控制系統(tǒng)仿真與CAD-實(shí)驗(yàn)報(bào)告

1、控制系統(tǒng)仿真與CA實(shí)驗(yàn)課程報(bào)告可編輯一、實(shí)驗(yàn)教學(xué)目標(biāo)與基本要求上機(jī)實(shí)驗(yàn)是本課程重要的實(shí)踐教學(xué)環(huán)節(jié)。實(shí)驗(yàn)的目的不僅僅是驗(yàn)證理論知識(shí)，更重要的是通過(guò)上機(jī)加強(qiáng)學(xué)生的實(shí)驗(yàn)手段與實(shí)踐技能，掌握應(yīng)用MATLAB/Simulink求解控制問(wèn)題的方法，培養(yǎng)學(xué)生分析問(wèn)題、解決問(wèn)題、應(yīng)用 知識(shí)的能力和創(chuàng)新精神，全面提高學(xué)生的綜合素質(zhì)。通過(guò)對(duì)MATLAB/Simulink進(jìn)行求解，基本掌握常見(jiàn)控制問(wèn)題的求解方法與命 令調(diào)用，更深入地認(rèn)識(shí)和了解MATLAB語(yǔ)言的強(qiáng)大的計(jì)算功能與其在控制領(lǐng)域的 應(yīng)用優(yōu)勢(shì)。上機(jī)實(shí)驗(yàn)最終以書(shū)面報(bào)告的形式提交，作為期末成績(jī)的考核內(nèi)容。二、題目及解答第一部分：MATLAB必備基礎(chǔ)知識(shí)、控制系統(tǒng)

2、模型與轉(zhuǎn)換、線性控制系統(tǒng)的計(jì) 算機(jī)輔助分析1.考慮著名的Rossol化學(xué)反應(yīng)方程組13 = _y_ 2J7 = X I ayE = 6 1- e)z選定口 =方=也2, c =且工i(0)=上0)=工式0) =0,繪制仿真結(jié)果的三維相軌途.并得出其在wy平面上的投影.f=inline(-x(2)-x(3);x(1)+a*x(2);b+(x(1)-c)*x(3),t,x,flag,a,b,c);t,x=ode45(f,0,1 00,0;0;0,0.2,0.2,5.7);plot3(x(:,1),x(:,2),x(:,3),gridfigure,plot(x(:,1),x(:,2),grid2.求

3、解T市的最優(yōu)化問(wèn)陋min (1； - 2jtj.十央4片士君,4Hlm口之0y=(x)x(1)A2-2*x(1)+x(2);ff=optimset;ff.LargeScale=off;ff.TolFun=1e-30;ff.TolX =1e-15;ff.TolCon=1e-20;x0=1;1;1;xm=0;0;0;xM=;A=;B=;Aeq=;Beq=;x,f,c,d =fmincon(y,x0,A,B,Aeq,Beq,xm,xM,wzhfc1,ff)Warning: Options LargeScale = off and Algorithm = trust-region-reflective

4、 conflict.Ignoring Algorithm and running active-set algorithm. To run trust-region-reflective, set LargeScale = on. To run active-set without this warning, useAlgorithm = active-set. In fmincon at 456Local minimum possible. Constraints satisfied.fmincon stopped because the size of the current search

5、 direction is less than twice the selected value of the step size tolerance and constraints are satisfied to within the selected value of the constraint tolerance.Active inequalities (to within options.TolCon = 1e-20):lower upper ineqlin ineqnonlin1.00001.0000-1.0000iterations: 5 funcCount: 20Isstep

6、length: 1stepsize: 3.9638e-26algorithm: medium-scale: SQP, Quasi-Newton, line-searchfirstorderopt: 7.4506e-09constrviolation: 0message: 1x766 char3.請(qǐng)將下面的傳遞函數(shù)模型輸入到MATLAB環(huán)境.卡 一如十2s3(sa+2)(sa+l)3+2j+5(h) H(z)=-/ + 0.568(a) s=tf(s);G=(sA3+4*s+2)/(sA3*(sA2+2)*(sA2+1)A3+2*s+5)G =sA3 + 4 s + 2sA11 + 5 sA9

