自动控制理论课程讲义automatic control theory

Lyapunov:BeijingInstituteofMarch12,AutomaticControlTheory(C

Lyapunov

March12, 1/AutomaticControlAutomaticControlTheory(CLyapunovMarch12,2/o???o?????? 1考虑g治系˙= ?x=[x1(t),x2(t),···,AutomaticControlTheory(CLyapunovMarch12, 3/f(x,t)=[f1(x,t),AutomaticControlTheory(CLyapunovMarch12, 3/o?????? fi(x,t)是x1,x2,···,xn? 有 连Y可 单值?数。程 表示x(t)=φ(t;x0,t0 x0=φ(t0;x0,t0 看成系?3n? 空间由初始 x0出发所 一;迹，称之?系 $?如J存3?所有时间t?满vx˙= xe，K把xe称? ?? ，显然有f(xe,t)= AutomaticControlTheory(C

Lyapunov

March12, 4/o?????? ?于非线5系?可能有不止一个?o?????? ?于线5?常系?，系?可由x˙Axe描述，??方程Axe=若系数矩阵非?异，K系?存3?一?? xe=0，否K，有?限?个??:2o亚普??稳定性定2XJ?uz???ε??A3,????δ(ε,t0)>0d?v AutomaticControlTheory(CLyapunovMarch12, 5/AutomaticControlTheory(CLyapunovMarch12, 5/AutomaticAutomaticControlTheory(CLyapunovMarch12, 6/o?????? ?? x0? $?φ(t;x0,t0)??k?o?????? kφ(t;x0,t0)−xek≤K?X(1 ??? xe?o??????e? o?o?????? AutomaticControlTheory(CLyapunovMarch12, 7/渐进渐进??如Jxe是系?x˙=f(x,t ???? ，而且满lim|φ(t;x0,t0)−xe|≤?中µ是任意? 量，K称??xe是渐进?◦大范大范?渐进??能实现渐进???大S(δ)球域，称?引力域，如J引力域充满G空间，称xe是大范?渐进????G，或称?全局渐进????G――??G ?一5。o?o?????? AutomaticControlTheory(CLyapunovMarch12, 8/o??????o?????? 3单值标量?f(xfx1···,xnXJ?u ?m ?":x??kf(x)>0?? x=0??kf(x)=0?K???f(x) XJ?u ?m ?":x??kf(x)≥0? x=0?kf(x)=0?K???f(x) ? XJ−f(x) ?K?f(x)?K ?XJ−f(x) ? K?f(x)?K? XJ? ?m?":x6=0,f(x)? ???K???"AutomaticControlTheory(CLyapunovMarch12,AutomaticControlTheory(CLyapunovMarch12, 9/o?????? 二次.o?????? Q(x)

αijxixj=xτP?P?nn阶实?称矩阵，α11

P α21

=.. .. ..

矩阵P称?二次.?数 权矩阵AutomaticControlTheory(AutomaticControlTheory(CLyapunovMarch12, 10/o?????? ??? P ??7?^?? ?o?????? ??? P?K ??7?^?? ?f??( >0;i?? <0;i??

??? P ? ??7?^?? cn− ?f?K ??"?∆1≥0;∆2≥0;···;∆n−1≥0;∆n=detP=??? P?K? ??7?^?? ?f??(≥0;i??∆ ≤ i??

,?∆n=detP=

AutomaticAutomaticControlTheory(CLyapunovMarch12, 11/o?????? 4李? ??基本?o亚普??第??{。设一个不受部?用系?，可由非线5方程描述。xn?G向量，f(x)n??数向量，它每个?x1···xn连Y可??数。把非线5?数向量f(x3??Gxxe附近?开成Taylor级数，有f(x)=

y+g(xe, AutomaticControlTheory(C

Lyapunov

March12, 12/o?????? ?中yxxe是no??????

