(完整版)MATLAB控制系统仿真计算

1、控制系统的模型与转换1.请将下面的传递函数模型输入到matlab环境。32s 4s 2z 0.568G(s) 23 H (z) 2，T=0.1ss3(s2 2)(s2 1)3 2s 5(z 1)(z20.2z 0.99) s=tf(s);G=(sA3+4*s+2)/(sA3*(sA2+2)*(sA2+1)A3+2*s+5); GTransfer function: sA3 + 4 s + 2sA11 + 5 sA9 + 9 sA7 + 2 sA6 + 12 $八5 + 4 $八4 + 12 $八3 num=1 0 0.56;den=conv(1 -1,1 -0.2 0.99);H=tf(num

2、,den,Ts,0.1)Transfer function:zA2 + 0.562.请将下面的零极点模型输入到zA3 - 1.2 zA2 + 1.19 z - 0.99matlab环境。请求出上述模型的零极点，并绘制其位置。G(s)8(s 1 j)(s 1 j)2_2s (s 5)(s 6)(s1)z=-1-j -1+j；p=0 0 -5 -6 -j j;G=zpk(z,p,8)Zero/pole/gain:8 *2 + 2s + 2)sA2 (s+5) (s+6)伊人2 + 1)pzmap(G)，、 (z 1 3.2)(z 1 2.6)H (z) 5一t , T=0.05s5z (z 8.2

3、)Pck!-Z*re M 邯i f1t r0M0.0 -1 O.G 0.6 0.4 z=0 0 0 0 0 -1/3.2 -1/2.6;P=1/8.2;Zero/pole/gain:zA5 (z+0.3125) (z+0.3846)亿-0.122)Sampling time: 0.05pzmap (H)线性系统分析1.请分析下面传递函数模型的稳定性。G(s)1332-s 2ss 2G(s)3s 1-2Z2 11s (300s600s 50) 3s 1 num=1;den=1 2 1 2;G=tf(num,den);eig(G)ans =-2.00000.0000 - 1.0000i0.0000

4、 + 1.0000i可见，系统有两个特征根在虚轴上, 一个特征根在虚轴左侧，所以系统是临 界稳定的。 num=3 1;den=300 600 50 3 1;1 PdB 工 5r 口 IVtip105 Iirf-L.fK fl rFde-ZEfuMBt&必0.1150.1。,&5S3-，n its由iO.HQ2r111界3JI111-1.5-1-05D0,Heel floas5G=tf(num,den);eig(G) ans =-1.9152-0.14140.0283 - 0.1073i0.0283 + 0.1073i可见，有两个特征根在虚轴右侧，所以系统是不稳定的。2.请判定下面离散系统的稳定

5、性。H(z) (z3 0.22产0.252 0.05)H(z)212.12z 2 11.76z 1 15.91z 5 7.368z 4 20.15z 321102.4z 2 80.39z1 340) num=-3 2;den=1 -0.2 -0.25 0.05;H=tf(num,den,Ts,0.1);eig(H) abs(eig(H)ans =-0.50000.50000.50000.50000.20000.2000可以看出，由于各个特征根的模均 小于1,所以可以判定闭环系统是稳定 的。 z=tf(z,0.1);H=(2.12*zA-2+11.76*zA- 1+15.91)/ :(zA-5-

6、7.368*zA-4-20.15*zA-3+102.4*zA-2+80.39*z-1-340);eig(H) abs(eig(H)ans =0000000000000000000000004.17244.17240.3755 + 0.1814i0.41700.3755 - 0.1814i0.4170-0.52920.5292-0.27160.27160.11930.1193Pule-Zelv Mb i；可以看出，由于4.1724这个特征根的模大于 1,所以可以判定闭环系统是不稳定的。3.设描述系统的传递函数为G(s)18 s7514 s65982 s536380 s4 122664 s3222

7、088 s2185760 s 40320s836 s7546 s6 4536 s522449 s4 67284 s3118124 s2 109584 s 40320具有零初始状态，请求出单位阶跃响应曲线和单位脉冲响应曲线。 num=18 514 5982 36380 122664 22088 185760 40320;den=1 36 546 4536 22449 67284 118124 109584 40320;G=tf(num,den)Transfer function:18 sA7 + 514 sA6 + 5982 sA5 + 36380 sA4 + 122664 sA3 + 2208

