東北大學(xué) 自動化 控制系統(tǒng)計算機輔助設(shè)計實驗

1、 控制系統(tǒng)計算機輔助設(shè)計 東北大學(xué)11控制系統(tǒng)計算機輔助設(shè)計_x2014_x2014_MATLAB語言與應(yīng)用控制系統(tǒng)計計算機輔助助設(shè)計第一部分1 functtion dx=llorennzeq(t,x)dx=-x(2)-x(33);x(1)+0.2*x(2);0.2+(x(1)-5.77)*x(3); x00=0;0;0; tt,y=ode445(llorennzeq,0,100,x0); pllot(tt,y) fiiguree;ploot3(yy(:,11),y(:,2),y(:,3),griid,2functtion y,yyeq=f2a(x)yeq=;y=4*xx(1)2+x(2)22

2、-4; Aeeq=;Beqq=;A=;B=; xmm=0;0; xM=;x00=0;0; f11=inlline(x(11)2-2*x(1)+xx(2); xx,f=fminncon(f1,xx0,A,B,Aeeq,Beeq,xmm,xM,f2aa);xx,fans = 11.00000 0f =-13（a） s=tf(s);G=(ss3+44*s+22)/s3/(ss2+22)/(s2+1)33+2*ss+5Transsfer funcctionn: ss3 + 4 ss + 22s11 + 5 s9 + 9 s7 + 2 s6 + 122 s55 + 44 s44 + 112 s3（b）

3、z=tf(z,00.1); H=(z22+0.5568)/(z-11)/(zz2-00.2*zz+0.999)Transsfer funcctionn: z2 + 0.568z3 - 1.22 z22 + 11.19 z - 0.999Samplling timee: 0.14 A=0 11 0;00 0 11;-5 -4 -13; B=0;00;2; C=1 00 0;00 0 00;0 00 0; D=0; G=ss(AA,B,CC,D); Ga = xx1 x2 x3 x11 0 1 0 x22 0 0 1 x33 -5 -4 -13b = u11 x11 00 x22 00 x33 2

4、2c = x11 x22 x33 y11 11 00 00 y22 00 00 00 y33 00 00 00d = u11 y11 00 y22 00 y33 00Contiinuouus-tiime mmodell. G=tf(GG)Transsfer funcctionn froom innput to ooutpuut 22 #1: s33 + 113 s2 + 4 s + 5 #2: 0 #3: 0 GGG=zpkk(G)Zero/polee/gaiin frrom iinputt to outpput 2 #1: (s+12.772) (s2 + 0.28366s + 0.3993

5、2) #2: 0 #3: 0根據(jù)微分方方程也可以以直接寫出出傳遞函數(shù)數(shù)模型： nuum=22; deen=11,13,4,5; G=tf(nnum,dden); GTranssfer funcctionn: 2s3 + 13 s2 + 4 s + 5 GGG=zpkk(G)Zero/polee/gaiin: 22(s+122.72) (s2 + 0.28836s + 0.39322)5 nuum=11,2; deen=11,1,00.16; H=tf(nnum,dden,Ts,1); HTranssfer funcctionn: zz + 22z2 + z + 0.116Samplling t

6、imee: 16functtion H=feeedbaack(GG1,G22,keyy)if naarginn=2;key=-1;eend,HH=G1/(symm(1)-key*G1*GG2);HH=simmple(H); ssyms J Kpp Ki s; gcc=(Kpp*s+KKi)/ss; g=(s+11)/(JJ*s22+2*ss+5); ggg=feeedbacck(g*gc,11) ggg=feeedbacck(g*gc,11)gg =(Ki + Kpp*s)*(s + 1)/(J*s3 + (KKp + 2)*ss2 + (Kii + KKp + 5)*ss + KKi)7（a

7、） s=tf(s); G=(2111.87*s+3117.644)/(ss+20)/(s+94.334)/(s+0.16844); Gcc=(1669.6*s+4000)/ss/(s+4); Hss=1/(0.011*s+11); GGG=feeedbacck(G*Gc,HHs)Transsfer funcctionn: 3359.33 s33 + 33.7322e0044 s22 + 11.3999e0055 s + 12770560.01 s6 + 2.185 s5 + 1442.1 s4 + 24444 ss3 + 4.3389e0004 ss2 + 1.3999e0005 s + 12

