控制系统仿真与CAD实验报告.docx
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控制系统仿真与CAD实验报告
《控制系统仿真与CAD
实验课程报告
、实验教学目标与基本要求
上机实验是本课程重要的实践教学环节。
实验的目的不仅仅是验证理论知识,更重要的是通过上机加强学生的实验手段与实践技能,掌握应用MATLAB/Simulink求解控制问题的方法,培养学生分析问题、解决问题、应用知识的能力和创新精神,全面提高学生的综合素质。
通过对MATLAB/Simulink进行求解,基本掌握常见控制问题的求解方法与命令调用,更深入地认识和了解MATLAB语言的强大的计算功能与其在控制领域的应用优势。
上机实验最终以书面报告的形式提交,作为期末成绩的考核内容。
二、题目及解答
第一部分:
MATLAB必备基础知识、控制系统模型与转换、线性控制系统的计算机辅助分析
1.
考虑碧名的Rossol化学反应方程组
选定d=c=且T1(O)=r2(0)=x3(0)=0.绘制仿育勢果的三维相轨:
卜并傅出其在脣丫平面上的投彭.
>>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)),grid,figure,plot(x(:
1),x(:
2)),grid
2.
求解下囲的最优化问题”
(a)min-2jti十ar空)
(4谥+鬭*''
Xfirtn<
I工1.jra>0
>>y=@(x)x(1F2-2*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:
OptionsLargeScale='offandAlgorithm='trust-region-reflective'conflict.
IgnoringAlgorithmandrunningactive-setalgorithm.Toruntrust-region-reflective,setLargeScale='on'.Torunactive-setwithoutthiswarning,useAlgorithm='active-set'.
>Infminconat456
Localminimumpossible.Constraintssatisfied.
fminconstoppedbecausethesizeofthecurrentsearchdirectionislessthantwicetheselectedvalueofthestepsizetoleranceandconstraintsaresatisfiedtowithintheselectedvalueoftheconstrainttolerance.
Activeinequalities(towithinoptions.TolCon=1e-20):
lowerupperineqlinineqnonlin
2
x=
1.0000
0
1.0000
f=
-1.0000
c=
iterations:
5funcCount:
20
Issteplength:
1
stepsize:
3.9638e-26
algorithm:
'medium-scale:
SQP,Quasi-Newton,line-search'firstorderopt:
7.4506e-09
constrviolation:
0message:
[1x766char]
3.
请持下面的传递函數欖理输入到MATLAB环境"
闾*)=科晶舞FT⑹恥)=时备瞎而护秒
(a)>>s=tf('s');G=(sA3+4*s+2)/(sA3*(sA2+2)*((sA2+1)A3+2*s+5))
sA3+4s+2
sA11+5sA9+9sA7+2sA6+12sA5+4sA4+12sA3Continuous-timetransferfunction.
(b)
>>z=tf('z',0.1);H=(zA2+0.568)/((z-1)*(zA2-0.2*z+0.99))
H=
zA2+0.568
zA3-1.2zA2+1.19z-0.99
Sampletime:
0.1seconds
Discrete-timetransferfunction.
4.
假设描述系统的常微分方程为期⑶⑹+13y(t)十4讥站+=请选择一组状态变量,
并将此方程在MATLAB工作空闻中表示出来.如果想得到系统的传递菌数和零极点模型:
我V将如何求取?
得出的结果又是怎样西?
由徴分方程模型能否直接写岀系统的传递函数模型?
>>A=[010;001;-15-4-13];B=[002]';C=[10
0];D=0;G=ss(A,B,C,D),Gs=tf(G),Gz=zpk(G)
x1
x2
x3
x1
0
1
0
x2
0
0
1
x3
-15
-4
-13
b=
u1
x1
0
x2
0
x3
2
c=
x1x2x3
y1100
d=
u1
y10
Continuous-timestate-spacemodel.
Gs=
sA3+13sA2+4s+15
Continuous-timetransferfunction.