8、8 sA2 + 185760 s + 40320 sA8 + 36 sA7 + 546 $八6 + 4536 $人5 + 22449 $人4 + 67284 $人3 + 118124 sA2 + 109584 s + 40320 step(G10) impulse(G,10)单位阶跃响应:单位脉冲响应:线性系统Simulink仿真应用1 .请分析下面传递函数模型阶跃响应。G(s)1s3 2s2 s 2利用Simulink建模，建立系统仿真模型如下:单击启动仿真按钮，双击示波器得到系统的阶跃响应如下:2 .请分析下面离散系统的脉冲响应。、3s3 4s2 1G (s)-22s2 (300s2 60

9、0s 50) 3s 1利用Simulink建模，建立系统仿真模型如下:单击启动仿真按钮，双击示波器得到系统的脉冲响应如下:3 .对离散采样系统进行分析，并求出其阶跃响应其中：G(s)s 3 二2 ZZs 7 s 3s 2利用Simulink建模，建立系统仿真模型如下:Step单击启动仿真按钮，双击示波器得到系统的阶跃响应如下:4 .设计控制器，使得下列系统稳定。(s 1)(s 2.3)(s 3)(s 2)(s 1.2)利用Simulink建模，未连入控制器时，仿真模型和响应如下:25利用Simulink建模，设计控制器:Time offset: 0从响应输出图形可以看出，连入控制器后系统稳定，

10、性能明显提高。四、基于MATLAB 的PID控制器设计设计题目：1.应用Ziegler Nichols算法设计PID控制器，实现系统的闭环稳定，并比较对各个系统的控制效果。G(s)1(s 1)3未连入PID控制器时的系统仿真及其性能指标如下:StepS*1可见，未调节时的系统性能有待提高,需设计输入: num=1;den=conv( 1,1,conv( 1,1,1,1);Step(num,den);K=dcgain (num,den)得出:K =1根据图形，得出:L=1.86T=4.41a. 3LIBM0.2D0510Tirnc offset 0PID控制器连入。15 如利用自定义的 Zieg

11、ler_std函数求出 Kp、Ti、Td输入： K=1;L=1.86;T=4.4;num,den,Kp,Ti,Td=Ziegler_std (3,K,L,T)得出：num =2.64002.83871.5262den =10Kp =2.8387Ti =3.7200Td =0.9300根据得出的Kp、Ti、Td值，设计PID控制器，并利用利用Simulink仿真建模。可见，加入PID控制器调节后，系统性能明显改善。G(s)1(s 1)5未连入PID控制器时的系统仿真及其性能指标如下:号一-0与输入： num=1;den=conv(1,1,conv(1,1,;conv( 1,1,conv( 1,1

12、,1,1 ); Step(num,den);K=dcgain (num,den)得出：K = 1根据图形，得出：L=3.4 T=6.8利用自定义的 Ziegler_std函数求出 Kp、Ti、Td 输入： K=1;L=3.4;T=6.8;num,den,Kp,Ti,Td=Ziegler_std (3,K,L,T)得出：num =4.08002.40000.7059den =1 0Kp =2.4000Ti =6.8000Td =1.7000根据得出的Kp、Ti、Td值，设计PID控制器，并利用利用 Simulink仿真建模。 仿真模型及其响应如下：Time offsM: 0可见，加入PID控制器

13、调节后，系统性能明显改善。-1.5s 1G(s) 3(s 1)3利用Simulink建模，未连入控制器时，仿真模型和响应如下:输入：num=-1.5 1;Time (sec)den=conv( 1,1,conv( 1,1,1,1);Step(num,den);K=dcgain (num,den)得出：K =1根据图形，得出：L=1.8T=5.7利用自定义的 Ziegler_std函数求出Kp、Ti、Td输入： K=1;L=1.8T=5.7;num,den,Kp,Ti,Td=Ziegler_std (3,K,L,T)得出：num =3.42003.80002.1111den =1 0Kp = 3

14、.8000Ti =3.6000Td =0.9000根据得出的Kp、Ti、Td值，设计PID控制器，并利用利用 Simulink仿真建模。 仿真模型及其响应如下：可见，加入PID控制器调节后，系统性能明显改善。五、模糊控制器设计设计任务：试设计一个模糊控制器，实现对室内温度的控制的模拟。参考输入：(1)温度18-40 C范围内分为七个论域，NB NM NS ZE PS PM PB ;隶属度函数满足高斯分布；(2)温度变化率-2 2c范围内分为七个论域，NB NM NS ZE PS PM PB ;隶属度函数满足高斯分布；参考输出：变频空调输出的控制信号。在一定范围内分为七个论域，NB NM NS

