自動控制理論課程講義automatic control theory

Lyapunov:BeijingInstituteofMarch12,AutomaticControlTheory(C

Lyapunov

March12, 1/AutomaticControlAutomaticControlTheory(CLyapunovMarch12,2/o???o?????? 1考慮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 連Y可 單值?數(shù)。程 表示x(t)=φ(t;x0,t0 x0=φ(t0;x0,t0 看成系?3n? 空間由初始 x0出發(fā)所 一;跡，稱之?系 $?如J存3?所有時間t?滿vx˙= xe，K把xe稱? ?? ，顯然有f(xe,t)= AutomaticControlTheory(C

Lyapunov

March12, 4/o?????? ?于非線5系?可能有不止一個?o?????? ?于線5?常系?，系?可由x˙Axe描述，??方程Axe=若系數(shù)矩陣非?異，K系?存3?一?? xe=0，否K，有?限?個??:2o亞普??穩(wěn)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/漸進漸進??如Jxe是系?x˙=f(x,t ???? ，而且滿lim|φ(t;x0,t0)?xe|≤?中μ是任意? 量，K稱??xe是漸進??大范大范?漸進??能實現(xiàn)漸進???大S(δ)球域，稱?引力域，如J引力域充滿G空間，稱xe是大范?漸進????G，或稱?全局漸進????G――??G ?一5。o?o?????? AutomaticControlTheory(CLyapunovMarch12, 8/o??????o?????? 3單值標量?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階實?稱矩陣，α11

P α21

=.. .. ..

矩陣P稱?二次.?數(shù) 權(quán)矩陣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亞普??第??{。設一個不受部?用系?，可由非線5方程描述。xn?G向量，f(x)n??數(shù)向量，它每個?x1···xn連Y可??數(shù)。把非線5?數(shù)向量f(x3??Gxxe附近?開成Taylor級數(shù)，有f(x)=

y+g(xe, AutomaticControlTheory(C

Lyapunov

March12, 12/o?????? ?中yxxe是no??????

?f

?Jacobian矩陣。3xxe時記?A,A=AutomaticControlTheory(CAutomaticControlTheory(CLyapunovMarch12, 13/

nn常數(shù)矩陣o?????? A所有A征根具有負實部 o?????? 進? A有至少一個正實部根時 非線5系?3? 不? A征根?不具有正實部， 至少有一個A征值 部?零，?法應用1一方法 ????5。~非線5系?方程1AutomaticControlTheory(CLyapunovMarch12, 14/AutomaticControlTheory(CLyapunovMarch12, 14/o?????? ?中α,β,γ均大于零，輸入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?????? 顯然 0時 0，系?具有負實部Aγ根，漸進?? u<0時cos(arcsinαu)<0，線5系?不?，從 非線5系?3xe附近不??o亞普??第?{o亞普??穩(wěn)定 定n設不 部? 系?$?方程??? 如下(˙=xe=

如J有連Y 一階? 純量?數(shù)V(x,t)存3，并且滿v以下^件：AutomaticControlTheory(CLyapunovMarchAutomaticControlTheory(CLyapunovMarch12, 17/o?????? V˙(xo?????? ?V

<

?x nK : ?? 是漸進?

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

確??? ??5) :是系 ?一?? ，?取一個正?純量 數(shù)V(x?V(xx2x2 AutomaticControlTheory(CLyapunovMarch12, 19/AutomaticControlTheory(CLyapunovMarch12, 19/o??o?????? (x 又由于kxk時，V(x)→∞，因此系?3??:是大范漸進? o亞普??第?{o亞普??穩(wěn)定 定n?于系?(10)，若可以 單值標量?數(shù)V(x)，且V(x)可??如JV(x?V˙(x滿vV(x,t) AutomaticControlTheory(CLyapunovMarch12, 20AutomaticControlTheory(CLyapunovMarch12, 20/o?????? K : ?? 是? ?如JV(x?V˙(x滿vV(x,t) K : ?? 是不? 如下系 ??5"

xe= ?1AutomaticControlTheory(C

Lyapunov

March12, 21/o?????? (1)試?正?李? ?數(shù)V(x)=2x2+x2>0， 2(x)2不?，? ??? ??5(?理是充分不必 (2)?李? ?數(shù)V(x)=x2+x2>0， 2(x)2x222負半?，所以??:是? ，尚?法 是否漸進??AutomaticControlTheory(C

