计算机控制实验2.docx
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计算机控制实验2.docx
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计算机控制实验2
练习2-1.给定系统的传递函数如下:
求该系统的阶跃响应曲线,记录超调量、上升时间、过渡过程时间。
程序如下:
%%%%%%%%%%%%%%%
clearall;clc;
num=[25];den=[1425];
s=tf(num,den);
step(s)
%%%%%%%%%%%%%%%
练习2-2.已知系统的开环传递函数为:
求出该系统在单位负反馈下的阶跃响应曲线,记录超调量、上升时间、过渡过程时间。
程序如下:
%%%%%%%%%%%%%%%%%%%%%%%
clearall;clc;
num=[20];den=[1836400];
[num,den]=cloop(num,den,-1);
s=tf(num,den);
step(s)
%%%%%%%%%%%%%%%%%%%%%%%
练习2-3已知系统的传递函数为:
+
——
1求系统的阶跃响应;
2阶跃响应曲线线型用“*”号表示;
3阶跃响应图应加上横坐标名、纵坐标名和标题名,并加上网格线。
程序如下:
%%%%%%%%%%%%%%%%%%%%%%%%%
clearall;clc;
num1=[6.3233];den1=[1];
num2=[11.4235];den2=[10];
num3=[11.4235];den3=[10];
num4=[1];den4=[1650];
num=conv(num1,conv(num2,conv(num3,num4)));
den=conv(den1,conv(den2,conv(den3,den4)));
[num,den]=cloop(num,den,-1);
s=tf(num,den);
step(s)
%%%%%%%%%%%%%%%%%%%%%
练习2-4求T1、T2、T3系统的阶跃响应;
1将T1、T2、T3系统的阶跃响应图画在同一窗口内;
2T1、T2、T3系统的阶跃响应曲线分别用不同的线形和颜色表示;
3将‘T1、T2、T3’分别标注在对应的曲线上。
程序如下:
%%%%%%%%%%%%%%%%%%%%%%%
clearall;clc;
num1=[2];den1=[122];
num2=[42];den2=[122];
num3=[1];den3=[2331];
t1=tf(num1,den1);t2=tf(num2,den2);t3=tf(num3,den3);
holdon;
step(t1);
step(t2);step(t3);
holdoff;
%%%%%%%%%%%%%%%%%%%%%%%%
练习2-5一个系统的状态空间描述如下:
①
求出G(S)=Y(S)/U(S);
4绘制该状态方程的单位阶跃响应曲线。
程序如下:
%%%%%%%%%%%%%%%%%%%%%%
clearall;clc;
A=[-1-1;6.50];B=[11;10];
C=[10;01];D=[00;00];
[num,den]=ss2tf(A,B,C,D,2);
printsys(num,den);
step(num,den)
%%%%%%%%%%%%%%%%%%%%%%
练习2-6典型二阶欠阻尼系统的传递函数为:
极点位置:
式中:
①设ωa=1,σ=0.5,1,5,求阶跃响应;
②设σ=1,ωa=0.5,1,5,求阶跃响应;
③设:
求阶跃响应;
5
设求阶跃响应;
6阶跃响应对应的时间:
t=0至t=10,分析参数变化(增加、减少与不变)对阶跃响应的影响。
1.
%%%%%%%%%%%%%%%%%%
clearall;clc;
wa=1;se=0.5;
num2=[(wa)^2+(se)^2];den2=[12*se(wa)^2+(se)^2];
step(num2,den2);
%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%
clearall;clc;
wa=1;se=1.5;
num2=[(wa)^2+(se)^2];den2=[12*se(wa)^2+(se)^2];
step(num2,den2);
%%%%%%%%%%%%%%%%%%%%
2.
%%%%%%%%%%%%%%%%%
clearall;clc;
wa=0.5;se=1;
num2=[(wa)^2+(se)^2];den2=[12*se(wa)^2+(se)^2];
step(num2,den2);
%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%
clearall;clc;
wa=1.5;se=1;
num2=[(wa)^2+(se)^2];den2=[12*se(wa)^2+(se)^2];
step(num2,den2);
%%%%%%%%%%%%%%%%%
3.
%%%%%%%%%%%%%%%%%%%
clearall;clc;
wn=2^0.5/2;e=1/2^0.5;
num1=[wn^2];den1=[12*e*wnwn^2];
step(num1,den1);
%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%
clearall;clc;
wn=2^0.5;e=1/2^0.5;
num1=[wn^2];den1=[12*e*wnwn^2];
step(num1,den1);
%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%
clearall;clc;
wn=5/2^0.5;e=1/2^0.5;
num1=[wn^2];den1=[12*e*wnwn^2];
step(num1,den1);
%%%%%%%%%%%%%%%%%%
4.
%%%%%%%%%%%%%%%%%%
clearall;clc;
seta=30/180*pi;
wn=2^0.5;e=cos(seta);
num1=[wn^2];den1=[12*e*wnwn^2];
step(num1,den1);
%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%
clearall;clc;
seta=45/180*pi;
wn=2^0.5;e=cos(seta);
num1=[wn^2];den1=[12*e*wnwn^2];
step(num1,den1);
%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%
clearall;clc;
seta=60/180*pi;
wn=2^0.5;e=cos(seta);
num1=[wn^2];den1=[12*e*wnwn^2];
step(num1,den1);
%%%%%%%%%%%%%%%%%%
5.由上述实验结果可得ωa不变时;σ的增加导致系统由二阶震荡变为一阶系统。
超调量消失,稳定时间减小。
σ不变时,ωa增加时,导致系统调节时间减少,但出现超调量,系统变为二阶震荡;ζ不变时,Wn的增加导致调节时间降低。
θ增加时,调节时间边长,且出现超调量,超调量变大。
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