∂f

?Jacobian矩阵。3xxe时记?A,A=AutomaticControlTheory(CAutomaticControlTheory(CLyapunovMarch12, 13/

nn常数矩阵o?????? A所有A征根具有负实部 o?????? 进? A有至少一个正实部根时 非线5系?3? 不? A征根?不具有正实部， 至少有一个A征值 部?零，?法应用1一方法 ????5。~非线5系?方程1AutomaticControlTheory(CLyapunovMarch12, 14/AutomaticControlTheory(CLyapunovMarch12, 14/o?????? ?中α,β,γ均大于零，输入u?o?????? 。˙1,˙2"xe

arcsinγuα0γ

# 方程(

y1=x1−arcsinαy2=1

ye= AutomaticControlTheory(CLyapunovMarch12, 15/˙2AutomaticControlTheory(CLyapunovMarch12, 15/o????o?????? y1=y2= A α αA征方程det(λI−A)=λ2+βλ+αcos(arcsinγu)=AutomaticAutomaticControlTheory(CLyapunovMarch12, 16/o?o?????? 显然 0时 0，系?具有负实部Aγ根，渐进?? u<0时cos(arcsinαu)<0，线5系?不?，从 非线5系?3xe附近不??o亚普??第?{o亚普??稳定 定n设不 部? 系?$?方程??? 如下(˙=xe=

如J有连Y 一阶? 纯量?数V(x,t)存3，并且满v以下^件：AutomaticControlTheory(CLyapunovMarchAutomaticControlTheory(CLyapunovMarch12, 17/o?????? V˙(xo?????? ∂V

<

∂x nK : ?? 是渐进?

，如Jkxk→∞有V(x,t)→∞，K : ??AutomaticControlTheory(CLyapunovMarch12, 18AutomaticControlTheory(CLyapunovMarch12, 18/o????o??????

确??? ??5) :是系 ?一?? ，?取一个正?纯量 数V(x?V(xx2x2 AutomaticControlTheory(CLyapunovMarch12, 19/AutomaticControlTheory(CLyapunovMarch12, 19/o??o?????? (x 又由于kxk时，V(x)→∞，因此系?3??:是大范渐进? o亚普??第?{o亚普??稳定 定n?于系?(10)，若可以 单值标量?数V(x)，且V(x)可?♠如JV(x?V˙(x满vV(x,t) AutomaticControlTheory(CLyapunovMarch12, 20AutomaticControlTheory(CLyapunovMarch12, 20/o?????? K : ?? 是? ♠如JV(x?V˙(x满vV(x,t) K : ?? 是不? 如下系 ??5"

xe= −1AutomaticControlTheory(C

Lyapunov

March12, 21/o?????? (1)试?正?李? ?数V(x)=2x2+x2>0， 2(x)2不?，? ??? ??5(?理是充分不必 (2)?李? ?数V(x)=x2+x2>0， 2(x)2x222负半?，所以??:是? ，尚?法 是否渐进??AutomaticControlTheory(C

Lyapunov

March12, 22/o?????? (3)?李? ?数V(x)=(x1+x2)2+2x2+x2>0， 2˙˙2)1˙12˙ =−2(x2+x2)< 负?，kxkV(x)→∞，所以??:是大范?渐进? 系???5(xe˙1˙2AutomaticControlTheory(C

Lyapunov

March12, 23/o?????? ?正?李? ?数V(x)=x2+x2>0， ˙˙=2x2+2x2> 正?，所以? 是不? o亚普??第?{o亚普??稳定 定n?于系?(10)，如J可以 可 标量?数V(x)，并满v V˙(x)?K? V˙(x)3??(10 ?") $?;?????"AutomaticControlTheory(C

Lyapunov

March12, 24/o?????? K?? 是渐进? 如下非线5系??? ??51˙=−β(1+x

—x (β> 显然xe=0是系 ?? 。?J李? 数V(xx2x2>0 ˙˙2=−2β(1+x2)2AutomaticControlTheory(C

Lyapunov

March12, 25/o?????? 显然 x2=0,−1时V˙(x)o?????? 。下 于零(1x2(t0以及x1任意。3此种情/下，只有??(2)x2(t)≡−1以及x1任意。 出x1(t)≡0,˙1(t)=−1 矛?结J，所以?种情/不会发生3方程(10) $?;迹上！所以，非零 ;迹上V˙(x)不??零。所以xe=0是系 AutomaticControlTheory(CLyapunovAutomaticControlTheory(CLyapunovMarch12, 26/Lyapunov?{3?5?~X? A^1线5?常系 渐进?系? 方程可表示 A?常系数矩阵。?该系 ?数V(x)=xτ AutomaticControlTheory(C

