( s) H ( s) ? (s ? p1 ) ?( s ? pn )
? G(s) H (s) ? ?K? K s ? z1 ? s ? z m s ? p1 ? s ? pn ?1

s ? p1 ? s ? pn s ? z1 ? s ? z m

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K (s ? z1 )?( s ? zm ) G( s) H ( s) ? (s ? p1 ) ?( s ? pn )

? argG( s) H ( s) ? ? ?( s ? z j ) ?? ?( s ? pi )
j ?1 i ?1

m

n

? (2k ? 1)? , k ? 0,?1,?2,.....

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j?

si
?1

p1
?3
?2

z1
?4

z2
p2

0

?

?1 ? ? 2 ??3 ?? 4 ? (2k ? 1)? , k ? 0,?1,?2,..... return

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5.2.2 Rules for Drawing Root Locus
Rule 1 : The Symmetry of the Root Locus The root loci must be symmetrical with the respect to the horizontal real axis because complex roots must appear as pairs of complex conjugate roots.

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Rule 2: The Starting Points and the End Points of the Root Locus The locus begins at the open-loop poles (the closed-loop poles for K=0), and ends at the open-loop zeros (the closed-loop zeros for K=?). The number of branches approaching to infinity is n-m.

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Proving:
? G ( s) H ( s) ? K ? ( s ? z j ) z j ? ? open ? loop zero j ?1 ? ( s ? pi )
i ?1 n m

pi ? ? open ? loop pole
? ( s ? pi ) ? K ? ( s ? z j ) ? 0
i ?1 j ?1 n m

1 ? G ( s) H ( s) ? 0

At the starting point of the root locus: K=0 ?? (s ? pi ) ? 0, s ? pi ; (i ? 1, 2, ?, n)

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At the end point of the root locus: K ? ? and the characteristic equation can be written as m 1 n ? ( s ? pi ) ? ? ( s ? z j ) ? 0 j ?1 K i ?1

when

K ??

s ? zj

( j ? 1, 2, ?, m)

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Rule 3:The Root Locus on the Real Axis The root locus on the real axis has odd number of poles and zeros on its right, complex poles or zeros and poles and zeros on its left have no effect.

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It can be ascertained by the angle criterion.

s1 ? p3

j

p2

180 ?

j 180 ?

p3

p2 s1 p1 s1 ? p1 s1 ? p2

p4

s1

p1

p3

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Rule 4 : The Asymptotes of the Root Locus The loci proceed to the zeros at infinity along asymptotes centered at ? a and with angles ? . The point on the ? pi ? ? z j ?a ? real axis given by n?m The Angle of the asymptotes with respect to the real axis is
2k ? 1 ?? ? n?m (k ? 0, 1, ?, n ? m ? 1)

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j
60 ?

?a

?2

?1

n?m 0 ? (?1) ? (?2) ? ? ?1 3?0

? pi ? ? z j ?

0
2k ? 1 ?? ? (k ? 0, 1, ?, n ? m ? 1) n?m ?? / 3; k ? 0 ? ? ?? ; k ? 1 ?? ? / 3; k ? ?1 ?

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Rule 5 : The Breakaway Point of the Root Locus The root locus left the real axis at a breakaway point. The breakaway point on the real axis can be evaluated graphically or analytically. The locus leaves the real axis where there are a multiplicity of roots, typically two.

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j
j
60 ?

?2

?1

0

?2

?1

0

Breakaway point

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Assume the breakaway point s=d:
(1) d K ds ( GH ( s) ? s ?d ? 0
n

K ? (s ? z j )
j ?1

m

? ( s ? pi )
i ?1

n

? ?1 ;

N (s) ?K ? ? ?? ) D( s) ? (s ? z j )
i ?1 m j ?1

? ( s ? pi )

? N ( s) D' ( s ) ? N ' ( s ) D( s) ? 0
n 1 1 (2) ? ?? j ?1 d ? z j i ?1 d ? pi m

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j

K GH ( s) ? s( s ? 1)

-1

-0.5

0

d d (1) K s ?d ? [? s( s ? 1)]s ?d ? ?(2d ? 1) ? 0 ds ds ? d ? ?0.5
n 1 1 1 1 (2) ? ?0?? ? ? d d ?1 j ?1 d ? z j i ?1 d ? pi m