Gz=
(s+12.78)(sA2+0.2212s+1.174)
Continuous-timezero/pole/gainmodel.
5.
已知某系统的差分方程摸型为+2)+y(k+1.)十0】切的=卷依+1.)v2u(k).试将其输入到MATLABX作空间"
设采样周期为0.01s
>>z=tf('z',0.01);H=(z+2)/(zA2+z+0.16)
H=
zA2+z+0.16
Sampletime:
0.01seconds
Discrete-timetransferfunction.
6.
假设某单位负反馈系统申,=(3十+劄+5)’Gc(3)=(KFS七瓦订2、试用MAT1-A13推导岀闭环系统的传递函数模型。
>>symsJKpKis;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.
从二面给出西典型反馈控制系统结构子模型中一求出总系统的狀态方程与传递函数模型’并得出各个模型的零极点模型表示”
(b)G(z~[
169.6.-十」IIII
“(g十4;i
211.87s+317.64
357S6.7--1十108444
(「1十」住一1十勿)心一1十74.04]
+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=
359.3sA3+3.732e04sA2+1.399e05s+127056
0.01sA6+2.185sA5+142.1sA4+2444sA3+4.389e04sA2+1.399e05s+127056
Continuous-timetransferfunction.
Gd=
a=
x1
x2
x3
x4
x5
x6
x1
-218.5
-111.1-29.83
-16.74
-6.671
-3.029
x2
128
0
0
0
0
0
x3
0
64
0
0
0
0
x4
0
0
32
0
0
0
x5
0
0
0
8
0
0
x6
0
0
0
0
2
0
b=
u1
x1
4
x2
0
x3
0
x4
0
x5
0
x6
0
x1
x2
x3
x4
x5x6
y1
0
0
1.097
3.559
1.6680.7573
y1
u1
0
Continuous-timestate-spacemodel.
Gz=
35933.152(s+100)(s+2.358)(s+1.499)
(sA2+3.667s+3.501)(sA2+11.73s+339.1)(sA2+203.1s+1.07e04)Continuous-timezero/pole/gainmodel.
(b)设采样周期为0.1s>>z=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=
-108444zA5+1.844e04zA4+1.789e04zA3
1.144e05zA5+2.876e04zA4+274.2zA3+782.4zA2+47.52z+0.5
Sampletime:
0.1seconds
Discrete-timetransferfunction.
Gd=
x1
x2
x3
x4
x5
x1
x2
x3
-0.2515-0.00959-0.1095-0.05318
0.25
0.25
0.125
x4
-0.01791
0.03125
x5
b=
u1x11x20x30x40x50
x1
x2
x3
x4
x5
y10.39960.63490.10380.050430.01698
u1
y1-0.9482
Sampletime:
0.1seconds
Discrete-timestate-spacemodel.
Gz=
-0.94821zA3(z-0.5)(z+0.33)
(z+0.3035)(z+0.04438)(z+0.01355)宫2-0.11z+0.02396)
Sampletime:
0.1seconds
Discrete-timezero/pole/gainmodel.
8.
lL知系统的方框图如图匸3:
斤几试推导岀从输入信号到输出倍号以"的总杀统模型。
>>s=tf('s');g仁1/(s+1);g2=s/(sA2+2);g3=1/sA2;g4=(4*s+2)/(s+1)A2;g5=50;g6=(sA2+
2)/(sW+14);G仁feedback(g1*g2,g4);G2=feedback(g3,g5);GG=3*feedback(G1*G2,g6)
GG=
3sA6+6sA5+3sA4+42sA3+84sA2+42s
sA10+3sA9+55sA8+175sA7+300sA6+1323sA5+2656sA4+3715sA3+7732
sA2+5602s+1400
Continuous-timetransferfunction.
9.
is-l)a(sa十2s十400)
砾传递函数模型&⑶7仆亠"汕二亠]十皿对不同采擇周取“
0.01,0.1和秒对之进行高散化,比较原系统的廉跃响应与各离散系统的阶跃啊应曲域.