Lyapunov

March12, 27/LyapunoLyapunov?{3?5?~X?A^=xτAτPx+xτ=xτ(AτP+PA)x=−xτAutomaticControlTheory(CLyapunovMarch12, 28/如J?称矩阵Q是AutomaticControlTheory(CLyapunovMarch12, 28/LyLyapunov?{3?5?~X?A^定定n线5系?(13 ?? xe=0?渐进? 充分必要件是：?任意给 正?矩阵Q，存3正?矩阵P?矩阵方AτP+PA= 解证明(必要5)：如J系?(13)是渐进? ，KG=移阵limeAt=0 E时变矩AutomaticControlTheory(CLyapunovMarch12, 29AutomaticControlTheory(CLyapunovMarch12, 29/Lyapunov?{Lyapunov?{3?5?~X?A^ E(0)= 解。?(17)1一式两?从t= t=∞积分ZE(∞)−E(0)=

ZE(t)dt E(t)dt 令ZP E(t)dt

ZeAτtQeAtdt 0 limeAt=0→??AutomaticControlAutomaticControlTheory(CLyapunovMarch12, 30/Lyapunov?{3?5?~X? A^K可满−Q=AτP+P任取n?非零常向量x0 矩阵P| 二次 正?5ZxτPx xeAτtQeAtx 0Z

:=

xτ(t)Q0由于eAt是非?异矩阵，x(tn?非零向量。由于Q正?5，K被积?数是正?二次.?数，3积分区间[0,∞上?取正值。所以(19)积分是正，K矩阵P是正?。AutomaticControlTheory(C

Lyapunov

March12, 31/AutomaticAutomaticControlTheory(CLyapunovMarch12, 32/Lyapunov?{3Lyapunov?{3?5?~X?A^程(15)可 AτP+PA= 系系?方程x˙1=x˙2=−x1−??5?系?3??Lyapunov?{3?5?~X? A^!P具有?称/P

，K由式 p11 p12 p22=1，即P

。容 ?1 12阵P正?。因此系? : ?? 是大范?渐进?相 ?数V(x)=xτP

13x2+2x1x2+可求出

=2 V˙(x)=−(x2+ AutomaticControlTheory(C

Lyapunov

March12, 33/LyLyapunov?{3?5?~X?A^AutomaticControlTheory(CLyapunovMarch12, 34/确确?使系?渐进? O益K 范?LyapuLyapunov?{3?5?~X?A^x˙1=x˙2=−2x2+x˙3=−0.5Kx1−

X? ? A

AutomaticControlTheory(CLyapunovMarch12, 35/ 空 :是该系AutomaticControlTheory(CLyapunovMarch12, 35/000Q000,001Lyapunov?{3?5Lyapunov?{3?5?~X?A^3

˙(x)x˙3(t)≡0\→x1(t)≡0,x2(t)≡0，可见只有 空:满vV˙(x)? 于零。所以半负?Q矩阵可1，由矩阵方程AτP+PA=−Q可以解出K2+ P 48−

0

由Sylvester?理，满vP?正 ^件AutomaticControlTheory(CAutomaticControlTheory(CLyapunovMarch12, 36/Lyapunov?{3?5?~X? A^所以K满v上述^件时，系?渐进??推 系?所有A征 实部 于负常数 充分必要^件是?任意给 正?矩阵Q，?存3正?矩阵P满v矩阵方AτP+PA+2σP= 2线5时变系 李? ??5分设系?方程˙(t)xe=

AutomaticControlTheory(C

Lyapunov

March12, 37/Lyapunov?{3?5?~X? A^系?3??:xe=0大范?渐进?? 充分必要^件是：?任意给? 连Y正??称矩阵(positivedefinitesymmetricmatrices)Q(t)，存3 Y?称正?矩阵P(t)使V(x(t),t)=xτ(t)P(t)

