MATLAB控制系統(tǒng)各種仿真例題包括simulink解法

1、一、 控制系統(tǒng)的模型與轉(zhuǎn)換1 請(qǐng)將下面的傳遞函數(shù)模型輸入到matlab環(huán)境。 ，T=0.1s s=tf(s); G=(s3+4*s+2)/(s3*(s2+2)*(s2+1)3+2*s+5); G Transfer function: s3 + 4 s + 2-s11 + 5 s9 + 9 s7 + 2 s6 + 12 s5 + 4 s4 + 12 s3 num=1 0 0.56; den=conv(1 -1,1 -0.2 0.99); H=tf(num,den,Ts,0.1) Transfer function: z2 + 0.56-z3 - 1.2 z2 + 1.19 z - 0.992 請(qǐng)

2、將下面的零極點(diǎn)模型輸入到matlab環(huán)境。請(qǐng)求出上述模型的零極點(diǎn)，并繪制其位置。 ，T=0.05sz=-1-j -1+j; p=0 0 -5 -6 -j j; G=zpk(z,p,8) Zero/pole/gain: 8 (s2 + 2s + 2)-s2 (s+5) (s+6) (s2 + 1)pzmap(G) z=0 0 0 0 0 -1/3.2 -1/2.6; p=1/8.2; H=zpk(z,p,1,Ts,0.05) Zero/pole/gain:z5 (z+0.3125) (z+0.3846)- (z-0.122) Sampling time: 0.05pzmap（H）num=0,7.

3、1570,-6.4875 ;den=1,-2.2326,1.7641,-0.4966;sysd=tf(num,den,0.05,variable,z-1) Transfer function: 7.157 z-1 - 6.487 z-2-1 - 2.233 z-1 + 1.764 z-2 - 0.4966 z-3 Sampling time: 0.05二、 線性系統(tǒng)分析1. 請(qǐng)分析下面?zhèn)鬟f函數(shù)模型的穩(wěn)定性。 num=1;den=1 2 1 2;G=tf(num,den);eig(G)ans =-2.0000 0.0000 - 1.0000i 0.0000 + 1.0000i可見，系統(tǒng)有兩個(gè)特征

4、根在虛軸上，一個(gè)特征根在虛軸左側(cè)，所以系統(tǒng)是臨界穩(wěn)定的。 num=3 1;den=300 600 50 3 1;G=tf(num,den);eig(G)ans = -1.9152 -0.14140.0283 - 0.1073i 0.0283 + 0.1073i可見，有兩個(gè)特征根在虛軸右側(cè)，所以系統(tǒng)是不穩(wěn)定的。2. 請(qǐng)判定下面離散系統(tǒng)的穩(wěn)定性。 num=-3 2; den=1 -0.2 -0.25 0.05; H=tf(num,den,Ts,0.1); eig(H) abs(eig(H) ans = -0.5000 0.5000 0.5000 0.50000.2000 0.2000可以看出，由于

5、各個(gè)特征根的模均小于1，所以可以判定閉環(huán)系統(tǒng)是穩(wěn)定的。 z=tf(z,0.1); H=(2.12*z-2+11.76*z-1+15.91)/;(z-5-7.368*z-4-20.15*z-3+102.4*z-2+80.39*z-1-340); eig(H) abs(eig(H) ans = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4.1724 4.1724 0.3755 + 0.1814i 0.4170 0.3755 - 0.1814i 0.4170 -0.5292 0.5292 -0.2716 0.2716 0.1193 0.1193

6、可以看出，由于4.1724這個(gè)特征根的模大于1，所以可以判定閉環(huán)系統(tǒng)是不穩(wěn)定的。3. 設(shè)描述系統(tǒng)的傳遞函數(shù)為，假定系統(tǒng)具有零初始狀態(tài)，請(qǐng)求出單位階躍響應(yīng)曲線和單位脈沖響應(yīng)曲線。 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 s7 + 514 s6 + 5982 s5 + 36380 s4 + 122664 s3 + 22088 s2 + 185760 s + 40320-

7、s8 + 36 s7 + 546 s6 + 4536 s5 + 22449 s4 + 67284 s3 + 118124 s2 + 109584 s + 40320 step(G,10) impulse(G,10)單位階躍響應(yīng)：?jiǎn)挝幻}沖響應(yīng)：三、 線性系統(tǒng)Simulink仿真應(yīng)用1. 請(qǐng)分析下面?zhèn)鬟f函數(shù)模型階躍響應(yīng)。 利用Simulink建模，建立系統(tǒng)仿真模型如下：?jiǎn)螕魡?dòng)仿真按鈕，雙擊示波器得到系統(tǒng)的階躍響應(yīng)如下：2. 請(qǐng)分析下面離散系統(tǒng)的脈沖響應(yīng)。 利用Simulink建模，建立系統(tǒng)仿真模型如下：?jiǎn)螕魡?dòng)仿真按鈕，雙擊示波器得到系統(tǒng)的脈沖響應(yīng)如下：3. 對(duì)離散采樣系統(tǒng)進(jìn)行分析，并求出其階躍

