neu控制系统仿真CAD作业Word文档下载推荐.docx
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x=
1.0000
0
f_opt=
-1
3.
(a)
s=tf('
s'
);
G=(s^3+4*s+2)/(s^3*(s^2+2)*((s^2+1)^3+2*s+5))
Transferfunction:
s^3+4s+2
------------------------------------------------------
s^11+5s^9+9s^7+2s^6+12s^5+4s^4+12s^3
(b)
z=tf('
z'
0.1);
H=(z^2+0.568)/(z-1)/(z^2-0.2*z+0.99)
z^2+0.568
-----------------------------
z^3-1.2z^2+1.19z-0.99
Samplingtime:
0.1
4.手工运算整理得
令x1=y;
x2=dx1/dt;
x3=dx2/dt;
dx3/dt=2*u-5*x1-4*x2-13*x3;
A=[0,1,0;
0,0,1;
-5,-4,-13];
即可得出状态方程:
[dx1/dt;
dx2/dt;
dx3/dt]=A*[x1;
x2;
x3]+[0;
2];
将上述状态方程输入到MATLAB命令窗口;
调用tf(G),zpk(G)指令即可得出传递函数和零极点模型
A=[010;
001;
-5,-4,-13];
B=[0;
C=[100];
D=0;
G=ss(A,B,C,D)
a=
x1x2x3
x1010
x2001
x3-5-4-13
b=
u1
x10
x20
x32
c=
y1100
d=
y10
Continuous-timemodel.
tf(G)
2
----------------------
s^3+13s^2+4s+5
zpk(G)
Zero/pole/gain:
----------------------------------
(s+12.72)(s^2+0.2836s+0.3932)
当然可以由微分方程直接得出传递函数;
用拉普拉斯变换法可直接输入:
G=tf(2,[11345])
5.采用z变换方法;
设采样周期为1s
1);
G=(z+2)/(z^2+z+0.016)
z+2
---------------
z^2+z+0.016
1
6.
symsGJGcKpKis
G=(s+1)/(J*s^2+2*s+5);
Gc=(Kp*s+Ki)/s;
GG=feedback(G*Gc,1)
GG=
(s+1)*(Kp*s+Ki)/(J*s^3+2*s^2+5*s+Kp*s^2+s*Ki+Kp*s+Ki)
7.
G=tf([211.87,317.64],conv(conv([1,20],[1,94.34]),[1,0.1684]));
Gc=tf([169.6,400],[1,4,0]);
H=tf(1,[0.01,1]);
GG=feedback(G*Gc,H)
359.3s^3+3.732e004s^2+1.399e005s+127056
----------------------------------------------------------------------------------
0.01s^6+2.185s^5+142.1s^4+2444s^3+4.389e004s^2+1.399e005s+127056
zpk(GG)
35933.152(s+100)(s+2.358)(s+1.499)
--------------------------------------------------------------------------
0.02(s^2+3.667s+3.501)(s^2+11.73s+339.1)(s^2+203.1s+1.07e004)
(b)
z=tf('
G=(35786.7*z^-1+108444)/(z^-1+4)/(z^-1+20)/(z^-1+74.04);
Gc=1/(z^-1-1);
H=1/(0.5*z^-1-1);
-108444z^6+1.844e004z^5+1.789e004z^4
-------------------------------------------------------------------------
1.144e005z^6+2.876e004z^5+274.2z^4+782.4z^3+47.52z^2+0.5z
zpk(GG)
-0.94821z^4(z-0.5)(z+0.33)
-------------------------------------------------------------
z(z+0.3035)(z+0.04438)(z+0.01355)(z^2-0.11z+0.02396)
8.
g1=tf(1,[1,1]);
gc1=tf([1,0],[1,0,2]);
h1=tf([4,2],[1,2,1]);
G=feedback(g1*gc1,h1)
s^3+2s^2+s
--------------------------------------
s^5+3s^4+5s^3+11s^2+8s+2
g2=tf(1,[1,0,0]);
h2=50;
Gc=feedback(g2,h2)
--------
s^2+50
H=tf([1,0,2],[1,0,0,14]);
GG=3*feedback(G*Gc,H)
3s^6+6s^5+3s^4+42s^3+84s^2+42s
-----------------------------------------------------------------------------------------------------
s^10+3s^9+55s^8+175s^7+300s^6+1323s^5+2656s^4+3715s^3+7732s^2+5602s+1400
9.