V(x(t),t)?系 ?数y??? ?数V(x(t),t)=xτ(t)P(t)AutomaticControlTheory(C

Lyapunov

March12, 38/Lyapunov?{3?5?~Lyapunov?{3?5?~X?A^˙(t)=[A(t)x(t)]τP(t)x(t)+xτP˙(t)+xτ(t)P(t)[A(t)x(t)]=xτ(t)Aτ(t)P(t)=xτ(t)[Aτ(t)P(t)+P˙(t)+P(t)

=−xτ(t)Q(t)由上式可求

AutomaticControlTheory(CLyapunovMarch12, 39/AutomaticControlTheory(CLyapunovMarch12, 39/Lyapunov?{3?5?~X? A^上式 J方程(Riccati A殊情况，?解ZP(t)=Φτ(t0,t)P(t0)Φ(t0,t) Φτ(τ,t)Q(t)Φ(τ, 式中Φ(τ,t)――系?˙(t)=A(t) =移矩阵 Q(t)=IZP(t)=Φτ(t0,t)P(t0)Φ(t0,t) Φτ(τ,t)Φ(τ, 显然 ?取Q=I时，可以?L系 =移矩阵Φ(τ,计算矩阵P(t，并根据矩阵P(t是否连Y、?称、正?5来分?线5时变系 ??5AutomaticControlTheory(C

Lyapunov

March12, 40/Lyapunov?{3?5?~X? A^3离散系 李? ??5分设离散系?方程x(k+1)=Ax(k)xe=0

A?常系数非?异矩阵。离散系?3?? xe=0?渐进 充分必要^件?：给?任一正??米A矩阵(Hermitianmatrix)或是?称矩阵Q，存3一个?米A矩阵或是?称矩阵P，AτPA−P= AutomaticControlTheory(C

Lyapunov

March12, 41/Lyapunov?{3?5?~X? A^而且纯量?数xτPx是系 ?数。如∆V(x(k))=−xτ(k)Q?任一 S列不 于零，KQ可取正半?矩阵y??? ?数V[x(k)]=xτ(k)P式中P?正??米A或实?称矩阵。V[x(k+1V[x(kO线5连Y系? ∆V(x)=V[x(k+1)]−AutomaticControlTheory(C

Lyapunov

March12, 42/LyLyapunov?{3?5?~X?A^∆V[x(k)]=V[x(k+1)]−=xτ(k+1)Px(k+1)−xτ(k)P=[Ax(k)]τP[Ax(k)]−xτ(k)P=xτ(k)AτPAx(k)−xτ(k)P=xτ[AτPA−PAutomaticControlTheory(CLyapunovMarch12,AutomaticControlTheory(CLyapunovMarch12, 43/

Lyapunov?{3?5?~X? A^根据渐进? ^件，Q?正?矩阵是离散系?渐进? 分^件。根据给 Q验算由矩阵方AτPA−P=−Q→确P是否?正?，矩阵正?5是系3??G渐进?必要^件4线5时变离散系 李? ??5分设线5时变离散系?方程x(k+1)=A(k+1,k)x(k)xe=0

AutomaticControlTheory(C

Lyapunov

March12, 44/Lyapunov?{3?5?~X? A^K系?3?? 大范?渐进? 充分必要^件是：意给 正??米A或实?称矩阵Q(k)存3一个正 ?A或实?称矩阵P(k+1Aτ[(k+1),k]P(k+1)A(k+1,k)−P(k)= 并且V[x(k),k]=xτ(k)P(k)

是系 ?数。差分方程 解P(k+1)=Aτ(0,k+1)P(0)A(0,k+—

Aτ(i,k+1)Q(i)A(i,k+AutomaticControlTheory(C

Lyapunov

March12, 45/Lyapunov?{3?5Lyapunov?{3?5?~X?A^P(k+1)=Aτ(0,k+1)P(0)A(0,k+—AutomaticControlAutomaticControlTheory(CLyapunovMarch12, 46/

Aτ(i,k+1)A(i,k+X?5U?X?5U? 1?义V˙(x)与V(x)比 负值η,−V(x) ?于渐进??系?，η?取正值 大渐进? $?x(t)于?? 快。解AutomaticControlTheory(CLyapunovMarch12, 47/AutomaticControlTheory(CLyapunovMarch12, 47/