8、270566 zppk(GGG)Zero/polee/gaiin: 359933.1152 (s+1000) (s+2.358) (s+1.4999)(s2 + 33.6677s + 3.5001) (s2 + 111.733s + 339.1) (s2 + 2203.11s + 1.077e0044)（b） z=tf(z); G=(357786.77*z-1+10084444)/(z-11+4)*(z-1+200)*(zz-1+74.004); Gcc=1/(z-11-1); H=1/(00.5*zz-1-1); GGG=feeedbacck(G*Gc,HH)Transsfer funcct

9、ionn: -11084444 z6 + 1.8444e0004 z5 + 1.7889e0004 z41.1444e0055 z66 + 22.8766e0044 z55 + 2274.22 z44 + 7782.44 z33 + 447.522 z22 + 00.5 zz Samplling timee: unnspeccifieed zppk(GGG)Zero/polee/gaiin: -0.944821 z4 (z-00.5) (z+00.33)z (z+0.30035) (z+00.044438) (z+00.013355) (z22 - 0.111z + 0.022396)Samp

10、lling timee: unnspeccifieed8 s=tf(s);c1=feeedbaack(11/(s+1)*ss/(s2+2),(4*s+2)/(s+1)22);c2=feeedbaack(11/s22,50);G=feeedbacck(c11*c2,(s22+2)/(s33+14)Transsfer funcctionn: s6 + 2 s5 + s4 + 144 s33 + 228 s2 + 14 sss10 + 3 s9 + 555 s88 + 1175 ss7 + 3000 s66 + 11323 s5 + 26656 ss4 + 37115 s3 + 77732 ss2

11、+ 56002 s + 14400 9 s=tf(s); G=(s+11)2*(s22+2*ss+4000)/(ss+5)2/(ss2+33*s+1100)/(s22+3*ss+25000); G11=c2dd(G,00.01)Transsfer funcctionn:4.7166e-0005 z5 - 0.000013996 z4 + 9.5996e-0005 zz3 + 8.118e -0005 z2 - 0.000012889 z + 4.355ee-0055 z6 - 5.5592 zz5 + 13.26 zz4 - 17.06 zz3 + 12.58 zz2 - 5.0032 zz

12、+ 0.85211Samplling timee: 0.01steep(G11) G22=c2dd(G,00.1)Transsfer funcctionn:0.000039822 z55 - 00.000039199 z44 - 00.0000336 z3 + 0.00077842 z2 - 0.0007766 zz + 00.000032144z6 - 2.6644 zz5 + 4.0044 zz4 - 3.994 z3 + 2.5449 z2 - 1.0556 z + 0.20199Samplling timee: 0.1steep(G22) G33=c2dd(G,11)Transsfer

13、 funcctionn:8.6255e-0005 z5 - 4.488e-0005 z4 + 6.5445e-0006 zz3 + 1.2211e-005 z2 - 33.2999e-0006 z + 1.011ee-0077z6 - 0.00419 z5 - 0.070992 z4 - 0.000045449 z3 + 0.00024955 z22 - 33.3477e-0005 z + 1.125ee-0077 Samplling timee: 1 sttep(GG3)10（a） G=1/(ss3+22*s22+s+22);pzmmap(GG)系統(tǒng)極點均均在虛軸左左側(cè)，系統(tǒng)統(tǒng)穩(wěn)定（b） G

14、=1/(66*s44+3*ss3+22*s22+s+11); pzzmap(G)系統(tǒng)極點在在虛軸右側(cè)側(cè)側(cè)，系統(tǒng)統(tǒng)不穩(wěn)定（c） G=1/(ss4+ss3-33*s22-s+22); pzzmap(G)系統(tǒng)極點在在虛軸右側(cè)側(cè)側(cè)，系統(tǒng)統(tǒng)不穩(wěn)定11（a） z=tf(z,00.1); H=(-3*z+2)/(z3-0.2*z2-0.25*zz+0.005); pzzmap(H)系統(tǒng)所有極極點均在單單位圓內(nèi)，所所以系統(tǒng)穩(wěn)穩(wěn)定（b） z=tf(z,00.1); H=(3*zz2-00.39*z-0.09)/(z44-1.77*z33+1.004*z2+0.268*z+0.024); pzzmap(H)系統(tǒng)所