Lyapunov

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

Lyapunov

March12, 23/o?????? ?正?李? ?數(shù)V(x)=x2+x2>0， ˙˙=2x2+2x2> 正?，所以? 是不? o亞普??第?{o亞普??穩(wěn)定 定n?于系?(10)，如J可以 可 標量?數(shù)V(x)，并滿v V˙(x)?K? V˙(x)3??(10 ?") $?;?????"AutomaticControlTheory(C

Lyapunov

March12, 24/o?????? K?? 是漸進? 如下非線5系??? ??51˙=?β(1+x

—x (β> 顯然xe=0是系 ?? 。?J李? 數(shù)V(xx2x2>0 ˙˙2=?2β(1+x2)2AutomaticControlTheory(C

Lyapunov

March12, 25/o?????? 顯然 x2=0,?1時V˙(x)o?????? 。下 于零(1x2(t0以及x1任意。3此種情/下，只有??(2)x2(t)≡?1以及x1任意。 出x1(t)≡0,˙1(t)=?1 矛?結(jié)J，所以?種情/不會發(fā)生3方程(10) $?;跡上！所以，非零 ;跡上V˙(x)不??零。所以xe=0是系 AutomaticControlTheory(CLyapunovAutomaticControlTheory(CLyapunovMarch12, 26/Lyapunov?{3?5?~X? A^1線5?常系 漸進?系? 方程可表示 A?常系數(shù)矩陣。?該系 ?數(shù)V(x)=xτ AutomaticControlTheory(C

Lyapunov

March12, 27/LyapunoLyapunov?{3?5?~X?A^=xτAτPx+xτ=xτ(AτP+PA)x=?xτAutomaticControlTheory(CLyapunovMarch12, 28/如J?稱矩陣Q是AutomaticControlTheory(CLyapunovMarch12, 28/LyLyapunov?{3?5?~X?A^定定n線5系?(13 ?? xe=0?漸進? 充分必要件是：?任意給 正?矩陣Q，存3正?矩陣P?矩陣方AτP+PA= 解證明(必要5)：如J系?(13)是漸進? ，KG=移陣limeAt=0 E時變矩AutomaticControlTheory(CLyapunovMarch12, 29AutomaticControlTheory(CLyapunovMarch12, 29/Lyapunov?{Lyapunov?{3?5?~X?A^ E(0)= 解。?(17)1一式兩?從t= t=∞積分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可滿?Q=AτP+P任取n?非零常向量x0 矩陣P| 二次 正?5ZxτPx xeAτtQeAtx 0Z

:=

xτ(t)Q0由于eAt是非?異矩陣，x(tn?非零向量。由于Q正?5，K被積?數(shù)是正?二次.?數(shù)，3積分區(qū)間[0,∞上?取正值。所以(19)積分是正，K矩陣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?稱/P

，K由式 p11 p12 p22=1，即P

。容 ?1 12陣P正?。因此系? : ?? 是大范?漸進?相 ?數(shù)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/確確?使系?漸進? O益K 范?LyapuLyapunov?{3?5?~X?A^x˙1=x˙2=?2x2+x˙3=?0.5Kx1?

X? ? A

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

˙(x)x˙3(t)≡0\→x1(t)≡0,x2(t)≡0，可見只有 空:滿vV˙(x)? 于零。所以半負?Q矩陣可1，由矩陣方程AτP+PA=?Q可以解出K2+ P 48?