? d ? 1 ? d ? 0 ? d ? ?0.5

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K GH ( s) ? s( s ? 1)(s ? 2)
d1 ? ?1.577 ( to be given up ) d 2 ? ?0.423 ( breakaway point )

K ( s ? 2) GH ( s) ? s( s ? 1)
d1 ? ?0.586 (breakaway point ) d 2 ? ?3.414 ( breakaway point )

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Rule 6:The angle between the direction of emergence (or entry) of q coincident poles (or zeros) on the real axis(根轨迹 离开或进入实轴上q重极点(或零点)方 向之间的夹角)

(2k ? 1)? ?? q

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j

K GH ( s) ? 2 s ( s ? 1)

-1

0

K GH ( s) ? 3 ( s ? 1)

3 poles
-1

j

0

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Rule 7:The Angle of departure of the locus from Complex Poles and The Angle of arrival at Complex Zeros(using the phase criterion) The angle of departure from complex poles is given by (2k+1)π –Σ(angles of the vectors from all other open-loop poles to the poles in question) + Σ(angles of the vectors from the open-loop zeros to the complex pole in question).
? p ? (2k ? 1)? ? ? ?( pi ? p j ) ? ? ?( pi ? z j )
i

n

m

j ?1 j ?i

j ?1

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arg G ( s ) H ( s ) ? ? ?( s ? z j ) ?? ?( s ? pi ) ? (2k ? 1)?
j ?1 i ?1 s ? pi

m

n

lim arg G ( s ) H ( s ) ? ? ?( pi ? z j ) ?? ?( pi ? p j ) ? ? pi ? (2k ? 1)?
j ?1 n j ?1 j ?i m

m

n

? ? pi ? (2k ? 1)? ? ? ?( pi ? p j ) ? ? ?( pi ? z j )
j ?1 j ?i j ?1

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The angle of arrival at a complex zero may be found from the same rule and then the sign changed to produce the final result.
? z ? (2k ? 1)? ? ? ?( zi ? z j ) ? ? ?( zi ? p j )
i

m

n

j ?1 j ?i

j ?1

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arg G ( s ) H ( s ) ? ? ?( s ? z j ) ?? ?( s ? pi ) ? (2k ? 1)?
j ?1 i ?1

m

n

lim arg G ( s ) H ( s ) ? ? ?( zi ? z j ) ?? ?( zi ? p j ) ? ? zi ? (2k ? 1)?
s ? zi j ?1 j ?i m j ?1 n

m

n

? ? zi ? (2k ? 1)? ? ? ?( zi ? z j ) ? ? ?( zi ? p j )
j ?1 j ?i j ?1

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K ( s ? 5) K ( s ? 5) GH ( s) ? ? 2 s( s ? 4s ? 8) s( s ? 2 ? j 2)(s ? 2 ? j 2)
? ?2? j 2 ? 180? ? (?(?2 ? j 2 ? 0) ? ?((?2 ? j 2) ? (?2 ? j 2))
? ?((?2 ? j 2) ? (?5)) 2 ?1 4 ?1 2 ? 180? ? tg ? tg ? tg ?2 0 3 ? 180? ? 135? ? 90? ? 33? ? ?12?
?1

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Rule 8 : The Root Locus Crossing the Imaginary Axis The point where the locus crosses the imaginary axis, we have s=jω into the characteristic equation and solving for ω.

1 ? GH ( s) ? 0 ? Re[1 ? GH ( j? )] ? 0 Im[1 ? GH ( j? )] ? 0

1 ? GH ( j? ) ? 0

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K ( s ? 5) G( s) ? s( s 2 ? 4s ? 8)

s3 ? 4s 2 ? s( K ? 8) ? 5K ? 0

5 K ? 4? 2 ? 0 ( K ? 8) ? ? 2 ? 0

(5K ? 4? ) ? j[(K ? 8)? ? ? ] ? 0
2 3

K ? 32 ? ? ?6.32

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Note The actual point at which the root locus crosses the imaginary axis can also be readily evaluated by utilizing the Routh criterion.