num=conv(conv([1,1],[1,1]),[1,2,400]);
den=conv(conv(conv([1,5],[1,5]),[1,3,100]),[1,3,2500]);
G=tf(num,den);
step(G)
grid;
holdon;
G1=c2d(G,0.01);
step(G1)
G2=c2d(G,0.1);
step(G2)
G3=c2d(G,1);
step(G3)
10.(a)>
G=tf(1,[1,2,1,2]);
eig(G)
ans=-2.00000.0000+1.0000i0.0000-1.0000i
临界稳定
(b)>
G=tf(1,[6,3,2,1,1]);
ans=-0.4949+0.4356i-0.4949-0.4356i0.2449+0.5688i0.2449-0.5688i
不稳定
(c)>
G=tf(1,[4,1,-3,-1,2]);
ans=-0.8057+0.4664i-0.8057-0.4664i0.6807+0.3372i0.6807-0.3372i
11
(a)>
H=(-3*z+2)/(z^3-0.2*z^2-0.25*z+0.05);
abs(eig(H))
ans=0.50000.50000.2000
稳定
H=(3*z-0.39*z-0.09)/(z^4-1.7*z^3+1.04*z^2+0.268*z+0.024);
ans=
1.19391.19390.12980.1298
12.
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=[000030]'
;
rank(ctrb(A,B))
ans=5完全可控
13
显然,这个自治的微分方程组的解析解为x(t)=x(0)*expm(A*t)
A=[-5200;
0-400;
-32-4-1;
-320-4];
x0=[1201]'
symsst;
x=expm(A*t)*x0
-3*exp(-5*t)+4*exp(-4*t)
2*exp(-4*t)
18*exp(-4*t)-18*exp(-5*t)-18*t*exp(-4*t)+4*t^2*exp(-4*t)
10*exp(-4*t)-9*exp(-5*t)-8*t*exp(-4*t)
由于符号运算不能图解,将x分成四个表达式,用离散化的t向量为为自变量,绘图
t=[0:
0.02:
x1=-3*exp(-5*t)+4*exp(-4*t);
x2=2*exp(-4*t);
x3=18*exp(-4*t)-18*exp(-5*t)-18*t.*exp(-4*t)+4*t.^2.*exp(-4*t);
x4=10*exp(-4*t)-9*exp(-5*t)-8*t.*exp(-4*t);
plot(t,x1,t,x2,t,x3,t,x4),grid
数值解法:
f=@(t,x)[-5200;
0-400;
-32-4-1;
-320-4]*x;
[t,y]=ode45(f,[0,2],[1201]'
可见,数值解与解析解很接近。
14.
G=tf(conv([1,6],[1,-6]),conv(conv(conv([1,0],[1,3]),[1,4-4*j]),[1,4+4*j]))
rlocus(G)
增益Gain从0~+无穷,在右半平面都存在不稳定极点,因此不论开环增益为多大,系统始终不稳定。
不妨当增益为1时,求闭环系统的特征根。
GG=feedback(G,1);
eig(GG)'
-3.9404-4.2387i-3.9404+4.2387i-3.43230.3132(右半平面的极点,不稳定)
G=tf([1,2,2],[1,1,14,8]);
rlocus(G)
从根轨迹可以看出,所有根轨迹均在s左半平面,因此无论增益为多大,闭环系统始终稳定
15.