15、ZE PS PM PB ,隶属度函数为常数1。模糊推理过程，output=输入隶属度函数值*输出论域的中心值。注：本模糊程序采用PAM控制方式的压缩机，则其车出的转速范围为：010500转/分。控制规则：%if input is NB and errorinput is NB, then output isNB;%if input is NB and errorinput is NM, then output isNB;%if input is NB and errorinput is NS, then output isNB;%if input is NB and errorinput is

16、 ZE, then output isNM;%if input is NB and errorinput is PS, then output isNM;%if input is NB and errorinput is PM, then output isNM;%if input is NB and errorinput is PB, then output isNS;%if input is NM and errorinput is NB, then output isNB;%if input is NM and errorinput is NM, then output isNM;%if

17、 input is NM and errorinput is NS, then output isNM;%if input is NM and errorinput is ZE, then output isNM;%if input is NM and errorinput is PS, then output isNM;%if input is NM and errorinput is PM, then output isNS;%if input is NM and errorinput is PB, then output isNS;%if input is NS and errorinp

18、ut is NB, then output isNM;%if input is NS and errorinput is NM, then output isNS;%if input is NS and errorinput is NS, then output isNS;%if input is NS and errorinput is ZE, then output isNS;%if input is NS and errorinput is PS, then output isNS;%if input is NS and errorinput is PM, then output isZ

19、E;%if input is NS and errorinput is PB, then output isZE;%if input is ZE and errorinput is NB, then output isNS;%if input is ZE and errorinput is NM, then output isZE;%if input is ZE and errorinput is NS, then output isZE;%if input is ZE and errorinput is ZE, then output isZE;%if input is ZE and err

20、orinput is PS, then output isZE;%if input is ZE and errorinput is PM, then output isPS;%if input is ZE and errorinput is PB, then output isPS;%if input is PS and errorinput is NB, then output isZE;%if input is PS and errorinput is NM, then output isPS;%if input is PS and errorinput is NS, then outpu

21、t isPS;%if input is PS and errorinput is ZE, then output isPS;%if input is PS and errorinput is PS, then output isPS;%if input is PS and errorinput is PM, then output isPM;%if input is PS and errorinput is PB, then output isPM;%if input is PM and errorinput is NB, then output isPS;%if input is PM an

22、d errorinput is NM, then output isPS;%if input is PM and errorinput is NS, then output isPM;%if input is PM and errorinput is ZE, then output isPM;%if input is PM and errorinput is PS, then output isPM;%if input is PM and errorinput is PM, then output isPM;%if input is PM and errorinput is PB, then

23、output isPB;%if input is PB and errorinput is NB, then output isPS;%if input is PB and errorinput is NM, then output isPM;%if input is PB and errorinput is NS, then output isPM;%if input is PB and errorinput is ZE, then output isPM;%if input is PB and errorinput is PS, then output isPB;%if input is

24、PB and errorinput is PM, then output isPB;%if input is PB and errorinput is PB, then output isPB;1.输入为 程序为 x1 = (18:0.1:40);y0 = gaussmf(x1, 1 18);y1 = gaussmf(x1, 1 21);y2 = gaussmf(x1, 1 25);y3 = gaussmf(x1, 1 29);y4 = gaussmf(x1, 1 33);y5 = gaussmf(x1, 1 37);y6 = gaussmf(x1, 1 40);plot(x1,y0 y1 y

25、2 y3 y4 y5 y6)2 .误差图：程序为： x1 = (-2:0.1:2);y0 = gaussmf(x1, 0.5 -2);y1 = gaussmf(x1, 0.5 -1.3);y2 = gaussmf(x1, 0.5 -0.7);y3 = gaussmf(x1, 0.5 0);y4 = gaussmf(x1, 0.5 0.7);y5 = gaussmf(x1, 0.5 1.3);y6 = gaussmf(x1, 0.5 2);plot(x1,y0 y1 y2 y3 y4 y5 y6)3 .程序为；x=35;ex=-0.8;% define input type in fuzzy z

26、oney0 = gaussmf(x, 1 18);y1 = gaussmf(x, 1 21);y2 = gaussmf(x, 1 25);y3 = gaussmf(x, 1 29);y4 = gaussmf(x, 1 33);y5 = gaussmf(x, 1 37);y6 = gaussmf(x, 1 40);a=y0 y1 y2 y3 y4 y5 y6;b=max(a);% caculate input in fuzzy zone,get input_type and input_authorityvalueif x=18if b=a(1)type= NB ;authorityvalue=y0;elseif b=a(2)type= NM ;authorityvalue=y1;elseif b=a(3)type= NS ;authorityvalue=y2;elseif b=a(4)type= ZE ;authorityvalue=y3;el