7、+ 9 sA7 + 2 sA6 + 12 sA5 + 4 sA4 + 12 sA3Continuous-time transfer function.(b) z=tf(z,0.1);H=(zA2+0.568)/(z-1)*(zA2-0.2*z+0.99)H =zA2 + 0.568可編輯zA3 - 1.2 zA2 + 1.19 z - 0.99Sample time: 0.1 secondsDiscrete-time transfer function.4.假設(shè)描述系統(tǒng)的常微分方程為y十1的十州+ 5y=2女的，請(qǐng)選擇一組狀態(tài)變量， 并將此方程在MATLAB工作空間中表示出來(lái),如果想得到系統(tǒng)的

8、傳遞函數(shù)和零極點(diǎn)模型， 我門(mén)將如何求??？得出的結(jié)果又是怎樣的？由微分方程模型能否直接寫(xiě)出系統(tǒng)的傳遞函數(shù)模 型？A=010;001;-15-4-13;B=002;C=100;D=0;G=ss(A,B,C,D),Gs=tf(G),Gz=zpk(G)G =a =x1 x2 x3x1010x2001x3 -15-4 -13b =u1x1 0x2 0x3 2c =x1 x2 x3y1 100d =u1y1 0Continuous-time state-space model.Gs =2sA3 + 13 sA2 + 4 s + 15Continuous-time transfer function.Gz

9、=2(s+12.78)S2 + 0.2212s + 1.174)Continuous-time zero/pole/gain model.5.已知某系統(tǒng)的差分方程模型為式A+ 2)十以/+ 1.) +06孤幻= (/+ 1.)+2注泳)，玳將其輸入 到MATLAE工作空間，設(shè)采樣周期為0.01s z=tf(z,0.01);H=(z+2)/(zA2+z+0.16)H =zA2 + z + 0.16Sample time: 0.01 secondsDiscrete-time transfer function.6.假設(shè)某單位負(fù)反饋系統(tǒng)中，G)=(呂+*+ 5), Gc(3) = (KFS + Ki

10、)/3.試用 MATLA U推導(dǎo)出閉環(huán)系統(tǒng)的傳遞函數(shù)模型. syms J Kp Ki s;G=(s+1)/(J*sA2+2*s+5);Gc=(Kp*s+Ki)/s;GG=feedback(G*Gc,1)GG =(Ki + Kp*s)*(s + 1)/(J*sA3 + (Kp + 2)*sA2 + (Ki + Kp + 5)*s + Ki) 7.從二面紿出的典型反饋控制系統(tǒng)結(jié)構(gòu)子模型中一求出總系統(tǒng)的狀態(tài)方程與傳遞函數(shù)模型一并 得出各個(gè)模型的零極點(diǎn)模型表示.211.87s+ 317.64/（占）=169.6s + 400因(占十Ji(b)35786.7-1 十 IIIS444(第-1 H- 4)

11、(2 -1 十 20) i-1 十 74.04 j11公小一”日一+ 211)( + 94.34 )(s + II.16S4)(a)s=tf(s);G=(211.87*s+317.64)/(s+20)*(s+94.34)*(s+0.1684);Gc=(169.6*s+400)/ (s*(s+4);H=1/(0.01*s+1);GG=feedback(G*Gc,H),Gd=ss(GG),Gz=zpk(GG)GG =359.3 sA3 + 3.732e04 sA2 + 1.399e05 s + 1270560.01 sA6 + 2.185 sA5 + 142.1 sA4 + 2444 sA3 +

12、4.389e04 sA2 + 1.399e05 s + 127056Continuous-time transfer function.Gd =a =x1x2x3x4x5x6x1-218.5-111.1 -29.83-16.74 -6.671-3.029x212800000x30640000x40032000x5000800x6000020b =u1x1 4x2 0x3 0x4 0x5 0x6 0x1 x2 x3 x4 x5 x6y1 001.097 3.5591.668 0.7573d =u1y1 0Continuous-time state-space model.Gz =35933.15

13、2 (s+100) (s+2.358) (s+1.499)(sA2 + 3.667s + 3.501) (sA2 + 11.73s + 339.1) (sA2 + 203.1s + 1.07e04)Continuous-time zero/pole/gain model.(b)設(shè)采樣周期為0.1sz=tf(z,0.1);G=(35786.7*zA2+108444*zA3)/(1+4*z)*(1+20*z)*(1+74.04*z);Gc=z/(1-z);H=z/(0.5-z);GG=feedback(G*Gc,H),Gd=ss(GG),Gz=zpk(GG)GG =-108444 zA5 + 1.