>
G=(s-1)/(s+1)^5;
G.ioDelay=2;
G=pademod(G,1,2);
rlocus(G)为了方便看清晰临界稳定增益,此处截取了部分根轨迹,从图中可见,当增益Gain<
1时,闭环系统稳定
16.(a)>
G=8*(s+1)/s^2/(s+15)/(s^2+6*s+10);
nyquist(G)
bode(G),grid;
[gm,pm,wg,wp]=margin(G)
gm=30.4686pm=4.2340wg=1.5811Wp=0.2336
nichols(G)以下是局部Nichols图,上面反映了相位裕度和幅值裕度
相位裕量和幅值裕量均大于零,因此单位反馈构成的闭环系统稳定
step(feedback(G,1))
H=0.45*(z+1.31)*(z+0.45)*(z-0.957)/z/(z-1)/(z-0.368)/(z-0.99);
nyquist(H)
bode(H),gridon
[gm,pm,wg,wp]=margin(H)
Warning:
Theclosed-loopsystemisunstable.
gm=0.5434pm=-26.6555wg=0.8620wp=1.1975
相位裕量小于零,因此单位反馈构成的闭环系统不稳定。
step(feedback(H,1))
17
G=100*tf([1/2.5,1],conv([1/0.5,1,0],[1/50,1]));
Gc=1000*tf(conv([1,1],[1,2.5]),conv([1,0.5],[1,50]));
nichols(G*Gc)
gm=Infpm=63.2953wg=Infwp=18.8688
由以上图形和数据,可判断出单位负反馈系统稳定
以下是单位阶跃响应:
GG=feedback(G*Gc,1);
step(GG)
第二部分
sim('
simu2'
100),plot(tout,yout)
f=@(t,y)[y
(2);
y(3);
y(4);
-2*y
(1)-4*y
(2)-63*y(3)-5*y(4)+exp(-3*t)+exp(-5*t)*sin(4*t+pi/3)];
[t,y]=ode45(f,[0,100],[10.50.50.2]'
plot(t,y(:
1))
3
Simulink仿真模型
plot(tout,yout)
4换成一阶微分方程组的形式如下:
dx1/dt=t*sin(x2*exp(-2.3*x4));
dx2/dt=x1;
dx3/dt=sin(x2*exp(-2.3*x4);
dx4/dt=x3;
5系统仿真模型
[A,B,C,D]=linmod('
test5'
G=ss(A,B,C,D);
subplot(221),step(G),grid,subplot(222),bode(G),grid,subplot(223),nyquist(G),grid,subplot(224),nichols(G),grid
阶跃响应与频率响应曲线如下:
6
G=210*tf([11.5],conv(conv([11.75],[116]),conv([11.5+3*j],[11.5-3*j])));
[gm,gp,wm,wp]=margin(G)
gm=4.8921gp=60.0634wm=7.9490wp=3.9199
step(GG)
[gm1,gp1,wm1,wp1]=margin(GG)
Inwarningat26
Inlti.marginat67
gm1=Infgp1=-43.3034wm1=NaNwp1=24.6059
A=[0100;
0010;
-3123;
2100];
B=[1234;
0123]'
Q=diag([1234]);
R=eye
(2);
[K,S]=lqr(A,B,Q,R)
K=
-0.09781.21181.87670.7871
-3.8819-0.46682.67131.0320
S=
5.44000.6152-2.31630.0452
0.61521.8354-0.0138-0.7582
-2.3163-0.01381.9214-0.3859
0.0452-0.7582-0.38590.8540
eig(A-B*K)
-12.2563
-1.6786+0.9981i
-1.6786-0.9981i
-1.4627
G=ss(A-B*K,B,eye(4),0);
step(G)
8
A=[-0.20.5000;
0-0.51.600;
00-14.385.80;
000-33.3100;
0000-10];
B=[000030]'
C=[10000];
D=0;
P=[-1-2-3-4-5];
K=acker(A,B,P)
0.00040.0004-0.00350.3946-1.4433
-5.0000
-4.0000
-3.0000
-2.0000
-1.0000
step(ss(A-B*K,B,C,D),10)
p1=[-10-20-15-12-13]'
L=place(A'
C'
p1)'
[Gc,H]=obsvsf(G,K,L);
step(feedback(Gc*G,H))
grid
functiony=optfun_8(x)
assignin('
base'
'
Kp'
x
(1));
Ki'
x
(2));
Kd'
x(3));
[t_time,x_state,y_out]=sim('
opt1'
[0,21.000000]);
y=y_out(end);
ctrl_pars=
0.7922
0.9595
0.6557
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