提示:
后面将介绍,如果已知系统撲型为G,则Step(G[即可绘制出其阶跃响应曲线.
>>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));Gd仁c2d(G,T0),Gd2=c2d(G,T1),Gd3=c2d(G,T2),step(G),figure,step(Gd1),figure,step(Gd2),figure,step(Gd3)
Gd1=
4.716e-05zA5-0.0001396乙八4+9.596e-05zA3+8.18e-05zA2-0.0001289z+4.355e-05
zA6-5.592zA5+13.26乙八4-17.06乙八3+12.58乙八2-5.032z+0.8521
Sampletime:
0.01seconds
Discrete-timetransferfunction.
Gd2=
0.0003982zA5-0.0003919zM-0.000336zW+0.0007842zA2-0.000766z+
0.0003214
zA6-2.644zA5+4.044zA4-3.94zA3+2.549zA2-1.056z+0.2019
Sampletime:
0.1seconds
Discrete-timetransferfunction.
Gd3=
8.625e-05zA5-4.48e-05zA4+6.545e-06zA3+1.211e-05zA2-3.299e-06z+
1.011e-07
zA6-0.0419zA5-0.07092zA4-0.0004549zA3+0.002495zA2-3.347e-05z+
1.125e-07
Sampletime:
1seconds
Discrete-timetransferfunction.
辛」.==
O
■!
«.?
•£T-・Jr.T*lfi^F±
10.
判定下列逹续传递函数模型的稳定性。
(町嗣+2辭十应十2庆4十艮3+2辭十書十」J)曲十昇—3屛一总十2
(a)>>G=tf(1,[1212]);eig(G),pzmap(G)ans=
-2.0000
-0.0000+1.0000i
-0.0000-1.0000i
系统为临界稳定
(b)>>G=tf(1,[63211]);eig(G),pzmap(G)
ans=
-0.4949+0.4356i
-0.4949-0.4356i
0.2449+0.5688i
0.2449-0.5688i
^Dle-ZeTiMap
期4.1
有一对共轭复根在右半平面,所以系统不稳定
ar
JL4
L£-CQUS-JeE-Jff'E-
(c)>>G=tf(1,[11-3-12]);eig(G),pzmap(G)ans=
-2.0000
-1.0000
1.00001.0000
有两根在右半平面,故系统不稳定。
11.
判足F雹采样系统的伦足世
(a)坦二)=
—迟十2
d二0.2日二。
25匸I0.05’
(•>)卅(■=)=
3尹—0.3恥-0.09
-1.7z3\~lMz2+0.268c+0,02J
⑴>>H=tf([-32],[1-0.2-0.250.05]);pzmap(H),abs(eig(H'))ans=
0.5000
0.5000
0.2000
系统稳定
⑵>>H=tf([3-0.39-0.09],[1-1.71.040.2680.024]);pzmap(H),abs(eig(H'))ans=
1.1939
1.1939
0.1298
0.1298
系统不稳疋
12.
給出连续系统的状态方程模型,请判定系统的稳定性.
■-0.2
0
0
01
•01
0
-0.5
1.6
0
0
0
(a).=
0
0
—1-1.3
S5.8
0
x(t)十
0
ti(f)
0
0
0
—33.3
100
0
.0
0
0
0
-10-
.30J
(1)>>A=[-0.20.5000;0-0.51.600;00-14.385.80;000-33.3100;0000-10];B=[000030]';C=zeros(1,5);D=0;G=ss(A,B,C,D),eig(G)
x1
x2
x3
x4
x5
x1
-0.2
0.5
0
0
0
x2
0
-0.5
1.6
0
0
x3
0
0
-14.3
85.8
0
x4
0
0
0
-33.3
100
x5
0
0
0
0
-10
b=
u1
x10ans=
-0.2000
-0.5000
-14.3000