X?5U? ?中x0,t0分别?系 初始 ?初始时刻。?方便?论?η=min−V˙

ηm代入

R—t

V(x)≤ = 显然ηm给出了V(x)趋于?? 速 计。1/ηm是表征李 ?数V(x)趋于??:快 ?大时间常数，该常数?用传?控制理论计算 系?g由响应时间常 一半AutomaticControlTheory(C

Lyapunov

March12, 48/X?5U? 2线X?5U? 2设线5?常系? x˙=Ax，矩阵 所有A征值?具有负部，即线5系?渐进??。KV(x)=xτPV˙(x)=−xτQ式中P?正??米A或实?称矩阵，Q=−(AτP+PA)ηm=

−˙AutomaticControlTheoryAutomaticControlTheory(CLyapunovMarch12, 49/

=

xτQxτP

X?5U?X?5U? ∂η= xτQ

=2QxxτPx−2xτQxPx= xτPxτQλ=xτPx，代入45

(xτP(Q−λP)xmin= (QP−1−λI)xmin= 由于x?非零向量，所以λ必?Q 一个A征值。此 于Q A征值λmin 系?方程

# AutomaticControlTheory(AutomaticControlTheory(CLyapunovMarch12, 50/

−1

X?5U? 试求取系 ?数，以及从封闭曲线V(x)=150边界一 封闭曲线V(x)=0.06内一:响应时 上界)?1)Q=I，根据AτP+PA=−Q求矩阵P， # # 0

p12

1

−1 pP p

12

12 12AutomaticControlTheory(C

Lyapunov

March12, 51/X?5U? ?数及 数V(x)=xτPx

1(3x2+2x1x2+ V˙(x)=−xτQx=−(x2+ 因此有

2η=2

2(x2+

令dη=0，由dη=0，

3x1+2x1x2+x2−x1x2−x2= x1=1.618x2,x1= AutomaticControlTheory(C

Lyapunov

March12, 52/X?5U? 把以上结J代入(50)中， ηm1=0.553,X?5U? ηm2>ηm1，ηm=ηm1=0.553ηm也可由QP−1矩 A征值求" 根据|QP−1−λI|=0，考 Q=I,P 3/21/2， 21−2

2=02=2 12求出矩阵 两个A征值?λ1=1.447,λ2=0.553，AutomaticControlTheory(CAutomaticControlTheory(CLyapunovMarch12, 53/X?5U? 求取从封闭曲线V(x)= V(x)=X?5U? 因V(x,t)≤V(x0,所以t−t0≤−1

V(x, =

1

= 从曲线V(x)=150上出 任一;迹，进入V(x)=0.06所包区域内，所I时间不超L14.148个时间单?。如J1/ηm是李氏?数收敛时间常数，Kg由响应时间常数上限AutomaticControlTheory(CLyapunovMarchAutomaticControlTheory(CLyapunovMarch12, 54/X?5U? ?X?5U? V(x,t)=x2+x2+···+x2,V(x0,t0)=x2+x2+···+

x2+x2+···+x2≤(x

+

??论方便， 2−ηm(

—1ηm(t−tx1≤

x1≤x10e

1 2−ηm(

—1ηm(t−t

ηm=x2≤

x2≤x20e

=

2−η(t−t

—1η(t−t

AutomaticControlTheory(CLyapunovAutomaticControlTheory(CLyapunovMarch12, 55/

xn≤

2 X?5U?X?5U? 3设线5系? 方程 ?中系?矩阵 素依赖于可?参数α。参数α?K是使二次.积分指标J

∞xQAutomaticControlTheoryAutomaticControlTheory(CLyapunovMarch12, 56/X?5U? ，?中Q?正?或正半?常数矩阵X?5U? 系?渐进??，由5能指标中给?正?或正半?矩阵Q可L矩阵方程Aτ(α)P+PA(α)=解出正 ?参数 矩阵P(α)。5能指标(55)可化ZJ xτQxdt

Z −˙ =xτ(0)P(α)x(0)−xτ(∞)P(α)????→x(∞)=AutomaticControlTheory(CLyapunovMarch12, 57/=xτ(AutomaticControlTheory(CLyapunovMarch12, 57/