0

由Sylvester?理，滿vP?正 ^件AutomaticControlTheory(CAutomaticControlTheory(CLyapunovMarch12, 36/Lyapunov?{3?5?~X? A^所以K滿v上述^件時，系?漸進??推 系?所有A征 實部 于負常數(shù) 充分必要^件是?任意給 正?矩陣Q，?存3正?矩陣P滿v矩陣方AτP+PA+2σP= 2線5時變系 李? ??5分設系?方程˙(t)xe=

AutomaticControlTheory(C

Lyapunov

March12, 37/Lyapunov?{3?5?~X? A^系?3??:xe=0大范?漸進?? 充分必要^件是：?任意給? 連Y正??稱矩陣(positivedefinitesymmetricmatrices)Q(t)，存3 Y?稱正?矩陣P(t)使V(x(t),t)=xτ(t)P(t)

V(x(t),t)?系 ?數(shù)y??? ?數(shù)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殊情況，?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) =移矩陣 Q(t)=IZP(t)=Φτ(t0,t)P(t0)Φ(t0,t) Φτ(τ,t)Φ(τ, 顯然 ?取Q=I時，可以?L系 =移矩陣Φ(τ,計算矩陣P(t，并根據(jù)矩陣P(t是否連Y、?稱、正?5來分?線5時變系 ??5AutomaticControlTheory(C

Lyapunov

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

A?常系數(shù)非?異矩陣。離散系?3?? xe=0?漸進 充分必要^件?：給?任一正??米A矩陣(Hermitianmatrix)或是?稱矩陣Q，存3一個?米A矩陣或是?稱矩陣P，AτPA?P= AutomaticControlTheory(C

Lyapunov

March12, 41/Lyapunov?{3?5?~X? A^而且純量?數(shù)xτPx是系 ?數(shù)。如?V(x(k))=?xτ(k)Q?任一 S列不 于零，KQ可取正半?矩陣y??? ?數(shù)V[x(k)]=xτ(k)P式中P?正??米A或?qū)?稱矩陣。V[x(k+1V[x(kO線5連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^根據(jù)漸進? ^件，Q?正?矩陣是離散系?漸進? 分^件。根據(jù)給 Q驗算由矩陣方AτPA?P=?Q→確P是否?正?，矩陣正?5是系3??G漸進?必要^件4線5時變離散系 李? ??5分設線5時變離散系?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?漸進? 充分必要^件是：意給 正??米A或?qū)?稱矩陣Q(k)存3一個正 ?A或?qū)?稱矩陣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 ?數(shù)。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?義V˙(x)與V(x)比 負值η,?V(x) ?于漸進??系?，η?取正值 大漸進? $?x(t)于?? 快。解AutomaticControlTheory(CLyapunovMarch12, 47/AutomaticControlTheory(CLyapunovMarch12, 47/

X?5U? ?中x0,t0分別?系 初始 ?初始時刻。?方便?論?η=min?V˙

ηm代入

R—t

V(x)≤ = 顯然ηm給出了V(x)趨于?? 速 計。1/ηm是表征李 ?數(shù)V(x)趨于??:快 ?大時間常數(shù)，該常數(shù)?用傳?控制理論計算 系?g由響應時間常 一半AutomaticControlTheory(C

Lyapunov

March12, 48/X?5U? 2線X?5U? 2設線5?常系? x˙=Ax，矩陣 所有A征值?具有負部，即線5系?漸進??。KV(x)=xτPV˙(x)=?xτQ式中P?正??米A或?qū)?稱矩陣，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 一個A征值。此 于Q A征值λmin 系?方程

# AutomaticControlTheory(AutomaticControlTheory(CLyapunovMarch12, 50/

?1

X?5U? 試求取系 ?數(shù)，以及從封閉曲線V(x)=150邊界一 封閉曲線V(x)=0.06內(nèi)一:響應時 上界)?1)Q=I，根據(jù)AτP+PA=?Q求矩陣P， # # 0

p12

1

?1 pP p

12

12 12AutomaticControlTheory(C

Lyapunov

March12, 51/X?5U? ?數(shù)及 數(shù)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? 把以上結(jié)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" 根據(jù)|QP?1?λI|=0，考 Q=I,P 3/21/2， 21?2