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Rule 9:The sum of the closed-loop poles If there are at least two more open-loop poles than open-loop zeros, the sum of the closed-loop poles is constant, independent of K, and equal to the sum of the open-loop poles.
(如果开环极点比开环零点至少多2个,闭环极点 的和为一不依赖于 K 的常数,且等于开环极点的 和。)

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Rule 10:The gain at a selected point s1 on the locus

The gain at a selected point s1 on the locus is obtained by joining the point to all open-loop poles and zeros and measuring the length of each line st ? pi , st ? z j . The gain is n given by ?s? p
K? ? s ? zj
j ?1 i ?1 m i

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At the breakaway point s=-2.6, Gain K is

K?

? s ? pi ? s ? zj
j ?1 i ?1 m s ? ?2.6

n

?

? 2.6 ? 2 ? 2.6 ? 3 ? 2.6 ? 1 ? j 2 ? 2.6 ? 1 ? j 2

? 0.04

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Summary of the Root Locus plotting rules (steps * are optional): Step 1: Mark the poles and zeros with “x” and “o” on the complex plane. Step 2: Draw the locus on the real axis to the left of odd number of real poles and zeros. Step 3: Draw n-m asymptotes centered at ? a and leaving at angles ? .
?a

? p ??z ?
i

j

n?m

2k ? 1 ?? ? n?m

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Step 4: Compute departure angles from poles “x” and arrival angles at zeros “o”. Step 5*: Using s=jw find the locations where the locus crosses the imaginary axis. Step 6*: Find any multiple roots on the locus by solving dK ( s )

ds

?0

the solutions s0 are the poles multiple of locations for s. If these roots are on the real axis, these are either break away or break-in points. Step 7: Complete the root locus using the facts developed in steps 1-6.

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return

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5.3 Examples for drawing root locus
Example 1: Plot the root locus of the following system

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Step one: find the poles and zeros of GH and plot them.

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Step two: draw locus on the real axis

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*Step three: calculate asymptote angles and center of asymptotes:

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*Step four: compute departure angles at poles and arrival angles at zeros *Step five: find locations on the imaginary axis

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Step six: find multiple roots, especially on the real axis.

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Q: How can we determine the value of the gain K for a particular pole location?

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K ( s ? 2) Example: G0 ( s) ? s( s ? 3)(s 2 ? 2s ? 2)

Drawing root locus
1、p1 ? 0, p2 ? ?3, p3, 4 ? ?1 ? j , z1 ? ?2 2、realaxis: (??,?3), (?2,0) 0 ? (?3) ? (?2) ? (?2) 3、? ? ? ?1 3 (2k ? 1)? ? ? ?? ? ? ,? , 3 3 3 4、? p3 ? 450 ? (1350 ? 900 ? 26.60 ) ? 1800 ? ?26.60

? p 4 ? 26.60

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5、s( s ? 3)(s ? 2s ? 2) ? K ( s ? 2) ? 0
2

? s ? 5 s ? 8 s ? (6 ? K ) s ? 2 K ? 0
4 3 2

令s ? j? ?? ? 0 ?? ? ?1.61 ? ? 4 ? 8? 2 ? 2 K ? 0 ?? ,? ? 3 ?? 5? ? (6 ? K )? ? 0 ?K ? 0 ? K ? 7

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Root Locus

3

2

?26.60
1

Imag Axis

0

-1

-2

-3 -5 -4 -3 -2 Real Axis -1 0 1

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Root Locus

3

2

1

Imag Axis

0

-1

System: sys Gain: 6.98 Pole: -0.00305 - 1.61i Damping: 0.00189 Overshoot (%): 99.4 Frequency (rad/sec): 1.61

-2

-3 -5 -4 -3 -2 Real Axis -1 0 1

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K Example: G0 ( s) ? s( s ? 1)(s ? 2)

Drawing root locus
1、p1 ? 0, p2 ? ?1, p3 ? ?2 2、realaxis: (?1,0), (??,?2) 0 ? (?1) ? (?2) 3、? ? ? ?1 3 (2k ? 1)? ? ? ?? ? ? ,? , 3 3 3 4、breakawayp int : d1 ? ?0.423, d 2 ? ?1.577(?) o

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5、s ( s ? 1)(s ? 2) ? K ? 0 ? s ? 3s ? 2s ? K ? 0
3 2