15、有極極點不全單單位圓內(nèi)，所所以系統(tǒng)不不穩(wěn)定12 A=-0.2 0.5 0 0 0;0 -00.5 11.6 00 0;00 0 -14.33 85.8 0;0 0 0 -333.3 100;0 0 0 0 -10; B=0;00;0;00;30; C=0 00 0 00 0; G=ss(AA,B,CC,0)a = x1 x2 xx3 x44 x5 x11 -0.2 00.5 0 00 0 x22 0 -00.5 1.6 00 0 x33 0 0 -14.3 85.88 0 x44 0 0 0 -33.33 100 x55 0 0 0 00 -10b = u11 x11 00 x22 00 x3

16、3 00 x44 00 x55 300c = x11 x22 x33 x44 x55 y11 00 00 00 00 00d = u11 y11 00Contiinuouus-tiime mmodell. pzzmap(G)系統(tǒng)所有極極點均在虛虛軸左側(cè)，所所以系統(tǒng)穩(wěn)穩(wěn)定或 eiig(G)ans = -00.20000 -00.50000 -144.30000 -333.30000 -100.00000極點均在左左半軸，所所以系統(tǒng)穩(wěn)穩(wěn)定13數(shù)值解： ff=(tt,x)-5*xx(1)+2*x(2);-4*x(2);-3*x(1)+22*x(22)-4*x(3)-x(44);-33*x(11)+2

17、*x(2)-4*xx(4); t_finaal=100; x00=1 2 0 0; tt,x=ode445(f,0,tt_finnal,x0); pllot(tt,x)解析解functtion Ga,Xa=ss_aaugmeent(GG,cc,dd,XX)G=ss(G);AAa=G.a;Caa=G.cc;Xa=X;Baa=G.bb;D=GG.d;if (llengtth(ddd)0&sum(abs(dd)1e-55), iif (aabs(ddd(4)1ee-5), AAa=AAa ddd(2)*Ba,ddd(3)*Ba;. zzeross(2,llengtth(Aaa),dd(11),-dd

18、d(4);dd(4),ddd(1); CCa=CCa ddd(2)*D ddd(3)*D;XXa=XXa;1;0;BBa=BBa;0;0; eelse, AAa=AAa ddd(2)*B;zeeros(1,leengthh(Aa) ddd(1); CCa=CCa ddd(2)*D;XXa=XXa;1;Ba=B;00; eendendif (llengtth(ccc)0&sum(abs(cc)1e-5),MM=lenngth(cc); AAa=AAa Baa zerros(llengtth(Aaa),M-1);zzeross(M-11,lenngth(Aa)+1) eeye(MM-1);zero

19、os(1,lenggth(AAa)+MM); CCa=CCa D zeroos(1,M-1);Xaa=Xaa;cc(1);ii=11; ffor ii=2:MM,ii=ii*ii;Xa(lenggth(AAa)+ii)=ccc(i)*ii; eend,endGa=sss(Aa,zeroos(siize(CCa),Ca,D); ccc=2;dd=-3,0,2,2;xx0=11;2;00;1; A=-5,2,0,0;0,-4,00,0;-3,2,-4,-1;-33,2,00,-4;B=0;0;0;1;C=2 1 0 0;D=00; G=ss(AA,B,CC,D); GGa,xxx0=sss_auu

20、gmennt(G,cc,ddd,x00);Gaa.a,xxx0ans = -5 2 0 0 0 0 0 0 -4 0 0 0 0 0 -3 2 -4 -1 0 0 0 -3 2 0 -4 0 2 1 0 0 0 0 -3 -2 0 0 0 0 0 2 -3 0 0 0 0 0 0 0 0ans = 1 2 0 1 1 0 2 syyms tt;y=GGa.c*expmm(Ga.a*t)*xx00; laatex(y); llatexx(y)ans =-6,e-5,tt+110,e-4,tt14（a） s=tf(s); g=(s+66)*(ss-6)/s/(ss+3)/(s+44+4i)/(s+