14、844e04 zA4 + 1.789e04 zA31.144e05 zA5 + 2.876e04 zA4 + 274.2 zA3 + 782.4 zA2 + 47.52 z + 0.5Sample time: 0.1 secondsDiscrete-time transfer function.Gd =a =x1x2x3 x4x5x1 -0.2515 -0.00959 -0.1095 -0.05318 -0.01791x20.250000x300.25000x4000.12500x50000.031250b =u1x1 1x2 0x3 0x4 0x5 0c =x1x2x3x4x5y1 0.39

15、960.6349 0.1038 0.05043 0.01698d =u1y1 -0.9482Sample time: 0.1 secondsDiscrete-time state-space model.Gz =-0.94821 zA3 (z-0.5) (z+0.33)(z+0.3035) (z+0.04438) (z+0.01355) (zA2 - 0.11z + 0.02396)Sample time: 0.1 secondsDiscrete-time zero/pole/gain model.8.Li如系統(tǒng)的方程圖如四事3所示，試推導(dǎo)出從輸入信號(hào)r(i)到輸出信號(hào)y(t)的總系統(tǒng)模型.可

16、編輯s=tf(s);g1=1/(s+1);g2=s/(sA2+2);g3=1/sA2;g4=(4*s+2)/(s+1)A2;g5=50;g6=(sA2+2)/(sA3+14);G1=feedback(g1*g2,g4);G2=feedback(g3,g5);GG=3*feedback(G1*G2,g6)GG =3 sA6 + 6 sA5 + 3 sA4 + 42 sA3 + 84 sA2 + 42 ssA10 + 3 sA9 + 55 sA8 + 175 sA7 + 300 sA6 + 1323 sA5 + 2656 sA4 + 3715 sA3 + 7732sA2 + 5602 s + 14

17、00Continuous-time transfer function.9.已知傳遞函數(shù)模型G(S)=is- l)a(sa 十 2s + 400)(5 + 5)2(s2 抬十 100) *(32 十A十 2500),對(duì)不同采槨周期T =0.01,0.1和T = i秒對(duì)之進(jìn)行離散化，比較原系統(tǒng)的階躍喃應(yīng)與各離散系統(tǒng)的階躍響應(yīng)曲線. 提示：后面將介紹，如果已知系統(tǒng)模型為G,則用BSp(G)即可繪制出其階躍響應(yīng)曲線.s=tf(s);T0=0.01;T1=0.1;T2=1;G=(s+1)A2*(sA2+2*s+400)/(s+5)A2*(sA2+3*s+100 )*(sA2+3*s+2500);Gd1

18、=c2d(G,T0),Gd2=c2d(G,T1),Gd3=c2d(G,T2),step(G)figure,ste p(Gd1),figure,step(Gd2),figure,step(Gd3)Gd1 =4.716e-05zA5 - 0.0001396zA4 + 9.596e-05zA3 + 8.18e-05zA2 - 0.0001289z +4.355e-05zA6 - 5.592 zA5 + 13.26 zA4 - 17.06 工3 + 12.58 zA2 - 5.032 z + 0.8521Sample time: 0.01 secondsDiscrete-time transfer f

19、unction.Gd2 =0.0003982zA5 - 0.0003919zA4 - 0.000336zA3 + 0.0007842zA2 - 0.000766z +0.0003214zA6 - 2.644 zA5 + 4.044 zA4 - 3.94 zA3 + 2.549 zA2 - 1.056 z + 0.2019Sample time: 0.1 secondsDiscrete-time transfer function.Gd3 =8.625e-05 zA5 - 4.48e-05 zA4 + 6.545e-06 zA3 + 1.211e -05 zA2 - 3.299e-06 z +1

20、.011e-07zA6 - 0.0419 zA5 - 0.07092 zA4 - 0.0004549zA3 + 0.002495zA2 - 3.347e-05z + 1.125e-07Sample time: 1 secondsDiscrete-time transfer function.10.判定下列連續(xù)傳遞函數(shù)模型的穩(wěn)定性。出3 + 24內(nèi)十2( 二4十左號(hào)十2理十十U)囪4+四 $+2(a) G=tf(1,1 2 1 2);eig(G),pzmap(G)ans =-2.0000-0.0000 + 1.0000i-0.0000 - 1.0000io1.?修T-m.TL f 百=一系統(tǒng)為臨