令s ? j? ? ? j? ? 3? ? j 2? ? K ? 0
3 2

? K ? 3? 2 ? 0 ? ? ? ? 2 ? ?1.414, K ? 6 ? 3 ?? ? ? 2? ? 0

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Root Locus

4 3 2 1 System: sys Gain: 0.385 Pole: -0.423 - 3.37e-009i Damping: 1 Overshoot (%): 0 Frequency (rad/sec): 0.423

Imag Axis

0 -1 -2 -3 -4 -6 -5 -4 -3 -2 Real Axis -1 0 1

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Root Locus

4 3 2 1

System: sys Gain: 5.97 Pole: 0.000569 + 1.41i Damping: -0.000404 Overshoot (%): 100 Frequency (rad/sec): 1.41

Imag Axis

0 -1 -2 -3 -4 -6 -5 -4 -3 -2 Real Axis -1 0 1

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Generalized root locus 1. Zero-degree root locus
System contains inner positive-feedback- loop

2. Parameter root locus
Equivalent unity feedback transform

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Zero-degree root locus The close-loop characteristic equation:

G( s) H ( s) ? 1 1 ? G( s ) H ( s ) ? 0 magnitude requirement

G(s) H (s) ? 1
angle requirement

No change!

argG( s) H ( s) ? ?G( s) H ( s) ? 2k? , k ? 0,?1,?2,..... Changed!

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Rules for drawing zero-degree root locus
1. Number of root locus branches is the number of characteristic roots. No change!

2. The symmetry of root locus.
3. The start point and end point.

No change! No change!

4. The locus on the real axis has even number of real poles and zeros on its right. Changed!
5. The Asymptotes of the Root Locus .

?a

? p ??z ?
i

j

n?m

2k ?? ? (k ? 0, 1, ?) n?m

No change!

Changed!

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6. The breakaway point of the root locus

No change
7. The angle of departure of the locus from complex poles and the angle of arrival at complex zeros.

? p ? 2k? ? ? ?( pi ? p j ) ? ? ?( pi ? z j )
i

n

m

j ?1 j ?i

j ?1

? z ? 2k? ? ? ?( zi ? z j ) ? ? ?( zi ? p j )
i

m

n

Changed

j ?1 j ?i

j ?1

8. The root locus crossing the imaginary axis.

No change

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K Example: G0 ( s) ? s( s ? 1)(s ? 2)

Drawing the zero-degree root locus
1、s1 ? 0, s2 ? ?1, s3 ? ?2 2、realaxis: (0,??), (?2,?1) 0 ? (?1) ? (?2) 3、? ? ? ?1 3 2k? 2? 2? ?? ? 0, ,? 3 3 3 4、breakawayp int : d1 ? ?0.423(?), d 2 ? ?1.577 o

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Root Locus

5 4 3 2 1 System: sys Gain: 0.385 Pole: -1.58 Damping: 1 Overshoot (%): 0 Frequency (rad/sec): 1.58

Imag Axis

0 -1 -2 -3 -4 -5 -4 -3 -2 -1 0 1 2 3 4 5 Real Axis

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10(1 ? K h s) Example: G0 ( s) ? s( s ? 2)

Drawing the parameter root locus.
chracteris equation: s 2 ? 2s ? 10K h s ? 10 ? 0 tic ? equivalentopen- loop transfer: 10K h s G 0 (s) ? 2 ? ?1 s ? 2s ? 10 1、z1 ? 0, p1, 2 ? ?1 ? j 3
'

2、realaxis: (??,0) 3、breakawayp int : d1 ? 3.16(?), d 2 ? ?3.16 o 4、? p1 ? 198.40 ? ?161.60

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Root Locus 3

?161.60

2 System: sys Gain: 0.432 Pole: -3.16 Damping: 1 Overshoot (%): 0 Frequency (rad/sec): 3.16

1

Imag Axis

0

-1

-2

-3 -5 -4 -3 Real Axis -2 -1 0

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Root Locus 3

2

1

Imag Axis

0

-1

-2

System: sys Gain: 0.116 Pole: -1.58 - 2.74i Damping: 0.5 Overshoot (%): 16.3 Frequency (rad/sec): 3.16

-3 -5 -4 -3 Real Axis -2 -1 0

return

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5.4 Relationship between Performance and the distributing of closed-loop zeros and poles 一、闭环极点决定阶跃响应的类型 1、闭环零、极点表示的阶跃响应表达式(零极点形 式)
K C ? (s ? z j )
j ?1 m

GC ( s) ?