2=02=2 12求出矩陣 兩個A征值?λ1=1.447,λ2=0.553，AutomaticControlTheory(CAutomaticControlTheory(CLyapunovMarch12, 53/X?5U? 求取從封閉曲線V(x)= V(x)=X?5U? 因V(x,t)≤V(x0,所以t?t0≤?1

V(x, =

1

= 從曲線V(x)=150上出 任一;跡，進入V(x)=0.06所包區(qū)域內(nèi)，所I時間不超L14.148個時間單?。如J1/ηm是李氏?數(shù)收斂時間常數(shù)，Kg由響應時間常數(shù)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

+

??論方便， 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設線5系? 方程 ?中系?矩陣 素依賴于可?參數(shù)α。參數(shù)α?K是使二次.積分指標J

∞xQAutomaticControlTheoryAutomaticControlTheory(CLyapunovMarch12, 56/X?5U? ，?中Q?正?或正半?常數(shù)矩陣X?5U? 系?漸進??，由5能指標中給?正?或正半?矩陣Q可L矩陣方程Aτ(α)P+PA(α)=解出正 ?參數(shù) 矩陣P(α)。5能指標(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? 顯然，5能指標是 ?數(shù)，?極 充分必要^件?J=

>?α ="0#" ="0#" ,"#="10Ax

確? 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??矩陣，所以J=xτ(0Px(0，P可由矩陣方AτP+PA=?Q確?， #

#

p12

2

?1 解矩陣方程

" 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能指標 極值 √1+AutomaticControlTheory(AutomaticControlTheory(CLyapunovMarch12, 60/X?5U? 4 反 設~?系? 方程

# "x˙1

x1

0 ?1 ? ?數(shù)?V(x)=x2+x u=0時，V(x)'于時變化率

V˙(x)=2x1x˙1+2x2x˙2=2x1x2?2x1x2≡所以系?不是漸進? ，僅僅是? AutomaticControlTheory(C

Lyapunov

March12, 61/X?5U? 有控制輸入即u6=0時，VX?5U? V˙(x)=2x1x2+2x2(?x1+u)=若取u=?kx2,k>022是負半 ，而且3解曲線上V(x)不??零，所以系?是漸AutomaticControlTheoryAutomaticControlTheory(CLyapunovMarch12, 62/??5X o??????5? 考慮不 部?考慮不 部? 非線5系 f(0)=?取李???數(shù)k˙k2'于時 數(shù)V˙(x)=f˙τ(x)f(x)+fτ(x)fAutomaticControlTheory(C Lyapunov March12, 63/??5??5Xo??????5?

?f

F(x)

?fn

是系?(60 ?可比矩陣。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? 非線5?常系?,有f(0)=且f(x)?xi(i=1,2,···,n)?可?. Fτ(x)+F(x)負?時,系?? xe=0即 空 :是漸進? 有f(x)τf(x)→∞K系 ?? 是大范?漸進? 斯基?理只是漸進?? 充分^件。如JFτ(x)+F(x)不是負? ，不能 ?系?不是漸進?? 。?u?5X??n?????^AutomaticControlTheory(C

Lyapunov

March12, 65/??5X o??????5? ~?應 斯基方法證明系 ?? 是大范?漸? x˙1=?5x1+x˙=x?x?

y??(1)檢驗

是否?負值?f1(x)=?5<

=?1?3x2< 兩個 數(shù)均?負值，可嘗 斯基方法AutomaticControlTheory(C

Lyapunov

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

2 ?1?2Fτ(x)+F(x)V(x)

2 ?2?2

(負?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?漸進? ~

x˙1=?3x1+x˙=

?3x1+?f(x)

??f

= x x1

xe=?可比矩陣?角線存3 素，不能應 斯基?理AutomaticControlTheory(C Lyapunov March12, 68/??5X o??????5? 2變量F?法――Variable設不 部? 非線5系x˙=f(x, ?? 是 空 :。設 ?系?漸進? ? ?數(shù)V(x)是 顯?數(shù)，而不是 顯?數(shù)。?數(shù)?時 數(shù)可表示

?Vx˙2+

?V

=(grad AutomaticControlTheory(C

Lyapunov

March12, 6