21、界穩(wěn)定(b) G=tf(1,6 3 2 1 1);eig(G),pzmap(G)ans =-0.4949 + 0.4356i-0.4949 - 0.4356i0.2449 + 0.5688i0.2449 - 0.5688i尸 KPLD:qs.3en.JffE-有一對(duì)共腕復(fù)根在右半平面，所以系統(tǒng)不穩(wěn)定(c) G=tf(1,1 1 -3 -1 2);eig(G),pzmap(G) ans =-2.0000-1.00001.00001.0000有兩根在右半平面，故系統(tǒng)不穩(wěn)定。11.打定二套*樣品統(tǒng)的已定吧(a)田:)=一舐十23 二口.24二025丁 I 0.05,3# - 0.39 - 0.09門(mén)

22、 二L7產(chǎn) 門(mén)簿公2十0.268二十0.021(1) H=tf(-3 2,1 -0.2 -0.25 0.05);pzmap(H),abs(eig(H) ans =0.50000.50000.2000系統(tǒng)穩(wěn)定(2) H=tf(3 -0.39 -0.09,1 -1.7 1.04 0.268 0.024);pzmap(H),abs(eig(H) ans =1.19391.19390.12980.1298系統(tǒng)不穩(wěn)定。12.治出連續(xù)系統(tǒng)的狀態(tài)方程模型，請(qǐng)判定系統(tǒng)的穩(wěn)定性.-0.215000 0 10-0.51.6000.T(i)=00-11.3S5.80制力十0000-33.31000.0000-10-

23、,30 J(1) A=-0.2 0.5 0 0 0;0 -0.5 1.6 0 0;0 0 -14.3 85.8 0;0 0 0 -33.3 100;0 0 0 0 -10;B=0 00 0 30;C=zeros(1,5);D=0;G=ss(ABC,D),eig(G)G =a =x1x2x3x4x5x1-0.20.5000x20-0.51.600x300-14.385.80x4000-33.3100x50000-10b =u1x1 0ans =-0.2000-0.5000-14.3000-33.3000-10.0000x2 0x3 0x4 0x5 30c =x1 x2 x3 x4 x5y1 00

24、000d =u1y1 0Continuous-time state-space model.系統(tǒng)穩(wěn)定。13.請(qǐng)求出下面自治系統(tǒng)狀態(tài)方程的解析解-320-4并和數(shù)值蕤得出的曲線比較. A=-5 2 0 0; 0 -4 0 0; -3 2 -4 -1; -3 2 0 -4; A=sym(A);syms t;x=expm(A*t)*1;2;0;1 x = 4*exp(-4*t) - 3*exp(-5*t)2*exp(-4*t)12*exp(-4*t) - 18*exp(-5*t) + 3*t*exp(-4*t) - 4*tA2*(exp(-4*t)/(4*t) + exp(-4*t)/(2*tA2)

25、+ 8*tA2*(exp(-4*t)/2 - exp(-4*t)/(2*t) - 16*t*(exp(-4*t) - exp(-4*t)/(2*t)6*exp(-4*t) - 9*exp(-5*t) - 8*t*(exp(-4*t) - exp(-4*t)/(2*t) G=ss(-5 2 0 0; 0 -4 0 0; -3 2 -4 -1; -3 2 0 -4,1;2;0;1,eye(4),zeros(4,1);tt=0:0.01:2;xx=;for i=1:length(tt)t=tt(i); xx=xx eval(x);end y=impulse(G,tt); plot(tt,xx,tt,

26、y,:)解析解和數(shù)值解的脈沖響應(yīng)曲線如圖所示，可以看出他們完全一致14.武繪制下列開(kāi)環(huán)系統(tǒng)的根軌跡曲線，并大致確定使單位負(fù)反憒系統(tǒng)稔定的K值范圍.K十6)(呂- 6) = r十小十之 4 十-4j)(反4 4 - 4j)/ + 看3 十 1&避 + 8s(a) s=tf(s);G=(s+6)*(s-6)/(s*(s+3)*(s+4-4j)*(s+4+4j);rlocus(G),grid*4 (Mcnndfl不存在K使得系統(tǒng)穩(wěn)定(b) G=tf(1,2,2,1 1 14 8 0);rlocus(G),gridQ.ii .jw；iMt ijii6oit irsj_ , ,.cj-. .j_l_7.

