7_6_状态反馈中的稳态误差分析

State Space: Analyze and Design

CHAPTER 7

By Hui Wang hwang@http://www.77cn.com.cn FTP:10.13.21.29/168

Outline of Chapter 7Review and Introduction Controllability and Observability Linear transformation and Canonical forms State Feedback for SISO System State-Variable Feedback: steady-State Error Analysis State Observer for SISO System………..2

Outline of Section 5State-Variable Feedback: steady-State Error AnalysisBasic concept Step input: r(t)=R0u-1(t), R(s)=R0/s Ramp input: r(t)=R1u-2(t), R(s)=R1/s2 Parabolic input: r(t)=R2u-3(t), R(s)=R2/s3 Use of steady-state error coefficients

State feedback: steady-State Error Analysis 1. Basic concept A steady-state analysis of a stable system is an important design consideration (see Chap. 6 of this textbook). The three standard inputs are used (step, ramp and parabola) for a steady-state analysis of a state-variable-feedback control system. In analysis following, the phase-variable representation is used, in which the necessary values of the feedback coefficients are determined. And the results only apply when phase variables are used.

State feedback: steady-State Error Analysis2. Step input: r(t)=R0u-1(t), R(s)=R0/s Noted that developments are based on phase-variable representation, therefore, view back to P440, Eq.(13.76), the statevariable-feedback control system ratio isK G ( s w+ c w 1s w 1++ c1s+ c0 ) Y ( s)= R( s ) s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2++ (a0+ K G k1 ) K G ( s w+ c w 1s w 1++ c1s+ c0 )+ (a0+ K G k1 )

(13.76)

Y ( s)= R( s) R(s)= R0 s

s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2+

Applying the final-value theoremK G c0 R0= a0+ K G k1

For zero steady-state error

y (t ) ss

R0=

K G c0 R0 a k1= c0 0 a0+ K G k1 KG

where c0≠0

(13.80)5

State feedback: steady-State Error Analysis2. Step input: r(t)=R0u-1(t), R(s)=R0/s Thus, zero steady-state error with a step input can be achieved with state-variable-feedback, even if G(s) is Type 0. For an all-pole plant (i.e. the system in which G(s) has no zeros) as followingK G c0 Y (s) G(s)== n R ( s ) 1+ G ( s ) H ( s ) s+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2++ (a0+ K G k1 )

In such an all-pole plant c0=1, and for a Type 1 or higher plant, a0=0, in that case, the required value is k1=1. Therefore, the feedback coefficient k1 is determined by the plant parameters once zero steady-state error for a step input is specified. This specification is maintained for all state-variable-feedback control system designs throughout the remainder of this text.6

State feedback: steady-State Error Analysis2. Step input: r(t)=R0u-1(t), R(s)=R0/s

From Eq.(13.76), the system characteristic equation iss n+ (a n 1+ K G k n ) s n 1+ ( a n 2+ K G k n 1 ) s n 2++ ( a0+ K G k1 )= 0

(13.81)

A necessary condition for all clo

sed-loop poles of Y(s)/R(s) to be in the LH s plane (stable system) is that all coefficients in Eq.(13.81) must by positive. (Why?) So, constant terma0+ K G k1= K G c0

must be positive.

State feedback: steady-State Error Analysis3. Ramp input: r(t)=R1tu-2(t), R(s)=R1/s2 IF the input of the state-variable-feedback control system is ramp or parabolic, the error function E(s) must first determined. From the Eq.(13.76), the state-variable-feedback control system output isY ( s)= R( s) s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2+ K G ( s w+ c w 1s w 1++ c1s+ c0 )+ (a0+ K G k1 )

Then the error function E(s) isE ( s)= R( s) s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2+ s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2++ (a0+ K G k1 ) K G ( s w+ c w 1s w 1++ (a0+ K G k1 )+ c1s+ c0 )

(13.82)

k1= c0 E ( s)= R( s)

a0 KGs n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2++ (a1+ K G k 2 ) K G ( s w+ c w 1s w 1++ (a0+ K G k1 )+ c1s )

s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2+

(13.83)8

State feedback: steady-State Error Analysis3. Ramp input: r(t)=R1tu-2(t), R(s)=R1/s2 For a ramp input, the steady-state error isR (s)= R1 s2

(Applying the final-value theorem)e(t ) ss= lim sE ( s )=s→0

a1+ K G k 2 K G c1 R1 K G c0

(13.84)