8、響應(yīng)。 其中：利用Simulink建模，建立系統(tǒng)仿真模型如下：?jiǎn)螕魡?dòng)仿真按鈕，雙擊示波器得到系統(tǒng)的階躍響應(yīng)如下：4. 設(shè)計(jì)控制器，使得下列系統(tǒng)穩(wěn)定。 利用Simulink建模，未連入控制器時(shí)，仿真模型和響應(yīng)如下：利用Simulink建模，設(shè)計(jì)控制器：從響應(yīng)輸出圖形可以看出，連入控制器后系統(tǒng)穩(wěn)定，性能明顯提高。以下只做了解四、 基于MATLAB的PID 控制器設(shè)計(jì)設(shè)計(jì)題目：1. 應(yīng)用Ziegler Nichols算法設(shè)計(jì)PID控制器，實(shí)現(xiàn)系統(tǒng)的閉環(huán)穩(wěn)定，并比較對(duì)各個(gè)系統(tǒng)的控制效果。 未連入PID控制器時(shí)的系統(tǒng)仿真及其性能指標(biāo)如下： 可見，未調(diào)節(jié)時(shí)的系統(tǒng)性能有待提高，需設(shè)計(jì)PID控制器連入。輸入

9、： num=1; den=conv( 1,1,conv( 1,1,1,1 ); Step(num,den); K=dcgain (num,den)得出：K =1根據(jù)圖形，得出：L=1.86 T=4.4利用自定義的Ziegler_std函數(shù)求出Kp、Ti、Td輸入： K=1; L=1.86; T=4.4; num,den,Kp,Ti,Td=Ziegler_std (3,K,L,T)得出：num = 2.6400 2.8387 1.5262den =1 0Kp =2.8387Ti =3.7200Td =0.9300根據(jù)得出的Kp、Ti、Td值，設(shè)計(jì)PID控制器，并利用利用Simulink仿真建模。

10、仿真模型及其響應(yīng)如下：可見，加入PID控制器調(diào)節(jié)后，系統(tǒng)性能明顯改善。未連入PID控制器時(shí)的系統(tǒng)仿真及其性能指標(biāo)如下： 可見，未調(diào)節(jié)時(shí)的系統(tǒng)性能有待提高，需設(shè)計(jì)PID控制器連入。輸入： num=1; den=conv(1,1,conv(1,1,;conv( 1,1,conv( 1,1,1,1 ); Step(num,den); K=dcgain (num,den)得出：K = 1根據(jù)圖形，得出：L=3.4 T=6.8利用自定義的Ziegler_std函數(shù)求出Kp、Ti、Td輸入： K=1; L=3.4; T=6.8; num,den,Kp,Ti,Td=Ziegler_std (3,K,L,T)

11、得出:num = 4.0800 2.4000 0.7059den = 1 0Kp = 2.4000Ti = 6.8000Td =1.7000根據(jù)得出的Kp、Ti、Td值，設(shè)計(jì)PID控制器，并利用利用Simulink仿真建模。仿真模型及其響應(yīng)如下：可見，加入PID控制器調(diào)節(jié)后，系統(tǒng)性能明顯改善。利用Simulink建模，未連入控制器時(shí)，仿真模型和響應(yīng)如下：可見，未調(diào)節(jié)時(shí)的系統(tǒng)性能有待提高，需設(shè)計(jì)PID控制器連入。輸入:num=-1.5 1; den=conv( 1,1,conv( 1,1,1,1 ); Step(num,den); K=dcgain (num,den)得出:K =1根據(jù)圖形，得出

12、：L=1.8 T=5.7利用自定義的Ziegler_std函數(shù)求出Kp、Ti、Td輸入: K=1; L=1.8 T=5.7; num,den,Kp,Ti,Td=Ziegler_std (3,K,L,T)得出:num = 3.4200 3.8000 2.1111den =1 0Kp =3.8000Ti =3.6000Td =0.9000根據(jù)得出的Kp、Ti、Td值，設(shè)計(jì)PID控制器，并利用利用Simulink仿真建模。仿真模型及其響應(yīng)如下：可見，加入PID控制器調(diào)節(jié)后，系統(tǒng)性能明顯改善。以下可以不看五、 模糊控制器設(shè)計(jì)設(shè)計(jì)任務(wù)：試設(shè)計(jì)一個(gè)模糊控制器，實(shí)現(xiàn)對(duì)室內(nèi)溫度的控制的模擬。參考輸入：（1）溫