? (s ? s )
i i ?1

n

Y ( s) ? R( s )

1 R(s) ? s

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Y ( s ) ? GC ( s ) R( s ) ?

KC ? (s ? z j )
j ?1

m

? (s ? s )
i i ?1

n

1 ? s

A0 An A1 ? ? ??? s s ? s1 s ? sn A0 n Ai ? ?? s i ?1 s ? si

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式中:
A0 ? K C ? (s ? z j )
j ?1 m n s ?0

? (s ? s )
i i ?1

?

K C ? (? z j )
j ?1

m

? (? s )
i i ?1

n

Ai ?

K C ? (s ? zi ) s? ( s ? s i )
i ?1 i ?1 n

m

(s ? s j )

s?s j

?

K C ? (s j ? zi ) s j ? ( s j ? si )
i ?1 i? j i ?1 n

m

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L :y(t ) ? L [Y (s)] ? A0 ? ? A j e
?1 ?1 j ?1

n

s jt

结论:y(t)与zi和sj分布有关。 2、闭环零极点分布与阶跃响应之间的定性关系 1)闭环极点在S左半平面,则系统稳定。负实数, 则单调衰减;负实部共轭复根,则振荡衰减。 2)闭环极点远离虚轴,则相应分量衰减快,快 速性好。

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3)闭环复极点应配置在S平面中与负实轴 成450夹角线附近,则系统平稳性好(最 佳阻尼比)。 4)闭环极点之间的距离要大。 5)闭环零点应靠近离虚轴近的闭环极点, 用以抵消极点对瞬态响应过程的影响, 改善快速性。 6)原则性结论,为利用闭环零极点直接对 系统动态过程性能进行定性分析提供依 据。

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二、闭环零点对阶跃响应的影响 三、除闭环主导极点以外的其他极点对阶跃响应 的影响 1、主导极点:距离虚轴最近;附近没有闭环零 点;其它闭环极点距虚轴较远(5~10倍)。对 系统动态过程性能影响最大,起主导作用。 2、利用主导极点的概念,可使系统分析简化。 在高阶系统设计中,常希望系统具有一对共轭 复数主导极点,对这类高阶系统的分析,可利 用二阶系统时域分析的结论。 3、利用主导极点估算系统的性能指标 4、非主导极点对阶跃响应的影响:工程上认为 忽略不计。离虚轴越远,影响越小。

Automatic Control Theory Chapter 5

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四、闭环偶极子对阶跃响应的影响 1、偶极子:一对靠得很近的零点、极点。 2、若偶极子间的距离<1/10到其它零极 点的距离,则它们对阶跃响应的影响可 忽略。 3、可在系统中有意识加入适当的零点, 以抵消对动态过程影响较大的不利闭环 极点,使系统的动态性能获得改善。 4、举例。 return

Automatic Control Theory Chapter 5

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5.5 Compensation by using Root Locus Method

r(t) + e(t) Gc(s) -

u(t)
Gp(s)

y(t)

H(s)
compensator

H(s)=1

Plant

Automatic Control Theory Chapter 5

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Adding a single zero moves root locus to the left achieves the higher stability. Note: location of the zero.

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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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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

Automatic Control Theory Chapter 5

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Root Locus 4

stable

3

2

K ( s ? 1.5) G( s) ? 2 s ( s ? 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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Root Locus 10 8 6 4 2

unstable

K ( s ? 3) G( s) ? 2 s ( s ? 2)

Imag Axis

0 -2 -4 -6 -8 -10 -3 -2.5 -2 -1.5 Real Axis -1 -0.5 0

Automatic Control Theory Chapter 5

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Shortcomings

Difficult to realize

The noise is amplified, esp. the higher freq. noise.