In order to obtain e(t)ss=0, it is necessary thate(t ) ss= lim sE ( s )=s→0

a1+ K G k 2 K G c1 R1= 0 K G c0

a1+ K G k 2 K G c1= 0a1 ) KG

For an all-pole plant c0=1 and c1=0; thus k2=-a1/KG, in that cases n+ (a n 1+ K G k n ) s n 1+ ( a n 2+ K G k n 1 ) s n 2+

k 2= c1 (

(13.85)

+ ( a1+ K G k 2 ) s+ ( a0+ K G k1 )= 0 (13.81)

=0

State feedback: steady-State Error Analysis3. Ramp input: r(t)=R1tu-2(t), R(s)=R1/s2 We can see, the coefficient of s in the system characteristic equation (13.81) is zero, and the system is unstable. Therefore, zero steady-state error for a ramp input cannot be achieved by state-variable-feedback for an all pole stable system.s n+ (a n 1+ K G k n ) s n 1+ ( a n 2+ K G k n 1 ) s n 2++ ( a1+ K G k 2 ) s+ ( a0+ K G k1 )= 0 (13.81)

If a1+ K G k 2> 0 in order to obtain a positive coefficient of s in the system characteristic equation (13.81), the system may be stable but a finite steady-state error exists. The error in an all-pole plant ise(t ) ss= lim sE ( s )=s→0

a1+ K G k 2 a R1= (k 2+ 1 ) R1 KG KG

State feedback: steady-State Error Analysis3. Ramp input: r(t)=R1tu-2(t), R(s)=R1/s2Y ( s)= R( s) s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2+ K G ( s w+ c w 1s w 1++ c1s+ c0 )+ (a0+ K G k1 )

When a plant contain at least one zero, so that c1>0, it is possible to specify the valu

e of k2 given byEq.(13.85) and to have a stable system.k 2= c1 ( a1 ) KG

(13.85)

In that case, zero steady-state error with a ramp input can be achieved.s n+ (a n 1+ K G k n ) s n 1+ ( a n 2+ K G k n 1 ) s n 2++ ( a1+ K G k 2 ) s+ ( a0+ K G k1 )= 0 (13.81)K G c1> 0

Thus, a stable state-variable-feedback system having zero steady-state error with both a step and a ramp input can be achieved only when the control ratio has at least one zero.11

State feedback: steady-State Error Analysis4. Parabolic input: r(t)=(R2t2/2)u-3(t), R(s)=R2/s3 For a parabolic input, the all-pole and pole-zero plants are considered separately. For an all-pole plant, we haveE ( s)= R( s)

k1= c0

a0 KG

(13.80)+ c1s+ c0 )

=0 (13.82)

s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2+

s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2+

+ (a0+ K G k1 ) K G ( s w+ c w 1s w 1++ (a0+ K G k1 )

E (s)=

R2 s3

s n+ ( a n 1+ K G k n ) s n 1+ ( a n 2+ K G k n 1 ) s n 2+

s n+ ( a n 1+ K G k n ) s n 1+ ( a n 2+ K G k n 1 ) s n 2+

+ ( a1+ K G k 2 ) s+ ( a 0+ K G k1 )

e(t ) ss= lim sE ( s )=∞s→0

It is concluded that an all-pole plant cannot be follow a parabolic input.12

State feedback: steady-State Error Analysis4. Parabolic input: r(t)=(R2t2/2)u-3(t), R(s)=R2/s3 Under the condition that e(t)ss=0 for both step and ramp inputs to a polezero stable system (a plant containing at least one zero), we havek1= c0 a0 KG

(13.80)

k 2= c1 (

a1 ) KG

(13.85)+ c1s )++ (a0+ K G k1 )+ c2 s 2 )

E ( s)= R( s)

s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2+ s+ (a n 1+ K G k n ) sn n 1

+ (a1+ K G k 2 ) K G ( s w+ c w 1s w 1+n 2

+ (a n 2+ K G k n 1 ) s

(13.83) (13.88)

E ( s)= R( s)

s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2+

s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2+

+ (a 2+ K G k 3 ) K G ( s w+ c w 1s w 1++ (a0+ K G k1 )

e(t ) ss= lim sE ( s )=s→0

a2+ K G k3 K G c2 R2 K G c0

State feedback: steady-State Error Analysis4. Parabolic input: r(t)=(R2t2/2)u-3(t), R(s)=R2/s3 In order to obtain e(t)ss=0, it is necessary thate(t ) ss= lim sE ( s )=s→0

a2+ K G k 3 K G c2 R2= 0 K G c0

a2+ K G k 3 K G c2= 0a2 ) KG

k 3= c2 (

(13.90)

With this value of k3, the s2 term in Eq.(13.81) (system characteristic equation) has the positive coefficient KGc2 only when w>=2. Thus, a stable system having zero steady-state error with a parabolic input can be achieved, even if G(s) is Type 0, only for a pole-zero system with two or more zeros.