27、_Ii-J 41.5,5鼻.Am (aven4s1)4J5434,J54J1$5心4 n3UMJS y Wn 0Gar 5 S3 叫-Bcrifr-ii + 口1 6S1D 0-W3hWtrr S FfdE. I W9-山 16二3打3放大根軌跡圖像，可以看到，根軌跡與虛軸交點(diǎn)處，K值為5.53,因此，0K tau=2;n,d=paderm(tau,1,3);s=tf(s);G=tf(n,d)*(s-1)/(s+1)A5,rlocus(G)G =-1.5 sA2 + 4.5 s - 3sA8 + 8 sA7 + 29.5 sA6 + 65.5 sA5 + 95 sA4 + 91 sA3 + 5

28、5.5 sA2 + 19.5 s + 3Continuous-time transfer function.可編輯將匕nEW7?-*-*e： 0 SrPWS。時(shí)35 *。解電Rflnw拳口 G立4 甲Dh-nilooi 禺F iiK- n 內(nèi) F .Wt a.Ki二 * 1.13g中即伸口 3 MiZdgn口 -Q.CDlZIZ n。9nMe (Wj 1/ FnKuen： j.丘 HdJ、： 3 fr由圖得0Ks=tf(s);G=8*(s+1)/(sA2*(s+15)*(sA2+6*s+10);bode(G),figure,nyquist(G),figure,nichols(G),Gm,y,w

29、cg,wcp=margin(G),figure,step(feedback(G,1)Gm =30.46864.2340 wcg =1.5811 wcp =0.2336R*hI Ayr.ijrimih. aiar系統(tǒng)穩(wěn)定(b)z=tf(z);G=0.45*(z+1.31)*(z+0.054)*(z-0.957)/(z*(z-1)*(z-0.368)*(z-0.99);bode(G),figure,nyquist(G),figure,nichols(G),Gm,y,wcg,wcp=margin(G),figure,step(feedback(G,1)Warning: The closed-loop

30、 system is unstable. In warning at 26In DynamicSystem.margin at 63Gm =0.9578-1.7660wcg =1.0464wcp =1.0734系統(tǒng)不穩(wěn)定。17.假設(shè)受控對(duì)象模型為G(s)=，.津丫產(chǎn)石； 并假設(shè)由某種方法設(shè)計(jì)出串聯(lián)控制墨模s (1 + s/11.5J (1 + 3/&U)型為=當(dāng)歸=察丑，試用頻域響城的方法判定閉環(huán)系統(tǒng)的忖能.并用時(shí)域響 (s 十 0.5) (.5 十 50)應(yīng)檢驗(yàn)得出的結(jié)論。 s=tf(s);G=100*(1+s/2.5)/(s*(1+s/0.5)*(1+s/50);Gc=1000*(s+1)

31、*(s+2.5)/(s+0.5)*(s+50);GG=G*Gc;nyquist(GG),grid,figure,bode(GG)figure,nichols(GG),grid,figure,step(feedback(GG,1)o-l-35十工,三由奈氏圖可得，曲線不包圍（-1,j0）點(diǎn)，而開(kāi)環(huán)系統(tǒng)不含有不穩(wěn)定極點(diǎn)，所以根據(jù)奈氏穩(wěn)定判據(jù)閉環(huán)系統(tǒng)是穩(wěn)定的。用階躍響應(yīng)來(lái)驗(yàn)證，可得系統(tǒng)是穩(wěn)定的第二部分：Simulink在系統(tǒng)仿真中的應(yīng)用、控制系統(tǒng)計(jì)算機(jī)輔助設(shè)計(jì)、控制工程中的仿真技術(shù)應(yīng)用2.2考慮新單的戰(zhàn)勝微分方程y十5V+6,+ 4$十=日-+吧-也就口（盤(pán)+73）,且方程的初 值為期=1,（0=訓(xùn)