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Compensator

Pc be far away from the origin. Effect of the pole of the compensator on the root locus can be neglected 1. Expected dominant pole:

Automatic Control Theory Chapter 5

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(not on the original root locus) The angle condition:

2. Phase lead angle provided by Compensator:

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3. Proper choices of pc and zc assure the right phase lead angle.

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4. Selection of pc and zc (1) Infinite number of solutions for a required phase lead angle

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Different pairs (pc/zc) correspond to different open loop gain.

Relation between pc(zc) and OL gain
Before compensation

After the compensation

? Kv

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(2) Different cases:

When pc( zc) moves to the left

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Obtaining pc( zc) with the largest OL gain

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4 Example: Given G p ( s) ? s( s ? 2)

Design a compensator such that the desired CL poles have

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Solution: (i) Uncompensated system analysis

Automatic Control Theory Chapter 5

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(ii) Desired CL dominant poles

CL poles are not on the desired places. Adjusting K only cannot obtain the right CL poles sd is on the left to the original root locus. Phase lead comp. is required

Automatic Control Theory Chapter 5

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(iii) Calculate the required phase lead angle

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(iv) Determination of pc and zc

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By bisector method

(v) Determine OL gain
4( s ? 2.93) G p ( s)Gc ( s) ? s( s ? 2)(s ? 5.46)

Automatic Control Theory Chapter 5

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(vi) Check of the compensated system

CL poles:

The resulted closed loop system meets the prospect requirements.

Automatic Control Theory Chapter 5

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5. Procedure (for reference only) (i) Select desired dominant poles from
(ii) Calculate the required phase lead angle from the dominant poles. (iii) Calculate pc and zc from required phase lead angle. (iv) Determine the open loop gain. (v) System performance check and adjustment.

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Feature: ?Reduces the steady-state error. ?Moves the root locus to the left. ?Slows down the response speed.

Automatic Control Theory Chapter 5

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6. Improvements in Steady-State Error Type 1 system (as example)

Automatic Control Theory Chapter 5

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Let then
G p ( s)Gc ( s) ?
m

K ? ( s ? zi ) s? ( s ? p j )
j ?1 i ?1 n

m

s ? zc ? s ? pc

K ?
' v

K ? zi
i ?1

?p
j ?1

n

?

zc pc

? Kv Kc

j

Automatic Control Theory Chapter 5

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Steady-state error coefficient will increase by a factor Kc.

Kc ?
Selecting Pole and Zero

zc pc

(1) pc、 zc should be very close to each other, so root locus shifts very little.

Automatic Control Theory Chapter 5

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Automatic Control Theory Chapter 5

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pc?zc , The root locus will not change. pc?zc Minor change and dominant poles shift little only. pc?zc can guarantee that the root locus is almost unchanged in the areas around the dominant poles.

Automatic Control Theory Chapter 5

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pc、zc should be very close to the origin.

pc<<zc

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Example: Given

Design a phase lag compensator, the compensated system: Kv ? 5?1 s , sd unchanged.

Solution: (i) Uncompensated system analysis

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(ii) Required additional gain

Automatic Control Theory Chapter 5

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A lag compensator is required for raising Kv and attaining sd unchanged.

(iv) Check of the compensated system

Kv ? 5.3 ? 5s ?1

return

Automatic Control Theory Chapter 5

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5.6 Summary
?The relationship between performance and roots location ?The concept of Root Locus ?The RL drawing procedure ?The generalized RL(zero-degree and parameter RL) ?A powerful tool for the analysis and design of control system return

Automatic Control Theory Chapter 5

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The relative stability and the transient response performance of a closed-loop control system are directly related to the location of the closed-loop roots of the characteristic equation. ? The root locus method is utilized to investigate the movements of the characteristic roots on the splane as the system parameters are varied. ? The root locus method is a graphical technique and can be used to obtain an approximate sketch in order to analyze the initial design of a system and determine suitable alterations of the system structure and the parameter values.

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? A computer is commonly used to calculate several accurate roots at important points on the locus. ? The utilization of the root locus method has been extended to the design of several parameters for a closed-loop control system. ? The sensitivity of the characteristic roots was investigated for undesired parameter variations by defining a root sensitivity measure. ? For the design of modern control systems, the root locus method will continue to be one of the most important procedures of control engineering.

return

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The End of Chapter 5


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