21、4-4ii); rllocuss(g)（b） s=tf(s); G=(s22+2*ss+2)/(s44+s33+14*s2+8*s); rllocuss(G)K s=tf(s); G=(s-11)/(ss+1)5; G.ioDeelay=2Transsfer funcctionn: ss - 11exp(-2*s) * ss5 + 5 ss4 + 10 s3 + 100 s22 + 5 s + 1 rrlocuus(paade(GG,2)16（a） s=tf(s); G=8*(ss+1)/s2/(s+115)/(s2+6*s+10); nyyquisst(G), nyyquisst(G),gr

22、iid boode(GG) fiiguree;niccholss(G),gridd ggm,pmm,wg,wp=marggin(GG)gm = 300.46886pm = 44.23440wg = 11.58111wp =0.23336 GGG=feeedbacck(G,1)Transsfer funcctionn: 8 s + 8s5 + 21 s4 + 1000 s3 + 150 s2 + 8 s + 8 pzzmap(GG)通過以上的的分析，可可以看出該該系統(tǒng)是穩(wěn)穩(wěn)定的。采用階躍響響應(yīng)進行驗驗證如下圖圖： pzzmap(GG) sttep(GGG)（b） Z=-1.31;-0.0554;

23、0.957; P=0;11;0.3368;00.99; G=zpk(Z,P,0.455,Tss,0.1)Zero/polee/gaiin:0.45 (z+11.31) (z-0.9557) (z+0.054) z (z-11) (zz-0.999) (z-0.368)Samplling timee: 0.1 nyyquisst(G);griid boode(GG) niicholls(G),griid ggm,pmm,wg,wp=marggin(GG)gm = 00.95778pm = -11.76660wg = 100.46441wp = 100.73440 GGG=feeedbacck(G

24、,1); pzzmap(GG) sttep(GGG)17 z=-2.5;pp=0;-0.55;-500; G=zpk(z,p,100/2.5*0.5*50); z=-1;-2.55;p=-0.5;-550Gc=zpk(z,p,10000); GGG=feeedbacck(G*Gc,11) Zero/polee/gaiin: 100000000 (ss+1) (s+22.5)2(s+1) (s2 + 4.999s + 6.2339) (s2 + 955.01ss + 11.0022e0066) boode(GGG)通過bodde圖可以以看出該系系統(tǒng)是穩(wěn)定定的。驗證如下： sttep(GGG)第二

25、部分2 Y=dsollve(D4y+5*D33y+6*D2y+4*Dyy+2*yy=expp(-3*t)+eexp(-5*t)*sinn(4*tt+pi/3),.)y(0)=1,Dy(0)=11/2,D2yy(0)=1/2,D33y(0)=0.22); laatex(Y)3Simullink仿仿真圖形：系統(tǒng)輸出曲曲線：系統(tǒng)誤差曲曲線：4x1(t)=sinn(x2(t)*eexp(-2.3*x4(tt)x2(t)=x1(t)x3=siin(x22*expp(-2.3*x44)x4=x335Simullink仿仿真圖形：階躍響應(yīng)曲曲線：6 s=tf(s);g=2210*(s+1.5)/(s+11.75)*(s+16)*(s+11.5+33*j)*(s+11.5-33*j); gcc=52.5*(ss+1.55)/(ss+14.86); sttep(ffeedbback(g,1) sttep(ffeedbback(g*gcc,1)通過兩個圖圖比較可以以看出，原原系統(tǒng)超調(diào)調(diào)量過大，震震蕩嚴(yán)重，加加入控制器器后，系統(tǒng)統(tǒng)變得不穩(wěn)穩(wěn)定。7 A=0 11 0 00;0 00 1 00;-3 1 2 3;2 1 0 0; B=1 00;2 11;3 22;4 33; Q=diagg(1,2,3,4); R=eye(2); KK,S=lqr(A,B,Q