32、0） = l/2,y（0） = 02 試用Simulink搭放起系統(tǒng)的仿真模型，井繪 制出仿真結(jié)果曲線.由第2章介紹的知識(shí)，讀方程可以用微分方程數(shù)值解的形式進(jìn)行分析. 試比較二者的分析結(jié)果. syms y t;y=dsolve(D4y+5*D3y+6*D2y+4*Dy+2*y=exp(-3*t)+exp(-5*t)*sin(4*t+pi/3),y(0)=T,Dy(0)=1/2,D2y(0)=1/2,D3y(0)=1/5);tt=0:.05:10; yy=;for k=1:length(tt)ti=tt(k);yy=yy subs(y,t,ti);endplot(tout,yout,tt,yy,

33、:)3.建立起如圖57所示非線性系統(tǒng)網(wǎng)的Sinmlink框圖，并觀察在單位除跋信號(hào)輸入下系統(tǒng)的輸 出曲線和誤差曲線.圖齊了習(xí)題5的系統(tǒng)方框圖輸出曲線及誤差曲線4.已知某系統(tǒng)的Siniuli成傳真樞國(guó)如圖5/5所示，江由該梅圖當(dāng)出系統(tǒng)的數(shù)學(xué)模型公式可編輯m妃grataeIniegrat-or3*ro(lu A,B,C,D=linmod(part2_5);G=ss(ABC,D)* i/sInfegrjcac IiiiegratcriWarning: Using a default value of 0.2 for maximum step sizeT. he simulation step siz

34、e will be equal to or less than this value.You can disable this diagnostic by setting Automatic solver parameter selection diagnostic to none in the Diagnostics page of the configuration parameters dialog. In dlinmod at 172In linmod at 60a =x1x2x3x4x5x6x1000000x20-1000000x31300-100000x40200-0.88-100

35、00x50000-1000x6000294.1-29.41-149.3x70100-0.4400x8-27.5600001.045e+004x9000100-100x1000000x7x8x9x10x101.400x20000x30000x411.76000x501.400x60019.610x70000x80-6.66700x90000x100000b =u1x1 0x21x3 0x40x5 0x6 0x7 0x8 0x9 0x100c =x1x2x3x4x5x6x7x8x9x10y1 130000000000d =u1y1 0Continuous-time model.subplot(22

36、1),step(G),grid,subplot(222),bode(G),grid,subplot(223),nyquist(G),grid,subplot(224),nichols(G),grid階躍響應(yīng)和頻率響應(yīng)曲線-一工 Irili，口 i - 串Hi隼IH倬Cy 口市HCMlCfntl OlATl6.假設(shè)系統(tǒng)的對(duì)象模型為G(s) = -一八I：1” 并已知我們可以設(shè)什一個(gè)控 卜 十 Lf5J(一十 1d)(s + 1.5 j3)制器為&=W 請(qǐng)觀察在該控制器二系統(tǒng)的動(dòng)態(tài)特性：比較原系統(tǒng)和校正后 系統(tǒng)的幅值和相匝崎箝給出進(jìn)一步改進(jìn)系統(tǒng)性能的建議.s=tf(s);G=210*(s+1.5)

37、/(s+1.75)*(s+16)*(sA2+3*s+11.25);Gc=52.5*(s+1.5)/(s+14.86);GG=feedback(G*Gc,1);step(feedback(G,1),figure,step(GG),xlim(85 95) Gm,garma,wcg,wcp=margin(G)Gm =4.8921garma =60.0634wcg =7.9490wcp =3.9199 Gm,garma,wcg,wcp=margin(G*Gc)Warning: The closed-loop system is unstable. In warning at 26In DynamicS

38、ystem.margin at 63Gm =0.8090 garma =-6.0615wcg =17.1659wcp =18.90297.若系統(tǒng)的狀態(tài)方程模型為選擇加權(quán)矩陣。=/電(1,2,14)及冗=心，則設(shè)計(jì)出這一線性二次型指標(biāo)的最優(yōu)控制器及在 最優(yōu)控制下的閉環(huán)系統(tǒng)極點(diǎn)位置，并繪制出閉環(huán)系統(tǒng)各個(gè)狀態(tài)的曲線.A=0 1 0 0;0 0 1 0;-3 1 2 3;2 1 0 0;B=1 0;2 1;3 2;4 3;Q=diag(1 2 34);R=eye(2);K,P=lqr(ABQ,R),eig(A-B*K)K =-0.09781.21181.87670.7871-3.8819-0.46682.67131.0320P =5.440