X?5U? 显然，5能指标是 ?数，?极 充分必要^件∂J=

>∂α ="0#" ="0#" ,"#="10Ax

确? ξ>0使5能指ZJ xτ(t)Q Q0

0 (µ>0AutomaticControlTheory(C

Lyapunov

March12, 58/X?5UX?5U? )?由于A是??矩阵，所以J=xτ(0Px(0，P可由矩阵方AτP+PA=−Q确?， #

#

p12

2

−1 解矩阵方程

" p ξP

1+ 4 1+AutomaticControlTheory(AutomaticControlTheory(CLyapunovMarch12, 59/

X?5UX?5U? J=xτ(0)Px(0) ξ+1+ x2(0)+x1(0)x2(0)+1+µ4 x1(0)=1,x2(0)=0代入上式

4 J=ξ+1+4令∂J=0 使5能指标 极值 √1+AutomaticControlTheory(AutomaticControlTheory(CLyapunovMarch12, 60/X?5U? 4 反 设~?系? 方程

# "x˙1

x1

0 −1 ? ?数?V(x)=x2+x u=0时，V(x)'于时变化率

V˙(x)=2x1x˙1+2x2x˙2=2x1x2−2x1x2≡所以系?不是渐进? ，仅仅是? AutomaticControlTheory(C

Lyapunov

March12, 61/X?5U? 有控制输入即u6=0时，VX?5U? V˙(x)=2x1x2+2x2(−x1+u)=若取u=−kx2,k>022是负半 ，而且3解曲线上V(x)不??零，所以系?是渐AutomaticControlTheoryAutomaticControlTheory(CLyapunovMarch12, 62/??5X o??????5? 考虑不 部?考虑不 部? 非线5系 f(0)=?取李???数k˙k2'于时 数V˙(x)=f˙τ(x)f(x)+fτ(x)fAutomaticControlTheory(C Lyapunov March12, 63/??5??5Xo??????5?

∂f

F(x)

∂fn

是系?(60 ?可比矩阵。代入

V˙(x)=fτ(x)[Fτ(x)+F(x)]f 由(65)可以看出：如Jx6=0f(x)6=0^件成立，KV˙(xAutomaticControlTheory(CLyapunovMarch12, 64/?AutomaticControlTheory(CLyapunovMarch12, 64/??5X o??????5? Krasovskii??(60 不 部? 非线5?常系?,有f(0)=且f(x)?xi(i=1,2,···,n)?可?. Fτ(x)+F(x)负?时,系?? xe=0即 空 :是渐进? 有f(x)τf(x)→∞K系 ?? 是大范?渐进? 斯基?理只是渐进?? 充分^件。如JFτ(x)+F(x)不是负? ，不能 ?系?不是渐进?? 。?u?5X??n?????^AutomaticControlTheory(C

Lyapunov

March12, 65/??5X o??????5? ~?应 斯基方法证明系 ?? 是大范?渐? x˙1=−5x1+x˙=x−x−

y??(1)检验

是否?负值∂f1(x)=−5<

=−1−3x2< 两个 数均?负值，可尝 斯基方法AutomaticControlTheory(C

Lyapunov

March12, 66/??5X o??????5? E?可比矩阵F(x)=∂fK

2 −1−2Fτ(x)+F(x)V(x)

2 −2−2

(负?AutomaticControlTheory(C

Lyapunov

March12, 67/??5X o??????5? hV(x)=fτ(x)f(x) −5x+

ix−x−

−5x1+

x−x−2=(−5x1+x2)2+(x1−x2−x3)2→∞2所以系?3?? 是大范?渐进? ~

x˙1=−3x1+x˙=

−3x1+⇒f(x)

⇒∂f

= x x1

xe=?可比矩阵?角线存3 素，不能应 斯基?理AutomaticControlTheory(C Lyapunov March12, 68/??5X o??????5? 2变量F?法――Variable设不 部? 非线5系x˙=f(x, ?? 是 空 :。设 ?系?渐进? ? ?数V(x)是 显?数，而不是 显?数。?数?时 数可表示

∂Vx˙2+

∂V

=(grad AutomaticControlTheory(C

Lyapunov

March12, 6