State feedback: steady-State Error Analysis4. Parabolic input: r(t)=(R2t2/2)u-3(t), R(s)=R2/s3 IN a system having only one ze

ro (w=1), the value c2=0 and k3=-a2/KG produces a zero coefficient for the s2 term in Eq.(13.81) (system characteristic equation), resulting in an unstable system. Therefor, zero steady-state error with a parabolic input cannot be achieved by a state-variable-feedback pole-zero stable system with one zero. If k3≠ -a2/KG and a2+KGk3>0 in order to maintain a positive coefficient of s2 term in Eq.(13.81), a finite error exists. The system can be stable, and the steady-state error ise(t ) ss= lim sE ( s )= (s→0

a2+ K G k3 ) R2 K G c0

(13.91)

State feedback: steady-State Error Analysis Brief review(1) In analysis steady-state error, consider the minimun-phase transfer function G(s) is K ( s w+ c s w 1++c s+c ) Y ( s)G ( s)= U (s)=G w 1 1 0

s+ an 1sn

n 1

+ an 2 s

n 2

+

+ a1s+ a0

where kG>0, w<n, and the coefficients ci are all positive. The state-variable-feedback control system ratio isK G ( s w+ c w 1s w 1++ c1s+ c0 ) Y (s)= R( s ) s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2++ (a0+ K G k1 )

(13.76)

For an all-pole plant,

G (s)=

K G c0 Y (s)= n U ( s ) s+ an 1s n 1+ a n 2 s n 2+

+ a1s+ a0

IF ci=0 (i>0), it is a pole-zero plant.16

State feedback: steady-State Error Analysis Brief review(2) For a step input, zero steady-state error can be achieved with statevariable-feedback, even if G(s) is Type 0, and the feedback coefficient k1 is determined by the plant parameters. For a ramp input, zero steady-state error cannot be achieved by statevariable-feedback for an all pole stable system. The system may be stable but a finite steady-state error exists. The zero steady-state error with a ramp input can be achieved only when the control ratio has at least one zero. For a parabolic input, the all-pole plant cannot be follow the input. The zero steady-state error can be achieved, only for a pole-zero system with two or more zeros.

State feedback: steady-State Error Analysis5. Use of steady-state coefficients For the example represented by Fig.13.7a in P437, the feedback coefficient blocks are manipulated to yield the Heq block diagram as R UX1= Y

GH eq

The system control ratio isM (s)= Y (s) G (s)= R ( s ) 1+ G ( s ) H eq ( s )

(13.92)

kX

Fig. 13.7d

Similar, a Geq unity-feedback block diagram can be obtained, and the system control ratio is.G eq ( s ) Y (s) M (s)== R ( s ) 1+ G eq ( s )

R

U

G eq

X1= Y

kX

(13.93)

Fig. 13.818

State feedback: steady-State Error Analysis5. Use of steady-state coefficients Solving for Geq from Eq.(13.93)G eq ( s ) Y (s) M (s)== R ( s ) 1+ G eq ( s )

Noted that Eq.(13.76)

G eq ( s )=

M (s) 1 M (s)

K G ( s w+ c w 1s w 1++ c1s+ c0 ) Y (s)∵ M (s)== R ( s ) s n+ (a n 1+ K G k n ) s n 1+ (a n 2+ K G k n 1 ) s n 2++ (a0+ K G k1 )K G (s w+ cw 1s w 1++ c1s+ c0 )

(13.76)

∴ G

eq (s)=

s n+ (an 1+ K G k n )s n 1+

+ (a1+ K G k 2 )s+ (a0+ K G k1 ) K G (s w+ cw 1s w 1+

+ c1s+ c0 )

(13.95) This is a Type 0 transfer function, the step-error coefficient isK p= lim Geq (s)=s→0

K G c0 a0+ K G k1 K G c0

k1= c0 (a0/ K G )

K p=∞ e(t ) ss= 0

with a step input.19

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