13、度18-40范圍內(nèi)分為七個(gè)論域，NB NM NS ZE PS PM PB；隸屬度函數(shù)滿足高斯分布；（2）溫度變化率-2 2范圍內(nèi)分為七個(gè)論域，NB NM NS ZE PS PM PB；隸屬度函數(shù)滿足高斯分布；參考輸出：變頻空調(diào)輸出的控制信號(hào)。在一定范圍內(nèi)分為七個(gè)論域，NB NM NS ZE PS PM PB，隸屬度函數(shù)為常數(shù)1。模糊推理過程，output=輸入隸屬度函數(shù)值*輸出論域的中心值。注：本模糊程序采用PAM控制方式的壓縮機(jī)，則其輸出的轉(zhuǎn)速范圍為：010500轉(zhuǎn)/分?？刂埔?guī)則：% % if input is NB and errorinput is NB, then output is

14、NB;% % if input is NB and errorinput is NM, then output is NB;% % if input is NB and errorinput is NS, then output is NB;% % if input is NB and errorinput is ZE, then output is NM;% % if input is NB and errorinput is PS, then output is NM;% % if input is NB and errorinput is PM, then output is NM;%

15、% if input is NB and errorinput is PB, then output is NS;% % if input is NM and errorinput is NB, then output is NB;% % if input is NM and errorinput is NM, then output is NM;% % if input is NM and errorinput is NS, then output is NM;% % if input is NM and errorinput is ZE, then output is NM;% % if

16、input is NM and errorinput is PS, then output is NM;% % if input is NM and errorinput is PM, then output is NS;% % if input is NM and errorinput is PB, then output is NS;% % if input is NS and errorinput is NB, then output is NM;% % if input is NS and errorinput is NM, then output is NS;% % if input

17、 is NS and errorinput is NS, then output is NS;% % if input is NS and errorinput is ZE, then output is NS;% % if input is NS and errorinput is PS, then output is NS;% % if input is NS and errorinput is PM, then output is ZE;% % if input is NS and errorinput is PB, then output is ZE;% % if input is Z

18、E and errorinput is NB, then output is NS;% % if input is ZE and errorinput is NM, then output is ZE;% % if input is ZE and errorinput is NS, then output is ZE;% % if input is ZE and errorinput is ZE, then output is ZE;% % if input is ZE and errorinput is PS, then output is ZE;% % if input is ZE and

19、 errorinput is PM, then output is PS;% % if input is ZE and errorinput is PB, then output is PS;% % if input is PS and errorinput is NB, then output is ZE;% % if input is PS and errorinput is NM, then output is PS;% % if input is PS and errorinput is NS, then output is PS;% % if input is PS and erro

20、rinput is ZE, then output is PS;% % if input is PS and errorinput is PS, then output is PS;% % if input is PS and errorinput is PM, then output is PM;% % if input is PS and errorinput is PB, then output is PM;% % if input is PM and errorinput is NB, then output is PS;% % if input is PM and errorinpu

21、t is NM, then output is PS;% % if input is PM and errorinput is NS, then output is PM;% % if input is PM and errorinput is ZE, then output is PM;% % if input is PM and errorinput is PS, then output is PM;% % if input is PM and errorinput is PM, then output is PM;% % if input is PM and errorinput is

22、PB, then output is PB;% % if input is PB and errorinput is NB, then output is PS;% % if input is PB and errorinput is NM, then output is PM;% % if input is PB and errorinput is NS, then output is PM;% % if input is PB and errorinput is ZE, then output is PM;% % if input is PB and errorinput is PS, t

23、hen output is PB;% % if input is PB and errorinput is PM, then output is PB;% % if input is PB and errorinput is PB, then output is PB;1.輸入為：程序?yàn)椋?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)

24、;y6 = gaussmf(x1, 1 40);plot(x1,y0 y1 y2 y3 y4 y5 y6)2.誤差圖：程序?yàn)椋?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?yàn)?；x=35;e

25、x=-0.8;% define input type in fuzzy zoney0 = 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=18 if 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;elseif b=a(5) type=PS; authorityvalue=y4;elseif b=a(6) type=PM; author